Graphs and hypergraphs (North-Holland mathematical library)

$(n,p)

=

- 1, let

na 2

- - 1.

+

Let p > 2; let (no,n,, ..., n,) be a ( p 1)-tuple that maximizes f. If n, > 1 andn, > 1 w i t h j - i > 1, then the value offcan be increased either by taking n; = ni - 1 and n; = n j 1, or by taking ni = n, 1 and n; = n, - 1, since

+

f ( n , , ...)n, k 1,

= f ( n o , ...)n,)

..., n j T 1, ... n,) )

+

=

+ 1 k (nt-1 + + ni,l) n1

T (nj-1

It follows that there exists an index k, with 1

i Z k }>.,=1. i # k + l

+ n, + n j + l ) .

< k
- 2 such that

72

GRAPHS

Therefore,

P-1

=-

2

By replacing nk

1

+ P - 3 + 2 (nk.+ n,,, +

-

+ n k + , + 1 by n - ( p - 2), we obtain na Pa - P - 2. - n(P - 2) + $(4P) = 2 2

Since this equality is also true for p

=

-1 . 2

2, the theorem follows.

Q.E.D. Theorem 7 (Ghouila-Houri [1960]). The maximum value for the diameter of a strongly connected loopless l-graph with n vertices a,nd m arcs is

cp ( m , n )

I

if

=n-1

=[n

n < m <

n2+n-2 2

1-4

+ -

if

n2+n-2

<m

< n(n

- 1).

Let G be a strongly connected loopless graph with n vertices and m arcs; then m 2 n because each vertex is the initial endpoint of at least one arc. Furthermore, m < n(n - l), since the graph cannot have more arcs than a complete symmetric 1-graph. If m = n(n - l), the proof is immediate; therefore, we may assume m < n(n - 1) and m 2 n. If conditions (1) and (3) of Lemma 2 are fulfilled, then from Lemmas 1 and 2 there exists a strongly connected 1-graph with 6(G) 2 p . Conversely, if conditions (1) and (3) are not satisfied, then no such graph exists. Thus, the greatest possible value for the diameter is the greatest value cp(m, n) for an integer p satisfying (I) and (3). Inequality (1) is equivalent to

Then, by elementary calculus, we obtain the above value cp(nt, n) for p .

Q.E.D.

For a simple graph G

=

(X,E ) , we define the undirected radius p*(G) as

PATHS, CENTRES AND DIAMETERS

73

the radius p(G*) of the graph G* = (X,U ) obtained from G by replacing each edge in G by two oppositely directed arcs. Similarly, we define the undirected diameter 6*(G) of a simple graph G by

6*(G) = 6(G*). The following theorem is due to Camille Jordan [1869].

Theorem 8. r f G is u tree, and if S*(G) is even, then G has a unique centre, and all the elementary chains of maximum length pass through it; furthermore,

I f 6 * ( G ) is odd, then G has exactly two centres (which are adjacent vertices) and all the elementary chains of maximum length pass through them; furthermore, 1 p*(G) = 3 (6*(G) 1 ) .

+

The theorem is obvious for trees of order < 3. Let n > 3, and suppose that the theorem is true for trees of order < h ; we shall show that the theorem is true for a tree G = (A', E ) of order n. Let B be the set of pendant vertices of tree G. From Theorem (2, Ch. 3), we have I B 1 2 2, and hence, the subgraph G X - Bis a tree of order < n. Clearly, P*(G.Y-B) = p*(G) -

3

6*(Gx-B) = 6*(G) - 2 . Each centre of G X - Bis a centre of G, and vice versa. Each elementary chain of maximum length in G x - Binduces in graph G an elementary chain of length 6*(G). Since the theorem is true for G X - B by , the induction hypothesis, the result follows. Q.E.D. Remark. The two equalities of Theorem 8 can be summarized by:

where [r] denotes the greatest integer

< r.

For simple graphs, several results regarding p * ( G ) and S*(G) have been obtained: Vizing [1967] has determined the maximum number of edges in a simple graph with p*(G) = r. Murty [1968] has determined the minimum number of edges in a simple graph with 6*(G) < k such that the diameter remains < 1 after the elimination of any s vertices. The diameter of a planar 3-connected graph has been studied by M. Balinski [1966], B. Grunbaum [1967], Bollobas [1968], Bosik, Kotzig, Zndm [1968].

14

GRAPHS

5. Counting paths This section presents results about the number of different paths between each pair of vertices x, y in a graph G = ( X , V ) . It will be necessary to use matrix products as defined in linear algebra. The principal result for path counting is: Theorem 9. Consider two graphs G = ( X , U ) and H = ( X , V ) with the same vertex set, and let A = ((a:))and B = ((b:))denote respectirely their associated matrices. The matrix product A B corresponds to a graph G . H with rertex set X , and an arc from x to y for each distinct path from x to y composed of an arc of U followed by an arc of V. The graph G . H is called the composition product of G and H.

The number of distinct paths of the form [xi,xk, x j ] with (xi,xk) E U, (xk, x j )E V , equals af,b:. Thus, the total number of different paths from x1 to x j formed from an arc of U followed by an arc of k is n

a:b:=(ai,bi), k=l

where ( a', bj) denotes the scalar product of the row vector a' and the column vector bj, which is also the general coefficient of the matrix A . B.

Q.E.D. Corollary 1. If G is a graph and A is its associated matrix, the general coeficient p: of the matrix P = A k (the product of A with itselfk times) equals the number of distinct paths of length k from xi to x, in G. The theorem is true for k = 1. Let k > I , and suppose the theorem is true for the power k - 1. Then it is also true for the power k , since A k = A(Ak-') shows the number of paths of length 1 + (k - 1) = k from xi to x, (by Theorem 9).

Q.E.D. Corollary 2. A graph G possesses a path of length k if, and only $ A k # 0. G possesses no circuits if, and only if, A k = 0 for k suficiently large.

The proof follows immediately from Corollary 1. APPLICATION. The power index of a participant in a partial tournament.

Let G = ( X , U ) be a I-graph of a partial tournament with n participants. Let arc (x,y ) E U if participant x has beaten or had a draw with participant y. It is tempting to choose as the winner the most dominating participant, i.e., the vertex x whose outer demi-degree is as large as possible.

PATHS, CENTRES AND DIAMETERS

75

However, there is a good chance that this participant x has beaten a large number of very weak participants, and would lose to a participant y who has beaten only a few very strong players. If p:’(k) is the general coefficient of matrix Ak, i.e., the number of paths of length k from x to xi then a better estimation of the power of participant xi is the sum

+

+ +

pi@) = p‘l(k) p l ( k ) - * * p&). Consequently, the power index of participant xi can be defined as

Note that by virtue of the Perron-Frobenius Theorem, this limit always exists.

EXERCISES 1. Let S*(G) = sup d(x, y ) be the diameter of a simple connected graph C. Show that the following conditions are equivalent: (1) Any two vertices of G are connected by at most one elementary chain of length

<

6*(G).

(2) G has no cycles of length < 2 S*(G). (3) Either there exist n o cycles in G, or the length of the shortest cycle of G is 2 6*(G) + I . 2. Given two integers h and 6, the number n of vertices of a simple graph regular of degree h with diameter S satisfies d

n

<

I

+ h 1(h - 1 y - 1 : i=l

A simple graph with n vertices, regular of degree h, with diameter 8, for which the equality holds is called a Moore graph of type (h, 6). Show that the Petersen graph (Fig. 10.1 1) is a Moore graph of type (3, 2). Show that the Moore graphs of type (h, 1) are the (h 1)-cliques. Show that the cycles without chords of length 2 8 1 are Moore graphs of type (2, S). Show that there exists a Moore graph of type (7, 2) and that it is unique (A. Hoffman, R. Singleton [1960]). (Hoffman and Singleton have characterized the Moore graphs of types (h, 2) and (h, 3) with the exception of type (57, 2) for which no example is known.) 3. Show that a simple connected graph G satisfies the equivalent properties in Exercise 1 if, and only if, it is either a tree or a Moore graph. (J. Boshk, A. Kotzig, S. Z n l m [1968]) 4. Show that for a simple connected graph, two elementary chains of maximum length always have common vertices. Also show that if I, is the length of a longest elementary chain, then each vertex is the initial endpoint of an elementary chain of length 2 [+(lo 111. 5. Show that a simple graph G of order n > 1 which is isomorphic to its complementary (H. Sachs [1962]) graph G has a diameter equal t o 2 or 3.

+

+

+

CHAPTER 5

Flow Problems

1. The maximum flow problem

Consider a graph G with arcs denoted by 1, 2, ..., m, and consider a set of numbers bl, b2, ..., b,, cl, c2, ..., c, in Z such that

+

- co d bi d ci d 00. AJlowl in G is defined as a vector q = (ql,q z ,..., q',) E Z"such that: (1) qi E Z for i = 1, 2, ..., nz. (The integer qi is called an arcflow, and may be regarded as the number of vehicles travelling through arc i along its direction if pi 2 0 or against its direction if (pi < 0.) (2) For each vertex x, the sum of the arc flows entering x equals the sum of the arc flows leaving x, i.e., C qi = q j (x E X ) . i E w-(x)

j

E o+(x)

In other words, there is a conservation of arc flows at each vertex (Kirchhoff's Law). We shall study the following problems:

The compatible flow problem. Given a graph G with an interral [bi,ci] associatedivith each arc i,jind a j o w p such that

bi

< qi < ci

(i = 1, 2,

..., m).

c, is called the capacity of arc i, and represents the maximum number of vehicles that can use the arc i along its direction.

The maximum flow problem. Giceiz a graph G with an interval [bi,c,] associated with each arc &find a frow 40 such that

(i = 1, 2 , ..., m), (1) bi < 'pid ci (2) the arc Pow q 1 is as large as possible. Note that, for both of these problems, we may assume without loss of generality that G is a 1-graph. 76

77

FLOW PROBLEMS

The maximum flow problem occurs frequently with the following additional conditions : for i = 1,2, ..., m, (2) c, > 0 for all i, and cl = +a ( 3 ) arc 1 is an arc joining a vertex b, called the sink, to a vertex a, called the source, where (1) bi = 0

w - ( a ) = (1, 0 , 0, ... O), w'(b) = (1, 0, 0, ..., 0). (4) G is an anti-symmetric I-graph. )

In other words, only arc i = 1 enters vertex a, and only arc i = 1 leaves vertex b. This arc (b, a ) , generally omitted in the illustrations, is called the return arc. (This arc has no function other than the maintenance of the conservation of flow at vertices a and b.) The most important floy problem is to find the maximum number of vehicles that can be sent from a to b without violating the arc capacities. A graph G with a capacity ci associated with each arc i and which satisfies the conditions (l), (2), (3), (4) is called a transportation network and is often denoted by R = ( X , U,c(u)). Below are some examples that reduce to maximum flow problems in a transportation network.

EXAMPLE 1. Maritime trafic. Let the seaports a,, a2, ..., up,b,, b2, ..., b,, be represented by vertices, and suppose that bananas are ready for shipment at ports a l , a2, ..., up to ports bl, b2, ..., b,. Let s, denote the quantity available at a,, and let dj denote the quantity demanded at b,. Shipping routes can be represented by arcs of the form (a,,b j )with a capacity equal to the shipping capacity between the two seaports. Is it possible to satisfy all the demands? How should the bananas be shipped? To answer these questions, create a source a, and join a to each a, by an arc with capacity c(a, a,) = si . Next, create a sink b, and join each vertex b, to b by an arc with capacity-

c(bj, b) = dj . A maximum flow for this transportation network yields the number of bananas to ship along each route in order to satisfy all demands, if this is possible.

EXAMPLE 2. The battle of the Marne. The towns a,, a2, ..., a, each have

78

GRAPHS

motor cars to be sent to town b. If there is a relay route from town a, to town a,, let ci, denote the number of motor cars that can leave a, for a, each time period. Let ti, denote the traverse time from a, to at, let st denote the number

e

Source U

Sink Fig. 5.1

of motor cars initially available at a,, and let c, denote the number of motor cars that can park at a; simultaneously. How can we direct the transport of these motor cars so that as many as possible arrive at b within T time periods? This type of problem has been studied by R. Fulkerson under the name of “dynamic networks”. The problem can be reduced to a maximum flow problem in a transportation network with vertices ai(r), where i = 1, 2,..., n,and t = 0, 1, ..., T.

+

Vertex ai(t) and vertex a,(t 1) are joined by an arc of capacity c,. If there is a route from a, to a,, it is represented by an arc from a,(t) to a,(t + t,,) with a capacity of c,,. Add a source a, sink 6, arcs (a, a,(O)) with capacity s, and arcs ( b ( t ) , b) with capacity co. The maximum flow in this transportation network determines the optimal routes.

79

FLOW PROBLEMS

EXAMPLE 3. The selection of representatii>es.A set X of residents belong to various clubs C1,C,, ..., C, (which are not necessarily disjoint subsets of X) and t o various political parties PI,Pz,..., P, (which are disjoint subsets of A'). Each club must choose one of its members to represent it, and no person can represent more than one club, no matter how many clubs he belongs to. HOWshould one choose a system of distinct representatives A = { al, a,, ..., a, }, such that the numbers of representatives belonging to each party Pisatisfies bj


n P j l G cj

A solution to this problem is given by a maximum flow in the transportation network shown in Fig. 5.2 (for q = 4 and r = 3 ) .

fl

Source

h

Sink

Auxiliary vertex Fig. 5.2

The arc capacities are marked in parenthesis. It is left to the reader to verify that a maximum flow that saturates all source arcs determines a set of distinct representatives as required.

80

GRAPHS

General maximum flow algorithm (when a compatible flow is known) (Ford and Fulkerson [1956]). Consider an anti-symmetric 1-graph G with arc 1 = (b, a). Suppose that a compatible flow q satisfying b, < q, < c, is known. We shall augment successively the value of the arc flow q 1 by the following labelling procedure:

RULE1. Label vertex a, the terminal endpoint of the arc 1, with the index

+ 1.

RULE2. If x is labelled and y is unlabelled, label y with the index + k if (x,y ) is arc k , and if 4%

Y) < 4x9 Y ) *

RULE3. If x is labelled and y is unlabelled, label y with the index - k if ( y , x) is arc k , and if

4po,, x ) > MY, x)

*

I f sink b is labelled b y this procedure, we shaN show that rhe arcflow q1 can be augmented, i.e., Hie shall construct a new flow cp' such that cp; > q,. Let

P

=

b,ai, U 2 , ..., Uk, bj

be a chain from a to b in which each vertex ai+l.has been labelled from its predecessor a,. (1) If the edge [a,,a,,,] is directed from a, to a,,,, then we have q(ai, ai+d <

~((113

ai+b

9

Then, put q'(at, a i + l >= P(ai, a i + J + 1 * (2) If the edge [a,, a,,,] is directed from a,,, to a,, then we have

d a j + 1 , aj> > N a j + 1 9 aj)

-

Then, put q'(aj+1 9

aj)

= d a j + 1, aj)

- 1;

(3) For arc 1 = (6, a), put

rp'(b, a ) = q ( b , a)

+ 1 = p1 + 1 .

(4) For all other arcs, put

cp'(x, Y ) = cp(x, Y ) *

81

FLOW PROBLEMS

In this way a new flow cp' is constructed. Flow cp' is compatible because only the flow around a cycle p' = p [b, a] has been changed, and at each vertex of this cycle, the conservation of flow has been maintained. In fact, the new flow can be expressed as the vectorial sum

+

Q' = Cp

-k p'

if the algorithm cannot label sink

It remains t o show that

b, then the arc

j o w cpl is maximum.

Lemma 1. In graph G = ( X , U ) , let A ' C X be a set with a E A and b .$A. Then, for eachflow cp, such that bi < (pi < ci, 'PI

G

C

i

1

c i -

i# 1 iew-(A)

E a, + ( A )

bi .

Since the algebraic sum of the flow entering set A equals the algebraic sum of the flow leaving set A , it follows that

Hence,

C

C P ~ =

1

qi-

iEO+(A) i

E

pi<

i#l o-(A)

1

c i iEo+(A)

1

bi.

if1 iEo-(A)

Q.E.D. Lemma 2. if the Ford atid Fulkerson algorithm cannot label sink b, theti cpl is maximum. Let A denote the set of vertices labelled by the algorithm. Clearly, a E A and b $ A. Since no more labelling is possible, an arc ( x , y ) with x E A and y $ A satisfies q(x,y ) = c(x, y ) , and an arc ( y , x ) with y $ A and x E A satisfies cpb,X ) = b(y, 4 . Thus,

From Lemma 1, it follows that cppl is maximum.

Q.E.D. Theorem 1. (Ford, Fulkerson [1957]). In a graph G n f t h numbers hi, ci where - 03 d b, < ci < 03, the maximum value of a compatibleflow in arc I is maxcp, = min C ci -

+

cp

,4311

Adb

i

iem+(A)

The proof follows from Lemma 1 and 2.

i# 1 i

E

(u - ( A )

82

GRAPHS

In a transportation network R = ( X , U,c(u)), a cut between a and b is defined to be a set of arcs of the form w + ( A ) with a e ; A and b $ A . The capacity of a cut is defined to be the sum of the capacities of its arcs, i.e.,

Maximum flow theorem (Ford, Fulkerson [1957]). In a transportation network, the maximum value of the arcjow qr equals the minimum capacity of a cut. EXAMPLE. Consider the transport network in Fig. 5.3.

b

a

'.X-'

'.-//

Y Fig. 5.3

The return arc (6, a) has been deleted to simplify the figure. The capacity of each source arc and sink arc is 2. The capacity of each intermediate arc (arcs that are neither source'arcs nor sink arcs) is 1. A compatible flow cp is easily found, and the saturated arcs are indicated by heavy print. Its arc flows are cp(a, c) = d a , 4 = 2 ,

d a , e) = d e ,

= cp(d,f) = cp(d, g> = = q(c9.f) = d e ,

df,b) = c p k , b) = 2

s> = P(h, b> =

1

9

*

This flow is maximal in the sense that there are no paths from a to b composed entirely of unsaturated arcs. However, the flow is not maximum because the labelling algorithm can be used to construct a flow cp' with 'pl = 40; + 1. For example, the vertices can be labelled as follows: a ( + 1) ; e ( +

4 ; f(+ 4 ); s ( + eg) ;

c ( - cf) ; d ( - dg) ;

h ( + ch) ; b ( + hb).

83

FLOW PROBLEMS

This determines the chain : p[a, b] =

+ ae + ef - cf + ch + h b .

Let

W,f) +

e)

+

d C , f )

-

1 1

1 , etc.

@(a, el = d a ,

cp’(e,f> = cp’(c,f) =

The reader can easily verify that the new flow cp‘ is maximum. Note that for a network of this type, a maximum flow can be obtained directly by using the following rule: during the sequential flow augmentation, send an additional unit of flow toward the vertex that has the greatest capacity to receive flow. When this flow has been constructed, it will be maximum. However, this simplification is possible only for very special networks. In fact, the Maximum Flow Theorem simplifies even fu,rther when the network R is biparlite, i.e., the vertices form four disjoint scts, X,Y, { a }, { b } and the arcs are of the following types: rype 1. ( x ,y ) with x E X,y E Y, rype 2. (a, x ) with x E X, type 3. (y, 6) with y E Y , type 4. (b, a), the return arc, denoted by 1. To simplify, let

d ( y ) is called the demand at vertex y. If B c Y , then the demand of set B is defined to be =

c

Y

~

4Y)

*

B

If B c Y, let F(B) denote the maximum quantity offlow thar can be sent into B, i.e., the maximum flow for a network R’ obtained from R by changing the capacities in the following way:

b) = c’cy, b) = C’(X, Y ) = C’b,

+ 00 0 C(X,

Y)

ifyE B ifyE Y - B for all other arcs (x, y).

We now have the notation needed to state: Theorem 2. In a bipartite transportation network R = ( X , Y, U ) , the maximum value of a compatibleJlow in arc 1 is

84

GRAPHS

p1 = d(Y)

+ minY [F(B) B c

1. Consider a set B c Y, and construct from R the network R‘ as described above. The Ford and Fulkerson Theorem states that

F(B) = min c’[:w-(S)] SP h

S$i

We may restrict our attention to sets S containing B because, otherwise, + 03. Without changing in the right side of the above equation, S can be replaced by S - { b }. Since we are only interested in the minimum, we may also restrict our attention to sets S of the form S = A u B, where A c X. Thus F(B) = min c‘[o-(A u B)] = min c [ o - ( ~u B)] .

c’[o-(S)] =

A c X

A c X

2. If P and Q are two disjoint sets of vertices, in order to simplify, let

q E Q

Consider a set S with b E S, a f S. Let

SnX=A, SnY=B. Then,

+ c(X - A , B ) + c(Y - B , b) = c(cB-(A u B ) ) + d(Y - B) = c ( w - ( A u B ) ) - d(B) + d(Y) .

c[w-(S)J’ = c(a,A)

From the Ford and Fulkerson theorem, the maximum value of a flow in arc 1 equals q1 = minc[o-(S)]

= min min

[ d ( ~+) c ( o - ( ~u B ) )

- d(B)]

B c Y A c X

Seb

d ( Y ) + min (F(B) - d(B)) . B c Y

This proves the theorem.

Q.E.D. Corollary (Gale [1958]). A bipartite transportation network R = ( X , Y, U) has a compatible j l o ~that ’ saturates all the sink arcs i f , and only i f ,

F(B) 2 d(B)

( B c 9.

FLOW PROBLEMS

85

In fact, this condition is equivalent to min (F(B) - d ( ~ )=) 0. B c Y

Q.E.D. Algorithm for planar networks If the transportation network R is planar, the following algorithm may be used : 1. Place arc (b, a) horizontally in the plane, with vertex a at the left with vertex b at the right. Draw the graph above arc (b, a) such that the arcs do not cross one another. In this topological planar graph, find a superior path p' from a to b by always choosing the left-most arc (and back tracing when an impasse occurs). 2. Reduce the capacity of each arc in the superior path p1 from a to b by the amount

= min ci

.

iep'

Eliminate from the graph all arcs with zero capacity.

3. Similarly determine the superior path p z in the new network. Reduce the capacity of each of its arcs by the amount c2 = miqci . iepz

4. Repeat this process until there are no more paths from a to b. The required flow is obtained by sending &k units along each path p k . To validate this algorithm, it is first necessary to establish the following lemma. Lemma. Let A be a set of certices such that a E A , b 4 A and such that w + ( A )is a minimal cut. The superior path p1 encounters w + ( A )exactly once. Clearly, path p1 encounters the cut w + ( A )at least once. Suppose that the path encounters the cut at two arcs, say i andj, in that order. We shall show that there exists a path p from a to b that meets o + ( A ) only at arc i. Colour arc i red, the other arcs of o + ( A ) green, and colour all other arcs black. Since o + ( A ) is a minimal cut, arc 1 is not contained in any green and black cocycle in which all black arcs are similarly directed. By the Arc Colouring Lemma, (Ch. 2), it follows that arc 1 belongs to a red and black cycle with all black arcs having the same direction.

86

GRAPHS

Thus, this cycle is a circuit, and it induces the required path p . Similarly, there exists a path v from a to b that meets o + ( A )only at arcj.

Fig. 5.3

Since p1 is the superior path, the portion of path p following arc i and the portion of Y preceding arc j have a common vertex x . This implies that x E A and x 4 A , which is a contradiction.

Q.E.D. Proposition. The J ~ O W cp obtained by the Algorithm for planar networks is maximum. Let A. denote the set of vertices that can be reached by a path from a. Then aEAo,

b $A,.

Consequently, w + ( A , ) is a cut. Let w + ( A ) be a minimal cut contained in w + ( A o ) From . the lemma, the k-th step reduces the capacity of the cut w + ( A ) by exactly E ~ and , the procedure terminates with a zero cut capacity. Thus, 'p1

=

Ek = k

ci

= c(m+(A))

i€@+(A)

By Lemma 1 to Theorem 1, flow cp is maximum.

Q.E.D. 2. The compatible flow problem Clearly, there always exists a compatible flow in a transportation network R since the flow cp = 0 is compatible. However, there may not always exist a

87

FLOW PROBLEMS

flow compatible with the constraints bi ,<

'pi

< ci.First, we have:

Lemma. A necessary condition for the existence of a conipatibleflow is that for each cocycle o ( A ) ,

'c

i Em+(A)

ci-

c

ieo-(A)

biaO.

Clearly, if a compatible flow cp exists, then

Q.E.D. The sufficiency of this condition will be demonstrated later.

Compatible flow algorithm (J. C. Herz [1967]). Let G = ( X , U ) be a graph with numbers bi and ci associated with each arc i, By using an iterative procedure similar to the Ford and Fulkerson algorithm, we shall construct a compatible flow starting with any non-compatibie flow q. Let the distance of'pifiom the interval [ b i , ci] be defined to equal

[

=

o

di(pi) = bi - 'pi = pi - ci

if if

pi E [bi, ci]

if

pi > ci

'pi<

9

bi,

.

The algorithm successively reduces the value of n

If d(q) = 0, flow cp is compatible and the procedure stops. If d(cp) > 0, there exists an arc i with d,(cpi)> 0. Suppose for example that d,(cp,) > 0, and that cpl < b,. Then, sequentially, label the vertices according to the following rules: RULE1. Label vertex a, the terminal endpoint of arc 1 with the index

+ 1.

RULE2. If x is labelled and y is not labelled, label y with the indexj if < cp

(x, y ) is arc j and if cp,

RULE3. If x is labelled and y is not labelled, label y with the index - j if ( y , x) is a r c j and if 'p, > b,. If vertex 6, the initial endpoint of arc 1, is labelled, then a new flow cp' such that d(cp') < d(q) can be constructed by using the method of the Ford and

88

GRAPHS

Fulkerson algorithm. If the initial endpoint of arc 1 cannot be labelled, then the set A of labelled vertices satisfy a E A and b $ A . Since (pl < b l , it follows that

o=

1

c

cpi-

ierv+(A)

iEW +

c

cpi> (A)

iEW

ci-

-(A)

1

bi.

LEO - ( A )

From the lemma, no compatible flow exists, and the problem has no solution. This algorithm establishes the following result :

Compatible flow theorem (Hoffman [1958]). For a graph G with arc numbers b, and c, such that - 00 < b, < cl < + co for all i, a necessary and suficient condition that there exists aJow cp with bt < 'pi < ct for all i is that, for each set A c X ,

The necessity of the condition follows from the lemma. The condition is sufficient because each flow cp with d(cp) > 0 can be improved by using the preceding algorithm, until we obtain a flow cp' with d ( q ) = 0. Q.E.D. The following algorithm is in general more efficient.

Second compatible flow algorithm Let G be a graph with vertices x l , x2, ..., x , and with arc numbers b f , ci. The following rules construct a flow cp in G such that

bi

< cpi f

ci

'

( i = 1 , 2 ,..., m).

RULE1. Construct a transportation network R' from G by adding a source a, asink b, a return arc (6, a), and the various source and sink arcs.

RULE2. If b(x, y ) 2 0, let the capacity of arc ( x , y ) in network R' equal CYX,

Y ) = 4x9 y> - 4 x 9 Y> ,

and create a sink arc ( x , b) with capacity c'(x, b) = b(x, y )

.

Also, create a source arc (a, y ) with capacity

m,v) = W , Y )

*

RULE3. If b(x, y ) < 0, let the capacity of arc ( x , y ) in network R' equal

89

FLOW PROBLEMS

CYX,

Y ) = 4 x 9 u> -

m,Y>,

and create a source arc (a, x ) with capacity c’(a, x) =

- b(x, y ) .

Also, create a sink arc ( y , b) with capacity CYY,

b) =

- 0 ,v) -

We shall now show that a compatibleflow exists in G i f , and only i f , in transportation network R’, there exists a maximumjow cp‘ that saturates all source arcs and sink arcs. If we let q ( x , y ) = cp’(x,y ) + b(x, y ) , then clearly cp is a flow in G because for each vertex x in G,

C

bm+W

pi

- C

pi =

iE(D-(X)

C

( ~ f bi) ++

IErO+(X)

bi 3 0

Flow cp is compatible in G, because

0 < pi

< C; - bi

(i=1,2

,..., m),

and consequently,

b,

(i = 1 , 2,

< 401 < ci

..., m)

Conversely, it is easily seen that each compatible flow cp in G corresponds in R’ to a flow cp’ that saturates all source and sink arcs. This completes the proof. Q.E.D.

3. An algebraic study of flows and tensions Each flow considered above is a vector in Z“.Flows could also be considered in any ring R, with s , t ~ R OE R SER s , t ~ R

=-

s + t ~ R , (zero element), * --sER, => s.tER.

The space Z” is not a “vector space” on Z, because Z is not a “field”, but a “module” on Z, and

90

s, t E b ,

E

n,s E n m

-

GRAPHS

=

s

+ t = (sl + t l , ...)s, + tm)E Z",

is = ( h 1 , ..., Asm)E n r n .

Consequently, the set @ of all flows in graph G constitutes a subiiiodule of

Z",i.e., we have: cp'+$€@,

=>

'P',cp2E@ S€h,cp€@

,

*

SCpE@.

Theorem 3. Let G = ( X , U ) be a connected graph. Let H = ( X , V ) be a spanning tree of G. Denote the arcs of U - V by 1, 2 , ..., k, and denote the associafed cycles of'H by p', p2, ..., pk. A J o w cp is uniquely defined from the i d u e s cpl, cp2, ..., (Pk by 9 =

p1

+

'p2

p2

+

f

(Pk

pk.

Consider the vector k

This vector cp is a flow, since it is a linear combination of flows. Clearly cp takes only zero values outside of tree H . Let W c Vbe the set of arcs ifor whichcp, # 0. We shall show that W = 0 i.e., each connected component C of the partial graph (X,W ) reduces to a single vertex. If not, C is a tree, and from Theorem (2, Ch. 3), C necessarily contains a pendant vertex a. If, for example, the pendant arc is incident into a , then

c

O#

pi=

c

cp;=o,

I E m'(a)

i€m-(.)

which is a contradiction.

Q.E.D. Corollary. A rrecessary and suflcient condition that a rector cp he a J o w is that it is of the form Cp =

$1

p1

+

$2

p2 f ' * '

+

sk

pk 3

where sl,s,, ..., sk E Z and pl,p2, ..., pk are elementary cycles.

Since a cycle is a flow, this linear combination of cycles is a flow. The converse follows immediately from Theorem 3, because graph G may be assumed to be connected (otherwise, each connected component could be considered separately). Q.E.D. In particular, Theorem 3 shows that a cycle p can be obtained by the addi-

91

FLOW PROBLEMS

tion of all the cycles p*whose "out of tree" arc is used by p, In this addition, a cycle is preceded by a sign if the cycle is in the direction of p. Otherwise, the cycle is preceded by a - sign. Hence, Theorem 3 reduces the number of unknowns for the determination of a flow from 112 to m - n + 1.

+

Theorem 4. A necessary and su#icient condition that a rector cp be a flow uith no negatire components is that it is of the form cp = s1 p1

+

where sl,s,, ..., s k E Z,sl, s,, ..., s k

s2

p2

+ + sk p k "*

7

> 0 and pl, p2, ..., p k are circuits.

Clearly, a vector cp of the indicated form is a flow 2 0. Conversely, consider a non-zero flow cp 2 0, and let C be the graph obtained from G by removing all arcs i with 'pi = 0. Graph C contains no cocircuits, and therefore by the arc colouring Lemma, it contains at least one circuit pl. Let s, > 0 be the smallest flow in an arc of pl. The vector cpl = cp - s1 p1 is a flow 2 0 that has more zero.components than cp. If cp' is not a zero vector, repeat this process, etc. ... Finally, a zero flow of the form cpk

= cp - s1 p1 - s2 p2

- ... - s k p k = O ,

is obtained with sl,s2, ..., sk E Z and sl,s2, ..., sk 2 0.

Q.E.D. A tension (or potential difference) is defined to be a vector 8 = O,,J E Z" such that, for each elementary cycle p ,

(el,O,, ...,

This equality can be restated by saying that the .scalar product ( p, 0 ) = p i di is zero, Let 0 denote the set of all tensions. Note that 0 is a submodule of Z", i.e.,

01,02E@ S€Z,0€0

=. *

O1

+02€@,

*SO€@.

Theorem 5. A cector 0 = (01,02, ..., 0,) is a tension and only if, there exists a function t ( x ) defned on the certex set X it.ith ralues in Z such that, f o r each arc i = (a, h), 8, = t(b) - t(a). The function t ( x ) is called a potential attached to tension 0. 1. If 8 is a vector defined by a function t ( x ) , then consider the cycle p = (il, i,,

..., ik)

92

GRAPHS

that successively encounters the vertices a,6, c, Pi,

..., z. Then, we may write

= t(b) - t(a)

Oil

. . . . . .=. . t(4 . . . .-. . 0 . . .) .

Pi,.Bi,

Pi,

ei, = t(a) - t ( z ) .

Adding the above equations yields

c ei- c ei=o.

isp+

isp-

2. It is easy to calculate successively the coefficients t ( x ) for a given tension 8, by the following rules: RULE1. Take any vertex xo, label xo and set 0. RULE2. If x is labelled and y is unlabelled, and if i t(x,) =

= (x, y )

is an arc, put

rb) = t ( x ) + oi .

If i

= (y, x) is

an arc, put

tb)= t

( ~) ei

.

In this way, all the vertices of a connected graph will be labelled. (If the graph is not connected, each connected component can be treated separately.) Each coefficient is uniquely defined by this process. Otherwise, there would exist two chains p1 and p2 from xo to x such that < P ' , e > # ,

and consequently, (pl

- p 2 , e ) # 0.

Since p1 - p2 is a flow, it is a linear combination of elementary cycles by virtue of Theorem 3. Therefore, there exists an elementary cycle p such that ( L O >

+ 0,

which contradicts the definition of a tension.

Q.E.D. Consequence. This theorem clearly shows that a cocycle w(A) is a tension, since we may let t(x) =

Hence, for an arc i = (a, b),

0 1

if if

XEA x$A.

93

FLOW PROBLEMS

I

t(b) - t(a) =

+1 -1 0

if if if

I

~EO+(A) i E o - ( A ) = o,(A) i$o(A)

.

Theorem 6. Let G = (X,U ) be a connected graph. Let H = (A’, V ) be a spanning tree of G with arcs 1, 2, ..., I, let 02,02, ..., m1 be the cocycles associated with H. A tension 8 = ( G , , e2, ..., 6,) is uniquely defined from the values GI, G2, ..., O1 by:

+ e2 m2 + + 8, .

e = el

+.-

The vector 8’ = 0 -

el o1

- e2 w 2 - ... - e1 0

1

is a tension that has zero value on each arc of the tree H. If 8‘ corresponds to a potential t’(x), then

t’(xl) = t‘(x2) = ... = t’(x,,). Consequently, 8’

=

0.

Q.E.D. Corollary. A necessary and suficient condition that a vector 0 be a tension is that it is of the form

e = s1 0 1 + s2 m2 + ..I + sk m k ,

where sl,s2, ...,sk E Z,and d,02, ._., okare elementary cocycles. Clearly, each linear combination of elementary cocycles is a tension, and conversely, each tension is a linearly combination of elementary cocycles, from Theorem 6. Q.E.D. Theorem 6 shows that a tension

e = (el, 02, ..., em), can be determined from A(G)

=

n

- 1 unknowns.

Theorem 7. A necessary and suficient condition thai 8 be a tension 2 0 is that

e = s1 0 1 + s,

0 2

+ -..+

-k,

where sl, s2, ..., s k E Z, sl,s2, ...,s k 2 0 and ol, w2, ..., wk are elementary cocircuits. Clearly, any linear combination of cocircuits with all coefficients 2 0 is a tension 2 0.

94

GRAPHS

Conversely, consider a tension 8 # 0, 8 > 0. We shall show that there exists an elementary cocircuit o1and an s1 > 0 such that the vector

e - s1 o1 has more zero components than the vector 8. Let i = 1 be an arc with el=min(8i/8i#0, 1 ~ i ~ n z ) . Put 0, = s1 > 0. Colour black all arcs i such that 0, > 0. Colour red all arcs i such that 0, = 0. There cannot exist a red and black elementary cycle with all black arcs in the same direction that contains arc 1. Thus, by virtue of the Arc Colouring Lemma, arc 1 is contained in a black elementary cocycle o' with all arcs in the same direction. Thus, o1is a cocircuit, and the vector

6 - s,

o1

has more zero components than the vector 6. This vector is also a tension > 0. If 6 - s1 o1# 0, this reduction process can be repeated until a tension - .-. - s k U k = 0. 8 - s1 d - s2

is obtained.

Q.E.D. Theorem 8. A vector cp E Em is apow if, and only if, it is orthogonal to each vector of 0.A rector 0 E Em is a tension if, and only if, it is orthogonal to each vector of 0. (Hence, 0 and @ are two orthogonal submodules of Em.) 1. We shall show that if cp E @ and

@ € 0,then

i.e., m

21 v i e i = ( c P , =~ )0 .

i=

For each elementary cycle p, (p,Q> From Theorem 3, cp is of the form

Thus,

=o.

6 and cp are orthogonal,

95

FLOW PROBLEMS

2. Let cp be a vector such that (cp, 0 ) = 0 for all 0 E 0.Vector cp is a flow, because if we take 0 = o ( x ) for some vertex x , then

C

2

'pi-

isw+(x)

~ i = ( ~ ( x ) , ~ ) = o -

iEa-(x)

<

3. Let 0 be a vector such that cp, 0 ) = 0 for all cp E CD. Vector 0 is a tension, because for every elementary cycle p, we have (P,Q)

=

0. Q.E.D.

Additional algebraic results for flows and tensions appear in Berge and Ghouila-Houri [1962], and Slepian [1968].

4. The maximum tension problem Consider a graph G whose arcs are denoted by, 1, 2, ..., in. Let kl, k2, ..., k,, 11, I, ..., I,, E Z,such that

-

CL)

< k, < I i < + 0 3 .

Several problems, similar to the above problems can be stated for tensions. The compatible tension problem. For graph G, find a tension 0 such that

ki

< O1 < I,

(i = 1,2,

... , m ) .

The maximum tension problem. For graph G , find (I tension 0 such that (1) k, G e, < zi (i (2) 8, is maxiinum.

=

1,2,

..., m),

EXAMPLE 1. The shortest path problem (Ch. 4, 5 2) is a special case of the maximum tension problem. Let a and b be two vertices of a graph G. Let I(x, y ) denote the length of arc ( x , y). Add to G an arc 1 = (b, a) and let t(x) denote the length I(p[a, X I ) of a shortest elementary path p[a, x] from a to x. For each arc ( x , y ) of G , let Nx, Y) = tw

-t(4

*

Consequently,

- 03 < e(x, v) G ~ ( xY). , We wish to maximize 8(b, a) = t(u) - t(b) = - t(b) , since t(b) must represent the length of the shortest path from a to b.

96

GRAPHS

EXAMPLE 2. Sequencing problems: The construction of a factory requires the performance of various distinct tasks designated by xl,x2,..., x,. Before starting a task x,, it is usually required that another task xi be sufficiently under way. Let k t jdenote the amount of time by which the start of task i must precede the start of taskj. Furthermore, a task xi cannot begin before a fixed time k; and has a known duration kr. Given these constraints, when should each task begin so that the construction project finishes as soon as possible? Consider a graph G whose vertices correspond to the tasks. An arc (xi,x j ) is present if task x j can start only after task xi is sufficiently under way. Add a source a and an arc (a, x) for each task x that cannot begin before a specified time. Finally, add a sink b and an arc (x,b) for each task x. Denote by t ( x ) the starting time of task x. Let t(a) =

0.

We wish to minimize t(b), i.e., to maximize the tension B(b, U ) = t ( ~-) t(b)

in the “return arc” (b, a). The constraints are

kf

< t(xi)

-0

k i j < t ( x j ) - t(xi) k:

< t(b)

+ co = B(xi,xj) < + co = B ( a , x i ) =g

- t ( x i ) = B ( x i , b) =g

$

00

if

(a,xi)eU

if

(xi, x i ) E U

if

(xi, b)

EU

.

The solution of sequencing problems as potential problems has been notably developed by B. Roy [1965]. Lemma. A necessary condition for the existence of a compatible tension is that,for each cycle p)

In fact, if such a tension 0 exists, then for each cycle p

O=(p,0)=(p+,0)-(1l-,0)2

C

ki-

iep+

C

li.

iep-

Q.E.D. The sufficiency of this condition will be demonstrated later. Compatible tension algorithm (J. C. Herz [1967]) Let G be a graph with the numbers ki 6 I, associated with each arc i. For a given tension 0, compatible or not, define the distance of 0, from the interval

97

FLOW PROBLEMS

[ki, lil

by

Pi, lil,

if if if

di(Bi) = ki - 8, I ==OBi - I ,

Bi < k , , 8, > I,.

We shall successively reduce the quantity m

d(e) =

C1 di(eJ

*

i=

If d(8) = 0, the tension 8 is compatible, and the procedure stops. Otherwise, there exists an arc i with d,(ei) > 0; let 1 = @,a) and suppose that

-=

el kl . d,(e,) = k , - o l , To construct a tension 8’ such that d(8’) < d(B), successively label the vertices of G in the following way: RULE1. Label vertex a, the terminal endpoint of arc 1, with the index

+ 1.

RULE2. If x is labelled and y is unlabelled, label y with the index + i if (x,y ) = i and if Bi < ki . RULE3. If x is labelled ar,d y is unlabelled, label y with the icdex - i if

( y , x) = i and if 0, b li .

If vertex b, the initial endpoint of arc 1, cannot be labelled by this procedure, then the tension 8 can be improved. In fact, the set A of labelled vertices satisfies UEA, b#A.

+ 1, and 6’ compatible, because The tension 8’ = 8 - o ( A ) satisfies 0; = i E o + ( A )implies Bi > k, and 0; 2 k,. Similarly, i E o - ( A ) implies 0, < and 0; < l i . If this labelling procedure labels vertex b, then there exists a chain v = [a, a,,

a2,

ee.7

4

9

in which each vertex has been labelled from the preceding one. Thus iEvf i E vP =

< k,, ei 2 I,.

0,

40,b] + [b, a] is a cycle. Since Ol < k l , we have

98

GRAPHS

From the preceding lemma, it follows that no compatible tension exists. The algorithm yields the following result:

Compatible tension theorem (Ghouila-Houri [1960]). Giren a graph G and numbers k, and li where - co < k, < I, < + cc for i = 1,2, ..., m, a necessary and suficient condition that there exists a tension 0 = (el,&, ..., 0,) with k, < Oi < I, for all i, is that for each cycle p,

C+lii e p

C

ki>O*

lop-

The necessary part of the theorem follows from the preceding lemma. The condition is sufficient because, then, each tension 0 with d(0) > 0 can be improved using the preceding algorithm until we obtain tension 8’ with d(8’) = 0. Q.E.D.

Corollary 1 (Roy, [1962]). There exists a tension 8 uith 0, 2 kifor all i, K and only if, for each circuit p,

+ OD for all i. There exists a tension 8 suih that Oi < I, for all i, if, and only i f ,

The proof is achieved by taking 1,

=

Corollary 2. f o r each circuit p,

CI,20. iEl(

Corollary 3. A vector Y

=

( Y , Y , ... Y,) 3

9

9

is called a subtension if there exists a tension 8 such that yi < 9, for all i. A necessary and sufJicient condition that a rector y is a subtension is that (Cp,Y)
((PEG,

cp20).

Furthermore, a subtension y and a flow cp 2 0 satisfv { cp, y ) = 0, if, and only if, y is a tension on the partial graph generated by the arcs i with cpi > 0. 1. If y satisfies the above inequality, then by taking for cp a circuit p, we obtain

99

FLOW PROBLEMS

From Corollary 1, y is therefore a subtension. 2. Conversely, let y be a subtension. If cp E @, cp 2 0, then m

m

Furthermore, equality holds if, and only if, ‘pi

>o

yi =

ei.

Clearly, if ( cp, y ) = 0, then y is a tension on the partial graph generated by the arcs i such that qpi> 0. Finally, if y is a tension on the partial graph generated by the arcs i with ‘pi > 0, then

C

YiVi=O,

Vi’O

and, consequently,

(cP,Y>=

C

W’O

yiqi+

C

yiqi=O*

v1=0

Q.E.D. Maximum tension algorithm We shall construct a tension that maximizes the value 8, of the tension in arc 1 = (b, a). Starting with any compatible tension 8, achieves the labelling procedure for the compatible tension Problem. If vertex.b cannot be labelled, the set A of labelled vertices satisfies a E A , h $ A , and the tension 8’ = 8 - w(A) satis1. Thus, the value of the tension in arc 1 can be improved. fies 0; = 0, If vertex b can be labelled, we shall show that the value 0, of the tension 8 in arc 1 is maximum.

+

For each compatible tension 8 and for each cycle p that uses arc I along its direction, we have

If the algorithm labels vertex b, then there exists a chain v[a,b] such that i e v+ => Oi = k i , i e v* 6, = l i .

100

GRAPHS

Hence, for the cycle p

=

+ [b, a ] , we have

v[a, b]

From inequality (l), we see that 0, has reached its maximum value. Theorem 9. For a graph G with arcs numbers k , and I,, the maximum calue of a compatible tension in arc I is

m a x e l = min e

(i$-

li

C

-

P

k).

iEP+

1 Ep+

iZ1

If 8 is a compatible tension with

it can be improved by using the above algorithm. The theorem follows.

Q.E.D.

EXERCISES 1. Let

.T = (TI, Tz, ..., Tn) be a partition of a finite set X , and let

Y

sz,...,Sm)

= (Sl ,

be a family of subsets of X . Show that if every union of k of the Ti contains at most k of the S, (for k = 1, 2, ..., n), then there exist indices i,, ia, ..., i, such that T t p n S p# 0

p = 1 , 2,...,m .

for

Hint: This result is easily shown by constructing the appropriate transportation network. 2. Let

Y = (TI, Tz,..., Tn) be a partition of a finite set X , and let Y

= (Sl, SZ,

...,Sm)

be a family of subsets, and let c,, ca, ..., cn be positive integers. Show how to construct a set of representatives A = { a 1 , a * ,..., a m } . such that al

E

S,,a2 E Sz, ..., etc., and such that IAnTjISq

Dour

j=1,2,

..., n

101

FLOW PROBLEMS

By reducing this problem to a flow problem, show that such a set of representatives exists if

for all I c { 1, 2, ..., m } and all J

C

{ 1, 2,

..., n}.

3. Let

9-= (Tl, T2, ..., T n ) be a partition of a finite set X , and let

Y = (Sl,S2, ...,Sm) be a family of subsets, and let b l , b z , ..., b, be positive integers. Show how to construct a set of representatives A

with al

E

=

{ a l , a z ,..., a m } ,

S1,u2 E Sz,..., and such that IAnTjIgbj

for

j=l,2

,..., n .

Show that such a set of representatives exists if

for all I c { 1, 2,

..., m } and for all J

c { 1, 2,

..., n }.

4. Use the Compatible Flow Theorem to prove the following!

Let R be a transportation network with source a and sink b. Associated with each source arc i E w + ( u ) are two numbers b, and c, such that 0 < b, < c,. Associated with each sink arc i E w - ( b ) are two numbers b; and c; such that 0 < b; < c;. A necessary and sufficient condition that there exists a flow 'p satisfying b~ < qt < C I bJ < pj < c)

for for

iE

w+(a),

j E w-(b) ,

is that both of the following conditions hold: (1) There exists a flow 'pl with (p:

3 br

pf

< cJ

for for

iEw+(a)

for for

i e w+(a) j e w-(b) .

j E

w-(b).

(2) There exists a flow 'p2with 2

< CI

2

2 b;

qi

(pj

CHAPTER 6

Degrees and Demi-Degrees

1. Existence of a p-graph with given demi-degrees

For a graph G = ( X , U ) , the outer demi-degree d $ ( x ) of a vertex x is defined to be the number of arcs having x as their initial endpoint, i.e.

Similarly, the inner demi-degree d ; ( x ) o f a vertex x to be the number of arcs having x as their terminal endpoint, i.e. &(x) =

c mE(x, Y ) .

Y E X

Finally, we define the degree dG(x) of a vertex x is defined to be the integer d,(x) = d G + ( X )

+ dG(X) .

Thus, a loop at vertex x increases the degree of vertex x by 2.

A p-graph is a graph with m,+(x,y ) 6 p for all vertices x, y. Given integers r l r r 2 , ..., r , , sl,s2,..., s,, we may ask if there exists a p-graph G with vertices xl, x2, ..., xn such that dC+(xk)= rk

(k = 1 , 2 , ..., n)

di(Xk) =

(k = 1, 2 ,

sk

n)

In this case, the pairs (rk, sk) are said to constitute the demi-degrees of a p graph. Theorem 1. Let (rl, sl), (r2,s2), ..., (r,,, s,) be pairs of integers with si 2

The pairs

( r k , sk)

n

s2

2 *.. 2 S " .

constitute the denii-degrees of a p-graph if, and only (fi k

102

103

DEGREES AND DEMI-DEGREES n

Construct a transportation network R with vertices x l , x 2 , ..., x,, ,TI,,T2,

..., X,, and with a source a and a sink b. Join vertices x, and f , by an arc with capacity c(x,, Xj) = p . Join vertices a and x, by an arc with capacity c(a, x i ) = r,. Join vertices 2, and b by an arc with capacity c(F,, b e s,. Any flow that saturates the source and sink arcs of R defines a p-graph (X,U ) having demi-degrees ( r k , s,). Conversely, each p-graph (X,U ) with demi-degrees ( r k ,sk) defines a flow in R that saturates the source and sink arcs. From Theorem (2, Ch. 5), a necessary and sufficient condition for the existence of such a flow is that condition (2) hold and that, for each set B = { Fil, Xiz, ..., Xi* ), the total flow that can enter B is greater than the total demand at B, i.e., -

-

-

F(xi,, xi2, ..., xi*)

-

-

> d ( x i , , x i r , ..., Xi,)

(1

< il

< i2 <

*a.

< ik

< n) .

This is equivalent to n

(1’)

CI min { r i , p k } > si, + si2 + ... + sir

i=

(1 < il < i2

<

< ik

< n) .

-

Clearly (1’) * ( I ) and, since the s, are indexed in decreasing order, (1) (1‘). Q.E.D. For p = 1, these necessary and sufficient conditions can be restated in a simpler form. Let rl > r2 > - * * ’ > r, be a n-tuple of integers. Corresponding to this n-tuple, associate the sequence (r:, r t , ...) where r z denotes the number of.r, that are greater than or equal to the integer k. Hence, r: > 1; 2 r: > * - To visualize the numbers r z , we can construct a diagram, called the Ferrers diagram, as in Fig. 6.1. Hatch the first r l squares in the i-th column in the positive quadrant. Then it is easily seen that rT equals the number of hatched squares in thej-th row in the positive quadrant. By counting, in two different ways, the number of hatched squares, we obtain: n

Cri= Cr:. i= 1

k>l

The sequences (Ti) and (r;) are called conjugates. Using these definitions, several corollaries follow from Theorem 1.

104

GRAPHS

j’k - - -

Fig. 6.1. Ferrers diagram for the sequence (r,) = (5, 4, 2, 2, 1, 1). Here, Cr:) = (6,4,2,2, 1)

Corollary 1. Given n pairs

(Ti,

s1

sJ of integers such that

2

... 2 s,,

s2

there exists a p-graph with d,C(x,) = r i , d;(xt) = s1for i only if,

c

=

1,2,

..., n if,

and

k

r: 2

i>1

c sI

(k = 1 , 2 , ...,n

- 1)

j=1

j=1

By counting in two different ways the number of hatched squares in the first pk rows of the associated Ferrers diagram for (rl , r 2 , ..., r,) we find that It

C min { ri, p k } = i=1

pk i=1

r:

.

Corollary 2 (Ryser [1957]; Gale [1957]). Consider npairs ( T i . such that s1

2

s2

There exists a I-graph G with d$ (xJ k

2 . * - 2 s,. =

ri and &(xJ

= st

if, and only if,

k

r: 2

j= 1

sI

Q.E. D. st) of integers

(k = 1 , 2 , ..., n

- 1)

105

DEGREES AND DEMI-DEGREES

This follows immediately from Corollary 1.

Corollary 3. Giuen a sequence (rl , r 2 , ..., r,,), there exists a 1-graph G with dG+(x,)= r, and d;(xi) = 1for all i if, and only if, t r i = n . i= 1

This condition is necessary because if such a graph Gexists, then n

n

c ri c =

i=l

=

&(Xi)

n

i= 1

Conversely, assume that the condition holds; by Theorem 1, we have only to show that, for all k < n, n

2 min { k, ri } 2 k .

i= 1

If there exists an ri 2 k , the above inequality is satisfied. Otherwise, each rl is less than k, and consequently, n

n

i= 1

i= 1

C min { k , ri } =

C

n 2 k

ri

.

Thus there exists a graph G with the desired properties.

Q.E.D. We shall now present necessary and sufficient conditions for the pairs ( r l , rl), ( r 2 , r2), ..., (rn,r,,) to constitute the demi-degrees of a symmetric pgraph, i.e. a p-graph G with d ( x , Y> = m a y , x>

(x, Y E X )

*

Theorem 2. Let Go be a symmetric graph such that each odd cycle contains a vertex with one or more loops, and let r l , r2, ..., r,, be positive integers. If Go has apartialgraph H with d i ( x i ) = d;;(xi) = rifor each vertex x i , then Go has a symmetric partial graph G with d i ( x i ) = d G ( ~ i= ) ri for all i. We shall assume that the theorem is true for any graph Go of order < n, and we shall show that the theorem is true for a graph Go of order n. Given a graph H, we shall construct from H the symmetric graph G. 1. Suppose first that there exists a vertex xo such that

m&o, Y> = m a y , xo>

(Y EX).

106

GRAPHS

Consider the subgraph Go of order tz - 1 obtained from Go by removing with given demi-degrees Fi = vertex xo. Graph Go has a partial graph ri - mH+(x,,xi), which can be transformed to a symmetric graph G with demi-degrees Ji (by the induction hypothesis). This yields graph G. 2. We may now suppose that for each vertex x of Go,there exists a vertex y adjacent to x with

-

m,+(x,Y> > d ( Y , x> Since d$(x) = d;(x), there exists a vertex z adjacent to x with

m&, z ) < m,+(z,x). Clearly x , y and z are distinct vertices. 3. Choose any vertex xl, and let x2 be any vertex such that

m 2 x l , x2> >

.

m&2

XI)

Let x3 be any vertex such that

, x3) > 4 x 3 xz> In this way, a ,sequence xl,x2, x3, ... is defined and, since the graph is m&2

9

finite, an elementary cycle =

[xp

9

x p + k - l ~x p + k

xp+l,

=

xpl

will be formed. 4. If cycle ,u is even, transform H to H ' by removing an arc between each of the pairs: (xp, x p + l ) ,

(xp+2, xp+3),

(xp+k-2,

as.9

Xp+k-l)

and adding an arc between each of the pairs: (xp+2,

xp+A ( x p + 4 ,

xp+3),

.-.,(Xp+k,

Xp+k-l).

This does not alter the demi-degrees, and produces a graph H ' with

5. If cycle ,u is odd, and if H has a loop at one of its vertices, say x, then transform H into H ' by removing an arc between each pair ( x p , x p ) , ( x p + l , Xp+2)9 ( x p + 3 , x p + 4 ) 9

...) ( x p + k - 2 ,

xp+k-l),

and adding an arc between each pair (XP+l,

x p ) , ( x p + 3 ,x p + 2 ) ,

*",

( X P + k , Xp+k--l)*

This does not alter the demi-degrees, and the graph H ' satisfies (1).

107

DEGREES A N D DEMI-DEGREES

6. Finally, if cycle p is odd and if H has no loops attached to p , then there exists a vertex of p, say x,, that is incident with a loop in Go, Transform H into H ’ by adding an arc between each pair ( x p , xp)? ( x p + 2 3 x p + l > ? ( x p + 4 ,

xp+3),

..., ( X p + k - l ,

Xp+k-2),

and removing an arc between each pair (xp? x p + l > ,

(xp+Z,

xp+3),

.--)( X p + k - l ,

1 Xp+k)*

Again, this does not alter the demi-degrees, and graph H’ satisfies (1). After repeating this procedure a finite number of times, a symmetric graph G is obtained. Q.E.D. Corollary. Given integers rl 2 r2 2 .*.2 r,, a necessary and suflcient condition that there exists a symmetric p-graph with &(xi) = & ( x i ) = ri for all i, is that

The result follows by applying Theorem 2 to the p-graph G,, with vertices x l , x 2 , ..., x, a n d p arcs going from xi to x, for all i and allJ, and then, invoking Corollary 1 to Theorem I . The following consequence of Theorem 1 is used to characterize tournaments.

Theorem 3. (Landau [1953]; Moon [1963]). There exists a coniplefe antisymmetric 1 -graph with outer demi-degrees rl

< r2 < -..< r,,

il; and only il;

I where

I:(

ri = i= 1

(3

denotes the binomial coefficient

P! 4!(P-4)!’

The condition is necessary because, in a complete anti-symmetric I-graph the number of edges joining the vertices of the set { x1 ,x 2 , ..., xk } is less than or equal to the number of arcs leaving the vertices of the set.

108

GRAPHS

The condition is suficient. To prove sufficiency, let si

= (n

- 1) - ri

(i = 1 , 2,

..., n) .

Thus, s1

p

2

s2

2

s3

2

*..

2

s,.

1. First we shall show that the conditions 6f Theorem 1 are satisfied for 1. Note that

= n

n

C sI = n(n - 1) - C ri = n(n - 1) - n(n

- 1) - n(n - 1) =

2

2 Furthermore, by letting rk denote the number of r, that are < k, i=1

i= 1

C ri.

i= 1

Note that, for any integers k and t ,

2[(:)+(!j’-k(r-l)]=r(t-

l)+k(k-l)-2kt+2k =

(k

- t ) 2 + ( k - r) z o

,

Thus,

k

2 k(n - 1) -

C ri = i=1

c k

si

y

i=l

and the conditions of Theorem 1 are satisfied.

2. From Part 1, there exists a 1-graph G & ( x i ) = ri , & ( x i ) = si

=

( X , U ) such that

(i = 1,2,

..., n) .

If this 1-graph G has neither loops nor two oppositely directed arcs joining the same pair of vertices (call such arcs multiple edges), then G is a complete anti-symmetric 1-graph because the number of arcs is case, the theorem has been verified.

xr=lri

=

DEGREES AND DEMI-DEGREES

109

Otherwise, we shall alter G and decrease the number of loops and multiple edges without changing its demi-degrees. Suppose that at vertex xi there are p(xi) loops and q(xi) multiple edges. The number of vertices that are not adjacent to X I is

1

1

(n - 1) - T G ( X J =(n - 1) - p i i-si -q(xJ-2 p(x*)]=q(xJ +2 P(Xi) . Colour red each multiple edge and loop, and insert a green edge between each pair of non-adjacent vertices; then the number of red edges incident to x equals the number of green edges incident to x. Suppose we travel through the coloured edges of the graph without using two edges of the same colour in succession and without using the same edge twice. Then, after arrival a t a vertex x , we can always leave x except perhaps if x is the initial vertex of the tour. In other words, if there exists red edges, then there exists a cycle

bl, Y 2 ,

Y11 > with alternately red and green edges such that **-)

Y2k3

u, 4. u d v 3 , Y 4 ) , dv4, Y 3 ) E u , Oll,YZ),dYZ,Y1)E

02,

Y3), (Y39

Y2)

s....................

@Zk,

Y l ) , @I

9

Y2k)

$

u-

Thus G may be altered by removing arcs ( y 2 , y l ) ,(y4, y3), etc... and by adding arcs ( y 2 ,y3),( y 4 ,y5), ..., ( y Z ky, l ) , without changing any demidegree. This process decreases the number of red edges. It can be repeated until no more red edges are present. Q.E.D.

2. Existence of a p-graph without loops and with given demi-degrees Consider the following problem: Given a graph Go = ( X , U),construct a partial graph H with gitlen demidegrees d; (x) and d; (x). For A c X and B c X,let m& ( A , B ) denote the number of arcs in Go whose initial endpoint is in A and whose terminal endpoint is in B. If an integer ri is associated with each xi E X then, for each A c X , let

Theorem 4. For a graph Gowith certices x1 , x2 , ..., x,, and integers ri, si, for i = 1, 2, ..., n, a necessary and suflcient condition that Go hare a partial subgraph H with

110

GRAPHS

i= 1

i= 1

_ -

Consider a bipartite transportation network R = ( X , X , U ) with vertex s e t s X = { x , , x , ,..., x , , ) a n d X = ( x , , f , ,..., x-,,},withasourceaanda sink 6, and with arcs ( 4 9

Xi)

(a, xi)

(F,,b)

(for 1 (for 1

with capacity mc',(xi,xi) with capacity r, with capacity s, .

< i < n) < j < n)

If condition (2) is satisfied, the desired partial graph H exists if, and only if, network R has a maximum flow that saturates the sink arcs; from Theorem (2, Ch. 5), this is equivalent to

(Ac X),

F(A) a d(A)

(1')

where F ( z ) is the maximum amount of flow that can enter

Z,i.e.

n

W) = i C min { T i , d 0 ( x i , 1)}, = 1 and d ( Z ) is the total demand of set 2,i.e.

x-

d(A) = XJ

E

SJ

=

s(2) .

A

Clearly, condition (1') is equivalent to condition (1).

Q.E.D.

Corollary 1. Given pairs of integers ( r l , s,), ( r 2 , s2), ..., (r,,, sn), a necessary and sufficient condition f o r ;he pairs to constitute the demi-degrees of a p-graph H without loops, is that n

(1)

C1 min { ri, P I A -

i= n

{ x i }

I)

( A c X)

S(A)

n

We apply Theorem 4 to a complete symmetric p-graph Go without loops. Clearly,

I

m&(xt, A ) = P A - { x i

The result follows.

1I

*

Q.E.D.

DEGREES AND DEMI-DEGREES

111

These conditions simplify considerably when rl s1

> r2 2 2 r, > s2 z . * z s,. ***

Consider a sequence ( r l , r 2 , ..., r,) of positive integers such that rl 2 r2

Z

.-*

> r,.

Let Fk denote the number of indices i such that i < I$ and ri 2 k - 1 plus the number of indices i such that i > k and ri z k . The sequence (rk) is called the corrected conjugate of sequence ( r J . The numbers r, can be visualized on the corrected Ferrers diagram (Fig. 6.2), formed by dividing the positive quadrant into three parts: hatched, dotted or empty. All squares on the principal diagonal are dotted. In the i-th column, the first r, squares not on the diagonal are hatched. All other squares are empty.

t

hFig. 6.2. Corrected Ferrers diagram for the sequence (7, 5, 5 , 5 , 5 , 3, 2, 1). The corrected conjugate sequence is (7, 6 , 5, 4, 4, 5 , 1, 1)

Clearly, the number of hatched squares in the k-th row of the diagram equals Fk, and therefore r i = C rk.

1

iZ1

k3l

112

GRAPHS

Corollary 2. Let (r,, si) be pairs of integers with

> > s2 >

2 rn ,

rl 2 r2 s1

***

> s,.

Let (Fi) be the corrected conjugate of ( T i ) . There exists a 1-graph H without loops with d i ( x i ) = ri and d;; ( x i ) = si for all i, and only if, (1) n

C-

Ti=

i31

C si. i= 1

1. If such a I-graph exists, then n

C1 min { ri, 1 A

i=

- {xi } 1 2 s ( 4

(A

= XI.

By taking A = { x l , x z , ..., XI, >,condition (1) follows. Condition (2) obviously holds. 2: Conversely, suppose that conditions (1) and (2) are satisfied. For any set A of cardinality k, n

c min { ri, I A -

i= 1

min{r,,k- 1)

= xl€A

+ C

{Xi

1

> = k

k

min{ri,k)

xiex-A

From Corollary 1, there exists a graph H without loops with dH+(xi)= ri and d;(x,) = si for all i. Q.E.D. We shall now consider necessary and sufficient conditions for the pairs ( r l , rl), ( r 2 , r2), ..., (r,, r,) to constitute the demi-degrees of a symmetric p graph without loops, i.e. a p-graph with: mcf(x, Y> = m a y , XI

(x, Y E X ) .

First, we shall prove the following very general result : Theorem 5 (Fulkerson, Hoffman, McAndrew [1965]). Let Go = ( X , V )be a symmetric graph of order n without loops, such that any two rertex-disjoint elementary cycles of odd length are joined by an edge. Let r l , r z , ..., r, be integers whose sum is eren. If Gohas a partial graph H with d i (xi) = d; ( x i ) = ri for all i, then Go has a symmetric partial graph G with

113

DEGREES AND DEMI-DEGREES

(3) If the numbersf(x, y ) are all even, then the graph G defined by 1 mc+(x,Y > = +,

Y)

is the required graph. Otherwise,-fl(x, y ) determines a non-empty set El c Y 2 ( X )defined by

EI = { [x, YI / 1x3 YI E 9 2 ( f l , f i ( x , Y )

1

(mod. 2)

}*

Let H , = ( X , El) denote the simple graph having El as its edge set. Note that in H , all degrees are even, because

dHI(xk= )

C

fl(x,, y )

= 2 rk

=0

(mod. 2)

.

Y E rH,(xk)

2. We shall now successively construct a sequencef,,f,, ...,f, of functions of two variables that satisfy (l), (2), and (3) and define (as above) sets of edges

E2 2 2 Eq and simple graphs H I , H,, ..., H p . When we encounter a set E, El

2

=

(i.e., a functionf,(x, y ) with even values that satisfy conditions (l), (2), and (3)), we shall obtain the required graph G, defined by

1

Y).

mc+(x, Y ) = #’

3. If the simple graph H1 has an edge, then it has a cycle (because all its degrees are even). Let p denote such a cycle. If cycle p is even, define a function f,(x, y ) , that differs from fi(x,y ) only on the edges of p, by adding alternately 1 and - 1 to f l ( x , y ) while traversing cycle p. Since cycle p is even, f i ( x , y ) also satisfies conditions (l), (2) and (3), and determines a new set of edges E, = El - p. The simple graph H z = ( X , E,)

+

114

GRAPHS

defined in this way has only even degrees. If H1 contains an even cycle p', we can similarly define H3 = (X,E3), etc. Repeat this process until a simple graph H , = (X, E,) without any even cycles has been found. 4. If the graph H, has an edge, it contains a cycle p (because all its degrees are even). This cycle p is necessarily elementary (because otherwise the edges of p would contain an even cycle which contradicts the definition of H,) and odd. Let p =

[a17 a 2 r

a2k+l3 a2k+2

=

'11

be this elementary odd cycle of H,.Then 2k+l

c f,(ai, a i + l )= 1

(mod. 2)

i= 1

and

n

(mod. 2)

= x r i = O

.

i= 1

This shows thatf, has at least one odd value on an edge not in p, and that H , contains an edge not in p. Thus H , has another odd elementary cycle v # p , and v has no common vertices with p (since, otherwise, they would form an even cycle). From the assumptions on Go, these two cycles are joined by an arc of Go. For example, let

I

p = a2, v = Cbl, 6 2 , (a1 Y b , ) E u

a2k+ 1,

.*.,b,+,b l l t

1

Edge [al b,], which is not in H,, satisfies f,(a,, b,)

=0

(mod. 2)

.

,

Iff&, b,) = 0, the function f,+ is obtained by lettingfp+,(a,, b,) = 2, and by adding alternatively - 1 and 1 to the edges [u,, a,], ..., [azk+ a,], and also to the edges [b, b,], ..., [b,t+ b,]. Iffp(al, 6,) # 0, let f,+l(al, b,) = fp(ul,6,) - 2 and add alternately 1 and - 1 to the edges [a,, a,], ..., [ a z k + ,all , and to the edges [b,, b,], ...,

+

+

[b21+17

bll*

Function f,+, also satisfies conditions (l), (2) and (3), and it yields

DEGREES A N D DEMI-DEGREES

115

the simple graph H,+, = (X, E P + , ) ,where E,,, = E, - ( p u v). Since E is finite, we obtain in a finite number of steps a set Eq = 0. Q.E.D. Corollary. Given the integers p 2 I, r1 2 rz >...a r, 2 1 such that condition for the existence of a symmetric p-graph G without loops whose vertices satisfy dC+(x,)= d;(xi) = r.I 2 is that

2ri is ecen, a necessary and suficient n

The proof follows by applying Theorem 5 to a graph Go with vertices x l , x 2 , .. ., x, and with p arcs from xi to x j for each pair (xi,x j ) with xi # x j . 3. Existence of a simple graph with given degrees

This section describes necessary and sufficient conditions for a sequence of integers dl 2 d2 2 ... 2 d, to constitute the degrees of a simple graph. Theorem 6 (Erdos, Gallai [1960]). Let dl 2 d2 2 * * * >d, be a decreasing sequence of n integers with 2 di ecen, and let (di) denote its corrected conjugate sequence. The following conditions are equit'alent: (1) There exists a simple graph G n!hose tiertices x, satisfy dG(xi)= di; k

k

( k = 1 , 2 , ..., n ) ; (3)

C di G k(k - 1) + C i=1

min { k , d, }

( k = 1 , 2 ,..., n ) .

j=k+l

(1) A (2) Condition (1) implies the existence of a I-graph G* without loops such that d,+,(xi)= d,;(xi) = di, for all i; this implies (2), by Corollary 2, Theorem 4. (2) 5 (1) Condition (2) implies the existence of a I-graph H without loops such that d,f(xi) = dG(xi) = d i , from Corollary 2, Theorem 4. Since 2 diis even, this implies from Theorem 5 the existence of a symmetric 1-graph G* without loops such that d&(xi) = &*(xi) = di . (2)

3

(3) Consider the corrected Ferrers diagram for the sequence (di),and denote by xrC the number of empty squares in [0, k ] x [0, k ] ; condition (3) is equivalent to k

(3')

k

116

GRAPHS

Since ctk > 0, condition (2) implies condition (3’), which implies condition (3).

(3)

2

(2) Suppose condition (3’) is satisfied and there exists adinteger k with

We shall show that this results in a contradiction. Clearly k > 1. Since only one square in [0, 11 x [0, 11 is dotted, we have a1 = 0 and, from condition (3’), dl < dl a1 = dl. Let q be the largest integer such that d, 2 k - 1. Then q < k (since ctk > 0). Hence

+

1 di > ic= di= q(k - 1 ) + k

k

1

i= 1

c k

di

+

i=qt 1

c n

di.

i=l+l

Thus n

a

2d i > q ( k -

di.

1)+

i= 1

i=k+l

On the other hand, from condition (3), we have 4

n

i= 1

i=q+ 1

1 di < q(q - 1) + C

min { di, 4 ) G n

L

G q(q

- 1)

+ 1

i=q+ 1

min { di, q }

< q(q - 1) + ( k - 4 ) 4 +

+ 1 i=k+

1

di <

n

c

a

i=k+ 1

Hence a

n

(ii)

Comparing (i) and (ii), we obtain the desired contradiction. Q.E.D. We shall now describe conditions for a sequence (di) to constitute the degrees of various types of simple graph. Theorem 7. Consider two sequences rl 2 r2 rp and s1 > s2 2 with p G q. There exists a simple bipartite graph G = (A’, Y, E ) on the sets > . a * >

... 2 s,,

117

DEGREES AND DEMI-DEGREES

X = { x l ,..., x p }

and

Y = { y l ,..., y , } ,

such that dG(xi)= ri dG(Yj) = sj

(i = 1 , 2,

,..,p ) ,

( j = 192, ...) 4 ) ,

and only i f i

(1 1 4

r: =

C

sj

j= 1

i2t

Clearly, a necessary and sufficient condition that there exists such a graph G is that there exists a I-graph whose demi-degrees are given by the pairs ( 0 , s,), (0. sz), ..., ( 0 , s,) ( r l ,s,+l>,( r z , s q + J 3..., ( r p ,sq+&

where Sqfl

= sq+2 =

..* = s q c p= 0 .

From Corollary 2, Theorem 1, such a l-graph exists if, and only if, we have both L

I,

1 ~ : aC i=

j= 1

1

sj

( k = 1 , 2 ,..., q + p ) , a n d

(2')

These conditions are equivalent to conditions (1) and (2) above.

Q.E.D. Theorem 8. The numbers dl , d 2 , ..., d,, constitute the degrees of a tree i f i and only i f , (1) di 2 1 ( i = 1 , 2 )..., n),

c di n

(2)

=

2(n - 1).

i= 1

From Theorem (19, Ch. 3), these conditions are equivalent to T ( n ; d , , d 2 , ..., d,) # 0 .

Q.E.D. Theorem 9. Let dl

> d2 >...> d,,

be a sequence of integers, n 2 2. A

118

GRAPHS

necessary and sufficient condition for the existence of a simple connected graph G with degrees d,(x,) = d,, is that

>1

d,

(1)

n

2 di is etren

(3)

i= 1 k

(4)

k

CdiG i= 1

Cdi i= 1

( k = 1 , 2 ,..., n ) .

Suppose these conditions are satisfied; then from conditions (3) and (4) there exists a simple graph with the given degrees. From (2), it has at least I I - 1 edges, and therefore, if this graph is not connected, it has a cycle (Theorem I , Ch. 2). Let [x,y ] be an edge of this cycle, and let [a, b] be an edge of a different connected component. This exists, from condition (1). If edges [x,y ] and [a, b] are replaced by two new edges [x, a] and [ y , b],the number o f connected components is reduced without changing any degrees. By repeating this operation as many times as needed, we obtain a connected graph. Concersely, suppose there exists a simple connected graph G with degrees dG(Xi)= d i . Then conditions (l), (3) and (4) are satisfied. Furthermire, if m denotes the number of edges in G, n

i= 1

n

d, =

dG(xi)= 2 m

> 2(n - 1)

i= 1

(since the cyclomatic number satisfies v(G) dition (2) is also satisfied.

=

m -n

+ 1 > 0). Thus conQ.E.D.

Recall that a graph of order n > 2 is defined to be 2-connected if it is connected and if it has no articulation vertex (a vertex whose removal disconnects the graph).

Theorem 10. Let dl 2 d2 d,, be a sequence of integers, n > 2. A necessary and sujicient condition for the existence of a simple 2-connected graph G with degrees dG(xi)= di is that > a * * >

(1)

d, 2 2

119

DEGREES A N D DEMI-DEGREES

It

(3)

C di is ecen i= I

I

k

1. Assume that conditions (I), ( 2 ) , (3) and (4) are satisfied. Then there exists from Theorem 9 a simple connected graph G with d,(x,) = di for all i. If xk is an articulation point, the subgraph G‘ generated by X - { x k } has p’ 2 2 connected components. At least one of these connected components has a cycle, since

y(G’)= m’ -

11’

+ p’ = ni - d,

-

(n - 1)

+ p’ 2

Let [ y , z ] be an edge of this cycle, and let [ t , u] be an edge of another connected component of G’ (which exists from condition (1)). If the edges [ y ,z ] and [t, u] are removed, and two new edges [ y , f ] and [z, 211 are added, the degrees are not altered, but the number of connected components of G‘ is reduced. Furthermore, for each vertex x, this operation does not increase the number of connected components of the subgraph Gx+) (supposing that [ t , u] has been selected in a cycle if its component in G ‘ is not a tree). Repeating this operation as many times as needed, we obtain a graph H such that HX-(,, has only one connected component (for each vertex x). Graph His the required graph. 2. Concersely, the degrees of a 2-connected graph G satisfy the required conhas m - dl edges, n - 1 vertices, and ditions, because Gx-,,,,

Thus

Hence, condition (2) holds. Conditions (l), (3) and (4) clearly hold.

Q.E.D.

120

GRAPHS

EXERCISES 1. For a multigraph G without multiple edges (but possibly with loops) let ~ G ( x = ) d&) if vertex x has no loops. Let 6,(x) = dG(x)- 1 if vertex x has a loop. In other words, a loop increases by 1 (not 2) the “corrected degree” 6&). Show from Theorem 2 that for a sequence dl > dz 3 *.. > d., there exists a multigraph G (without multiple edges) with n vertices xl,x 2 , ...,xn,such that S,(X;) = d; for i = 1, 2, ..., n, if and only if k

k

d: 3 1=1

1dj

(k = I , 2,

...,n)

i=l

(Ramachandra Rao [1969]) 2. Show that the pairs (r,, sl)are the demi-degrees of an arborescence with root x1 if, and

only if, rl = 0

rr = 1

(i # I )

fsc=n--l. I= 1

3. Show that the pairs ( r , , sl)are the demi-degrees of a strongly connected functional graph if, and only if, (i=l,2,

rt=si=l

...,n ) .

4. Show that a sequence d l , d z , ..., d, constitutes the degrees of

if,

2 dl is even.

a multigraph if, and only (Hakimi [1962])

5. A simple graph G is said to be k-edge-connectedif it is not disconnected by the removal of less than k edges. Show that if n > 1 and k > 1, then a sequence d l , d z , ..., d. con-

stitutes the degrees of a k-edge-connected graph if, and only if, (1) the dl are the degrees of a simple graph, (J. Edmonds [1964]) (2) dl > k for all i. 6. Let dl 3 dz p - . B d, be a sequence of integers with 2 dl even. Show that there exists a multigraph G with multiplicityp without loops with vertices xl of degree dG(xI)= dl if, and only if,

14n

Hint:This can be shown from the Corollary to Theorem 5 . 7 . Let dl 2 dz 2 > d,, be a sequence of integers. From its corrected conjugate sequence (a,, dz, a,, ...). Show that di 3 d2 > d, 3-, or that there exists an integer k such that -

-

-

> dr >, ... B db , = dii + 1 dtbz > dktJ > .*. di

.

dnbl 8. Let dl 3 dz B - . be a decreasing sequence, and let (d,) be its corrected conjugate sequence. Consider an integer k s n, and let lo be the largest integer 1 such that the number of hatched or dotted squares in the I-th column of the corrected Ferrers diagram is 3 k.

121

DEGREES AND DEMI-DEGREES

For 1, k k, show that n

di i=l

=

min

=

131. ( i i l

1

d,+hlo-k.

i=lo+ 1

Let ok denote the number of empty squares in the square [O,k] x [0, k] of the corrected Ferrers diagram. For lo < k, show that

i= 1

9. Show that if G is a 3-connected graph with degrees d1

(1)

dk

c

(3)

< d, <-<

d,,, then

33

di is even

i

(4)

(S. B. Rao and A. Ramachandra Rao [I9691 have shown that these necessary conditions are also sufficient for the existence of a 3-connected graph with degrees d,.) 10. Let d, 3 dz 2 ... 3 d, be a sequence of integers with 2 d, even. Show that there exists a multigraph G with multiplicityp without loops with vertices x i of degree d,(x,) = d, if, and only if,

id.
3

i=k+l

m i n { p k , d , ) ( k = 1 , 2,..., n)

(V. Chungphaisan) 11. Show, for the case k = 1, that there exists a simple graph with degree sequence

(dl , .. , d,) containing a regular partial graph of degree k if, and only if, (i) there is a simple graph with degree sequence (dl, _.., d"), (ii) there is a simple graph with degree sequence (d, - k, .... d, - k). (V. Chungphaisan has verified this conjecture of B. Griinbaum and S. Kundu has shown that the above statement is true for all k.)

CHAPTER 7

Matchings

1. The maximum matching problem Given a simple graph G = (X,E ) , a matching is defined to be a set E, of edges such that no two edges of Eo are adjacent. If Eo is a matching, and if El c E,, then E, is also a matching. We shall study the following problem: Finda matching Eo such rhat I E, 1 is maximum. A vertex x is said to be saturated by a matching Eo if an edge of Eo is attached to x. Let S(E,) denote the set of all saturated vertices. A matching that saturates all vertices of G is called a perfect matching. Clearly, a perfect matching is a maximum matching. In this chapter, we shall use dark lines to denote the edges of E,, and light lines to denote the edges of E - E,.

Truncated chessboard

Maximum matching of the corresponding graph Fig. 7.1

EXAMPLE1. Problem of the truncated chessboard. Consider an 8 x 8 chessboard whose upper left and lower right corner squares have been removed. (See Fig. 7.1.) We have 31 dominoes, each domino covering exactly two adjacent squares of the chessboard. Can we cover the 62 squares of the chessboard with the 31 dominoes? This problem is equivalent to finding a maximum matching in a graph whose vertices correspond to the squares of the truncated chessboard. In this 122

123

MATCHINGS

graph, two vertices are adjacent if they represent adjacent squares in the chessboard. (See Fig. 7.1.) It is easy to see that the matching in Fig. 7.1 is not perfect. A simple argument that does not use matching theory shows that no perfect matching is possible : Colour the squares of the chessboard black and white (as usual). Note that the truncated chessboard does not have the same number of black and white squares, since the two missing squares must have the same colour. Clearly, each arrangement of the dominoes covers the same number of black and white squares. Hence, no perfect matching can exist. If the chessboard had the same number of black and white squares, the number of dominoes needed to cover it would be difficult to calculate without using matching theory. EXAMPLE 2. The Battle of Britain. In 1941, the Royal Air Force consisted of planes requiring two pilots. However, certain pilots could not fly together because of language differences or training deficiencies. Given these restrictions, what is the greatest number of planes that can be airborne simultaneously? This problem is solved by finding the maximum matching in a graph whose vertices correspond to the pilots and whose edges join pilots who can fly together. EXAMPLE3. Personnel assignment problem. An office has p secretaries x l , x 2 , ..., x, and q jobs y l , y,, ...,y,. Each secretary is trained to perform at least one job. Is it possible to assign each secretary to a job for which she is qualified? Let T(xi)denote the set of jobs for which secretary xi is qualified. The problem reduces to finding a matching that saturates all the vertices of X in a bipartite graph ( X , Y, r). EXAMPLE 4. Dating problem. In a co-educational American college, each girl has k boy-friends, and each boy has k girl-friends. Is it possible to have a dance in which all students simultaneously dance with one of their friends ? Later, it will be shown that this is possible. Consider a matching Eo. An alternating chain is defined as a simple chain (i.e. a chain that does not use the same edge twice) whose edges are alternately in E,, and in Fo = E - Eo, i.e. alternating dark and light lines. Lemma. Let G = ( X , E ) be a simple graph, and let Eo and El be two matchings in G. Consider the partial graph G ’ with edge set (Eo - E l ) u

- Eo)

Each connected component of C’ is of one of the.foIlowing types: Type I . Isolated vertex. Type 2. Even elementary cycle whose edges are alternately in Eo and E l .

124

GRAPHS

Type 3. Elementary chain whose edges are alternately in Eo and El and whose endpoints are distinct and are both unsaturated in one of the two matchings. Let a E X . We have three cases: CASE1. If a $ S(Eo - E l ) and a $ S(E1 - E,), then a is an isolated vertex. CASE 2. If a E S(Eo - E l ) and a $ S(El - E,,), fhen a is the endpoint of an edge in Eo - E,. No other edge of Eo - El is attached to a (because Eo is a matching); no edge of El - Eo is attached to a (because a @ S(El - E,)). Furthermore, a $ S ( E 1 ) (because, otherwise, an edge of El attached t o a would belong t o El - E,).

CASE2. If a E S ( E , - E l ) and a E S ( E 1 - Eo), there exists a unique edge of Eo - El attached to a and a unique edge of El - E, attached t o a. Since these three cases are exhaustive, the maximum degree of the partial graph (K(Eo - El) (El - EON

"

is 2. This shows that the connected components must be of one of the types described above. Q.E.D. Theorem 1 (Berge [1957]). A matching Eo is maximum iA and on1-v i f , there exists no alternating chain between any two distinct unsaturated mrtices. 1. If E,, is a matching for which there exists an alternating chain between two unsaturated vertices, then by interchanging the dark and light edges 1. along this chain, we obtain a new matching El with I El I = I Eo I Thus, matching Eo was not maximum.

+

2. Suppose matching E,, satisfies the condition of the theorem, and let E, be a maximum matching. From 1, we know that matching El satisfies the condition of the theorem. Thus, I Eo - El I = I El - Eo I since the elementary chains of the partial graph ( X , ( E , - El) u (El - E,)) are necessarily even. Thus l E o l = IE, I ? and, therefore, Eo is a maximum matching.

Q.E.D.

Corollary 1. Let Eo be a maximum matching and consider the alternating chain p = (el, fl, e,, fi, ...), where ei E E, and f i E Fo = E - Eo. Let the

MATCHINGS

125

operation which interchanges dark and light edges in p be called a ‘‘transfer” on

CI. Each maximum matching El can be obtainedfrom Eo by a sequence of transfers along vertex-disjoint alternating chains that are, for the matching Eo, either alternating elementary cycles or alternating elementary even chains starting at an unsaturated vertex. It suffices to make the transfers along the connected components of the partial graph generated by (E, - E l ) U (El - E,), since the connected components are of the above two types by virtue of the lemma. Q.E.D.

Corollary 2. An edge is called “free” if it belongs to a maximum matching but does not belong to all maximum matchings. An edge e is free iL and only i f , for an arbitrary maximum matching E,, edge e belongs to an eren alternating chain beginning at an unsaturated vertex or to an alternating cycle. If e belongs to an alternating chain of this type, then clearly, e is free. Conversely, if e is free, suppose, for example, that e E Eo and e f El for some maximum matching E l . Thus e E ( E , - E l ) u (El - E,) and e belongs to a connected component of the partial graph generated by (E, - E l ) u (El - E,). Hence, for the matching E,, e belongs to an even alternating chain beginning at an unsaturated vertex or to an alternating cycle.

Q.E.D. Theorem 2 (Erdos, Gallai [1959]). The maximum number of edges in a simple graph of order n with a maxiinurn matching of q edges (n 2 2 q > 0) is

Note that in the case n

=

2 q , the graph K Z q ,a clique with 2 q vertices,

is clearly a graph of order n with m

=

(‘2”)

edges and with a maximum

matching of cardinality q. In the second case, the graph formed by the union of a ( 2 q + 1)-clique

126

K,

GRAPHS

+

and a set S, - (,

,

+ 1)

of n - (2 q

+ 1) isolated vertices is clearly a graph

of order n with a maximum matching of cardinality q and with m = (2q2+ l) edges. Finally, in the third case, take a q-clique K, and a stable set Sn-,and join in all possible ways the vertices of Kq with the vertices of S,,-,. Clearly, this graph of order n has a maximum matching of cardinality q since n - q > q and has m =

(3+

q(n - q) edges.

We shall now show that the given numbers represent the maximum possible number of edges. For the first case (n = 2 q), this is evident. Suppose that n22q+l. Let S(Eo)denote the set of unsaturated vertices in- a maximum matching Eo where I Eo I = q. Since n > 2 q , we have S(Eo) # @. Let El denote the edges in matching Eo that have one endpoint adjacent to setleral vertices of S(Eo).By Theorem 1, the other endpoint of such an edge cannot be adjacent to S(Eo),because then there would exist an alternating chain between two distinct unsaturated vertices, and Eo would not be a maximum matching. Let E, = Eo - El,and let q1 = I El I and q, = I E, I. Thus, q1 + q2 = q. For i = 1, 2, let X,denote the set of endpoints of the edges of Ei. Thus,

X, n X, = 0,X, n S ( E , )

=

0,X ,

n S(E,) = @ .

1. Two edges of El cannot generate a 4-clique because then there would be an alternating chain joining two vertices of S(Eo).Thus, the number of edges of Gjoining two vertices of X, satisfies

2. The number of edges of G joining Xl and S(Eo)satisfies

GE,))6 ql(n - 2 q ) .

mG(X1,

3. Let [x,, y,] be an edge of E 2 . If neither x2 nor y , is adjacent to the set X i of vertices of Xl that are non-adjacent to S(Eo),then mG({ x2,

y2 },

x - x2) 6 2 41 + 2 < 3 41 + 2 .

If the edge [x,,y 2 ] has an endpoint x, adjacent to Xi,(see Fig. 7.2), its other endpoint y 2 is not adjacent to S(Eo).Similarly, endpoint y , cannot be adjacent to two vertices of Xi.Thus,

MATCHINGS

n z , ( x z , X - X2) - X2)

m ~ ( ~X 2 ,

< 29, + I , < 9, + 1 ,

and, hence,

'"Gt{ x2, Y2

1, X - X2) < 3 q1 + 2 .

Finally, we obtain

E2 Fig. 7.2

Consequently, m = m,(x,,

x - X2) + ~ G ( X XZ Z, ) + m G ( x 1,S(E,))+ w , ( x , ,

3 q z q I +2q2

If nG7 Hence m

5q+3 7

c

I f n >-5 q + 2 m

(:)

+i212)+

then n - 3 9

ql(n - 2 q ) + ( 2 : l ) - ( y )

+-

2

=

=

+qFGO.

(2 q + 1) . 3 3

+ UOZ

then - 9)

+ (4 - ql)

v j

- n)

x,)

-

1 2ql(q - ql),

128

GRAPHS

Combining the above inequalities for all cases yields,

Note that

is equivalent to

or

Q.E.D. Maximum matching algorithm Consider a simple graph G = ( X , E ) with a matching E,, and associate with it a 1-graph = (A', U ) , where (x, y ) E U if there exists a vertex such that [x, 21 E E - Eo and [z, y ] E E,. For each even alternating chain p in G,

H

Fig. 7.3

129

MATCHINGS

that starts from an unsaturated vertex, there corresponds i n mentary path ,ii For . example in Fig. 7.3, jl

= [ a , b , c , e , d l corresponds to

c,

= [ a , c,

c a unique ele-

4.

For each elementary path ji in there corresponds a unique even chain fl of G, but this chain is not necessarily simple. For example, ,= i[ a , c , d , b] corresponds to p = [ a , b , c , e , d , c , b] (edge bc appears twice). A path in will be called legal if it corresponds to a single chain in G. Otherwise, it will be called illegal. Hence, the matching problem reduces to finding a legal path in that connects an unsaturated vertex (e.g. vertex a) to the neighbours of another unsaturated point (e.g. vertices b or g). One could use known algorithms for finding all elementary paths in (see Ch. 4,§ 1) and then each path of G that is illegal could be eliminated. In Fig. 7.3, the chain ji = [a, c, e, g] yields the desired alternating chain p = [a,6, c, d, e , f , g , h] in G . As noted by Jack Edmonds ([1962], [1965]), it is not necessary to explore each elementary path of starting at vertex a in order to reach an unsaturated vertex by a legal path. Modifications of Edmonds’ algorithm have been suggested (C. Witzgall and C . T. Zahn [1965]; M. Balinski [1970]; B. Roy [1969]).

c

c

c

c

2. The minimum covering problem

Given a simple graph G = (X, E ) , a coaering is defined to be a family F c E such that each vertex x E Xis the endpoint of at least one edge of F. The problem of finding a minimum cardinality covering has many similarities to the maximum matching problem. The covering problem is a more general case of a problem known in logic as “Quine’s Problem”. EXAMPLE. In the fort shown below (Fig. 7.4), there is a tower at the endpoints

Fig. 7.4. Minimum covering of a graph (in dark lines)

130

GRAPHS

of each wa!l. A guard stationed at a wall can watch both towers a t the end of his wall. What is the minimum number of guards needed to watch all the towers? Since the minimum covering of the corresponding graph is 7 edges, it follows that 7 guards will be required.

Theorem 3 (Norman, Rabin [1959]). In a simple graph G = ( X , E ) of order n, a maximum matching Eo and a minimum covering Fo satisfy

I Eo I f I Fo I = n. Giuen a maximum matching E,, a minimum couering

F1 = Eo u { e, 1 Y E S ( E d 1 is obtained by adding to Eo,for each unsaturated certex y , an edge e, of G that is incident to y. Giren a minimum corering F,, a maximum matching El is obtained by remotling successir7elyfrom F, edges that are adjacent to an unremored edge.

If Eo is a maximum matching, the set F~ = E o u ( e , / y E % E o ) ) is clearly a covering, and

IF1 I

=

IEO

I + (n - 2 I E O I )

=n

- IEOI.

Furthermore, if Fo is a minimum covering, the set El obtained by the successive elimination of edges of Fo that are adjacent to an (unremoved) edge of Fo is a matching. Since in G, the edge of Fo do not form chains of length 3, each removed edge creates exactly one unsaturated vertex of E l . Hence,

I Fo I - IE, I = I X - W

l

)

I = n - 2 I El I

and IF01

Since 1 El

1 < 1 Eo 1, IF,

=n-

IEII.

it follows that

I = n - l E o l < n - IE, I

=

I FoI.

Thus, the covering Fl is also a minimum covering. Since I Fl I = I Fo 1, it follows that I El 1 = 1 Eo 1, and, consequently, the matching El is a maximum matching. 1 Fo 1 = n. Finally, 1 Eo 1 Q.E.D.

+

This theorem shows that the minimum covering problem reduces to the maximum matching problem, which we shall study in the next section.

MATCHINGS

131

3. Matchings in bipartite graphs A graph G is said to be bipartite if its vertex set can be partitioned into two classes such that no two adjacent vertices belong to the same class. Theorem 4. For a graph G, the following conditions are equiralent: (1) G is bipartite, (2) G possesses no elementary cycles of odd length, ( 3 ) G possesses no cycles of odd length.

(1) =- (2) because if G is bipartite, we can colour the vertices red and blue such that two adjacent vertices have different colours. If G has an elementary cycle of odd length, then the vertices of the cycle cannot alternate in colour.

(2) 3 (3) Suppose that G possesses no elementary cycles of odd length, but there exists a cycle p = [ x o ,xl, ..., x,, = x o ] of odd length. If there are two vertices xt and xk in cycle p such that j < k and x j = x k , then the cycle can be decomposed into two cycles p [ x , , x k ] and p [ x o ,x j ] + p[Xk, x o ] . Furthermore, one of these cycles has odd length (otherwise, p would have even length). Clearly, each time that the cycle p is decomposed in this way, an odd cycle remains. When this decomposition terminates, there will remain an odd elementary cycle, which contradicts (2).

(3)

3

(1) We shall show that a graph without odd cycles is bipartite. Suppose that the graph is connected (otherwise, each connected component could be considered separately). Successively, colour the vertices using the following rules : RULE1. Colour an arbitrary vertex a blue. RULE2. If vertex x is blue, colour red all vertices adjacent to x. If vertex y is red, colour blue all vertices adjacent to y. Since the graph is connected, each vertex is coloured. A vertex x cannot be coloured both red and blue, since then vertices x and a would be contained in a cycle of odd length. This colouring determines a partition of the vertices into two classes and G is bipartite. Q.E.D.

Henceforth, a bipartite graph with vertex sets X and Y and with edge set E will be denoted by G = (A', Y, E ) . For any A c X u Y, the set of vertices adjacent to set A is denoted by T,(A).

132

GRAPHS

Konig's theorem [1931]. For a bipartite graph G number of edges in a matching equals

-AI

min (I

A c X

=

( X , Y, E), the maximum

+ I r G ( A ) I)

Consider the transportation network with vertices X u Y and a source a and a sink b. Source a is joined to each y j E Y by an arc of capacity c(a, y j ) = 1. Sink b is joined to each xi E X by an arc of capacity c(xl, b) = 1. Finally, y, is joined to xi by an arc of capacity 1 if y, E Tc(x,).

Fig. 7.5

For a set A c X,the total demand of set A equals d ( A ) = I A I. The maximum amount of flow that can be sent into A equals F ( A ) = I T,(A) I. A flow in this network defines a matching in the graph in which xi and y j are matched if a unit of flow traverses arc ( y j ,x,). Conversely, each matching defines a flow. The cardinality of a maximum matching is, therefore, equal to the value of a maximum flow between a and b. By Theorem (2, Ch. 3, max I E ,

I = d ( x ) + min ( F ( A ) - d ( A ) ) = A c X

€0

= IX

I

+ min

(I T,(A) I - I A I ) =

A c X

=

min (I X - A I

+ TG(A)).

A c X

Q.E.D.

133

MATCHINGS

For a graph G, a transversal set T is defined to be a set of vertices such that each edge has at least one endpoint in T.An equivalent formulation of Konig’s Theorem is: Corollary 1. For a bipartite graph G, the maximum number of edges in a matching equals the minimum number of vertices in a transversal set. Let Eo be a maximum matching, and let To be a minimum transversal set. Clearly, I To 1 2 I Eo 1 since To contains at least one endpoint of each edge in Eo. Furthermore, for each A c X , the set T = ( X - A ) u T,(A) is a transversal set of G, and from Konig’s theorem,

I E, I

= min

(I

X -A

A C X

I + I T,(A) I ) 2 I To I .

Hence, I Eo 1 = I To I.

Q.E.D. For a graph G, a stable set S is defined to be a set of vertices such that no edge has two distinct endpoints in S ; another formulation of Konig’s theorem is :

Corollary 2. For a bipartite graph G, the maximum number of vertices in a stable set equals the minimum number of edges in a cotiering. If T c X denotes a transversal set, and if E,, denotes a matching, then, from Corollary 1, max I Eo I = min I T I . T

Eo

If T is a transversal set, its complement S = ( X u Y ) - T is a stable set. If S is a stable set, its complement is a transversal set. Thus, max 1 S

I = IX 1

S

+ I Y I - min 1TI. T

From Theorem 3, min 1 F 1 = I X

I + 1 Y 1 - max I Eo 1 .

F

Eo

Hence, maxIS( =minIFI. S

P

Q.E.D. I f a bipartite graph G = ( X , Y , E ) has a matching that saturates all the vertices of X , then we say that X can be matched into Y. If this matching also saturates all the vertices of Y, we say that X can be matched onto Y.

134

GRAPHS

The following theorem is an easy consequence of the Konig theorem. G

Theorem 5 (P. Hall [1934]; “Konig-Hall Theorem”). In a bipartite graph = (X,Y , E ) , X can be matched into Y if, and only if,

I

rG(A)

I 2 IA I

(A

x)

-

From the Konig Theorem, X can be matched into Y if, and only if,

I x I = max I E,, I = min (1 x - A I + I rG(AjI). A c X

EO

This is equivalent to

Q.E.D. Corollary 1. In a bipartite multigraph G = ( X , Y, E ) with

IXl=p, IYI=q, index the certices x , E X and y j E Y such that

< dG(XJ

dG(x1) < (1IG(xz)< dGb1)

> dG(Yl)

2

**’

9

2 dG(yq) .

A sujicient condition that X can be matched into Y is that q 2 p and dG(xl)> 0 and

i

1-1

k-1

do(&)

’2 4 d Y J 1-1

(k = 2,3, ....PI.

Consider two subsets A c X and B c Y with k and k - 1 elements, respectively. From the above inequality, it follows that k

2

k- 1

dG(Y) =

mG(X,

B, *

YCB

Thus, the number of edges leaving A is strictly greater than the number of edges entering B. Hence, T,(A) C$ B for each set B of k - 1 elements; consequently 1 T,(A) I > k - 1 = I A 1 1. Finally,

-

I rG(A) I 2 I A 1 Therefore,

(A

x>.

X can be matched into Y. Q.E.D.

135

MATCHINGS

Corollary 2. If, in a bipartite multigraph G

mind,(X) 2 max dG(y)

=

and

(X,Y, E ) , we hare

I Y 1 2 IX I ,

Y E Y

XEX

then X can be matched into Y. Let min d G ( x ) = d,, max dG(y) = d2 . YEY

X P X

Thus d, 2 d,, and by indexing the vertices as described above, &(XI)

+ dJx2) + + dG(Xk) 2 kd, > ( k - I)dl 2 (k - I)d2 2 1..

2 dG(Y,)

+ dG(Y2) + ... + dG(Yk-1)

(k

=

2 , 3 , ..., P> *

Therefore, X can be matched into Y. Q.E.D. Corollary 3. I f G = (X,Y , E ) is a bipartite multigraph with no isolated vertices and Hith I Y I 2 I X 1, and such that for some vertex x1 E X,

min d G ( x ) 2 max d,(y) , Y € Y

xfx, xex

then X can be matched into Y. We may suppose that x1 has minimum degree (otherwise, the proof is immediate). Then

+

+ + dG(xk)2 dG(xl)+ ( k - 1) dl =-

d G ( x l ) dG(xz)

> (k - 1) d2 2 d G ( ~ 1 )+ ... + dG0.k-

I).

Hence, X c a n be matched into Y.

Q.E.D. Corollary 4. In a bipartite multigraph G = (X,Y, E ) , there exists a matching that saturates all the certices with maximum degree.

First, suppose that there exists a bipartite multigraph -

a,

G =(X,Y, with G = (X,Y , E ) as a subgraph suppose X c of degree h = max d , ( z ) .

1,Y

c

y, and G is regular

zeXuY

Prom Corollary 2, 1 can be matched onto L in G, since This matching saturates each vertex in G of degree h.

I XI

=

I Fl.

136

GRAPHS

We shall construct this graph Denote these h replicas by

G

= ( X , Y ,E ) ,

c by taking h replicas of multigraph

G.

G’ = ( X ’ , Y‘, E ’ ) , G” = ( X ” , Y ” ,E ” ) , ...

..., ~ ( h - 1 )= ( X ( h - l ) ,

E(h-1)

y(h-1) 7

1.

Let X consist of X U X‘ U--VF h - l ) together with some additional vertices. Similarly, let F consist of Y u Y’ u . . . u Y ( h - l )together with some additional vertices. These additional vertices are determined as follows : If xi E X and &(xi) < h, create, in F, h - &(xi) additional vertices, and join ~ X ~ X”, E ..., xIh-’)€X ( h - l ) , the each of these vertices to xi E X , x l X’, analogues of x i . Repeat this construction if y j E Y and &(y,) < h. In this way, a multigraph having G as a subgraph is constructed. Q.E.D.

c

Note that this result allows us to give an affirmative answer to the Dating Problem (Example 4, 0 1). Corollary 5. In a bipartite graph G = ( X , Y, E ) , X can be matched into Y if, and only i f ,

IX-

fG(B)I


(B c Y ) .

(1) If X can be matched into Y, and if B c Y, then from Theorem 5,

No vertex x E X

- T,(B)

is adjacent to B. Thus, T~(X c )Y - B ,

and, hence,

I X - ~ G ( BI )G I TG(X - TG(B))I < I Y - B I . (2) If X cannot be matched into Y, there exists a set A c X such that > I T,(A) I. Let B = Y - T,(A). Since no vertex of B is adjacent to A ,

IA I

A c X - I-,@). Hence,

IX -

I

~ G ( B ) 2I I A I > TG(A)I = I Y - B I . Q.E.D.

This corollary is in fact a reformulation of the Konig-Hall theorem that will be needed later. The next result is a reformulation of Bernstein’s theorem.

Theorem 6. In a bipartite graph G = ( X , Y, E ) , a necessary and suficient

137

MATCHINGS

condition that rhere exists a matching that simultaneously saturates X and B c Y is that (1) X can be matched into Y, i.e.

I T,(S) I 2 I s I

(S

=X);

(2) B can be matched into X , i.e.

I w-)Ia I T I

( T cB ) .

Clearly, conditions (1) and (2) are necessary. We shall show that if there exists a matching Eo from B into A c X , and that if there exists a matching El from X into Y , then there exists a matching saturating both X and B. We shall now construct from El a matching E; in which the saturated vertices of X remain saturated, and an unsaturated vertex b E Bin El becomes saturated in E ; . Since b E S(Eo),b is the end-point of a chain p of the form: Ab,

ZI =

( e , , e2, e37 e4r

with

...I

e l , e 3 , ... E E, - El , e 2 ,e4, ... E E , - E, .

Fig. 7.6

Suppose that p is as long as possible. Then, the last vertex z of this chain belongs to Y (otherwise, z belongs to A , and z is an endpoint of an edge of El - Eo that can extend chain p). If z E Y , then z $ B (otherwise z is an endpoint of an edge of Eo - El that can extend chain p). Thus, z E Y - B. Thus, E; = El u (p n Eo) - (p n El)is a matching saturating b ; besides, each vertex of X u B that is saturated in El remains saturated in E ; . By repeating this procedure as many times as needed, a matching saturating both X and B can be obtained. Q.E.D.

138

GRAPHS

Corollary. In a bipartite graph G = ( X , Y, E), a necessary and sufJicient condition that there exists a matching simultaneously saturating X and B c Y is that ( S c X) .

min{ITc(S)I,IXI-IB-J;;(S)I)21SI

Using Corollary 5 to Theorem 5, the two conditions of the above theorem can be written as

I TG(S>1 2 I s I

(1)

(2')

IB-rG(S)I
(S

= X)

(S

= XI

or

IxI

(2")

-

3

IB-rG(s)I

Is1

(S c X ) .

Q.E.D. This condition was obtained independently by linear programming methods by Hoffman and Kuhn [1956]. Lemma (Folkman, Fulkerson [1967]). Let G = ( X , Y , E ) be a bipartite multigraph with maximum degree < h and with I E I = m. Let m', mu,h' and h" be positive integers with m' m" = m , h' + h" = h .

+

The edges of G can be partitioned into two classes E' and E" (that form two partial rnultigraphs G' and G"), such that

I E' I = m' , I E" 1 = m" ,

max dG.(z) < h' , Z € X U

Y

max d,..(z)

< h" ,

Z € X U Y

if, and only if,

m' - h' I X - A I - h' m'' - h " I X - A I - h " I for all A c X and for all B c Y.

I Y - B I < mG(A, B ) , Y-Bl<mG(A,B),

1. Necessity. If there exist two partial multigraphs G ' and G " satisfying the above conditions, we have

m' = I E'

I < mG*(X - A , Y ) + mG,(X, Y - B) + mG*(A,B) < h' 1 X - A I + h' I Y - B 1 + rn,(A, B) .

2. Sufficiency. Consider the bipartite transportation network R obtained by joining each xi E X to a source a and by joining each y, E Y to a sink b, and by adding a return arc (b, a). Let ai = max

{ 0, dG(xi)- h" >, Bj = max { 0, dG(yj)- h" } ;

139

MATCHINGS

Let the interval of permitted flow values in arc u be

C%

I

h'l

CP j h'l 9

[b(u),c(u>]= [O, m,(xi, ; Cm', m'l

~ j ) ]

if if if if

u

=

( a , xi)

u = ( y j , b) u = (xi, Y j ) u = (b,a ) .

Each flow in R that is compatible with these intervals determines a partial graph G' with I E' 1 = m' and with

1 < dG.(xi)< h' max { 0, d G ( y j )- h" 1 < d G , ( y j )< h'

max { 0 , d,(xl) - h"

The partial graph G " dGW(si)

=

G - G ' satisfies I E"

=

t i G , , ( ~ ' j )=

t/G(xi)

I = m" and

- d G , ( x i ) < h"

d,(yj) - dG,(yi)

< k"

Thus the required partition of the edges of G has been found. Conversely, such a partition determines in graph R a compatible flow within the permitted intervals. From the Compatible Flow Theorem (Ch. 5 , Q 2), a necessary and sufficient condition for the existence of a compatible flow in R is that

b(o-(~))< c(w+(S))

(S c X u Y u ( a , b

1).

It is left to the reader to verify that the above condition is equivalent to that of the lemma. Q.E.D.

G

Theorem 7. (Dulmage, Mendelsohn [1961]). For a bipartite multigraph = (A', Y, E ) with I E I = m and with maximum degree < h, let

a = m a x ( n ~ - r n , ( A , ~ ) - ( / 1 - 1 ) ( 1 ~ - ~+ I I Y-BI)}, Ac X Bc Y

p = min Ac X Bc Y

{ m G ( A ,B ) + I x - A 1

+ I Y - B I} .

Then,

Furthermore, for each integer m' such that IS < m' < p , there exists a matching E' c E with I E' I = m' whose removal results in a multigraph with maximum degree < h - 1.

140

GRAPHS

1. If

[ 51 > p, then there exist sets A c X and B c Y such that

n? > h(m,(A, B)

+ IX

-A1

+I

Y - B ] )2

~m,(A,B)+m,(X-A,Y)+m,(A, Y - B ) = m , which is a contradiction. Thus,

2. If

[ t] < c,then there exist sets A c X and B c Y such that m -<m-mG(A,B)-(hh

1 ) ( ~ x - - ~ ~ + ~ Y - B ~ ) .

Hence,

m -

-i >in,(A, B ) + ( h in

h - 1 h

2 --m,(A,

B)

1)

(I

X -A

+ -h- (-hh1

1

+ 1 Y - B 1)

IX - A

2

I + h I Y - B I) 2

h -1 > --m, h which is a contradiction. (3) If m' satisfies c < m' have

< p, then, for all A

c

X and for all B c Y, we

2 in - i n & , B ) - ( h - I ) (1 x - A I + I Y - B ~ ~ ~ ' ~ ~ ? , ( A , B ) + I x - A I + I Y - B I .

\ in'

Let h'

=

1, h"

=

I)

h - 1, m" = m - m'. The above inequalities become

m" - h" I X - A I - h" I Y - B I in' - h'l X - A I - h' I Y - B I

< in,(A, B) < in,(A, B ) .

From the lemma, G can be decomposed into two partial bipartite multigraphs G' = (A', E') G" = ( X , E - E ' ) , and with

I E' I = in', max d,.(z) < 1, max dG,,(z)Q h

-

1.

Clearly, E' is the required matching.

Q.E.D.

141

MATCHINOS

4. An extension of the Konig theorem

In this section, we shall consider a multigraph with the properties: (1) there are no loops, (2) if p and p’ are two odd cycles without a common vertex, there exist two adjacent vertices x E p and x’ E p’. Such a graph is called semi-bipartite (see Ch. 6 , Q 2). We shall now study conditions for which a semi-bipartite graph possesses a perfect matching (i.e. a matching that saturates all vertices). Lemma 1. Let G = (X, r)be a symmetric semi-bipartite I-graph with I X I even; there exists a perfect matching i f , and only i f , there exists a partial graph H of G with (x 1 The proof follows from Theorem (5, Ch. 6). dH+(X) =

&(x)

=

EX)

.

Lemma 2. Let G = (X, r) be a 2-graph; there exists a set of elementary circuits of G that partition X i f , and only i f ,

I f(4 I 2 IA I

(A

=

x,

Associate with G a bipartite graph GI = ( X , E ) obtained by taking two replicas X and of set X,and joining xi E X to 2, E by an edge if x, E T(x,). In graph G, set Xcan be partitioned into. circuits if, and only if, the bipartite graph GI has a perfect matching, i.e., if, and only if,

x

I

fG,(4

I = I T(4 I > I A I

x

(A

=XI. Q.E.D.

Theorem 8. A semi-bipartite graph G = (X,E ) possesses a perfect matching i f , and only i f , (1) I X I is even, (2) I rG(A) I A ( A Apply Lemmas 1 and 2 to the symmetric semi-bipartite 1-graph G* obtained from G by replacing each edge by two oppositely directed arcs. Corollary. I f G is a semi-bipartite regular multigraph of degree h that has an ecen number of certices, then G possesses a perfect matching. Make two replicas X and of the vertex set of G, and construct a bipartite multigraph H = (X,1,E ) with m,(x, j ) = mG(x,y). For each A c X,we have

1A1

=

mH

((A,

rH(A))

mH

(x,r H ( A ) )

=

I rG(A) I

*

142

GRAPHS

Thus I f , ( A ) 1 2 I A I and, from Theorem 8, the simple graph G' obtained from G by collecting the multiple edges has'a perfect matching, which is also a perfect matching in G. Q.E.D.

5. Counting perfect matchings For certain graphs (particularly planar graphs) simple methods are available to count the number of distinct matchings. These methods are the result of work done independently by Fisher [1961] and by Kasteleyn [1961]. Kasteleyn's proof has been simplified by RCnyi [1966] and by Pla [1970]. We shall use Pla's proof. First, let us review some definitions from matrix algebra. If A = ((a:))is a square matrix of order n, the determinant of A is written as

where

2 42)

1 (r

=&I)

... ...

is a permutation of degree n, and E(O) = + 1 or - 1 if the permutation is even or odd, respectively. The permanent of A is defined to be the number Perm A = C

a&)

... a&) ,

0

Proposition 1. Let G = (X,U ) be a I-graph with certices x l , x2, ..., x,, and let A = ((a;))be a square matrix of order n defined by a: = 1 i f ( x , , x,) E U, and af = 0 $(xi, x j )6 U. Then the permanent of A equals the number of pairwise disjoint systems of circuits that partition X . Each non-zero term of the expansion of the permanent corresponds to such a system of circuits, and conversely. Q.E.D. A skew-symmetric matrix B = ((bj)) can be associated with an antisymmetric l-graph G = ( X , U ) by letting + 1 if ( x i , x j ) e U , b'. = - 1 if ( x j , x i ) U ~, 0 otherwise. Matrix B is called the adjoint matrix of G. If B = ((bj)) is a skew-symmetric matrix of even order n = 2 k , the pfafian of B is defined to be Pf B = CE(O,) bi: bf: ... b2t-l ,

4

n

143

MATCHINGS

where n = [ i l ,i 2 ] .[i3, ia]...[i2k-l, i z k ] is a permutation of degree n that decomposes into k cycles of length 2 (Le., x(i) = j impliesj # i and x(j) = i ) , such that i l < i3 < < i2k-1 il < i,, i3 < i4, ..., izk,

-=

and where cn is the permutation

Proposition 2. r f G = ( X , U ) is an anti-symmetric I-graph of even order n = 2 k , and if B = ((bj))is its adjoint matrix, then 1 Pf B I is less than or equal to the number of perfect matchings in G . Furthermore, the number of perfect matchings in G equals I Pf B I if, and only if, each term in the expansion of Pf B has the same sign. Each term in the expansion of the pfaffian corresponds to a perfect matching, and conversely. If two non-zero terms have opposite signs, then they cancel out one another. Q.E.D. Note that the pfaffian is easily calculated using the following well-known theorem from linear algebra: Proposition 3. I f B is a square skew-symmetric matrix of order n, then Det B = (Pf B)*

=o

i f n is even , i f n is odd.

The proof can be found in most comprehensive texts on linear algebra. Consider an anti-symmetric 1-graph G = ( X , U )with a perfect matching Wo c U. If p is a cycle of G, let p + denote the arcs of p that are directed in the direction of travel through the cycle, and let I p + I denote the cardinality of this arc set. If, for a family M = (pi / i E I ) of cycles, the numbers I p t I are all odd, G is said to be net1 directed with M . Finally, cycle p is said to be alternable if there exists a perfect matching in which p is an alternating cycle.

Theorem 9 (Kasteleyn [1961]). Let G = ( X , U ) be an anti-symmetric 1-graph of eren order 2 k ; let B be its adjoint matrix, and let W, be a perfect matching in G. The follorving three statements are equivalent: (1) AN the non-zero term in the expansion of Pf B hace the same sign, (2) G is t%3elldirected for the family of its alternable cycles, (3) G is well directed with the family of the alternating cycles for Wo.

144

GRAPHS

If W, is the only perfect matching in G, the result follows since G has no other alternable cycle. Thus we may assume that G has several perfect matchings. (I)

3

(2) For example, let p = [x,, x2, ..., x Z p x,] , be an alternable cycle, and let Wand W' be two perfect matchings i n G such that

-

(W

W') u (W' - W ) = p .

Suppose that the set of arcs common to W and W' is W n W' = {

(XZP+l, X Z p + 2 ) ,

.-.,( X Z k - 1 3

X,d>

*

The terms corresponding to Wand W' in the expansion of Pf B are respectively 1

b43 ... bii-' b;;:; ... b:i-'

8 = E(O)

b2

0'

biP b: b;

and =

~(0')

... bf,Pfi ... b i i - '

,

where

61

=(:.

2 2P

...

3 2

...

2p+ 1 2 p 1

2P '

+

2 p -1

... ...

Note that E(U) ~(0') = + I , because 2 p - 2 transpositions are required to pass from o to d.Thus, from (I),

+1

=

6 0' = - bi b j

... b;;-'

b:" .

Consequently,

Ip'l

E

Ip-I

=1

(mod2),

and for the alternable cycle p, graph G is well directed.

(2) 3 (3) This follows because each alternating cycle of W, is an alternable cycle. (3)

3

(1) Let W be a perfect matching different from W , . From Corollary 1 to Theorem I , a sequence of perfect matchings W, , W , , ..., W, can be constructed so that for each i the arcs of

(Wi- Wi+,>u ( W i + l

- Wi)

define an elementary cycle pi that is alternating in W,. Thus, from (3), I p; I is odd.

145

MATCHINGS

As above, we can show that the terms O(Wo)and O(W) corresponding to matchings W, and W satisfy O(Wo)B(W ) = + 1. Consequently, all nonzero terms in the expansion of Pf B have the same sign. Q.E.D.

Theorem 10. If G = (A’, U ) is an anti-symmetric I-graph, and i f M = (pi1 i E I ) is a family of linearly independent cycles, then, by reversing thl direction of certain arcs, a 1-graph G‘ = (X,U‘) that is well directedfor M can be obtained. Let si = number of arcs of G directed in the direction of travel in p i , s‘ = number of arcs of G directed against the direction of travel in pi. Let = 1 if ujepi ‘ij = o if uj+pi. Finally, let = 1 if the direction of arc u, should remain unchanged to obtain graph G‘, zi = 0 otherwise.

i

Graph G’ will be well directed for M if, and only if. for each i, rn

s1

+ j1 cij z j 3’1 =1

(mod 2).

Hence, G’ is obtained from a system of equations of m variables zl, z 2 , ...,

z, in the field of integers modulo 2: (i E I )

.

j= 1

If family M consists of independent cycles, then this system consists of principal equations and, therefore, it has a solution (zl, z2, ..., z,) in the field of integers modulo 2. Q.E.D. For example, consider the non-planar graph i n Fig. 7.7 with the perfect matching W , shown in dark lines. The alternating cycles are linearly independent since each possesses a distinct edge (marked with a cross in Fig. 7.8). Thus, from Theorem 10, it is possible, by directing the edges, to obtain a graph G = ( X , U ) in which the alternating cycles are well directed. The adjoint matrix of graph G = ( X , U ) is:

146

GRAPHS

0

6

5

1

1

0

4

Fig. 7.7

- 1

1 - 1

1

0

1

-1 0

-1

0

From Theorem 9, the number of distinct perfect matchings is

I Pf B I

= JDi5t B = J 3 . 3 . 2 . 2 = 6

We shall show that this method always works for planar graphs. Theorem 11 (Kasteleyn [1961]). Let G = (A', U ) be a connected planar graph with a perfect matchirig Wo that is well directed for the family of contours of its bounded faces. (From Theorem 10, one such orientation always exists.) Then G is well oriented for the family of its alternating cycles. Let p be an alternating cycle for the perfect matching Wo. Cycle p surrounds an even number of vertices because the vertices in the interior p are matched together by W , . Let H be the subgraph of G generated by vertices situated on p or in the interior of p. Clearly, H i s planar and connected. Suppose H has n vertices, m arcs and f finite faces, v l , v 2 , ..., v,. Successively traverse the cycles v l , v 2 , .. ., v, along their direction and then traverse p against its direction. While doing this, count the total number 5 of arcs traversed along their direction. It is clear that each arc of H will be traversed once in each direction; thus, ( = m.

147

MATCHINGS

Next, by summing over the cycles, we have f

t

=

c I :v

i= 1

I+

(w- I p + I) ,

where I@) is the length of p. Since G is well directed for the cycles vl, we have m=f+Z(p)+Ip+I.

The numbers n, m andfsatisfy the Euler relation for planar graphs (see Corollary 1, Theorem 2, Ch. 2), f = m - n 1. Hence

+

IpfI--m+f+/(p)=n+

= 1 + (n - I(p)) = 1 ,

1 +/(p)

since n - l ( p ) is the number of vertices in the interior of p, which is even. Thus, G is well directed for p. Q.E.D. Remark. T o calculate the number of distinct perfect matchings in a simple planar graph G, it suffices to give the edges of G a suitable anti-symmetric orientation, to determine the adjoint matrix B and to calculate 4 D e t B.

EXERCISES 1. Suppose that G = ( X , E ) is a connected graph without isthmi (an edge whose removal disconnects the graph), and each vertex of G has degree 3. Consider a maximum matching Eo. Show that there exists a chain whose edges belong alternately to Ea and E - Eo that uses exactly once each edge of E - Eo and uses each edge of Eo twice. (P.Medgyessy [1950]) 2. In a bipartite graph G = (A', Y, E), let

6(A)

, A I - I Tc;(A)

1,

60 = max & ( A ) . AcX

Show that

+

(1) &At U A z ) ~ ( A n I Az) > 6 ( A i ) f & A d . (2) Using the above inequality, show that the family

d -{ A / & A ) - - n } satisfies Ai,AzCd

7

Ai

u

A2,

AinA2Ed.

(3) If So > 0, the set of vertices of X not saturated by at least one maximum matching is

3. Consider a bipartite graph G = (A', Y, E ) for which there exists a matching of X into Y. Show that there exists an xo E Xsuch that, for each y E r G ( x o ) ,at least one maximum matching uses the edge [ X O , yl.

148

GRAPHS

Hint: If I TJS)

I > I S I for each

S #

@’,

any xo E X can be chosen. (M. Hall [1948])

4. Deduce from the preceding exercise that if the minimum degree dG(x) for x E X is equal to k. then there exist at least k ! distinct maximum matchings. (M. Hall [1948]) 5. In a bipartite graph (X,Y, r)with I XI = I Y 1, let k = max{ d d x d ddxd * - * 4-d ~ ( x d -ddvd - d&d - *.* - d ~ b - d .]

+

k

+

Show.that there exist k disjoint matchings of X into Y. (0.Ore [1955]) 6. Showfhat in a graph G with n vertices and minimum degree k , a maximum matching V satisfies

I V : 3 min ( k ,

[-;I 1.

(P. Erdos, L. P6sa [19621) 7. Show that if V is a matching and T is a transversal set, then min I TI < 2 max I V I . Show that the equality holds if, and only if, the connected components of the graphs are all cliques of odd cardinality. 8. If k = min do(x), if G is connected, and if max I V I < n--+ 2 , then 3 mini TI < 2max 1 VI - k . (Erdos, Gallai [1961])

9. If a graph remains connected after any k - 1 of its vertices are removed, and if n-I maxl VI < , then k < max I VI,and 2 min I TI < 2max I V I - k (Erdos, Gallai [1961]) Show that this bound is attained by a graph G formed from a k-clique K8 and 1 > k 1 cliques K z , +~I, with each vertex of K,,, + being joined to each vertex of K k . 10. Show that 2 mi n I V I < m + m a x I V l . Also show by an induction on m that the equality holds if, and only if, the connected components of the graph are 2-cliques or 3-cliques. 11. Show that 4 min I TI < 2 n nr - max I V I . If G is a 2-clique, 3-clique, 4-clique or two triangles joined by one edge, then the equality holds. Are there other connected graphs for which this equality holds? (Erdos, Gallai [1961]) 12. Let a tree with diameter < 3 be called a double star. Let f ( G ) denote the minimum number of double stars needed to cover all the edges of a graph G with n vertices. (1) If r is the cardinality of a maximum matching of G, show that f(G)< 2 r . (2) Show thatf(G) < n - 2 r . ~

+

+

(3) Using (1) and (2), show that f(G)< (L. Lovisz [1968])

149

MATCHINCS

13. In a simple graph G = ( X , E ) , consider a family of sets %?= { c1, c2, . . . I c, 1 , with (1) I Ct I odd (i = 1, 2, ...,p ) ( 2 ) For each [ x , y ] E €, there exists an index i such that

.

min { I Ci I, { x , y 1, let c(CJ = max { k, 1 }, and let

I Ci n { x , y } I

If

I C, I < 2 k

+

=

1I} .

P

C(W =

1 C(C0. I=1

Show that for each matching Eo,

1 EOI C min c(W. Y

Also show that max l E o l = m i n c ( a , Eo

v

(J. Edmonds [1964]) 14. Let G = (X,Y, r )be a bipartite 1-graph without isolated vertices (where Tis acorrespondence from X onto Y ) . Show that the minimum number of arcs that must be added to G to make a strongly connected graph is max { I X 1, I Y I }. Hint: Use Corollary 1 to the Konig theorem.

CHAPTER 8

c -Matchings

1. The maximum c-matching problem

Consider a multigraph G = ( X , E ) with vertices xl, x z , ..., x, and a ntuple c = (cl, c2, ..., c,) of integers with ( i = 1, 2 , ..., n). 0 < ci < &(xi) The set E0 c E is called a c-matching if for each i, the set Eo(x,) of edges of Eo incident to x, satisfies

I EO(Xi) I G ci

'

A vertex x, is said to be saturated in the c-matching Eo if I Eo(x,)1 = c,. In this section, we shall study the problem of constructing a maximum c-matching. The maximum matching problem (Chap. 7) is a special case of the maximum c-matching problem for c = (1, 1, ..., 1). To each multigraph G, there corresponds a simple graph G defined as follows: For each xi E X , define two disjoint sets A i = ( a f / e E E ( x i ) } and B i = ( b : / k = 1 , 2 , . . . , d G ( x i ) - c i } Let the vertex set of G be the union of U;=l At and of U;=l B,. For each i, join each vertex of A , to each vertex of B , . For each edge c = [ x , ,xi] of G , construct an edge 2 = [a;, a;] in Theorem 1. A maximum matching i?, in graph that saturates U;=l B, induces a maximum c-matching Eo in graph G,and comersely. 1. Let Eobe a maximum matching of G. We may assume that Eo saturates each b: (by interchanging if necessary the edges of Eo and of E - Eo along a chain [b:, a;, u!] of length 2). Matching Eo in G defines a set of edges E, in G, and this set Eo satisfy

c.

I

I

€,(xi) < dG(xJ - I Bi I = dG(xi) - (dG(xi) Thus, Eo is a c-matching in G ; furthermore,

Ci)

= ci

151

C-MATCHINOS

2. Now consider a maximum c-matching El c E of graph G. Set El defines in G a matching Elthat saturates each b:, as shown in Fig. 8.1.

-E - Eo

b Fig. 8.1

Consequently,

Since Eo is a c-matching in G, I Eo I

< I El 1,

and therefore

Since Eo is a maximum matching, 1 El1 = I ED1. Consequently, I El I = Hence, Eo is a maximum c-matching of G, and El is a maximum matching of G. Q.E.D.

I Eo I.

Remark. This theorem demonstrates that a maximum c-matching can be constructed by determining the maximum matching Eo in graph G and then saturating each vertex b: by an interchange along alternating chains of length 2. Consider a multigraph G with a c-matching Eo. From G, construct a multigraph R(Eo) by adding to G a vertex xo called the origin that is joined to each vertex xi by cI edges, for all i. Multigraph R(Eo)is called a transfer net work.

152

GRAPHS

Let the edges of Eo be represented by dark lines, and let ci - I Eo(xi)I edges from xo to xi for all i be also represented by dark lines. All other edges of R(Eo)are represented by light lines. An alternating chain is defined as a chain of R(Eo)with edges alternately dark and light that repeats no edge. A transfer along an alternating chain p is defiiled as the interchange of the dark and light colouring along p. Theorem 2. (Berge [1958]). A c-matching Eo of a multigraph G = ( X , E ) is maximum if, and only if, there exists no alternating chain in R(Eo) that joins xo to itself and has dark initial and terminal edges. As in Theorem 1, construct a graph G corresponding to the multigraph G and a matching Eo in G corresponding to Eo that saturates each vertex in (J Bi. An alternating chain with the above properties in R(Eo)corresponds in G to an alternating chain that connects two distinct unsaturated vertices (if Eo is properly chosen), and vice versa. In graph G, such a chain exists if, and only if, the matching Eo in G is not maximum (Chapter 7, Theorem l), or, from Theorem 1, if, and only if, the c-matching Eo in G is not maximum. Q.E.D. Theorem 3. I f Eo and El are two maximum c-matchings of a multigraph G = ( X , E), then El can be obtained from Eo by transfers along alternating cycles of R(Eo)that are pairwise edge-disjoint (but not necessarily elementary). From Corollary 1 to Theorem 1 (Ch. 7), each maximum matching El in G can be obtained from the maximum matching Eo by a series of transfers along vertex disjoint alternating chains. Each of these chains is either an alternating elementary cycle or an even elementary chain starting at an unsaturated vertex. In either case, this alternating chain corresponds in R(Eo)to an alternating cycle that is not necessarily elementary. Q.E.D. Theorem 4. A “free edge” is defined to be any edge of G = ( X , E ) that is contained in some maximum c-matching but not in every maximum c-matching. An edge e is,free if, and only if, gicen a maximum c-matching Eo of G, e lies on an alternating cycle of R(Eo).

1. If e is contained in an alternating cycle of R(Eo),then e is evidently a free edge, since a transfer can be made along this alternating cycle. 2. Let e be a free edge, and suppose at first that e is contained in the maximum c-matching Eo. There exists a maximum c-matching El that does not contain e. Thus, from Theorem 3, e is contained in an alternating cycle of W O ) .

If e is not contained in Eo,then e is contained in a maximum c-matching

153

C-M ATCHINGS

El obtained from Eo by a transfer along an alternating cycle of R(ED).Again, e is contained in an alternating cycle of R(Eo). Q.E.D. 2. Transfers In this section we shall consider a simple graph G and show how to obtain from G all the graphs with the same degrees as G by a sequence of transfers of a particular form. Lemma. If G = ( X , E ) is a simple graph such that each even cycle of length > 4 has a chord that divides the cycle into two euen cycles. Let E, c E be a set of dark edges. Then any transfer along an alternating cycle p can be obtained by a sequence of transfers along alternating cycles of length 4.

If the length of cycle p equals 4, the result is trivial. If the result is true for cycles of length < 2 k , then we shall show that it is also true for an alternating cycle p of length 2 k , say c1 = [a,, a27

**.,

a11

02k,

-

Since this cycle has a chord of the form [a,, a,,,,,,], cycles : Pl =

bi,

~ 1 2=

[a,,

ai+1, **., ai+2p+1,arl

we have two even

Y

ai, ai+2p+1,a i + 2 p + 2 ,

e a . 3

a2kr

a13

*

Relative to E D ,only one of these two cycles is alternating, say p l . Since the length of p 1 is < 2 k, the transfer along p l can be accomplished by a sequence of transfers along alternating cycles of length 4: Eb = Eo - (eel n E,) u (p1 - E,) . E, Relative to ED,cycle p2 is an alternating cycle of length < 2 k. The transfer -+

-+ ES = EL - ( p 2 n Eb) u ( p 2 - EA) Eb can be accomplished by a sequence of transfers along alternating cycles of length 4. Thus, the transfer

E,

+

Eg = E ,

- (p n E,)

u (p - E,)

has been obtained. Q.E.D. Theorem 5. Let HD= ( X , Eo) and H I = ( X , El) be two simple graphs, with the same vertex set, such that

dHo ( x i ) = dH1( x i ) = d,

(i = 1, 2, ... , n)

.

154

GRAPHS

Let a, 6, C, d be any four rertices in X with ac E E,, bd E Eo, ad 4 Eo and bc 4 Eo. A “direct transfer on H,” is defined as an operation that removes edges ac and bd and adds edges ad and bc. Then, graph H 1 can be obtained from graph Ho by a sequence of direct transfers.

Clearly, Ho and H I are two partial graphs of a complete graph of the form G = ( X , E). In G each even cycle [ a l ,a,, ..., a,] has a chord of the form [a,,a, + 31. From the above lemma and from Theorem 3, Ho can be transformed into fZl by transfers along alternating cycles of length 4. These transfers are direct transfers. Q.E.D.

Theorem 6. Let Ho graphs such that

=

( X , Y , E,) and H ,

=

( X , Y, El)be two bipartite 1, 2,

&&xi) = d H , ( x i )= ri

(i

dHo(Yj) = dH,(yj) = s j

( j = 192,

=

Let xl, x , E X be two distinct certices, and let y,, y , vertices with

x,Y,EEo,

X*Y2EEo,

XIY24E0,

E

... , p ) ,

..., 4) Y be two distinct

X2Yl$EO.

A “bipartite transfer on H,” is dejned to be an operation that remores edges x1 y , and x2y , and adds edges x1 y , and x2y , Graph H1can be obtained from graph Ho by a sequence of bipartite transfers.

.

Clearly, H o and H I are partial graphs of a complete bipartite graph of the form G = (X,Y, E ) in which each x E X is joined to each y E Y. From the lemma and from Theorem 3, Ho can be transformed into H I by a sequence of transfers along alternating cycles of length 4. These transfers are bipartite transfers. Q.E.D.

Theorem 7. Let G

=

(X,U ) and G‘

=

( X , U ) be I-graphs such that

dG+(xi) = dG+,(xi)= ri

(i = I , 2, ..., n) .

&(xi) = d&(xi) = si

(i = 1, 2, ..., n ) .

Let a, b, x, y be vertices of X with a # 6, x # y , (a, x) E U, (6, y ) E U, (a, y ) 4 U, (b, x) 4 U. An “oriented transfer” is dejined to be an operation that removes arcs (a, x) and (6,y ) and adds arcs (a, y ) and (b, x). Graph G’ can be obtained from Graph G by a sequence of oriented transfers.

C-MATCHINGS

155

Graph G = (X,U ) corresponds to a bipartite graph H = ( X , X,E ) where [ x i ,Z,] E E if, and only if, ( x i , x,) E U. The proof follows when Theorem 6 is applied to the bipartite graph H. Q.E.D. Remurk. Consider a 1-graph G such that d,+(x) = d;(x) for each vertex x . Such a graph defines a permutation of degree n, and vice versa. Thus, Theorem 7 generalizes the well known theorem from algebra: Ecery permutation is a product of transpositions.

3. Maximum cardinality of a c-matching Consider a multigraph G = (A', E ) with a c-matching Eo c E, and the corresponding multigraph R(Eo)with dark and light edges (see Section 1). Recall that the origin xo is joined to some of the vertices of X by dark and light lines, the edges of Eo are dark, and the edges of Fo = E - Eo are light. Consider the chains starting at the xo and beginning with a dark edge. If there exists an alternating chain ,u starting at xo and going to X E X , then orient each edge in ,u toward x. It is possible that an edge will be given two opposite directions. If vertex x is the endpoint of a dark edge oriented toward x and is not the endpoint of any light edge oriented toward x , then vertex x is called dark. The set of all dark vertices is denoted by Xd. If vertex x is the endpoint of a light edge directed toward x and not the endpoint of any dark edge directed toward x , then vertex is called light. The set of all light vertices is denoted by X', If a vertex x is the endpoint of a light edge directed toward x and also the endpoint of a dark edge directed toward x , then x is called a mixed vertex. The set of all mixed vertices is denoted by X".Finally, if a vertex x is not the endpoint of any edge directed toward x, then vertex x is called inaccessible. The set of all inaccessible vertices is denoted by X'. Vertex xo, the origin of the network R(Eo), is so far unclassified: it will be defined to be right if there exist no alternating chains from xo to xo with dark edges at each endpoint; otherwise, xo is defined to be mixed. Each vertex of R(E,) is either light, dark, mixed or inaccessible. Theorem 8. Two dark certices can only be joined by a dark edge. Two light certices can only be joined by a light edge. A mixed i3ertex and an inaccessible oertex cannot be joined by an edge. A dark uertex and an inaccessible vertex can only be joined by a dark edge. A light uertex and an inaccessible vertex can only be joined by a light edge.

This follows immediately from the definitions.

156

GRAPHS

These results are summarized below : TvDe of vertex

I

Liaht

Dark

Mixed

I

Dark edge

Light

Inaccessible

I

Dark edge

Light edge

Mixed Inaccessible

Dark edge

Light edge

Consider the subgraph of R(Eo)generated by the set Xm.Denote the connected components of this graph by M , , M 2 ,.... Chain p is said to enter M I via edge [a, b] if p is of the form p = [ x o , a,, a 2 ,..., ak = a ,

6 , = b , b 2 , ..., b,]

with xo, a l , a,,

..., ak # M 1

bi,

..., b l E M 1 .

b2, b 3 ,

Lemma 1. Let x o 6 M I and x E M I . If an alternating chain p[xo,x]enters M I via an edge [a, b] and ferminafes with a dark (respecticely, light) edge, then there exists an alternating chain p'[xo, x ] entering M I ria [a, b] and terminating with a light (respecticely, dark) edge.

Fig. 8.2

If x is a mixed vertex, then there exists an alternating chain v [ x o ,x ] that terminates with a light edge. Since xo 6 M 1 and x E M I , there exist in v [ x o ,x ] edges with one endpoint in X - M I and the other endpoint in M1. Let [c, d] be the last edge of this type in v [ x o ,XI.

C-MATCHINGS

157

Finally, let y be the first vertex of ,u that is on v[d,x ] (such a vertex always exists). If p[b,y ] and v[d,y ] terminate with edges of the same type, the theorem is true because of the alternating chain : P'

= P[X,,

Yl

+vb,XI.

I f p[b,y ] and v[d,y ] terminate with edges of different types, then [a, 61 = [c, dl since, otherwise, p[x0,vl + V [ Y ,c ] would be a simple alternating chain, and c would be mixed, which contradicts c $ M,. Thus, the chain P' = P h , bl

+ v[d, X I

is alternating and satisfies the requirements of the theorem. Q.E.D.

Lemma 2. Let xo6 M , , x E M1 ; let [a, b] be an edge incident to M1 and directed into M I . There exists an alternating chain p [ x o ,x ] from xo to x that enters M , via edge [a, 61. Let Y be the set of vertices of M I that are accessible by an alternating chain entering via [a, b]. Let Z be the set of other vertices of M , . Since b E Y, we have Y # @. Suppose that Z # 0, then there exists an edge [ y , 21 with y E Y and z E Z. We shall show that this leads to a contradiction. From Lemma 1, y is accessible by two alternating chains ,ul[xo,y ] and ,uz[xo,y ] entering M , via edge [a, b] and terminating respectively with a dark and light edge. Thus, z is accessible to an alternating chain entering via [a, 61. This contradicts z E Z. Q.E.D.

Theorem 9. (Gallai [1950]).Let Eo be a set of dark edges in G,. Let M I be a component of the subgraph of R(Eo)generated by the mixed vertices. r f x o $ M , , there exists exactly one edge incident to M I and directed into M1. rfxo E M I , there exists no edge incident to M I and directed into M I .

1. If xo $ M I , let p be an alternating chain going from xo to M,. Let b be the first vertex of p in M , . The alternating chain p [ x o ,b] enters M , via an edge [a, b] that is incident to MI and is directed into M I . Let [c, d ] be an edge other than [a,b] that is also incident to M 1 and directed into M , . From Lemma 2, we know that d E M 1 is accessible by an alternating chain entering via [a, 61; from Lemma 1, we may assume that this chain terminates with an edge of a type different from [c, d ] . Thus c is a mixed vertex, which contradicts c $ MI. Thus [a, 61 is the only edge that is directed into M,.

158

GRAPHS

2. I f xo E M , , then we can return to the preceding case simply by adding vertices a, and b,, a dark edge [a,, b,] and a light edge [b,, x,]. Let a, replace xo as the origin. The only edge entering M 1 is [b,, xo].Thus there exists no edge of the original graph that enters M , . Q.E.D.

Theorem 10. Let Eo be a maximum matching, and let M , be a component of the subgraph generated by the mixed vertices in the multigraph R(E,). There is a dark edge incident to M 1 and directed exclusively into M,. All other edges incident to M , are light and exclusively directed out of M I . 1. Since the matching Eo is maximum, x, is light, and x , $ M , . From

Theorem 9, there exists exactly one edge directed into M,. Denote this edge by [c, X I , where c $ M I , x E M I . Suppose [c, x ] is light: since x is mixed, there exists an alternating chain p from xo to x that terminates with a dark edge. Chain p necessarily uses edge [c, x ] (which is the only possible entrance edge to Ml), followed by a dark edge incident to x and terminates with another dark edge incident to x. But this is impossible, since there is only one dark edge incident to x. Thus, edge [c, x ] is dark. 2. Let [ y , b] be another edge incident to M , with y E M 1 and b 6 M,. From Theorem 9, [ y , b] is necessarily directed out of M , . Furthermore, [ y , b] cannot be dark because y, being a mixed vertex, can be reached by an alternating chain terminating with a dark edge. Q.E.D.

Corollary 1. If Eo is a maximum matching, and if M 1 is a component of the subgraph generated by X", then I M , I 2 3 and 1 M , I is odd. Let [c, x ] be a dark edge incident to M1 with x E M,. Then, I M 1 1 2 3, because, otherwise, x cannot be reached by an alternating chain terminating with a light edge. Furthermore, since M1 consists of vertex x and pairs o f vertices joined by dark edges, I M 1 I is odd. Q.E.D.

Corollary 2. I f Eo is a maximum matching, each non-mixed certex that is adjacent to a mixed vertex is a light vertex. I f [c,x ] is a dark edge incident to a component M , with c 6 M1, and M , , then c $ X i , c $ X d and c $ X". Consequently, c E X ' . If [a, b] is a light edge incident to M 1 with a $ M , , b E M,, then similarly, a 4 X', a $ Xd,and a $ X". Thus, a E XI. Q.E.D.

x

E

159

C-MATCHINGS

Corollary 3. vertex is light.

Eo is a maximum matching, each vertex adjacent to a dark

Let a E X d . If [x,a] is a dark edge, then x 4 X i(since x is the endpoint of a directed edge); x 4 X d (because, otherwise, a would be mixed), and x 4 Xm (from Corollary 2). Thus x E X'. If [x,a ] is a light edge, then x 4 X i , x $ X d , and x 4 X" (from Corollary 2). Thus, x E X'. Q.E.D. Theorem 11. Let Eo be a maximum matching and let Il be a connected component of the subgraph generated by the inaccessible vertices. Each edge incident to I, is light and undirected. Each vertex x $ I , that is adjacent to a vertex of Il is light. rfgraph G has no isolated vertices, then I I, I is even and 2 2. 1. If [c, x ] is an edge incident to Il with x E I,, then c $ Xm,c 4 Xi,and c 6 X d (from Corollary 3). Thus, c E X', and [c, x] is right. 2. Since Z, does not contain xo nor unsaturated vertices, it contains only pairs of vertices joined by dark edges. Thus, 1 Il I is even and 2 2. Q.E.D.

Theorem 12. (Berge, [1958]).Given a siniple connected graph G and a subset S of the vertices, let p,(S) denote the number of components of odd order in the subgraph generated by X - S. The number of unsaturated vertices in a maximum matching is

5

= max (pi(S) sc

x

-I S

I)

*

1. Consider a set S c X , and let C1, C,, ..., C, be the components of odd order in the subgraph generated by X - S. If a component ck has no unsaturated vertices, there is at least one dark edge going from c k to a vertex s k e S because I c k I is odd. Two distinct components c k correspond to two distinct vertices sk. Thus, no, the number of unsaturated vertices, satisfies

p,(S) - no < (number of

c k

without unsaturated vertices)

< 1 S 1,

Hence, Pi(S)

- I S I G no

( S c x)

2. We shall show that S can be chosen so that p,(S) - S = no (this establishes the theorem). Let Eo be a maximum matching, and let X' be the set of light vertices. From Theorem 10 and Corollaries 1, 2, and 3, pi(X') = (number of components in G y )

+ I Xd I .

160

GRAPHS

Furthermore, the dark edges of G define a bijection between the set X' and the components M , that contains no unsaturated vertices and no dark saturated vertices. Thus

I X 1I

= (number of components in

Gp) + 1 X d I - no.

Hence

no = p i ( X r ) - I X'

I.

Q.E.D. Corollary 1. In a simple connected graph G = ( X , E ) of order n, the number of edges in a maximum matching equals 1

- (n

2

-0,

where

The proof is immediate from Theorem 12.

Corollary 2. (Tutte [1947]). A necessary and suficient condition for a connected graph to possess a perfect matching is that PAS) G I S I The proof is immediate.

(S c X )

The following result generalizes the Petersen theorem [18911 for regular graphs of degree 3. Theorem 13. Let G = ( X , E ) be a connected multigraph that is regular of degree h, with an etlen number of vertices, without loops, and with (ScX; S#@,X). mG(S,X-S)>h-l Then G possesses a perfect matching. Furthermore, each edge of G is free (i.e. each edge of G belongs to at least one perfect matching but does not belong to all perfect matchings). 1. By using Corollary 2, we shall show that there exists a perfect matching. Let S be a non-empty subset of X , where S # X , and let C,,C2,... be the connected components of odd order in the subgraph generated by X - S. By hypothesis, m,(S, C,) 2 h - 1. However, m,(S, C,)> h - 1 , because, otherwise, the number of edges in C, would equal 1

1

mG(Cl,C , ) = 2 [ h I C, I - ( h - l)] = 2[h(l C , I - 1)

+ 11

161

C-MATCHINGS

which is not an integer, since I C, I is odd. Thus, mG(S,

c1) > h

Hence,

h

IsI2

mG(S,

x-s) 2 m,(s, c1 u c,?u "*) = 1 mG(S, ck) 2 hp,(S). k

Thus, pi(S) < I S 1 for all S # X, S # 0. This inequality remains valid when S = X because p i ( X ) = 0, or when S = 0 because p i ( @ ) = 0 (since the graph has an even number of vertices and is connected). Thus, from Corollary 2, there exists a perfect matching Eo.

2. To show that.an edge e E E is free, it is sufficient from Theorem 4 to show that e is on an alternating cycle relative to a perfect matching Eo. We may assume that e E E - Eo (because if e E Eo and if some edge of E - Eo that is adjacent to e is on an alternating cycle, then edge e will also be contained in an alternating cycle). We shall suppose that e is a light edge of G that does not belong to any alternating cycle, and we shall show that this leads to a contradiction. Contract the endpoints of edge e into a single vertex xo and define a transfer network R(Eo) with origin xo. Since edge e appears in no alternating cycle, vertex xo is light. Assume that there exist mixed vertices; let C , be a connected component of the subgraph of R(Eo) generated by the mixed vertices. Clearly, xo 6 C,. Since I C, I is odd from Corollary 1 to Theorem 10, we know from Part 1 of this proof that mG(X

- c1

9

c1)

>, h

a

Furthermore, from Theorem 10, there is a single dark edge of R(Eo) that leaves C , , and thus there are at least h - I light edges that leave C , . Now consider the network R(Eo)obtained from R(E0)by contracting each component C, of mixed vertices. Thus, C , becomes a dark vertex c,, and no edge has been oriented in two directions. From Theorem 10, c1 is incident to a dark edge directed into c1 and to light edges each directed out of cl. Let X d and 8' respectively denote the sets of dark vertices and light vertices in network R(Eo).Let d:(x) denote the number of light edges of R(Eo) directed out of x. Consequently, d ; ( x ) >, h - 1

(x E 2 ) .

Besides, the number d;(x) of light edges directed into x satisfies d;(x)

< d,(x)

- 1=

h

-1

(X

EX' - { x0 }) .

162

GRAPHS

Since the theorem is obvious for h

=

1, we may assume that h > 1 ; thus,

By counting in two different ways the number of dark edges of R(Eo) that have exactly one direction, we obtain Since we have obtained two incompatible relations, the proof is achieved. Q.E.D. Corollary (Errera [1922]). I f G = (X, E ) i; a simple connectedgraph regular of degree 3 such that all the isthmi of G are on the same elementary chain, then G possesses a perfect matching. Recall that an “isthmus” is an edge [a, b] whose removal disconnects the graph. In a connected graph, the removal of an isthmus creates exactly two Connected components. Since 3 I X I = 2 1 E 1, the number of vertices of G is even. Consider the different connected components created by the removal of all of the isthmi. From the hypothesis, each of these connected components is joined to the rest of the graph G by one or two isthmi. Suppose a connected component Ci is joined to the rest of the graph by two isthmi [x, c] and [y, c’], where c, c’ E Ciand x, y $ C , . Consider the graph G, obtained from graph Gc, by joining vertices c and c‘. Graph G, is connected, regular of degree 3 and has no isthmi. Therefore, from Theorem 13, graph G, possesses a perfect matching that contains edge [c, c‘]. This matching corresponds to a matching Eo(Ct)of graph Gc, that saturates all vertices except c and c’. Suppose a connected component Djis joined to the rest of the graph by only one edge [d, x ] with d~ D,,x $ D,.Consider the graph Hiobtained from G,, by removing vertex d and by joining the vertices dl and d2 of D that are adjacent to vertex d in graph G. From Theorem 13, graph H , has a perfect matching that does not contain edge [d,, 41, and that corresponds in graph GD,to a matching Eo(D,) that saturates every vertex except d E D,. Thus, a perfect matching for G can be formed from the union of sets U Eo(C,),U Eo(D,), and the set of isthmi of graph G. Q.E.D.

I63

C-MATCHINGS

EXERCISE 1. Show the following result: Given a multigraph G = ( X , E ) , there exists a partial multigraph H with dH(xt)= d(i) if, and only if, for each pair S,T of disjoint sets, the number q(S, T )of components C of the subgraph of G generated by X S - T with rnc(C, T ) d ( C ) odd satisfies

-

q(S, T ) < d(S)- &T)

Hint: Apply Theorem 5 to the graph

+ dC x - s ( T ) .

+

G constructed from G as shown in Theorem 1. (W. T. Tutte [1954])

CHAPTER 9

Connectivity

1. &Connected graphs

The connectiviry ti(G) of a connected graph G is defined to be the minimum number of vertices whose removal disconnects G or reduces G to a single vertex. If G is not a clique, there exist two non-adjacent vertices a and 6 ; thus, X - { a, b } is a set whose removal disconnects G, and, consequently, K(G)


{ a ,b

> I= n - 2 .

On the other hand, if G is the n-clique K,,, we have K(K,) = n - 1 .

Graph G is said to be h-connected if its connectivity K(G) is 2 h. An articulation set of G is any set of vertices of G whose removal disconnects G. Let d denote the family of all articulation sets of G. A graph G # K,, is hconnected if, and only if,

K(G) = min I A I 2 h . A€ d

Thus G is h-connected i f , and only i f , (1) h < n - 1, and (2) there is no articulation set A of G with I A

I = h - 1.

Theorem 1. For a simple connected graph G, K ( G ) < min d,(x) . XEX

If G has n vertices, then K(G) < n - 1. Let xo be a vertex of minimum degree h, and suppose that h < K(G).The set Tc(xo)has cardinality h < n - 1. Thus G is not a clique, and K ( G ) = min I A A E J ~

I < I T,(x,) I64

1 = h < K(G),

165

CONNECTIVITY

which is a contradiction.

Q.E.D.

Theorem 2 (Harary [1962]). For each m, n with 0 Q n - 1

< in <

(2) ,

the maximum connectiuity for a simple connected graph with n vertices and 2m m edges is -n-].

[

1. From Theorem 1, a connected graph G satisfies 2m = xsx

d,(x) 2 n min d,(x) 2 n / t i ( G ) . X€X

Hence,

and so ti(G) g[LnE]

2. Note that, if G is a simple connected graph, then

It now remains to show that for each n, m, satisfying these inequalities there exists a simple connected graph G(n, m) such that

2m CASE1. Suppose - is an integer. Consider two subcases: n 2m - = 2k is an even integer. Construct a graph n Gn,2kwith vertices x,,, xl, ..., x,-~.Join each vertex xi to the vertices of SUBCASE1'. Suppose

{ x , / j = i+_ k'(m0d.n); 1 < k ' c k } .

Since each vertex has degree 2 k, the number of edges in graph Gn,Pkis

166

GRAPHS

It can be easily shown that this graph is (2 k)-connected.'l) Thus, take

G(n, m) = G n . Z k . 2m SUBCASE1". Suppose that - = 2 k + 1 is an odd integer. Then n n(2 k + 1) = 2 m implies that n is even. Form graph C(n, m) by taking graph Gn,2kand by joining vertex xi to vertex x, if n j =i - (mod. n ) , 2 n j = O , l , ...,- - 1 . 2 Thus, for G = G(n, m), 1 1 m(G) = - d,(xi) = - ( 2 k + 1) n = m 2 2 It can be easily shown that this graph is (2 k l)-connected.(2)

+

+

2m CASE2. Suppose that - is not an integer. Let q n

=

[ -21nm

.

SUBCASE 2'. Suppose that n or q is an even number. Form graph G(n, m) by taking graph G,,qand adding m - nq edges arbitrarily. ( I ) We shall show that G n , , k is (2 k)-connected. From the symmetry of G, it is sufficient to show that xo and x u , a = 1,2, ..., n - 1, cannot be disconnected by less than 2 k vertices. Suppose that x,, and x can be disconnected by 2 k - 1 vertices, x i l , xll, . =x,2k- where

..

...

1 6 il < iz < < i2k-1 6 n- I . One of the two intervals [O,a]. [a,n] contains at most k - 1 of these indices. Suppose it is interval [0,a]. Then, two consecutive vertices of the sequence obtained by removing xil, x:* from x o , xlr...,x,, are joined by an edge (because the difference between their indices is < k 1). Consequently, there is a chain from xo to x u , which is a contradiction.

...

-

+

(a) We shall show that G,.Zk+ I is(2 k 1) connected. Suppose that x o and xu are disconnected by 2 k vertices x i l , x , z . . . . ,xizk where 1 < il < iz < i z k d n - 1. If one of the two intervals [O, a] [a,n] does not contain k consecutive indices, then a chain from x,, to xu can be constructed as shown above. Therefore, suppose that x o and x , are disconnected by x i , xi+ ..., x i + k where 1 6 i < a - k 1, and

...

+

x.+j, X r n t j i l ,

...,X . + j + k - l ,

where 16j
8=

[ a i i + j + k 2- n - I

I

, 8' = 0

+ !2!

(indices mod. n).

Then,

B E [a + j + k , n + i -

11,

8' 6 [i + k , a + j - 11.

There is a chain from xo to x B and from xa, to xa. Hence, there is a chain from xo to xu since [ x B ,xa*]is an edge of Gn, 2k + I .

167

CONNECTIVITY

+

SUBCASE 2". Suppose that both n and q are odd integers. Let q = 2 k 1. Form the graph G(n, m ) by adding to Gn,2kn edges of the form [xi,x,], where . n - 1 modulo n, and also m - nq arbitrary edges. The total number j=i+2 of edges now equals (rn - nq) n 2kn = m. Similarly, it can be shown that this graph G(n, m ) is q-connected. Q.E.D.

+ +

The following theorem for h-connected graphs was discovered first by K. Menger [I9261 and rediscovered by H. Whitney [1932].

Lemma If a and b are two non-adjacent rertices of a simple graph G, then the maximum number of t-ertex-disjoint chains between a and b equals $ ( a , b)

=

min I A I , A

where the minimization is taken over all vertex sets A that do not contain either a or b and whose removal disconnects a from b. To prove the lemma, we shall construct a transportation network R with source a and sink b in which the flow 4p that maximizes 4pI determines 4p1 vertex-disjoint chains from a to b in G. First, replace each edge in G by two oppositely directed arcs with the same endpoints. Replace each vertex x # a, b by two vertices x' and x" joined by an arc (x', x") with capacity 1. Replace each arc entering x by an arc entering x' with capacity 1, and replace each arc leaving x by an arc leaving x" with capacity 1. Thus, the maximum flow between a and b represents the maximum number of vertex-disjoint paths between a and 6. The minimum capacity of a cut between a and b represents the minimum number of vertices needed to disconnect a from 6. From Theorem (1, Ch. 9, these quantities are equal. Q.E.D.

Theorem 3 (Menger [1926]). A necessary and sufficient condition for a simple graph to be h-connected is that any two distinct vertices a and b can be joined by h certex-disjoint chains. Let h be a positive integer. Since the proof is obvious for h h 2 2.

= 1,

suppose

Necessity. Let G = (X,E ) be a h-connected graph. The lemma shows that two non-adjacent vertices can be joined by h vertex-disjoint chains. We shall show that this is also true for two adjacent vertices. Suppose this were not true. Let a and b be two adjacent vertices that cannot

168

GRAPHS

be joined by h vertex disjoint chains. Consider the graph G' obtained from G by removing edge [a, b]. Vertices a and b cannot be joined in G ' by h - 1 vertex-disjoint chains. Thus, from the lemma, there exists in G ' a set A c X - { a , b } of cardinality < h - 2 that disconnects a and b. Then

1X

- A [

=

I XI - 1.4 1

3 (h

+ 1) - (h - 2 ) =

3,

and there is a vertex c distinct from a and b in X - A . We shall show that it is possible to join vertices c and a in graph G' by a chain that avoids A . If c and a are adjacent, clearly this is possible. If c and a are not adjacent, they are joined by h disjoint chains in'G, and therefore, they are joined by h - I disjoint chains in G ' ; one of these chains avoids A , because I A I < h - 1 . Similarly, vertices c and b can be joined by a chain that avoids A . Finally, vertices a and b in G can be joined by a chain that avoids A , which contradicts the definition of A . Suficiency. If two arbitrary vertices are joined by h disjoint chains, graph G is connected. At most one of these chains has length 1, and the union of the vertices of the h - 1 other chains contains at least h - 1 distinct vertices other than a and b. Hence,

n 2 (h - 1)

+2 >h.

Thus, h < n - 1. Furthermore, G has no articulation set A with I A I < h because, otherwise, we could choose two vertices a and b in two distinct components of the subgraph generated by X - A ; each chain joining a and b passes through set A , but there cannot be h distinct vertices in A . Thus G is h-connected. Q.E.D.

Corollary 1. If G is h-connected, the partial graph obtained by remocing an edge is (h - 1)-connected. If two vertices can be joined by h vertex-disjoint chains in G, they can be joined by at least h - 1 vertex-disjoint chains in the new graph. Q.E.D.

Corollary 2. If G is h-connected, the subgraph obtained by removing a vertex is (h - I)-connected. If two vertices of the new graph can be joined by h vertex-disjoint chains in G, then they can be joined by at least h - 1 vertex-disjoint chains in the new graph. Q.E.D.

C0N N ECTI V I TY

169

Corollary 3. Let G be a simple h-connected graph. Let B = { b, ,b,, .. ., bh } be a set of vertices with I B I = h. I f a E X - B, there exist h vertex-disjoint elementary chains pi[a,b,]joining a and B. We may assume that none of the b, are adjacent to a because, if k vertices of B are adjacent to a, then h can be replaced by h - k in the theorem. Let graph G' be the graph obtained from graph G by adding a vertex z and joining it to each vertex b, E B. We shall show that graph G ' is h-connected. Let S be a subset of X with 1 S I < h - 1;we have then to show that the subgraph generated by X u { z } - S is connected. This is clearly true if z E S, since Gx-sis connected. If z $ S, then, since I S I < h, one of the bi does not belong to S, and since this bi is adjacent to z , the subgraph under consideration is connected. From the lemma, there exist, then, h vertex-disjoint chains in G ' between a and z. These chains induce in C the required chains p,[a, b,].

Q.E.D. Corollary 4. Let G be a simple h-connectedgraph with h 2 2. An elementary cycle passes through an arbitrary set of two edges el and e, and h - 2 vertices a13 a29 .-.)ah-2. 1. If h = 2, the graph G' obtained from G by adding vertices a and b in the middle of edges e, and e 2 , respectively, is again 2-connected because no vertex of G' can disconnect G'. From the Menger theorem, there exist two vertex-disjoint chains in G' that join a and b. These chains determine an elementary cycle passing through edges e , and e,, and the proof is achieved. 2. Suppose h > 2 and suppose that the theorem is true for all k-connected graphs with k < h. We shall show that it is also true for a given hconnected graph G. We may assume that a,, a,, ..., uh-, are not the endpoints of either el or e, since then the theorem would follow from the induction hypothesis. The subgraph obtained from G by removing vertex ah-, is (h - 1)connected, from Corollary 2. Hence by the induction hypothesis, there is a cycle p o passing through e l , e 2 ,a,, a,, ..., a h - 3 . Let B be the vertex set of cycle po. Clearly, I B I 2 h. From Corollary 3, vertex a,,-, and set B can be joined by h vertex-disjoint chains. Suppose that each of these chains encounters B at only one vertex (i.e. does not return to B). Denote these chains by

170

GRAPHS

Suppose, for example, that cycle p o encounters vertices b,, b,, ..., b,E B

in this order. Among the h segments of p o determined by two consecutive vertices of the sequence b,, b,, ..., bh,b,, there is at least one segment that does not contain in its interior any of the h - 1 elements a,, a2,..., ah -3, el, e,. For example, let po[bl,b,] be this segment. The cycle required for the proof is then :

C(Z[ah-2,621 + P o [ ~ z6h-J , + I*o[bh,1711 - pI[ah-2,b,]. Q.E.D.

Corollary 5 (Dirac [1960]).If G is a simple h-connected graph with h 2 2, then there is an elementary cycle passing through h arbitrary vertices. Select any set of h vertices a,, a2, ..., ah. Consider an edge [ah- ,, x] and an edge [ a h , y ] .From Corollary 4, there exists an elementary cycle passing through these two edges and through vertices a,, a,, ..., Therefore, there exists an elementary cycle containing vertices a,, a,, ...,a,, , Q.E.D. Theorem. (Halin [1969]). I f G is a simple h-connected graph, either there exists an edge such that the partial graph obtained by remoaing this edge is also h-connected or there exists a vertex of degree h. The proof, which is omitted, uses Theorem 3. For the cases h the theorem was first proved by Las Vergnas [1968].

=

3 and 4,

We shall now study conditions on the degrees of a graph that imply h-connectivity. The following result is a slight generalization of a result of Bondy [1969].

Theorem 4. Let G be a simple graph with n vertices x,,x,, . .., x, such that

&(xi) G dG(x2) 6 < dG(X,) = d . Let il < iz < be the sequence of indices i such that dG(xi)< i. For the I-th index of this sequence, let 1

=

[f c d,(x,,,] k=l

.

Then, the number p of connected components of G satisjies

Let c1

< cz < -.<

c, be the cardinalities of the vertex sets of the p con-

171

CONNECTIVITY

nected components of G. We may assume that p > I because, if p 6 I, the proposed inequality is obvious. Graph G contains at least c1 vertices of degree < c1 - 1, and c1 c2 c k vertices of degree < c k - 1. Hence,

+ +

a*.+

< ck -

dG(X,,+cl+...+Ck)

+ + + ck ( k = 192, P) numbers cl + c2 + .-.+ c k are all different. < c1

c2

*“

*

For k = I , 2, . . . , p , the ik, and so Hence, c1 c2 ck

+ +-+

Thus, from the definition of q(f), 1

1

dG(xiJ

2 lq(l) *

k=l

I, we have d,(x,,) 2 q(I). Hence, n 2 Iq(0 + ( p - 1 - 0 q ( 0 + d

Furthermore, k >

+ P,

and thus

n

+ q(l) - d q(1)

+1

’

which implies that

Q.E.D. EXAMPLE.Consider the graph G in Fig. 9.1 with the following degree sequence; 1, 1 , 3, 3, 4, 4, 4, 4,4. The vertices xi with &(xi) < i are circled.

For

I = 1, we have q(1) = 1, and pG[9+;-4~=3.

For 1 = 2, we have q(2) = 2, and

( [

p 6 max 2,

+

-

“1)

=2.

172

GRAPHS

Thus, Theorem 4 shows that G cannot have more than 2 components.

Fig. 9.1

Corollary 1 (Bondy [1969]). Let G be a simple graph with n uertices x l , x 2 , ...,x, such that dG(X1)

< dG(Xz) < < dG(Xn) = d . * * a

Ih for some integer q, dc(xk) > k

(k = 1 , 2,

..., 4).

then the number p of connected components of G satisfies

n+q-d p g

q + l

-

There exists an index i with dG(xi)< i because dG(x,) < n ; let il be the smallest such index. We have q(1) = dG(Xi,) 2 dG(XJ 2 4 . Hence

The proof follows by letting 1 = 1 in Theorem 4.

Q.E.D. The graph in Fig. 9.1 shows that Corollary 1 is weaker than Theorem 4. Corollary 2. Let G be a simple graph with n certices x l , x2, ..., x, such that dG(x1)

< dG(xz) < < dG(Xn) = d . *.*

IfdG(xk) > k for all k ,< n - d - I , then G is connected.

173

CONNECTIVITY

In Corollary 1 , let q=n-d-1.

We have

n +q - d=(q and, from Corollary 1,

+ d + 1) +

- d < 2(q

+ I),

Thus graph G is connected.

Q.E.D. Corollary 3 (Bondy [1969]). Let G be a simple graph with n vertices xl, x2, ..., x, such that &(xi) < dc(xz) < ... < dG(x,). I f f o r some integer h < n dG(xk)2 k f h - 1

(k

< n - dG(x,-h+l) - 1)

then G is h-connected.

Consider a set A C X with cardinality h - 1. We shall show that A cannot be an articulation set. Index the vertices XIof G’ = G X e A so that dG.(x;) < dG,(x;) <

If k k

< 1 X’ I - d‘

- 1,

< dC,(Xh-h+l)= d’ .

’*.

then

< (n - h + 1 ) - (dG(x,-h+l) - h + 1 ) - 1

=n

- ~ J X . - ~ + ~ -) 1 .

From the hypothesis, this implies that dG(xk) 2 k

+ h - 1,

which implies that dG,(x;)

> dG(xk) - 1 A I

2k

+ h - 1 - ( h - 1) = k .

Corollary 2 shows that Gx-a is connected. Since this is proved for each set A c X with cardinality h - 1, graph G is h-connected.

Q.E.D. Corollary 4 (Chartrand, Kapoor, Kronk [1968]). Let G be a simple graph of order n such that,,for some integer h < n, the,following two conditions hold: n - 11 (1) for each k < - , the number of certices with degree < k + h - 1 2 is less than k ;

174

GRAPHS

(2) the number of certices with degree <

n ~

+ h - I is less than n - h + I . 2

Then graph G is h-connected.

After indexing the vertices as in Corollary 3, conditions (1) and (2) become equivalent to

11

+h -1

2- 2

dC(X,-k+l)

n -h The inequalities k > -and k < n - d G ( x , - , + , ) - 1 cannot be 2 satisfied simultaneously because this would imply that

and hence

which is a contradiction. Thus, if k


- d G ( x , - , + , ) - 1, then k

-h < n___ 2

and, from condition

(1'L dG(xk) 2 k

+h - 1.

Corollary 3 shows that graph G is h-connected.

Q. E.D. Corollary 5 (Chartrand, Harary [1968]). Let G be a sirlzple graph of order n such that, f o r some positive integer h < n, (x EX).

Then G is h-connected. To prove the corollary, it is sufficient to show that conditions (1') and (2') of Corollary 4 are satisfied. If k d,(x)

-h < n,then, for each vertex x of graph G, 2

n-h >+h-l>k+h-l. 2

175

CONNECTIVITY

Thus condition (1') is satisfied; condition (2') is obviously satisfied. Q.E.D.

2. Articulation vertices and blocks

A vertex whose removal from the graph increases the number of connected components is called an articulation vertex. An edge whose removal from the graph increases the number of connected components of the graph is called an isthmus. Using these definitions, we can redefine 2-connectivity: A graph is 2connected i f , and only if, it is of order n 2 3, connected, and has no articulation cer rices. A set A of vertices in graph G that generates a subgraph G, that is connected and without articulation vertices and is maximal with respect to this

I Fig. 9.2. Cactus

property is called a block. Thus, subgraph G, is either 2-connected (if I A I > 2), or an isthmus of G (if I A I = 2), or an isolated vertex of G (if I A I = 1). It is left to the reader to verify that the graph in Fig. 9.2 has 9 articulation vertices, 6 isthmi, and 13 blocks. A chord of an elementary cycle is defined to be an edge that joins two nonconsecutive vertices of the cycle. A cycle of length 3 has no chords. A cycle of length 4 can have 0, 1 or 2 chords. A cactus is defined to be a connected graph in which every block is either an isthmus or an elementary cycle without chords (see Fig. 9.2). The principal characteristics of 2-connected graphs are described in the following theorem.

176

GRAPHS

Theorem 5. The following properties are equicalent in a graph G of order 2 3:

(1) G is 2-connected.

(2) Euery ( 3 ) Every (4) Ecery ( 5 ) Ei7ery ( 6 ) Every (1)

3

two uertices of G lie on a common elementary cycle. certex and edge of G lie on a common elementary cycle. two edges of G lie on a common elementary cycle. two edges of G lie on a common elementary cocycle. two adjacent edges of G lie on a common elementary cycle.

(4) This was established by Corollary 4 to Theorem 3.

(4) 3 (3) 3 (2) This proof is obvious. (1) This follows from Theorem 3.

(2)

=>

(4)

* ( 5 ) Let [a,b ] and [c, d ] be two edges, and let [b, a , a l , a 2 , ..., ak, c , d , 4 , 4 , ..., 4 , bl

be an elementary cycle that contains them. Let A = { a, a,, ..., a k , c } . Consider the subgraph G X - A ,and let B denote the connected component of Gx-a that contains b and d. Then cocycle w ( B ) in G is an elementary cocycle that contains edges [a, b] and [c,d ] .

( 5 ) => (6) Let [a, b] and [a, c] be two adjacent edges. Let w(A) be an elementary cocycle that contains these edges and that separates the graph into two connected components A and B. If a E A , then b E B and c E B. Consequently, there exists an elementary chain p[b, c] in B that joins b and c. Hence, P[b, C I + [c, a1 + [a, bl is an elementary cycle that contains edges [a, b ] and [a, c].

(6)

(1) We shall show that G is 2-connected. Otherwise, G has an articulation vertex a. Let B and C denote the two connected components of Vertex a is joined to B by an edge [a, b] and to the subgraph GX-{=). C by an edge [a, c ] ;clearly, no elementary cycle can contain both these edges. This contradicts (6). Q.E.D. In the following theorem Ramachandra Rao has characterized the simple connected graphs with n vertices and r articulation vertices that have the maximum number of edges. 2

Lemma. A connected graph of order n 2 2 has at least two vertices that are not articulation zertices. There are exactly two such zwrtices if, and only if, the graph consists of an elementary chain.

177

CONNECTIVITY

From Chapter 3, Theorem 2, we know that a spanning tree of G has at least two pendant vertices. Clearly, these vertices cannot be articulation vertices of G. If G has only two vertices that are not articulation vertices, then each spanning tree of G has exactly two pendant vertices. This implies that G is an elementary chain. Q.E.D.

Theorem 6 (Ramachandra Rao [1968]). In a simple connected graph of order n 2 2 with r articulation wrtices, the maximum nutnber of possible edges is

Recall from the lemma that n - r 2 2; thus this formula is meaningful. If G is such a graph with the maximum number of possible edges, then each block of G is a clique with at least two vertices. If there a r e p blocks, then p 2 r 1. Let n, 2 2 be the number of vertices in the i-th block. We see easily, by induction on the number of blocks, that

+

P

C ni = n + p - I .

i= 1

Thus, the number of edges of G is

= max par+ 1

= max p>r+l

(

i

n+p-l--2p+2 2

p - l +

((

11-p+l

j+p-

i1

1)

=(n;r)+r. Note that we can construct a graph with r articulation vertices and with the maximum number of possible edges by taking a clique Kn-r with n - r vertices and attaching to one of these vertices an elementary chain of length r. Thus, there are r articulation vertices, and the number of edges is

m

=

(I'

2 ') +

i'.

Q.E.D.

We shall now apply the preceding results to graphs in which each elementary cycle of even length has at least two chords. Trees and cacti with only odd cycles are examples of such graphs. Other less trivial examples will be encountered later.

Lemma 1. If each ecen elementary cycle has at least two chords, then each even elementary cycle generates a clique. If this were not true, then there would exist an even cycle p of minimum length that does not generate a clique. Let p = [a,, a,, ...,a,, a,]. Clearly, p 2 6 , since for p = 4, a quadrilateral with two diagonals is a clique. An even chord (respectively, odd chord) is defined to be a chord that divides p into two even chains (respectively, odd chains). There are only two cases to consider: (1) Cycle p has an odd chord [a,, at].Then the cycles

are even and of length less than I(p), the length of p. These cycles generate two cliques (see Fig. 9.3).

"h

Fig. 9.3

Vertices aj€p[a27ai-11 and a, € p [ a i + l , a P I are adjacent because the cycle

179

CONNECTIVI7Y

is a quadrilateral and therefore possesses two chords, one of which is necessarily [a,, ak]. Thus p generates a clique.

(2) Cycle p has two non-adjacent euen chords [al, a,] and [ a j ,a,]. Suppose that these chords cross one another; for example, suppose ajEp[aIrail,

akEp[aiTal],

as in Fig. 9.4. Consider the following two even cycles: ~1

"2

= P [ Q , v ajl = /l[aj, a i l

+ [aj, akl - P[ai, 0 x 1 + [a,, all + la,, - p[ak, a l l f [a,, ajl

9

a

A

Fig. 9.4

Fig. 9.5

At least one of the cycles v i has length I(v,) with 1

I(vi)

< 2 I(P) + 2

since, otherwise, Np) = I(\?,)

+ I(%)

-4

1

> 2- 1(p)

+ 2 + 21 / ( p ) + 2 - 4 = l ( p ) . -

For example, let v, be this cycle vi. Then, KV,)

< 51 K p ) + 2 < QP)

*

Consequently, v1 generates a clique, and there exists in p at least one odd chord. From Part (1) of the proof, p generates a clique. If the two chords do not cross one another, suppose, for example, ak E [a,, a,], and aj E [a,, a,], where k < j (see Fig. 9.5). Consider the cycle = lal,

ail

$- p[ai, akl

+ [%, aj] -k p [ a j ,

*

180

GRAPHS

Since cycle v is shorter than cycle p, it generates a clique, and cycle p contains an odd chord. From part (1) of the proof, p generates a clique. Hence, in all cases, p generates a clique. This contradicts the definition of p. Q.E.D.

Lemma 2. Ifeach even cycle has at least two chords, then an odd elementary cycle with at least one chord generates a clique. It is sufficient to show that if p = [a,, a,, ..., a,, a,] is an odd cycle and if a, - and [a,, a, + are also chords. It is

,]

[al, at]is one of its chords, then [a,, evident that one of the cycles ~1

= ~ [ a lail ,

+ [ai, all

or

~2

ail

=

+ ~ [ a ia11 ,

is even. Suppose p1 is even. From Lemma 1, we know that p, generates a clique and this implies that [a,, ai - E E. Since [a,, a, - ,] E E, consider the cycle

P’ =

at-11

+ p[ai-l,

011

*

This cycle is even, and so we know from Lemma 1 that [a,, a i + , ]E E. Q.E.D.

Theorem 7. (Dirac [1960]).If each e w n elementary cycle has at least two chords, then each block is either a clique or an odd cycle without chords. Let B be a block of G. If B is not a clique, then there exist two non-adjacent vertices b, and b, in B. Since G, is 2-connected, there is a cycle p o that contains 6 , and b,. This cycle p o is odd and has no chords (since, otherwise, 6 , and b, would be adjacent, from Lemmas 1 and 2). Let Bo c B denote the set of vertices on p,,. Suppose first that Bo # B. Let a E B - Bo. From Corollary 3 to Theorem 3, there are two vertex-disjoint elementary chains p,[a, x] and p,[a, y ] joining a and Bo, that contain no vertices of B, except x and y. One of the cycles plra, XI

+ p o [ x ,Yi - P2[a,Yi

& [ a ? XI

- Pob, X I

-

P2[Q,Y1

is even. Since this cycle generates a clique (from Lemma l), vertices x and y are adjacent. Since p , has no chords, x and y are consecutive vertices of the cycle. Therefore, the even cycle contains b, and b,. Thus b, and h, are adjacent, which is impossible. Thus Bo = B is an odd cycle without chords. Q.E.D.

181

CONNECTIVITY

3. k-Edge-connected graphs In this section we shall study, for a general graph G = ( X , U ) , the cardinality of a minimum cut between two vertices a and b, i.e. the number: cc+(u, b)

= min mc+(A,x - A) , A cX aEA

beX-A

where mb(A, B ) denotes the number of arcs going from A to B. We shall also study, for a multigraph G = (A’, E ) , the number: c,(a, b)

= min m,(A, X

- A) .

AcX asA beX-A

Theorem 8. 1. Let G = ( X , U)be a graph; then c; (a, b) equals the maximum number of arc-disjoint paths from a to b. 2. Let G = ( X , E ) be a multigraph; then c,(a, b) equals the maximum number of edge-disjoint chains from a to b. 1. If G = ( X , U)is a graph, form from G a transportation network R with a as a source CI and b as a sink. Let each arc U E U have capacity c(u) = 1, and add a return arc (b, a) = uo with infinite capacity. In network R, the maximum flow problem (Ch. 5, 5 1) is to find a flow cp, satisfying 0 < q(u) < 1 for all u in CJ, that maximizes the flow value in the return arc u,,. Theorem (1, Ch. 5 ) shows that

max rp(uo) = min r n , + ( ~X, 9

- A).

Ac X As0 A+b

The left side of this equality equals the maximum number of arc-disjoint paths from a to b in graph G. The right side is, by definition, c,+(a, b). This proves Part 1 of the theorem.

2. If G = ( X , E ) is a multigraph, consider the graph G* obtained from G by replacing each edge of G by two oppositely directed arcs. The maximum number of elementary chains in G between a and b with no common edge equals the maximum number of elementary paths in G* from a to b with no common arc. From part 1, this number equals &(a, b) = cJa, b) . Q.E.D.

182

GRAPHS

Nash-Williams Lemma. If G = ( X , E ) is a multigraph, it is always possible to construct a set E' of new edges that matches the certices of odd degree, and is such that graph G' = (A', E ' ) satisfies

Ajb

for all a E X , b E X . Let Y denote the set of vertices of odd degree, and let 2 denote the set of vertices of even degree; then

Thus 1 Y 1 is an even number, and it is always possible to match together the vertices of Y. The proof of this lemma is long, and the reader is referred to the original presentation (Nash-Williams [1960]).

A connected multigraph G is said to be k-edge-connected if it cannot be disconnected by the removal of less than k edges, i.e., if and only if c,(x, y ) 2 k for all x, y E X , x # y . A multigraph is I-edge-connected if, and only if, it is connected. A multigraph is 2-edge-connected if, and only if, it is connected and has no isthmus. The least k such that G is k-edge-connected is called the edge-connectivity. Lemma. In a connected multigraph G, an edge is an isthmus if, and only if, no elementary cycle of G contains this edge. If [a, b] is not an isthmus, the graph remains connected after the removal of [a,b] and there is an elementary chain p[a, b] of G that does not contain [a, b]. Thus, [a,b] together with p[a, b] form the required elementary cycle. Conversely, i f edge [a, b ] lies on a cycle, then the removal of [a, b] does not disconnect the graph, and so [a, b] cannot be an isthmus. Q.E.D.

Theorem 9. A connected multigraph is 2-edge-connected $, and only if: euery edge lies on a cycle. The proof follows from the lemma. Theorem 10 (Robbins [1941]). Given a simpre graph G, its edges can be and only if, G is 2-edgedirected to.form a strongly connected 1-graph H connected.

v,

1. If such a graph H exists, then G is 2-edge-connected because the removal of an edge cannot disconnect H and, consequently, cannot disconnect G.

I83

CONNECTIVITY

2. Conversely, if G is 2-edge-connected, each edge of G lies on an elementary cycle. Let A , denote the set of vertices on a first elementary cycle of G. Direct the edges of this cycle to form a circuit, and arbitrarily direct all the edges with both endpoints in Al. The subgraph generated by A , is strongly connected. If X = A,, the proof is achieved. If X - Al # 0, there is a vertex a2 4 A , that is adjacent to a vertex a, E A , (because the graph is connected). By hypothesis, edge [al,az]is on an elementary cycle that contains a chain leaving A , at vertex a, and returning to A , at some vertex b , , no edge of which is already directed. Direct this chain so that it becomes a path. Let A z denote set A , augmented by the vertices of this path. Arbitrarily direct all undirected edges with both endpoints in A z . The subgraph generated by A 2 is strongly connected. If X # A 2 , this procedure can be repeated, as many times as needed, to obtain a strongly connected graph H. Q.E.D.

Applications of this result to traffic problems are easily seen. If the streets of a city are represented by a graph, then the streets can be directed for one-way traffic without cutting off traffic between any two points, if, and only if, the graph is 2-edge-connected. The above theorem can be generalized: Theorem 11 (Nash-Williams [1960]). Theedgesof a multigraph G can be directed to form a graph H = ( X , U ) with

=

(X,E )

We may assume that multigraph G is connected; otherwise, each connected component could be considered separately.

CASE1. Each certex of G has euen degree. In this case, it is easy to see that there exists a cycle p that uses each edge exactly once. (This result, due to Euler, is one of the oldest results in graph theory; a proof is presented in Chapter 1 1, $1.) If the edges are directed along the direction of travel in p, then a graph H i s A # X,we have obtained. For each A c X,A #

a,

because a traveller through p exits set A as many times as he enters set A . Thus, rnG(A,

x - A ) = rn,+(A,x - A ) + m,(A, x - A ) = 2m,+(A,X - A ) .

184

GRAPHS

Consequently,

c i ( x , y) = min r n , f ( ~x,

- A ) = -1 min. r n , ( ~ , x - A ) = 31 c,(x,

A3x AlY

y> ,

A3x

A+Y

which establishes the theorem for this case.

CASE2. There exist uertices of odd degree. The proof for this case depends upon the Nash-Williams Lemma given above. Add to G a set E' of edges that matches the vertices of odd degree, as in the lemma. Let G G' = ( X , E u E'). Since each vertex of graph G + G' has even degree, its edges can be directed as shown in Case 1 to form a graph H H' = ( X , U u V ' ) where H = ( X , U ) is graph G = ( X , E ) with its edges directed, and H' = ( X , U ' ) is graph G' = ( X , E ' ) with its edges directed. If A is a set of vertices corresponding to a minimum cut in H between two given vertices a and 6, then

+

+

+ ( u , b) cH

=

m,+(A,x - A ) = rn,f+,.(A, x - A )

- rn,+,(A,X

- A).

From Case 1 and the Nash-Williams Lemma, we have

=

[ p1G ( a ,b ) ] . Q.E.D.

Note that if G is a simple 2-edge-connected graph, then CdX,

v) 2 2

(x, Y E

x ; x # u) .

Consequently there exists a graph H , with the same edges as G, such that c&,

y)

>1

(x, Y

EX; x # Y).

In other words, His strongly connected. This is another proof for Theorem 10.

CONNECTIVITY

185

EXERCISES 1. Show that a k-connected graph is k-edge-connected. 2. Show that a connected graph is 2-connected if, and only if, for every three vertices a, b, x , there exists an elementary chain p[u, b ] that contains x. 3. Let G be a simple regular graph of degree 3. Show that G is k-connected if, and only if, G is k-edge-connected. 4. Tutte has shown that: A graph G is 3-connected if, and only if, G is a “wheel” (an elementary cycle p and a vertex xo joined to each vertex in p) or can be formed from a wheel by a sequence of operations of the two following types: (1) the addition of a new edge, (2) the replacement of a vertex x of degree > 4 by two adjacent vertices x’ and x” of degrees 2 3 such that a neighbour of x in the original graph is a neighbour of exactly one of x‘ and x u in the new graph. Verify this result with some simple examples. (Tutte [1961]) 5. A simple 2-connected graph is defined to be rriininially 2-connected if the partial graph resulting from the removal of any edge is not 2-connected. If G is minimally 2-connected, show that: ( I ) No edge is the chord of a cycle. (2) If G # K:,, then G contains no triangles. (3) If [ x , y ] is an edge and z is an articulation vertex of G - [ x , y ] , then z lies on every cycle that contains [x, y ] . (4) Each chain that contains [x. y ] and z contains a vertex of degree 2 in its interior. ( 5 ) If G # K3, there is a cycle of G that contains two non-adjacent vertices of degree 2. (6) The set of all the vertices of degree 2 is an articulation set. ( M . Plummer [1968])

CHAPTER 10

Hamiltonian Cycles

1. Hamiltonian paths and circuits

In a graph G = (A', U),a hamiltonianpath is defined to be a path that meets every vertex exactly once. Similarly, a hamiltonian circuit is defined to be a circuit that passes through every vertex exactly once. In a simple graph G = (A', E ) , a hamiltonian chain and a hamiltonian cycle are defined similarly.

EXAMPLE1. Voyage around the world (Hamilton). Consider 20 cities ..., t , represented by the vertices of a regular dodecahedron (polyhedron with twelve pentagonal faces and 20 vertices). How can we travel to every city exactly once and return home using only the edges of the dodecahedron? In other words, can we find a hamiltonian cycle in the graph in Fig. 10.1? Hamilton solved this problem as follows. a When the traveller arrives at the endpoint of j an edge, he has the choice of taking the edge to S his right (denote this choice by R), or the edge r to his left (denote this choice by L), or by staying where he is (denote this choice by 1). In k the obvious way, define a product of these operations: for example, L2R denotes the Fig. 10.1 operation of going left twice and then right once. Finally, two operations are defined to be equal if after starting from the same vertex they terminate at the same vertex. Note that the product is not commutative (for example LR # RL), but it is associative (for example (LL)R = L(LR)). For the graph in Fig. 10.1, a, b, c,

@

,

R5 RL2 R LR2 L RL3 R LR3 L

= L' = I = LRL =

RLR

= L2

= 186

R2.

,

HAMILTONIAN CYCLES

187

Hence, 1 = R5 = R2 R3 = (LR3L ) R3 = (LR3)" = [L(LR3L ) RI2 = = (L2 R3 LR)" = [L2(LR3L ) RLR]" = [L3R3 LRLR]" = =LLLRRRLRLRLLLRRRLRLR. This sequence contains twenty operations and contains no partial sequence that equals 1. Therefore, it represents a hamiltonian cycle. Another hamiltonian cycle is obtained by reversing the sequence. It is left to the reader to verify that no other cycles exist. Note that the voyage around the world can begin (and terminate) at any of the twenty cities. Hamilton solved the problem with the additional constraint that the first five cities to be visited had to be a, 6, c, d, e in that order. He found four solutions, starting with edge [a, b ] : RLRLRLLLRRRLRLRLLLRR RLRLLLRRRLRLRLLLRRRL RLRLRRRLLLRLRLRRRLLL RLRRRLLLRLRLRRRLLLRL EXAMPLE 2. Open voyage around the world. Suppose that the traveller in Example 1 need not return home. In this case, each trip corresponds to a hamilton chain in-the graph in Fig. 10.1, and the number of distinct trips available to the traveller increases greatly. All the trips starting with an R are indicated below, All other possible trips can be obtained by interchanging the operations L and R in the sequences given below. RRRLRLRLLLRRRLRLRL RLRLLLRRRLRLRLLLRR RLRRRLLLRLRRRLLLRL RLLLRRRLRLRLLLRRRL RLRLLRRRLRLLLRRLRL RRRLLLRLRLRRRLLLRL RLRRRLLLRLRLRRRLLL RLRLRRRLLLRLRLRRRL RRRLRRLRLRLRLLRLLL RRLRLRLLLRRRLRLRLL RRRLLRLRRRLLLRLRLR RLRLRLLLRRRLRLRLLL RRRLRRLRLRLLLRRRLR RLLLRLRLRRRLLLRLRL RRLLLRLRLRRRLLLRLR RLRLRLLLRRRLRLLRRR RLLLRLRLLRRLRLRRRL RRRLLLRLRLRLRRLRRR RLRRRLLLRLRLRRLRRR RRRLRRLRLRLRLLLRRR RLRLLLRRLRLLRRRLRL RLRLRRRLLLRRLRLLLR RLLLRLRRLLLRRRLRLR RRRLLLRLRLRLRRRLLL RLLLRLRLLRRRLRLLLR RLLLRLRRRLLRLRLLLR RLLLRLRRRLLLRLRRRL

188

GRAPHS

EXAMPLE 3. Knight’s journey (Euler). How can a knight be moved on a chessboard so that he visits each square exactly once? This problem of finding a hamiltonian chain has interested many mathematicians: e.g. Euler, de Moivre, Vandermonde, etc. Many methods have been proposed. One method that seems to work in practice is the following: Move the knight to the square from which he will control the smallest number of unvisited squares.

Fig. 10.2. Examples of the knight’s journey

Another method is to move the knight on only half of the chessboard and then to have him repeat this pattern on the other half (see Fig. 10.2). This method depends on the special structure of the chessboard and does not work in more general situations. EXAMPLE 4. The problem of Mr. No. Mr. No, a mythical Japanese detective, lives in the upper left square of a chessboard. He wishes to visit Mr. Go who lives in the lower right square of the chessboard. Mr. No can move to any adjacent square but cannot move diagonally. Is it possible for him to visit each square of the chessboard exactly once en route to Mr. Go? Clearly, the problem of Mr. No is to find a hamiltonian chain between two

189

HAMILTONIAN CYCLES

given vertices. To visit each square exactly once, Mr. No must make 63 moves and 63 colour changes. Clearly, after 63 colour changes Mr. No stops on a square with a colour different from the colour of his home square. Since Mr. No and Mr. Go have home squares of the same colour, no such trip is possible.

Theorem 1. The number h(G) of hamiltonian paths in a 1-graph G = (X,U) has the same parity as the number of hamiltonian paths in the complementary 1-graph c = (X,X x X - U). We shall show that if G and G are two complementary 1-graphs, then h(G) = h(G) modulo 2. Suppose that the vertices of G are indexed from 1 to n. Given a subset V of the arcs of G, let h(V) denote the number of arrangements (il, iz, ..., in) of (1,2, ...,n)suchthateacharcof Visoftheform(ik,ik+l).Note that ifh(V) is not zero, then IVIGn-1, and the arcs of V form a family of pairwise disjoint paths. Furthermore, if h(V) # 0 and if I V I < n - 1, the connected components of the partial graph ( X , V ) consist of r disjoint paths, where r > 1, and consequently, h( V ) = r ! = 0. Hence, h(V)$O

*

(VI=n-l.

Now note that h(G) equals the number of arrangements ( i l , iz, ..., in) such that no arc of G is of the form ( i k , ik+l). Thus,

n ! - h(G) =

c

vc

h(V) il

c

h(Y)

+ .-.+

(-Qk-I

c h(V) +

**a

V C IJ

Therefore

h ( 5 ) = n ! - h(G) =

C

VCU I VI=n-l

Hence, h(c)= h(G) modulo 2.

h(V) = h(G) . Q.E.D.

Theorem 2 (C. A. B. Smith [1946]). 111 a simple regular graph of degree 3. the number of hamiltonian cycles that contain a given edge is even. Suppose’ that there exists a hamiltonian cycle (otherwise, the result is 3 trivially true). This cycle is of even length (since there are - n edges, which 2 implies that n is even). Thus the edges of the cycle can alternately be coloured with two colours c1 and /3 and the edges not on the cycle can be coloured y.

190

GRAPHS

Thus, there exists a partition { E,, E 8 , E, } of the edge set into three perfect matchings. Such a partition will be called here a 3-colouration of the edges. A 3-colouration of the edges with three colours a, fi, y determines a vector pa, = (pl,p2, ..., p") with p i = 1 if edge e, is c( or fi, and pi = 0 otherwise. Let p E M if vector p defines a family of vertex-disjoint even cycles that use all vertices. Each 3-colouration { E,, E,, E,} determines three vectors pa87 p a y , BY M , with (mod. 2). Pub + P U , + PbLpV 3 0 Furthermore, if a vector p E M consists of a family of k(p) pairwise disjoint cycles, then it corresponds to 2k(u)-1distinct 3-colourations. Summing the above identity over all 3-colourations { E,, E,, E, } gives

C

2k(p)-1

=o

(mod. 2).

p=o

(mod. 2).

p

lc=M

Hence,

C p

I k(d=

1

Since this sum is over all the hamiltonian cycles of G, the number of distinct hamiltonian cycles that contain a given edge is even. Q.E.D. Corollary 1. I f a simple regular graph of degree 3 has a hamiltonian cycle, it has at least three hamiltonian cycles.

Let e be an edge of the hamiltonian cycle. At least two distinct hamiltonian cycles p, and p2 contain e. Let x denote the first vertex at which these two cycles diverge after passing through e. If [x, y] is an edge of p, and not of p2, then at least two hamiltonian cycles pass through [x, y ] , and they are both different from p z . Q.E.D. Corollary 2 (N. J. A. Sloane [1969]). I f a graph G = ( X , E ) has two hamiltonian cycles without common edges, it has at least three hamiltonian cycles. Let p1 = [ x , , x z , ..., x,, x,] be a hamiltonian cycle of G , and let p2 be another hamiltonian cycle of G that does not contain any edge of pl. CASE1. Suppose n is even. A regular graph of degree 3 can be formed with all the edges of p 1 and some edges of p z . Corollary I established that this graph has three hamiltonian cycles. Therefore, the result is true for G. CASE2. Suppose n is odd and [xt, x t + J E E for all i. Then, there exists a third hamiltonian cycle, namely: p3

= [XI,X " ,

x2,

%,x4,

' S . 2

&-l, X l l

*

191

HAMILTONIAN CYCLES

CASE3. Suppose n is odd and [x,, xz] 4 E. Suppose that p2

=

[xi,

xj,xkl,

xkz7

* * * Y

x k , - , ~ xi]

i,j f

*

Form a new graph G by adding edge [x,. xz] and by removing vertex x, and all edges incident to xl. Graph has a hamiltonian cycle F1 = [xz,x3, ..., x,, xz]. The edges of cycle ,iil and the edges [xi,

Xkll,

[xki,

xkj],

Cxk4, x k s l ,

xi]

[xk,-j,

Y

form a regular graph of degree 3 that has two hamiltonian cycles that use edge [x,, x,] by Theorem 2. Therefore G possesses two hamiltonian cycles distinct from p z . Q.E.D. Theorem 3 (Boshk [1967]). A simple regular bipartite graph of degree 3 has an even number of hamiltonian cycles. Let G = ( X , Y, E ) be a regular bipartite graph of degree 3 with 1 X 1 = 1 Y 1 = 22’ and let h(G) denote the number of hamiltonian cycles in G. For n < 6 , the theorem is true since the complete bipartite graph K 3 , , contains exactly 6 hamiltonian cycles. Assume that the theorem is true for graphs of order < n ; we shall show that it is also true for a graph G = ( X , Y, E ) of order n > 6. Since G contains no triangles, an edge [x,,y,] is adjacent to four edges [xl,yzl, Ixl,y31, [yll~21,[Y,, x31such that the vertices xl, x z I x3,Y 1, y z , ~3 are all distinct. Let G’ and G” denote the two graphs obtained from G by the transformations shown in Fig. 10.3. C

.

G‘

G”

Y2

u n

Y3

x3

‘>( x3

Y3

Fig. 10.3

Let h,, h; and hl denote the number of distinct hamiltonian cycles in the graphs G, G’, and G ”using the vertices x, , xz, x3,y , ,y z , y , as shown in Fig. 10.4. From the induction hypothesis,

+ h; + h i + hi + hi = 0 h(G”) = h’; + hi + h’j + hi + h’; E 0 h(G’) = hi

(mod. 2) (mod. 2)

.

192

GRAPHS

Furthermore, it is evident that hi = h i , hz = h ; ,

hs = h:, Hence, h(G) = hi

h6

= h;,

hj

= hj,

hi = h i ,

h,+ =

hi

h; = h ; .

+ hz f + h, + h~ -k = h'; hi + hi + hi 4- h; + h; = h'; + h; + hi + h i + h'; + h; + (hi + h;) + (hi + h;)

=

h3

h;

+

h;

=0

h6

(mod. 2 ) .

Therefore, the number of hamiltonian cycles in G is even.

Q.E.D.

2. Hamiltonian paths in complete graphs

Theorem 4 (Camion [1959]). IfC = (A', f)is a strongly connected, complete 1-graph, then G has a hamiltonian circuit.

HAMILTONIAN CYCLES

193

Let p = [a,, a,, ..., ah- 1, ah,ah+, = a,] be a circuit of maximum length h. Suppose that circuit p does not encounter some vertex b. Then, u i Er ( b ) 3 b #T(ai-l) * ui-l E T ( ~ ) bET(a3 3 ai+1 #T(b) * b eT(at+l) U

Fig. 10.5

Therefore, the vertices not lying on circuit ,u can be divided into two classes B, and B, as follows: If b E B , , then each arc joining b and p is directed towards p . If b E B,, then each arc joining b and p is directed towards 6 . By hypothesis, B, u B2 # $3. Since G is strongly connected, B1 # $3, B, # $3,and there exists an arc from Bz to B,. Denote this arc by ( b 2 ,b,), where b, E Bz and b, E B,. Thus the circuit [al,a,, ...,a h ,b ,b,, all is longer than p, which contradicts the maximality of p . Q.E.D. Corollary. Let G = ( X , r) be a complete 1-graph. There exists a t’ertex xo such that for each vertex y # xo there is a path from xo to y . Such a vertex xo is calleda “root” of G. Each root is the initial endpoint of a harniltonianpath of G.

,

1. The existence of a root of G was established in Theorem (4, Ch. 4), since any centre of G is a root. 2. Let xo be a root of G. Consider the graph G’ obtained from graph G by adding a vertex z, an arc (z, xo), and the arc ( x , z ) for every x # x o . By Theorem 4, graph G’, which is complete and strongly connected, has a hamiltonian circuit, which corresponds in G to a hamiltonian path starting from x o . Q.E.D. Algorithm. This corollary provides a very simple algorithm to construct

a hamiltonian path in a complete 1-graph G = ( X , r)without loops:

194

GRAPHS

Let a, denote the vertex with the largest outer demi-degree d&J. Vertex a, is a centre (Theorem 4, Ch. 4). In the subgraph generated by F(al) - { a1 }, let a2 denote the vertex with the largest outer demi-degree. In the subgraph generated by r(a,) - { a , , a2 }, let a3 denote the vertex with the largest outer demi-degree, etc. Then [a,, a z , ...,] is a hamiltonian path. Theorem 5. I f G = (X,r)is a complete, anti-symmetric, transitive 1-graph, then G has exactIy one hamiltonian path. A hamiltonian path must start from a root. Since G is transitive, vertex x , is a root if, and only if,

r(x,)u(x,}=x. There is always a root in G, and since G is anti-symmetric, the above equation shows that the root is unique. Thus, the first vertex of a hamiltonian path must necessarily be this root xl. By the same argument, the second vertex of the hamiltonian path is the unique root of the subgraph generated by X - { x1 }, etc. ..., and the hamiltonian path is unique. Q.E.D. Theorem 6 (Ridei [1934]). I f C is a complete, anti-symmetric I-graph, then the number h(G) of distinct hamiltonian paths in G is odd. We shall show that if G is a complete, anti-symmetric graph, then reversing the direction of an arc does not change the parity of h(G). Since graph G can be obtained from a complete, anti-symmetric, transitive graph by making successive reversals of arc directions, this fact will establish the theorem. Let (a, b) be an arc of G , and let G ' be the graph obtained from G by reversing the direction of (a, 6). Let G, be the graph obtained from G by adding arc (b, a). Let G2 be the graph obtained from G by removing arc (a, b). We want to show that

h(G) G

= h(G')

G'

Fig. 10.6

(mod. 2 ) .

195

HAMILTONIAN CYCLES

Since graph G, is anti-symmetric, each hamiltonian path of G, (taken in its reverse direction) defines a hamiltonian path of GI,the complementary l-graph of G,, and vice versa. Therefore (from Theorem l), we have h(G2) = h(??~)= h(Gl)

(mod. 2)

.

Let h,(ab), h,(ba), h,(O) denote respectively the number of hamiltonian paths of G, that contain arc ab, arc ba, neither arc ab nor ba. Then,

hl(0) = h(G2) = h(G,) = h,(O) Thus, h,(ab)

= h,(ba) modulo 2, h(G) = h,(O)

+ h,(ab) + hl(ba) .

and consequently,

+ hl(ab) = hl(0) + hl(ba) = h(G’).

This shows that h(G) = h(G’) modulo 2, and completes the proof. Q.E.D.

3. Existence theorems for hamiltonian circuits

Clearly, the greater the demi-degrees d,+(x)and d;(x) of a 1-graph G are, the greater are the chances that G has a hamiltonian circuit. In this section, we present a theorem of Ghouila-Houri [1960] that gives conditions for the existence of a hamiltonian circuit in terms of the demi-degrees of the graph. First, a lemma is needed:

Lemma. Consider a circuit with m tlertices, each labelled with either a cross or a circle. suppose it contains exactly no # 0 circles and exactly n, # 0 sequences of q consecutive crosses. Then no+nl<m-q+l. Suppose that there are p maximal sequences of crosses of cardinality 2 q. Each such sequence is framed by two circles. By hypothesis, p 2 1. Let c denote the total number of crosses. Let mi denote the total number of sequences of q crosses contained in the i-th maximal sequence. Then the length of this maximal sequence is q a i - 1. Hence c 2 (q

+ al

+ - 1) + ( q + a,

- 1)

+ -.. + (q + ap - 1) 2 2 q + (a1 + a2 + ..* + ap) - 1 .

Hence, no+nl=m-c

+

P

Ca,<m-q+l. i= 1

Q.E.D.

196

GRAPHS

Theorem 7 (Ghouila-Houri [1960]). Let G be a strongly connected 1-graph for each vertex x, d,+(x) d,(x) z n , then G has a hamiltonian circuit. Let n 2 3. Suppose that the result is true for any graph of order < n, and that there exists a graph G = ( X , r)of order I X I = n for which the theorem fails. We shall show that this leads to a contradiction.

of order n and uithout loops.

v,

+

By hypothesis,

pm1+ pm1

- 1x1 > 0 ( X E X ) . (i) Let p = [ x o , x l , x,, ..., x , , - ~ ,x,] be an elementary circuit of G of maximum length. Since G is strongly connected, h 2 2, and since there are no hamiltonian circuits, m < n. Let Xo = { xo, x1, x2, x,-11. Let X I , X,, ..., X , denote the strongly connected components of the subgraph generated by X - X,. 1. We shalI show that each subgraph G,, , G,, , ...,Gx,contains a hamiltonian circuit. Let X E Xi, 1 < i 6 p . Then, f o r k < m, ..a,

xk

ri(x)

x&+1 6

fi(x)

(Otherwise, the circuit p could be lengthened.) Hence,

1~

( xn)x0 I Q I X , I -

I r c fn~x0 1 .

ForyEX,,j#i,O,

Y Erdx) * Y $T,+(x) (because X, is a strongly connected component different from Xi). Thus, for each j 2 0, j # i, rii(x) n X , Q I X j I - r , + ( x ) n X j 1 *

I

1

I

xo Fig.'l0.7

197

HAMILTONIAN CYCLES

Combining this result with (i) yields

1 T,+(x)n xi I + I

2

I - I Xi I >

nXi

G(X)

-

C (I

G ( x ) nX j

j#i

I

+I~

( xn )

xiI - I X , I) 2 o

Since this result is valid for each x E Xi, the subgraph Gxl has a hamiltonian circuit by virtue of the inductive hypothesis. 2. We shall show that there exists an X i , i # 0, such that and

r,+(Xo) n X i # fa

G ( X o ) n Xi # fa. It suffices to show that there exists an Xithat is joined in both directions to X,. Since graph G is strongly connected, there exists at least one X,such that

mxo, x,# 0.

Consider a vertex x directions to X,,then

+ X,

f-l

U

X , . If xk E X,,and if no Xiis joined in both

rJ(xk)

=>

x $ri(xk).

Hence,

I T,+(x,)

n ( X - xo - X,)

IG

< Ix -xO -xjI - I r i ( x k ) n (x-xO

If y

E

X,,then for each x + X, u X,, xErJ(Y)

*

-xj)I.

x$ri(Y)

Hence,

Im

y)

(7 ( X

- xo - X,)

Finally, for each y (ii)

I G(Y) n

(10

2

-

Ix,

E

I< < I X - X0 - X , I - lri(y)n ( X - x0 - xi)/.

X, U Xi,we have, by virtue of (i),

I

u x,>I + ~ ( y n)(x, u x,) I 2

I

I

U X , ~- (X - x0 - X,)n ~ , + ( y )-

I (x- x0 - X,)

nri(y)

1 + I x - x0 - xiI 2 I x0 u X , I .

Furthermore, since G is strongly connected, there exists a path from X , to

X, of the form [z,, zl,..., z ~ - z,], ~ , with zl,z2,..., z t d l $ X o u X,.Since X,n T;(X,) = 0, this path has length > 1. Consider the subgraph generated by X, u X,together with the arc (z,, zt). Since this graph is strongly connected and has fewer vertices than G, inequality (2) shows that it contains a hamiltonian circuit. This circuit necessarily contains arc (z, , zt). By replacing arc (z,, zt) with the path z,, zl,. .., zt , we obtain a circuit of G that is longer than p, which is a contradiction.

198

GRAPHS

Hence, there is a component Xi with the required property. Denote this component by X , . 3. Here we shall show that each y E X , satisfies T,+(y)n Xo # @ and G ( Y ) n xo z 0. We may assume 1 X , I > 1, because, if I X , I = 1, this is obviously true. Let yo be a vertex of X , that does not satisfy, for example, G(Y0)

n xo f:

a.

Let [ y o , yl, ..., yp-l, y o ] be a hamiltonian circuit of XI with length q, such a circuit exists.) Lets be the smallest index such that r ; ( y g ) n Xo # 0.From the definition of X , , such an s exists, and s # 0. 1

< q < m. (From part (l),

1 .'!,

Xm-YO -1

--4--

XO

YO

s-1

XI Fig. 10.8

Since graph G has no circuit of length > m, xk E T;(y,) implies that x ~ + ~ , XktZ, x k t q 4 ri(yS-l). Thus,

I G~Y,-A n x0I Q m - 4 .

For i # 0, 1,

~ ~ ~ ( y s - l ) n x i ~ + ~ ~ -~ (1 y~S ~ l1) n ~ x0i ~ since there are no double edges between y S - , and Xi. Hence,

199

HAMILTONIAN CYCLES

which contradicts (i). Hence r a y ) n x0 z By a similar argument,

o

G ( y ) n .U, # la 4. We shall show that for each y , E XI,

1 GAYJ

n

(M).

(iEI).

xoI + I r2(yS-dn xo I < m

-q

+1.

In fact, X0 is a circuit of m vertices that can be labelled with circles and crosses in the following way: If x, E ~ Z ( Y ~ -label ~ ) ,x j with a circle.Otherwise, label x, with a cross. We have seen in Part 3 that x k E ri (y,) implies the existence of a sequence Xk-Cl, X k f 2 ,

Xk+q

of q crosses. From 3, there are no # 0 circles and n, # 0 sequences of q crosses. From the lemma,

I G(YJ

n

xo I + I rc+(yS-l) n x0 I < nl + no < m - q + 1

5. We shall show that there exists a vertex y, E Xl with

I MYJ

n

xo I + I C ( y s - l ) n x0 ] z

m

-4

+2.

From inequality (i), each vertex y E X , satisfies

I G ( Y ) n xo I + I m y ) n xo I a Z I Xo I - (I r,'cv> n Xl I + I Gcv)

- I xi I) - C (I ~ C + ( Y ) n xj I + I T ~ ( . Yn)x j I - I Xj I) 2 j+o,1

z

m - [(4

A', I

- 1) + (q - 1) - q] - o = m

-4

+2 .

By counting in two different ways the number of arcs joining X,, and XI,we obtain

Therefore, there exists at least one ys that satisfies the above inequality. This contradicts Part 4. Q.E.D. Corollary 1. I f C = (A',r) is a 1-graph without loops much that &(x) dD(x) 2 n - 1 (XEX) then G has a hamiltonian path.

+

9

200

GRAPHS

Add to G a vertex xo and join it to each other vertex by two oppositely directed arcs. The new graph G' is strongly connected, and d&(x)

+ d,.(x)

3 n-1

+ 2 = n'

(x E X ' ) ,

Graph G' has a hamiltonian circuit .from Theorem 7. Thus graph G has a hamiltonian path. Q.E.D. Corollary 2. If G is a complete I-graph, then G has a harniltonian path. (This result follows also from the corollary to Theorem 4.)

Corollary 3. Let G = ( X , r) be a strongly connected I-graph of order n without loops. If the graph remains strongly connected afier the removal of any vertex, and if d i ( x ) -t d,(x) 2 n

+1

(XEX),

then for each pair a, b of distinct vertices, there exists in G a hamiltonian path with endpoints a and b.

Consider the graph G' obtained from G by contracting { a, b } into a single vertex c and by letting T$(c) = T:(b) - { a } , f&)

=

f,-(a) - { b } .

Graph G' is strongly connected because for x # a, b graph G has a path from x to a that avoids b, and a path from b to x that avoids a. Thus, G ' has a path from x to c, and a path from c to x. These properties hold also for the graph G" obtained from G by contracting

- 1 + I TJb) Thus at least one of the following two inequalities is satisfied:

c )

1 + I I--&) I 2 I X' I ,

G ( C )

I + I TG,(C) I 2 I X ' 1 .

Im I

-1

20 1

HAMILTONIAN CYCLES

Suppose, for example, that the first inequality is satisfied. Then graph G ‘ satisfies dZ,(x)

+ d&X)

2 I x I - 1 = I X’ I

(x EX’)

.

Thus, from Theorem 7, G‘ has a hamiltonian circuit that corresponds in G to a hamiltonian path between a and b. Q.E.D.

Corollary 4 (Nash-Williams [1969]). Zf G loops, of order n, such that

=

( X , r) is a I-graph without

then G has a hamiltonian circuit.

From the Ghouila-Houri theorem, it sufKces to show that G is strongly connected. More precisely, we shall show that a 1-graph G without loops with

is strongly connected. Suppose that G were not strongly connected. Then there exist in G several strongly connected components Xl, X,, ..., X,,and their contraction yields a graph without circuits (Theorem 12, Ch. 3). There exists at least one connected component XI with &(XI, X - Xl) = 0, and one connected component X, with m $ ( X - X,, X z ) = 0. Suppose, for example, that I XI I 2 I X z I. If xo E X z , then

This contradicts the hypothesis.

Q.E.D. Corollary 5 (Bermond [1970]). .ISG = ( X , ZJ is a 1-graph of order n without loops such that

where k is an integer, 0 < k < n - 1, then each elementary path of length k is contained in a hamiltonian circuit.

Let p o

= [ao,a,,

..., a], be a path of length k in G. Let A = {ao,a,,

..., a k } .

202

GRAPHS

Construct a 1-graph Gofrom G by removing A , adding an auxiliary vertex a, adding an arc ( x , a) for each x E X - A with x E r;(ao), and adding an arc (a, y ) for each y E X - A with y E r z ( a k ) . It suffices to show that Gohas a hamiltonian circuit. Graph Go has no = n - k vertices. For a vertex x # a, d:o(x) = dGo(X] =

&(X)

rn;(x, A

- &(X,

A

- { a, }) 2 &(x) - k , - { ak }) 2 &(X) - k .

Furthermore,

dio(a) = d;(ak) - m i (ak, A - { ak }) 2 d:(Uk) dGo(a) = dF(ak) - mF(ak, A - { ak }) 2 d,(ak) Thus, for each vertex x of graph Go,

and

d&)

2

-k -k

9

n $ .

Hence, from Corollary 4, Go has a hamiltonian circuit, that corresponds in G to a hamiltonian circuit containing ,uo. Q.E.D. Corollary 6. I f G

=

( X , r)is a 1-graph without loops of order n such that

then for each pair a, b of distinct vertices there exists a hamiltonian path from a to b. If a E T(b),let G’ = G; otherwise let G’ = G + (b, a). Graph G’ satisfies the above conditions. Therefore, from Corollary 5 , G ” has a hamiltonian circuit containing (b, a) that corresponds in G to a hamiltonian path from a to b. Q.E.D. The following two conjectures are due to Nash-Williams [1969]. Conjecture 1. If G is a 1-graph without loops of order n > 5, such that for all x E X , n n d,‘(x) Z and dG(x) 2 2 then G has two arc disjoint arc-disjoint hamiltonian circuits.

HAMILTONIAN CYCLES

203

If G is a 1-graph without loops of order n 2 5 such that for all (x) < k is < k, and the number of with d g ( y ) < k is < k , then G has a hamiltonian circuit.

Conjecture 2.

n the number of vertices x with d$ k < ?, certices y

When sufficient conditions for the existence of a hamiltonian circuit are difficult to find, necessary conditions can be found by using the concept of a dissection. A dissection of graph G is defined to be a set of elementary paths of G such that each vertex of the graph is contained in exactly one path. Call a path closed if it also defines a circuit, otherwise call it open. The value of a dissection is defined to be the number of open paths in the dissection. The value is always less than or equal to the number n of vertices since a single vertex is a path of length 0. A hamiltonian path is a dissection of value 1. A hamiltonian circuit is a dissection of value 0.

Dissection theorem. For a 1-graph G = ( X , f), the minimum calue of a dissection equals a,, = rnax (i s I - I r ( ~ I).) S C X

Consider a dissection a = (d, a2, ..., a”, B1, B2, ..., Bq) of value q, where CI‘ is a closed path, and B’ is an open path. Let A‘ denote the set of all vertices a‘, and let B’ denote the set of all vertices in PI. Make two copies Xand Xof X , and form the bipartite graph H = ( X , 1, E) where ( x i , 2,) E E if, and only if, x, E f ( x , ) . Each circuit CI‘ defines uniquely in H a matching of I A‘ I edges, and each open path defines in H a matching of I B j I - 1 edges. Therefore the dissection defines a matching Eo of cardinality IEOI = C I A ‘ l + C ( I B j I - 1 ) . i

i

Thus n = ~ I A ’ I + ~ I B ’ I = ~ I A+ixI ( I B ’ I - l ) + q = I E o 1 + 4 . i

i

i

i

This correspondence between the matchings and the dissections is a bijection. Therefore, from the Konig theorem (Chapter 7), the minimum value of a dissection equals n - max I Eo I = n - (n - 6,) = 6, . This gives the required formula. Q.E.D. Corollary 1. I f C has a hamiltonian circuit, then 6 , = 0. The proof is immediate.

204

Corollary 2.

GRAPHS

IJ’ G has a hamiltonian path, then 0 < do 6

1.

The proof is immediate. Corollary 3. IJ’ a I-graph has no circuits, then a necessary and suficient condition for the existence of a haniiltonian path is that 6, = 1. The proof is immediate.

4. Existence theorems for hamiltonian cycles

Without loss of generality, we shall assume throughout this section that G is a simple graph. The main existence theorem for a hamiltonian cycle (P6sa [1962]) was a generalization of a previous result of Dirac [1952]. A new proof for P6sa’s theorem, due to Nash-Williams, permitted J. A. Bondy [1969] to give a stronger result. Finally, Chvital [ 19721 has proved an even stronger theorem, that is in some sense the best possible. The following theorem is an extension of Chvital’s theorem that has been modified to generalize a result of Kronk [19691. Theorem 8. Let G = ( X , E ) be a simple graph of order n with degrees dl < d2 < ... < d,. Let q be an integer, 0 < q < 11 - 3. I f , for erery k with q < k < 3 (n + q), the following condition holds:

(A)

dk-q < k

* dn-k 2 n

-k +q

then for each subset F of edges with I F I = q such that the connected components of ( X , F ) are elementary chains, there exists a hamiltonian cycle of G that contains F. Furthermore, this result is the best possible in the following sense: each sequence of degrees that does not satisfy condition (A) is majorized by a sequence of degrees of a graph that does not haoe the desired property. 1. We shall first show that condition (A) implies: dl > q. Suppose that dl 6 q, and let k q<

Furthermore, since dl

=q

k = -Q 2

+ 1. Then

+2 + q< -q- + n 2 2

< k , it follows from condition an-q-

1+q=n-

(A) that 1.

This shows that there exist at least q + 2 vertices joined to all the other vertices of G ; therefore dl 2 q + 1, which is a contradiction.

205

HAMILTONIAN CYCLES

2. We shall show that if a simple graph G satisfies condition (A), the graph G ’ obtained from G by adding a new edge also satisfies condition (A). Let s k = { x / &(x) < k } . and Sk = { x / &*(x) < k } . Clearly, I Sk (A‘)

I
1. Note that condition (A) is equivalent to I sk I >k I s,-k+,-, I < ?I - k . s k

5

Since G satisfies (A’), JSkJ>k-q *

*

1Sk12k-q

I SA-k+q-l I

<

]Sn-k+q-ll
k. This shows that G ’ also satisfies condition (A). 3

H -

3. To prove the main part of the theorem, suppose that there exists a graph satisfying condition (A) that does not have the desired property. By adding new edges as many times as needed, we obtain a graph G = ( X , E ) satisfying condition (A), that does not have the desired property, and such that the addition of any new edge gives a graph with the desired property. Let F c E be a set with 1 F I = q that forms pairwise disjoint elementary chains, and that is not contained in any hamiltonian cycle of G. Since G is not complete, there exist two non-adjacent vertices y1 and yn ; choose y1 and yn with &(yl) + dG(y,) as large as possible and with dG(y1) 6 ddyn). There is a hamiltonian cycle p’ containing the edges of F i n graph G‘ = G + [yl, y,], and, necessarily, [ y , , y , ] E p ’ ; therefore, there is a hamiltonian chain p in G that contains F, of the form

P

= [ Y l , Y29

.-*,

Ynl

*

Let I={i/l i < n - 1,[vl,Yi+llEE,[v,,Yi+ll~F}. Set I is not empty, since (B) Moreover,

1I I

2 dG(y1) -

(C) ~ E + I because, otherwise, the cycle

> d1 -

> 0*

[~i,ynl$E

[yl, ~ 2 ..., , yi, Y n , Y n - 1 , ..*> Yi+l, ~ would be a hamiltonian cycle containing F.

1 1

206

GRAPHS

4. Let k = dG(y,). We shall show that

Since dl > q, the first inequality of (D) holds. From (C), it follows that ~G(Y,>= ( X - {YnH

-(Yi/i~Il.

Hence,

< ( n - 1) - [ ~ G ( Y-I )41 -

&(yn)

Thus ~ G ( Y ,+ ) dG(yn)G n

(El

+ q - 1.

Finally,

5. From (C), we know that, for each i E I, [Yi,Ynl # E

and ~G(Y,)+ d ~ ( ~
Thus, dG(yi) < k

Since I I

I

2 k - q, there are k

-

(i E I ) .

q vertices of degree

dk-,

< k.

Hence


which implies that dn-, 2 n - k + q .

This shows that there are at least k

+ 1 vertices of degree

'[YI~~I~E do(x) 2 n - k q.

\

+

Thus, n

+ q < dG(x) + dG(y1) < ~ G ( Y +I ) ~ G ( Y ,.)

This contradicts (E) and completes the proof of the theorem.

2 k. Hence

201

HAMlLTONIAN CYCLES

6. To show that the conditions of the theorem are the best possible, consider a non-decreasing sequence dl < d2 < -.-< d, and an integer k such that

[dn-k

< n - k + 9‘-

1.

Sequence (di)is majorized by sequence (d;),where

i:-

4‘= n - k i - q - 1 1

if if if

O
We shall show that there exists a graph G’ with the sequence (d;) for its degrees. Let the vertex set of G’ be the union of three disjoint sets A , By C, where A is a set Sk-aof k - q isolated vertices, B is a clique Kk of k vertices, and C is a clique Kn-ak+qof n - 2 k + q vertices. Join each vertex of B to each vertex of A U C. Graph G’ has order n, and d,(a) = k if a E A , d,*(b) = n - 1 if b E B, and dG(c) = n - k + q - 1 if c E C. Consider a set F of edges of G’ that forms an elementary chain of length q in By and suppose that there exists a hamiltonian cycle p that contains F. In sequence p, at most k - q elements of B are followed by an element of A U C. Since [ A I = k - q, there are exactly k - q elements of B followed by an element of A . Thus there is no element of B followed by an element of C, which is impossible, since I C I = n - 2 k + q > 0. Q.E.D. Corollary 1. Let G be a simple graph with degrees dl < d2 < < d,, Let < q < n - 3. I f the following condition holds ‘1.

q be an integer, where 0

l < i < j < n 4
I

d,+d,an+q,

then,for each set F of edges with I F I = q such that the components of ( X , F ) are vertex disjoint elementary chains, there exists a hamiltonian cycle that contains F.

Suppose that G is a simple graph satisfying this condition. We shall show that the graph G also satisfies the conditions of Theorem 8.

208

GRAPHS

If the conditions of Theorem 8 were not satisfied, there would exist an integer k such that

Let i

=

k - q and j

=

we have

n - k ; then i < j since, otherwise, k 3 (I .Thus, 2 +

1 6 i e j 6 n 4 < i + q

dj Q j

+

- 1,

and, consequently,

+ d, 2 n + q .

di

On the other hand, we have also di

+ d, < (k - q ) + q + (n - k ) + q - 1 = n + q - 1,

which yields a contradiction.

Q.E.D. Corollary 1 was proved directly by Bondy [1970]. It has been generalized by Las Vergnas [1970], who proved the following stronger result: Let G = ( X , E ) be a simple graph of order n 2 3. Let the vertices xi of G be indexed arbitrarily, and let q be an integer, where 0 < q 6 n - 1. if 1 9 i < j 6 n i + j 2 n - q

dG(x,) 6 i dG(xj) Q j [Xi,

!

+4 +q - 1

sl 4 E

then, for each F c E with

=s dG(XJ

I F I = q such

+ d,(Xj)

2 n

+4

that the connected components of

(A",F ) are elementary cycles, there exists a hamiltonian cycle that contains F. The following result is an easy consequence of Corollary 1.

209

HAMILTONIAN CYCLES

Corollary 2 (Kronk [1969]). Let G be a simple graph of order n, and let q be an integer, where 0 < q < n - 2. Suppose that (1) for each k sucli that q < k < 3 (n q - I), the set S, of vertices of degree < k has cardinality < k - q, (2) i f n + q is odd, and i f k = 3 ( n q - l), then I Sk I < k - q.

+

+

Thenfor each set F of edges with I F 1 = q such that (A’, F ) consists of rertex disjoint elementary chains, there exists a hamiltonian cycle that contains F.

If the degrees of G are indexed in increasing order, conditions (1) and (2) are equivalent to (1’)

dk-q > k

if

q < k < n+q2

(2’)

dk-Q+l> k

if

k

We have d k - q > k for all k <

+

2

=

-

1

51 (n + q - 1) . from (1’). For k =

n+q-1 2

3

we have from (2’):

In either case, the conditions of Theorem 8 are satisfied, and there exists a hamiltonian cycle that contains F. Q.E.D.

Corollary 3 (Chvital [1971]). Let G be a simple graph of order n 2 3 with degrees dl < d2 < *-. < d,,. If n d , < k < - * dn-,Bn-k, 2 then there exists a hamiltonian cycle in G. Furthermore, if the sequence (d,) does not satisfy the condition, there exists a graph G’, with a sequence of degrees majorizing (dt)that has no hamiltonian cycles.

This Corollary is established by setting q

=

0 in Theorem 8.

Note that Corollary 3 does not characterize all the sequences that imply the existence of a hamiltonian cycle. For example, Nash-Williams has shown that each regular graph of degree h with 2 h + 1 vertices has a hamiltonian cycle, and the corresponding sequence does not satisfy the above condition.

210

GRAPHS

Corollary 4 (Bondy [1969]). Let G be a simple graph of order n 3 3 with degrees dl < d2 < --.< d,,. If i#j d,
I

then G has a hamiltonian cycle.

This corollary is established by setting q = 0 in Corollary 1.

Corollary 5 (Chvsital [1972]).Let G such that I X I = I Y I = n 3 2, and

=

dG(X1) < dC(x2) < d d ~ 1 )< d d ~ 2 G)

( X , Y, E ) be a simple bipartite graph

< dG(xn) < d&n) .

If dG(Xk) < k < n

=>

dG(J',-k) 2 n - k

+ 1,

then G has a hamiltonian cycle. Let G ' be the graph obtained from G by joining each pair of vertices in Y. Since a hamiltonian cycle of G' cannot contain an edge of G' - G, it is sufficient to show that graph G' has a hamiltonian cycle. The degree sequence of G ' is dk = dG(x,) for k < n, and d; = dG(yk-,,) ( n - 1 ) for k > n. Clearly, this degree sequence satisfies the conditions of Corollary 3 and, consequently, G has a hamiltonian cycle. Q.E.D.

+

Corollary 6 (Chvsital [1972]).I f dl < d2 < * * * G d, are the degrees of a simple graph G of order n 3 2, and if n dk < k - 1 < - - 1 * d,,+l-k 2 n - k 2

then G has a harnilronian'chain. Let G' be the graph obtained from G by adding a vertex x, + and by joining this vertex to all the vertices of G. ( n + 1 ) - 1 , we have Let (d;)be the degree sequence of G ' ; for k < 2

-

5 d i = dk 4- 1 < k dA+l-k = dn+l-k + 1 3 n + 1 k . From Corollary 3, graph G ' has a hamiltonian cycle, and there'fore, graph G has a hamiltonian chain. Q.E.D.

21 1

HAMILTONIAN CYCLES

It is easy to verify that this result is the best possible in the same sense as Theorem 8. Besides, Las Vergnas [I9711 has extended this corollary to prove a sufficient condition for the existence of a spanning tree of G with at most h pendant vertices. The following corollaries are classical results : Corollary 7 ( P b a [1962]). Let G be a simple graph of order n 2 3, such that n-1 (1) for each k such that 1 < k < , the number of vertices of degree 2 < k is < k, n-1 n-1 (2) ( i f n is odd) the number of vertices of degree Q -is < 2 2 .

-

Then G has a hamiltonian cycle. This corollary is established by setting q

=

0 in Corollary 3.

Corollary 8. Let G be a simple graph of order n 2 3 with degrees dl 6 n d2 < ... < d,,. I f k < - implies dk > k, then G has a hamiltonian cycle.

2

Since k

n-1. k , the number of vertices of degree < k is

2

< k. From Corollary 7, G contains a hamiltonian cycle. Q.E.D. Corollary 9 (Erdos, Gallai [1959]). Let G be a simple graph of order n 2 3.

Let x1 be the vertex of minimum degree. IfdG(xl) 2 2, and i f d G ( x )2 ?for all 2 x # x l , then G has a hatniltonian cycle.

If the vertices are indexed as in Corollary 1, we have dG(xl) > 1. For n n 1 < k < 1, we have d,(x,) 2 - > k. From Corollary 8, there exists a hamil2 2 tonian cycle. Q.E.D. Corollary 10 (Ore [1961]).Let G = (A', E ) be a simplegraph of order n 2 3, such that 44x1 + d&) < n 5 [x,Y ] E E -

Then G has a hamiltonian cycle. n Let k From Corollary 2, it suffices to show that the set 2' { x / x E X,d,(x) < k } has cardinality < k. <-I.

s k

=

212

GRAPHS

From the hypothesis,

Sk

is a clique because k <

since the degrees of its vertices are

n

2 - Also I sk I < k

+ 1,

< k.

+

I Sk 1 # k 1 because, otherwise, s k would be a connected component of G , and two vertices x E s k and y E X - Sk would not be adjacent, implying dc(x)

+ d d y ) < k + (n - k - 2) = n - 2 < n .

This contradicts the hypothesis of the 'corollary.

I Sk I # k because, otherwise, each vertex of s k is adjacent to at most one vertex of X - s k , and the number of edges leaving s k is ma(Sk,

n

=k < 2

.

n this implies that there exists a vertex y E X 2 Let x E s k . Vertices x and y are non-adjacent, and

Since I X - Sk I > adjacent to

x - sk) < I s k I

sk.

dc(x)

-3

s k

non-

+ d&) < k + (n - k - 1) = n - 1 < n .

This contradicts the statement of the corollary. Thus I Sk [ < k, and G has a hamiltonian cycle.

Q.E.D. Theorem 9 (Ore [196 I]). Let G be a simple graph with n tlertices and m edges such that

1 m 2 - ( n - 1)(n - 2 ) + 2 . 2 Then G has a hamiltonian cycle. Furthermore, this inequality is the best possible. 1. From Corollary 10, it suffices to show that there do not exist two nonadjacent vertices a and b with

+

~G(u) &(b) < n - I . If two such vertices exist, graph G can be obtained from the complete graph K,,by removing at least

[(n - 2) - d&)]

+ [(n - 2) - dG(b)]+ 1 = - 3 - [dG(a)+ dG(b)]2 2 n - 3 - (n - 1) = n - 2

=2n

edges. Therefore, the number m of edges in G satisfies

m

< -21 n(n - 1)

- (n - 2)

which contradicts the hypothesis.

=

1 2

- (n - l)(n

- 2) +

1,

HAMILTONIAN CYCLES

213

2. Construct a graph Gowith n vertices by taking a clique K,,-l with n - 1 vertices and adding a vertex xo that is joined to Kne1by a single edge. The number of edges in Go equals 1 m(GJ = - (n - 1) (n - 2) + 1 . 2 Graph Go has no hamiltonian cycles. This shows that the inequality of the theorem is the best possible. Q.E.D. The following theorem is a slightly stronger formulation of a result due to Erdos and Chvatal. Theorem 10 (Erdos, Chvdtal [1972]). Let G = (X,E ) be a simple graph, and let h and q be two integers such that h 2 and 0 < q < 2. If graph G is hconnected, and if G contains no stable sets of h - q + 1 vertices, then,for each F c E with F I Q q, G has a hamiltonian cycle that contains F.

I

Recall that a set of vertices is called stable if no two of its vertices are adjacent. Let F be a set of edges with 1 F I < q. From Corollary 4 toTheorem (3, Ch. 9), there exists a cycle p that contains F. Assume that p is the longest 1 because Corollary 4 to cycle with this property. Then, I p I 2 h Theorem (3, Ch. 5) shows the existence of a cycle that contains any h - 2 vertices and any two edges not incident to these h - 2 vertices. If p is not a hamiltonian cycle, there exists a vertex a 4 p. Since I p I 2 h, Corollary 3 to Theorem (3, Ch. 9), shows that a can be joined to the vertices of p by h elementary chains pl[a, xl], p2[a, xz], .. .,ph[a, x h ] , that are pairwise vertex-disjoint (except at a). Let xi denote the unique vertex of p, in p, and let y, denote the vertex of p that immediately follows x, in p. Let I = { i / [xi,y i ] 6 F } . No yi,i E I, is adjacent to a, because, otherwise, there exists a cycle longer than p. Since I Z I 2 h - q, and since the set { y, / i E Z} u { a }cannot be stable (its cardinality is 2 h - q + l), there is an edge [y,,y j ] in G with i, j E Z and i < j . Consider the cycle

+

k[Yi, xrl - pda, x,l

+ pclb, XI3 - A x , , Yjl + [yr,nl

that contains F. The length of this cycle is greater than the length of p. This contradicts the maximality of p. Q.E.D.

+

Remark. The bound h - q 1 on the cardinality of a stable set is the best possible : For q = 0, the complete bipartite graph Kh, + is h-connected and has no hamiltonian cycle. For q = 1, the complete bipartite graph Kh,h

214

GRAPHS

augmented by one edge (between two vertices of the first class) has no hamiltonian cycle that contains this additional edge. Another example is the Petersen graph (Fig. 10.11). For q = 2, h 2 3, a counterexample is provided by the graph K h , h augmented by two edges joining vertices of the first class and one edge joining two vertices of the second class. No hamiltonian cycle can contain the first two additional edges. One can conjecture:

Conjecture. Let G = ( X , E ) be a simple graph with connectivity K ( G ) and with stability number a(G) such that ci(G) < K ( G ) - q. Let F c E be a set of edges with I F I = q such that the connected components of ( X , F ) are elementary chains. Then G has a hamiltonian cycle that contains F. By Theorem 10, this conjecture is true for q

=

0, 1,2.

In the following results, we shall only consider simple bipartite graphs.

Theorem 11 (Las Vergnas [1970]).Let G = (X,Y, E ) be asimple bipartite graph, where X = { xl, x2, ..., x,,}, Y = { y l , y 2 , ..., yn}, n 2 2; let q be an integer, 0 < q < n - 1. If fXj,

(4

Ykl

6

+

ddxj) 6 47(Y*) 6 k

+q

I

*

+ d&k)

2n

+4+ 1

then each set F of q edges that forms certex-disjoint chains is contained in a hamiltonian cycle of G. Suppose that there exists a bipartite graph that satisfies condition (A) but does not have the required property. By adding as many new edges to this graph as needed, we obtain a bipartite graph G = (A', Y , E ) satisfying condition (A), but not the required property, and such that the addition of any new edge gives a graph with the required property. Let F c E be a set of q edges that is not contained in any hamiltonian cycle of G. Letj(x) denote the index of vertex x E X , and let k ( y ) denote the index of vertex y E Y. 1. Let a, E X and 6 , E Y be two nonadjacent vertices that maximize j(a,) k(bl). The graph obtained from G by adding edge [al, b,] has a hamiltonian cycle that contains F and, therefore, graph G has a hamiltonian chain p that contains F. Let

+

P =

bz, a29 637 ..-,ai-,,bi, *..) an, b l ] .

215

HAMILTONIAN CYCLES

2. Let I

Then, 1 I

I2

=

2

{i/

dG(a1)

< i < n, b l , h1 E E,

- q. If i E I, then

[at-,, bi16

F).

[a,-,, b,] $ E because, otherwise,

bi, ai, bi+l, ..*>an, 61, at-,, bt-1, ...)bz, a11 would be a hamiltonian cycle containing F. Thus TG(bl)c X - { a i - , / i € Z } and so

dG(bd

< n - I 1 < n - dG(al)+ 4 ,

or

dAad + dAb1) < n + q . (B) 3. Letj(a,) = s, k(b,) = t , so that a, = xs,b, = y t . If i E I, it follows from Part 2 that [ai- b,] $ E, and, therefore, Hence Thus there are at least I I I vertices x withj(x) s 2

1I I

< s, and, therefore,

2 dc(al)- q .

Hence

dG(X8) ,i s

+q.

dG(Yd G t

+4

By the same argument, *

Since G satisfies condition (A), and since [x,, y t ]$ E, it follows that

ddx,)

+ d&t)

3n

+q + 1.

This contradicts (B).

Q.E.D.

Corollary 1 (Bondy [1969]). Let G = (X,Y, E ) be a simple bipartite graph with 1 X I = 1 Y I = n 2 2. r f the vertices of X and of Y are indexed with increasing degrees, and if

d~(xi>< j , MY& dk

&(xi)

+ d&d

2n

then G has a hamiltonian cycle, The corollary is established by setting q = 0 in Theorem 11.

+ 1,

216

I

GRAPHS

Corollary 2. Let G = (A', Y, E ) be a simple bipartite graph with I X I = Y I = n > 2, and let q be an integer, where 0 Q q < n, such that (1) for each integer j Q 4 , the set S, = { x / x E X , d G ( x )< j ) has +

2 cardinality < j - q, (2) for each integer k < -t- ' , t h e set Tk = { y / y € Y,dG(y)6 k } has 2 cardinality < k - q. Then for each set F of q edges that forms vertex disjoint elementary chains, there is a hamiltonian cycle that contains F. This follows easily from Theorem 1 1 . Corollary 3 (Moon, Moser [1963]). Let G = ( X , Y, E ) be a simple bipartite graph with I XI = I Y I = n 2 2. r f l s k 1 < kand I Tk I < k foreachinteger n k < * Then G has a hamiltonian cycle. 2 The corollary is established by letting q = 0 in Corollary 2.

-

Corollary 4. Let G = (X,Y, E ) be a simple bipartite graph with I X I = n 1 Y 1 = n 2 2. r f each certex z E X v Y satis$es dG(z) > , except for two 2 vertices, x1 E X with dG(xl) 2 2, and y , E Y with dG(y,) Z 2, then G has a hamiltonian cycle.

If k = 1, then I f k > I andk

I sk I = 0 < k ; I T k I n <-, then 2

=

0
ISkl
I YI =n 2

dG(x)

then G has a hamiltonian cycle. The corollary is established by letting q Corollary 6. Let G

=

[ x , y1

f dG(Y) 6

=

E

9

0 in Theorem 11.

(X,Y, E ) be a simple bipartite graph with

IA'I=IYI=n,

IEl=m.

IX1 =

217

HAMILTONIAN CYCLES

rf m 2 n2 - n

+2,

then G has a hamiltonian cycle. It suffices to show that there do not exist in G two nonadjacent vertices x and y with do(x) d&) < n. If two such vertices exist, then graph G can be obtained from the complete bipartite graph K,,%by removing a set of edges of cardinality at least

+

(n - &X))

+ (n - d,(y)) - I = 2 n - (dG(X)+ dG(y))- 1 2 n - I .

On the other hand, the number of edges eliminated from K,,,, is n2

- rn 6 n2 - (n2 - n + 2)

=n

- 2,

which is a contradiction. Q.E.D.

5. Hamilton-connected graphs Consider a simple graph G = ( X , E ) . Graph G is defined to be Hamiltonconnected if for each pair x , y of distinct vertices, there is a hamiltonian chain with endpoints x and y . In this section we shall study sufficient conditions for a graph to be Hamilton-connected. The above definition can be generalized. Let F be a set of q edges that form vertex-disjoint elementary chains. Let x , y be a pair of distinct vertices which are the endpoints of two distinct chains of ( X , F ) . Graph G is defined to be q-Hamilton-connected if for each such F and for each such pair x, y, there is a hamiltonian chain with endpoints x and y that contains F. For q = 0, a q-Hamilton-connected graph is a Hamilton-connected graph.

Lemma. For two integers q and n with 0 < q < n - 2, let Z ( n , q ) be a class of simple graphs of order n satisfying the two following conditions: (1) I f G E X(n,q), each set F of q edges forming a system of certex-disjoint elementary chains is contained in a hamiltonian cycle. (2) If G E Z ( n , q), the graph G' obtained from G by adding any new edge also belongs to 2 ( n , q). Then each graph of %(n, q) is (q - 1)-Hamilton-connected. Let G = (A', E ) be a graph of X ( n , q), and let F be a set of q - 1 edges of G such that the connected components of ( X , F ) are elementary chains. Let x, y be two vertices that are the endpoints of two distinct chains of ( X , F). If [x,y ] E E, let G' = G; otherwise, let G' = G [x,y ] .

+

218

GRAPHS

From condition (2), graph G ' belongs to Z ( n , q ) ; since F u { [x,y ] } constitutes a system of vertex-disjoint elementary chains of total length q, graph G ' has a hamiltonian cycle that contains F u { [x,y ] }. Therefore, F is contained in a hamiltonian chain of G with endpoints x and y. Q.E.D. dl

Theorem 12. Let G be a simple graph of order n 2 3 with degrees < d, < < d,, that satisfies the following condition:


4 - 1

1

Then graph G is Hamilton-connected. The graphs that satisfy this condition form a class A"(n, 1) by Theorem 8. ~. Q.E.D. Theorem 13. Let G be a simple graph of order n 2 3 such that (1) for each integer k , where 1 < k <

I sk I < k - 1, n (2) i f n is even, I S,/, I < - - 1 .

n2 , the set

Sk

of vertices of degree


has cardinality

2 Then, graph G is Hamilton-connected. The graphs that satisfy these conditions form a class A"(n, l), by Corollary 2 to Theorem 8. Q.E.D. Theorem 14. Let G = (X,Y, E ) be a bipartite graph, ulith I X I = n 2 2, in which the fiertices xi E X and yi E Y are indexed such that: dG(X1)

dGGGyI)

< d G ( X 2 ) < < &(%) < dGcv") < dGGCy2) ***

I Y1 =

7

*

Suppose that, i f j and k are the smallest two indices such that d~(xjG ) i

+ 1,

dGGGyk)

< k + 1,

(if they exist), we have

+

dG(xj) dG(yk)> n

+2.

Then, graph G is Hamilton-connected. From Theorem 11, the graphs that satisfy these conditions form a class A"(2 n, 1). Q.E.D.

219

HAMILTONIAN CYCLES

Theorem 15. Let G

n

=

( X , Y, E ) be a bipartite graph with I X

> 2 satisfying the two following conditions: ( 1 ) for each j < , the set of certices xi of 2

*

degree

<j

I=1

YI=

has cardinality

< j - 1,

(2) for each k < -I- , the set of vertices y , of degree 2
< k has cardinality

From Corollary 2 to Theorem I I , the graphs that satisfy these conditions form a class X(2 n, 1). Q.E.D.

Theorem 16 (Erdos, Gallai [1959]). Let G be a simple graph of order n 2 3 with dG(x) dG(y) > n for each pair x , y of distinct, non-adjacent vertices. Then, graph G is Hamilton-connected.

+

From the Lemma, it suffices to show that (1)

x

z Y,

d6.W

+4

*

Y )G n

[x, Yl E

implies (2) each edge of G is contained in some hamiltonian cycle. The proof is similar to the proof for the first part of Theorem 8. Let G be a simple graph that satisfies (I), but not (2). We may assume as before that the addition of any new edge causes the graph to satisfy property (2). Let e, be an edge of G that is not contained in any hamiltonian cycle. There exist two non-adjacent vertices a and b, since G # K,,. Since G + [a, b] has a hamiltonian cycle containing e,, there exists in G a hamiltonian chain containing e, that is of the form

~ [ abl,

= [a, ~

z

~

r

3

-..) , y n - i , b] .

Let y1 = a, and let Z = { i / i > 2 , [ a , ~ i l ~ [EY,i - l , ~ r I + e o ) .

Then,

I I I 2 dG(a)- 1 . For i E I, we have yr- $ T,(b) because, otherwise, [a, Y , , Y i + 1 , Yi+Z, - - * s ~

n ~ ,i - 1 ~ ,

i - 2 ,

a1

220

GRAPHS

would be a hamiltonian cycle containing e, . Thus, n - 1 dG(6) 2 I I ( > &(a) and d&) d&) < n .

-

-1 d

+

But, from (l), this implies that [a, b] E E which is a contradiction. Q.E.D. Theorem 17 (Ore [1963]). If G is a simple graph of order n with m edges such that ( n - 1) ( n - 2) m > 3, 2 then G is Hamilton-connected.

+

Let Ga= ( X , E ) be a graph that satisfies the inequality. Consider two nonadjacent vertices a and 6 . Let Eodenote the set of edges adjacent to neither a nor b. Then, I E I = dG(4 + d d b ) + I En I , and 1

IEOI <+-)("-3), 1 I E I 3 ~ ( -n l ) ( n - 2)

+3.

Thus, dG(a)

+ dG(b) = I I

- I EO

I

3 n - 2

2 T-[(n

- 1) - ( n -

Therefore, from Theorem 16, G is Hamilton-connected. Q.E.D.

The inequality presented by this theorem is the best possible: Consider the complete graph K,, with n vertices and remove all the edges incident to vertex a except for two edges. If n 2 4, this graph is not Hamilton-connected because the two vertices adjacent to a cannot be the endpoints of a hamiltonian chain. The number of edges of this graph equals 1

m = - n ( n - 1) 2

1 - (n - j) = -(n - l ) ( n - 2) + 2 . 2

Theorem 18 (Moon [1965]). The minimum number of edges in a simple Hamilton-connected graph of order n 2 4 is [+ (3 n I)].

+

221

HAMLTONIAN CYCLES

1. If m < [-) (3 n + l)], then a graph G with m edges contains at least one vertex a with &(a) < 2 because, otherwise,

which is impossible. Therefore, this graph G is not Hamilton-connected (because n > 4 and the vertices adjacent to a cannot be the endpoints of a hamiltonian chain).

+

2. For m = [-)(3 n l)], we shall show that there exists a Hamiltonconnected graph G, with n vertices and m edges. For n = 2 k,graph G, is shown in Fig. 10.9. For n = 2 k - 1, graph G, is shown in Fig. 10.10. *

a q p"T9. Sk-2

xk-1

yk-2

yk-I

G2k

Fig. 10.9

G2r-

1

Fig. 10.10

It is left to the reader to verify the existence of hamiltonian chains of the forms PI!', 1' , ~ [ axi], , ~ j ] , P[x~, yj] in graph GZk, and hamiltonian chains of the forms PIG,xi19 P[xi, xjl, PIX,, Y in graph G2k-1. The number of edges in Gzkequals

~I

222

GRAPHS

The number of edges in G,,-,

=

equals

3 k - 1 = [ 21 ( 6 k - 2 ) ] = [ 1 -(3n 2

+ l)]. Q.E.D.

Theorem 19 (Karaganis [ 19681). If G is a connected graph of order n 2 2, then its cube G 3 (the composition product G ‘ G . G ) is Hamilton-connected. Graph G3 has the same vertex set as G . Two vertices are adjacent in G3 if, and only if, their distance in G is < 3 (see Chapter 4). Consider a spanning tree H of G. We shall show by induction that H 3 is Hamilton-connected. Clearly, this is true for n = 2. Suppose that that is true for all trees of order < n and consider a tree H of order n > 2. Let a and b be two distinct vertices of H. We shall show that H 3 has a hamiltonian chain between a and b. Since H is a tree, it possesses a unique chain p[a, b] = [a, xl, ..., b] between a and b. If edge [a, x,] is removed from H , two trees Ha and H b are formed that respectively contain a and b. Let p [ a , a’] be a hamiltonian chain in (Ha)3between a and a neighbour of a in Ha (where a’ is distinct from a if Ha is not a single vertex). Let p[b‘, b] be a hamiltonian chain of (Hb)3between b and vertex b’, the neighbour of x1 in H b (where b’ is distinct from b if H b is not a single vertex). Vertices a‘ and b‘ are separated in H by at most three edges. Thus, they are adjacent in H 3 . It follows that p[a, a’] + [a’, b’] + p[b’,b] is the required hamiltonian chain between a and b in H 3 . Q.E.D. Theorem 19 shows that the vertices of a tree of order n can be numbered from 1 to n with numbers 1 and n assigned arbitrarily, such that any two vertices with consecutive numbers are separated by at most 3 edges.“’ “ ) A proof for Theorem 19 already appeatedin 1960 when Sekanina showed that the cube of a connected graph has a hamiltonian chain. (M. Sekanina, “O n an Ordering of the Set of Vertices of a Connected Graph,” Publ. Fac. Sc. Bmo, 412, 1960, pp. 137-141.) Sekanina also asked for graphs whose square contained a hamiltonian cycle. J. Nash-Williams conjectured that the square of a 2-connected graph has a hamiltonian cycle. This was shown to be true by Fleischner (H. Fleischner, “The Total Graph of A Block is Hamiltonian,” to appear. H. Fleischner, “ O n Line Critical Blocks,” to appear). More recently, Hobbs and Nash-Williams, using Fleisch. ner’s methods, has shown that the square of a 2-connected graph is Hamilton-connected (A. M. Hobbs, C. St. J. A. Nash-Williams, “The Square of a Block is Hamiltonian Connected,” to appear).

HAMILTONIAN CYCLES

223

6. Hamiltonian cycles in planar graphs (abstract) It has often been conjectured that each regular 3-connected graph of degree 3 possesses a hamiltonian cycle. The first known counter-example was the Petersen graph (see Fig. 10.11). In fact, the Petersen graph is the only graph of order < 10 without hamiltonian cycles, such that the removal of any vertex creates a graph with a hamiltonian cycle (R. Sousselier, in Berge [1963]).

Fig. 10.11. Petersen graph

Fig. 10.12. Tutte graph

The conjecture is false even for planar graphs. Tutte [I9461 constructed the first 3-connected planar graph regular of degree 3 that has no hamiltonian cycles (see Fig. 10.12) Barnette and JucoviE [1970] have shown that for a 3-connected planar graph without hamiltonian cycles, the smallest number of vertices is 11; an example with 11 vertices is given in Fig. 10.13. Clearly, since this graph is bipartite with an odd number of vertices, no hamiltonian cycle exists. M. Balinski [1961] conjectured the existence of a hamiltonian chain in all planar cubic 3-connected graphs. B. Griinbaum and T. S. Motzkin [I9621

Fig. 10.13. Herschel graph

224

GRAPHS

presented a counter-example by constructing a cubic, planar 3-connected graph with 944 vertices in which no elementary chain could contain more than 939 vertices (see also Brown [1961], JucoviE [1968]). The simplest example, given in Fig. 10.14, contains only 88 vertices (Zamfirescu, in Griinbaum [19701).

Fig. 10.14. Zamfirescu graph

Tutte [1956] has shown that each 4-connected planar graph has a hamiltonian cycle (see Ore [1968], for a proof). A graph is defined to be (k)-edge-connectedif it cannot be disconnected into two connected components, each containing a cycle, by the removal of less than k edges. In 1884, Tait conjectured that each simple (3)edge-connected planar graph regular of degree 3 has a hamiltonian cycle. The Tutte graph (Fig. 10.12) provides a counter-example to this conjecture. Moreover, (4)edge connected planar graphs without hamiltonian cycles have been constructed by different authors (Hunter [1962], Lederberg [1966], Tutte [1960]).

Fig. 10.15. Another Tutte graph

225

HAMILTONIAN CYCLES

The problem is more difficult for planar (5)-edges connected graphs. The first example (Walther [ 19651 of a planar (5)-edge connected graph without hamiltonian cycles had 114 vertices. Figures 10.15 and 10.16 present the two simplest known examples with 46 vertices (Kozyrev and Grinberg, in Sachs [19681) and 44 vertices (Tutte [1972]), respectively.

Fig. 10.16. Graph of Kozyrev and Grinberg

Planar graphs with or without hamiltonian cycles have been constructed by Bosak [1967], Culik [1964], Kotzig [1962] and Sachs [1967].

EXERCISES 1. Show that the complete bipartite graph K,, has no hamiltonian cycles if p Z q, and it has [p / 21 1 disjoint hamiltonian cycles if p = q.

-

2. Show using the theorem due to P6sa (Corollary 7 t o Theorem 8) that if G is a graph

with n vertices, m edges and of minimum degree k, and if

then G has a hamiltonian cycle. Show that this is the best possible result.

(Erdos [1962]) 3. Show that if G = (X,Y, -E ) is a bipartite graph with

with minimum degree k

<

:,

and satisfies

then G has a hamiltonian cycle.

1 XI

=

I Y I = n, I E I = m,

226

GRAPHS

4. Show that the Tutte graph (Fig. 10.12) has no hamiltonian cycles. Hint: Use reductions to show that if the graph consisting of the triangle edfand its interior had a hamiltonian chain, then this chain cannot have both e andffor endpoints. 5. Show that the graph in Fig. 10.15 has no hamiltonian cycles

6. Show that the graph of Kozyrev and Grinberg (Fig. 10.16) has no hamiltonian cycles. Hint: Suppose the existence of a hamiltonian cycle p. Denote byf; the number of faces with a contour of i edges that lie in the interior region of p. Denote byf,” the number of faces with a contour of i edges that lie in the exterior region of p. Show that

C(i-2)fi’

=

n-2

=

C(i-2)fi”.

Since f;= fi” = 0 for i # 5,8,9, then . 3f; 6fi 7f; = 3fJ 6fi Hence fi = f : mod 3, which contradicts fi fi = 1.

+

+

+

+

+ 7fi.

7. If G = (A’,U )is a complete symmetric graph whose arcs are divided into twa classes U’and U”,show that there exist two vertices a and b and a hamiltonian circuit p such

that the portion of p from a to b contains only arcs of U’, and the portion of p from b to a contains only arcs of U”. (H. Raynaud [1970]) 8. I n a 1-graph G = (A‘, r),a humiltoniun bi-path is defined to be two elementary paths such that each vertex belongs to exactly one of these two paths. Let €denote the complementary 1-graph of G. Let h,(G) denote the number of hamiltonian circuits in G , h,(G) denote the number of hamiltonian paths in G , and h,(G) denote the number of hamiltonian bi-paths in G. Show that ho(G) = ho(G)

+ hl(G) + h2(G)

(mod. 2)

.

Show that the number of hamiltonian bi-paths in a complete anti-symmetric 1-graph is odd. (Berge, [1967]). Hinr: Use the proof of Theorem 1 , replacing the word “arrangement” by “circular permutation of degree n”. 9. A graph G of order n is defined to be hypohamiltonian if it has no harniltonian cycles, but the subgraph G , obtained by removing any vertex x has a hamiltonian cycle. Prove the following: (1) If G is hypohamiltonian, then n > 3. (2) If G is hypohamiltonian, then d,(x) 2 3 for each vertex x. (3) If G is hypohamiltonian, and if y and z are consecutive vertices of an elementary cycle of length n - 1 in G,, then x is not adjacent to both y and z .

<

for each vertex x. 2 (5) If G is regular of degree h, then hn = 2 m. (6) Let x, a, b, a’, b’ be distinct vertices of G such that x is adjacent to a and to a’, b is adjacent to b’, and arcs (u, 6 ) and (a’,b’) are both contained in some harniltonian cycle of G,. Then, G has a hamiltonian cycle. (7) If G is hypohamiltonian, then n k 7. (8) If G is hypohamiltonian, then n # 7. (9) If G is hypohamiltonian, then n # 8. (10) If G is hypohamiltonian, then n # 9. (11) If G is hypohamiltonian of order 10, then G is regular of degree 3. (12) The Petersen graph is hypohamiltonian. (13) Each hypohamiltonian graph of order 10 is isomorphic to the Petersen graph. (For a complete proof, see Herz, Duby, ViguC [1967].) (4) If G is hypohamiltonian, then d,(x)

HAMILTONIAN CYCLES

221

10. In a simple, regular graph of degree 3, consider two edges incident to a vertex a. Show that the parity of the number of hamiltonian cycles that contain these two edges does not depend upon which two edges incident to a have been chosen.

11. Let G = (A', U)be a complete strongly connected graph of order n 3 3; show that each vertex lies on some circuit of length k, for k = 3,4, ..., n.

12. Let G be a complete strongly connected graph of order n 3 4; show that there exist two distinct vertices a and b such that the subgraphs Gx-co,and Gx-,a, are both strongly connected.

CHAPTER 11

Covering Edges with Chains

1. Eulerian cycles One of the oldest combinatorial problems, due to Euler, can be stated as follows: An eulerian chain (respectively, eulerian cycle) is defined to be a chain (respectively, cycle) that uses each edge exactly once. When does a multigraph have an eulerian chain or an eulerian cycle? EXAMPLE 1. Is is possible to trace out the graph in Fig. 11.1 without lifting your pencil from the paper and without repeating any edge? After several attempts, the reader will find that this is impossible. However, this is possible for the graph in Fig. 11.2, which has an eulerian chain.

w Fig. 11.1

Fig. 11.2

EXAMPLE 2 (Euler). The city of Konigsberg (today known as Kaliningrad) is divided by the Pregel River that surrounds the Island of Kneiphof. There are seven bridges in the city as shown in Fig. 11.3. Can a pedestrian traverse

h

C

d G Fig. 11.3 228

COVERING EDGES WITH CHAINS

229

each bridge exactly once? This problem puzzled the residents of Konigsberg until Euler showed in 1736 that no solution exists. Consider the multigraph G in Fig. 11.3 whose vertices represent the districts a, b, c, d of Konigsberg and whose edges represent the bridges of Konigsberg. Clearly, the bridge problem is solved by finding an eulerian chain in this graph. meorem 1 (Euler [1766]).A multigraph G has an eulerian chain i f , and only if, it is connected (except for isolated vertices) and the number of vertices of odd degree is 0 or 2.

1. Necessity. If there is an eulerian chain p in G, then G is clearly connected. Furthermore, only the two endpoints of p (if they are distinct) can be of odd degree. Thus, there can only be 0 or 2 vertices of odd degree.

2. Suficiency. We shall prove by induction: Ifthere are only two vertices a and b whose degree is odd, then there exists an eulerian chain that starts at a andjnishes at b ; if there are no points whose degree is odd, then the graph possesses an eulerian cycle. We shall assume this statement to be true for graphs with fewer than m edges, and prove that it also holds for a graph G with m edges. To be more definite, assume that G has two vertices a and b of odd degree. The required chain p will be defined by a traveller who traverses the graph from a and moves in such a way that he never uses the same edge twice. If he reaches a vertex x # 6, he will have been along an odd number of the edges incident to x , and therefore he will be able to leave x by an edge which has not yet been used; when this is no longer possible, he must necessarily have arrived at 6. However, it is possible that not all edges have been used. After removing the used edges, we obtain a partial graph G whose vertices are all of even degree. Let C1, C,,..., ck be the connected components of G’ that have at least one edge. By the induction hypothesis, these components have eulerian cycles, pl,p, ,.. .,p k . Since G is connected, chain p encounters each of the C,. Without loss of generality, suppose that p first encounters component C, at vertex xl, then encounters component Ca at vertex X a , etc. Consider the chain

Clearly, this chain is an eulerian chain from a to b.

Q.E.D. The reader can now see that no solution exists for the Konigsberg Bridge problem (Example 2).

230

GRAPHS

Similarly, it is possible to show that the edges of a connected multigraph with exactly 2 q vertices of odd degree can be covered with only q chains. AN EULERIAN CYCLE.Consider a conLOCALALGORITHM to CONSTRUCT nected multigraph G in which all vertices have even degree. The following rules construct an eulerian cycle in G : RULE1. Starting from any vertex a, follow a chain without using the same edge twice. RULE2. I f we arrive at a vertex x different from a after the kth step, never depart from vertex x along an edge that is an isthmus of the partial Graph G, generated by the unused edges, unless x is a pendant vertex of G,. RULE 3. If vertex a is revisited, depart from vertex a along any unused edge if it exists. If no unused edge exists, stop. We shall show that the chain generated by the algorithm is eulerian. 1. It is always possible to follow the rules. Upon arrival at a vertex x # a, there is always in GI, an edge incident to x, because dc(x) is even, and therefore d,,(x) is odd. If this edge is unique, it is a pendant edge in Gkand can be used, If this edge is not unique, there is at least one other edge that is not an isthmus, since, otherwise, there are two isthmi in Gk joining x to two distinct connected components C and D. Note that C contains at least one vertex of odd degree, because, otherwise, 0 = C dGk(x)= 1 2 m,,(C, C) mod 2, XE

c

+

which is impossible. Similarly, D contains at least one vertex of odd degree. Since Gk contains exactly two vertices of odd degree, and since one of them is x, the contradiction follows.

2. If the rules are followed, the chain determined during the first k steps is the beginning of an eulerian cycle. Graph G, is connected except for isolated vertices. Upon arrival at a vertex x # a, there are only two vertices x and a of odd degree in G,. Therefore, from Theorem 1, GI,has an eulerian chain from x to a. Q.E.D. We shall now apply the Euler theorem to the study of the factors of a graph. A factor is defined to be a system of vertex-disjoint elementary cycles such that each vertex is contained in exactly one cycle. Theorem 2 (Petersen [1891]). I f G = (A', E ) is a regular multigraph of even degree h = 2 k , then G has k edge-disjoint factors.

23 1

COVERING EDGES WITH CHAINS

1. First we shall show that G has a factor. From Theorem 1, each connected component of G has an eulerian cycle. By directing each edge along the direction of travel in this eulerian cycle, we obtain a graph H = (X,U) with &(x) = &(x) =

k

(X

EX).

The bipartite multigraph If = ( X , y, E ) obtained by making two copies X and of X , and by taking

x

rn,-(xi, i j )= r n i ( x i , x j ) , is regular of degree k. Thus, from Corollary 4 to Theorem ( 5 , Ch. 7). contains a perfect matching E,, that corresponds to a factor in G.

B

2. The theorem is obviously valid for k = 1. Assume it is valid for every regular multigraph of degree 2 k' < 2 k ; we shall show that the theorem is valid for a regular multigraph G = ( X , E ) of degree 2 k. From part (l), we know that G has a factor Eo c E. The partial multigraph ( X , E - Eo) o f degree 2(k - 1) has k - 1 disjoint factors El,E 2 ,..., Ek-.l by the induction hypothesis. Thus, G has k disjoint factors Eo, E l , ..., Ek-1. Q.E.D. Corollary. I f G is a regular multigraph of odd degree 2 k min rn,(S,X

s+x

+ 1 such that

- S) 2 2 k ,

S# 0

then G has k edge-disjoint factors. Without loss of generality, we may assume that G is connected. Clearly, G is a graph with an even number of vertices. Then, from Theorem (13, Ch. 8), G possesses a perfect matching. The edges not contained in the matching form a regular graph of degree 2 k , which from Theorem 2 can be decomposed into k disjoint factors. Q.E.D.

Remark. Babler [1938] extended this corollary as follows: A regular graph 1 contains q edge-disjoint factors, q < k, if

of odd degree h = 2 k

+

min rnG(S,X S#X

s+ 0

- S) 2 2 q .

232

GRAPHS

2. Covering edges with disjoint chains In this section we shall study the following problem : Given a simple graph G = (X,E), cover its edges with as few as possible edge-disjoint chains of a certain type. What is the smallest number of such chains needed to cover E? EXAMPLE1 (Kirkman). Each day, n knights meet at a round table. No knight wants to sit next to the same neighbour twice. How many days can they meet? The answer is the maximum number of disjoint hamiltonian cycles in a complete graph with n vertices. Later we shall show that this number is

['!!I.

EXAMPLE 2 (Lucas). Six boys a, b, c, d, e, f and six girls a, 6, E, a, b, f join hands and dance in a circle, How many dances can they dance so that no boy ever joins hands with the same girl twice and no boy ever joins hands with another boy? The problem is solved by finding the maximum number of disjoint hamiltonian cycles in the complete bipartite graph K 6 , 6 . There are three disjoint hamiltonian cycles in K 6 , 6 : n

a ii b b c c d d e e f J a acbicedfeaf ba n

Z bf c d F e C f d

o .

d Fig. 11.4

These cycles are found by rotating the labels of the tips of the star in Fig. 11.4. Since each vertex has degree 6, clearly, these cycles cover all the edges. Theorem 3. The maximum number of pairwise edge-hamiltonian cycles in the complete graph K,, is 1. First, suppose that n = 2 k

-1 + 1 is odd. We shall show that k = I17 I

disjoint hamiltonian cycles can be found. Number the vertices 0, 1, 2, . .., 2 k and place them on a circle as in Fig. 11.5. Then, the sequence [ 0 , 1 , 2 , 2 k , 3 , 2 X - - 1 , 4 ,..., k + 3 , k , k + 2 , k + 1 , 0 ]

COVERING EDGES WITH CHAINS

233

is a hamiltonian cycle. If 1 is added (modulo 2 k ) to the index of each vertex # 0, i.e. if the system of solid lines in Fig. 11.5 is rotated around point 0, then a new hamiltonian cycle is obtained. This cycle is disjoint from the preceding one because the sum of two successive vertices is 2 or 3 in the first cycle and 4 or 5 in the second cycle. Thus k - 1 rotations around point 0 can be made, and these rotations produce k disjoint hamiltonian cycles.

k;l Fig. 11.5

+

2. Suppose that n = 2 k 2 is even. Number the vertices of the graph 0, 1, 2, ..., 2 k , a, and place them on a circle as shown in Fig. 11.5, with vertex a placed on another plane above the centre of the circle. Similarly, k disjoint hamiltonian cycles can be produced by rotating the solid lines in Fig. 11.5 around point 0. Thus K,, contains k tonian cycles.

=

[2 '1 disjoint hamil.Q.E.D.

n-1 Corollary 1. Z f n is odd, the edges of graph K, can be covered by 2 disjoint hamiltonian cycles. Since the degree of each vertex is n - 1 cycles cover all the edges.

=

2 k, the k disjoint hamiltonian

Q.E.D. Corollary 2. If n is even, the edges of graph K, can be covered by a perfect matching and

[

~

2

] disjoint hamiltonian cycles.

234

GRAPHS

If the k hamiltonian cycles are removed, the remaining partial graph is regular of degree (n - 1) - 2 k = 1. Q.E.D.

n

Corollary 3. r f n is even, the edges of graph K, can be covered by - disjoint 2 hamiltonian chains.

The edges of K,,, can be covered by disjoint Hamiltonian cycles as shown in Corollary 1. If we remove one vertex, these cycles correspond to '1 disjoint 2 hamiltonian chains of K,. Q.E.D. Corollaries 1 and 3 are also consequences of the following general theorem: Theorem 4 (LovBsz [1968]). The edges of a simple graph G with n vertices can

12

always be covered by - elementary chains and cycles that are pairwise edge-

-

disjoint, or less.

1. We may assume that the graph G has no isolated vertices and no isolated edges. Otherwise, we could prove the theorem for the graph obtained by removing isolated vertices and edges.

2. Let m denote the number of edges in graph G. If 2 m

- n < 0,- then

n

Since d,(x,) 3 1 for all i, we have d,(x,) = 1 for all i. This contradicts the hypothesis that G has no isolated edges. Thus 2 m - n > 0.

3. We shall assume that the theorem is valid for each graph G' of order n' with

[g] [5] =

and

2 m ' - n' < 2 m

-n

and we shall prove the theorem by induction on 2 m - n. 4. First, suppose that there is a vertex of even degree. Let x be such that: rG(x)

= { a l , a 2 ,..., ak, b,, b2, ..., b l l ,

d,(aJeven

, d,(bj)

odd ,

k 2 1.

Let G ' be the graph obtained from G by removing edges [a,,x ] for i = 1, 2, ..., k. Graph G' satisfies 2m' - n' < 2 m - n, and contains a minimum disjoint elementary chains and covering M' = { p i , ph, ...} of less than cycles (by the induction hypothesis).

COVERING EDGES WITH CHAINS

23 5

Since vertices u l and 6, have odd degree in G‘, each of these vertices is the endpoint of an elementary path of M’ of length > 0. Since the covering M’ is minimum, each of these vertices is the endpoint of exactly one path of

M’. 5. Thus Vertex a, is the endpoint of a chain p’[ac,zt] E M’.Suppose that x E $[ai, zJ. Let y: denote the next to the last vertex encountered in the partial chain ,u’[u,, XI.Since vertex y: is necessarily a b,, it follows that y: is the endpoint of a chain p’[y:, z:] E M’. If x E p ’ b t , z:], define similarly a vertex y:, etc. In this way, a finite sequence 0

ai=Yi,

1 2 Yi, Yi,***

is determined for each i < k. It is easy to see that yf’ = yl implies p and i = j . Furthermore, the set Y of all yf (for all i and all p ) satisfies

{

/i
c Yc

=q

I‘&) .

G’

6. Now, we shall show that a covering M’ of G’ in the following way: If p; E M‘ has an endpoint in Y, let

cr,

=

01,)of G is obtained from covering

*

236

GRAPHS

Thus, p; is an elementary chain or an elementary cycle of G. If p;[yr,z,"] E M ' has in Y only one endpoint, say yp, let P s = Pul =

+ CYP, X I

{ pl

if

+ [yp, x] } - { bf'!,x]

x

# PICYP, 4

otherwise.

Thus, p, is an elementary chain of G. If p&$', yy] E M' has two endpoints yp and y: in Y, let Pt

=

,4 + CYP, xl + Ivy, x l

x 4 PL:CYf, Yjl

if

+

= { p i -I- [ y f , XI [uj, XI} - { CyP",

XI, [yj",

x]

Thus, in the first case, pt is an elementary cycle of G, and pi is an elementary chain of G.

}

otherwise.

in the second case,

To show that the set M of all the chains p, forms a covering of G, note first that - an edge of G that is not of the form bf,x ] is also an edge of G and is covered by a single chain of M , - an edge of the form [yp ,x] is covered by the only chain of M that comes from the chain p; (or p i ) that has yf as an endpoint. 7. Now, suppose that each vertex of G is of odd degree; consequently, n is even.

Replace G by a graph GIobtained by taking a vertex x of odd degree 2 3 and adding a vertex a, in the middle of some edge incident to x , say [x, 21. Since graph G; (obtained from graph GI as in Part 4) satisfies 2 m ; - n', = 2 m

GI can be covered by

- (n + 1)

-= 2 m - n ,

[ ?] = n5 disjoint chains. Each vertex of GI that is

distinct from a, is the endpoint of a covering chain. Hence, a, is not the endpoint of a covering chain. Thus, the !! covering chains of GIalso cover G. 2

Q.E.D. Corollary. Let G

=

( X , E ) be a simple graph of order n with only vertices

of odd degree; then its edges can be covered by :-edge disjoint elementary chains

2 so that each vertex of G is the endpoint of exactly one of these chains.

Since 2 d,(x)

=2m

is even, then n is even. Therefore, from Theorem 3,

231

COVERING EDGES WITH CHAINS

the edges of G can be covered by If disjoint. chains and cycles. Furthermore, 2 each vertex is necessarily the endpoint of at least one covering chain, because its degree is odd. Q.E.D. We shall now state three conjectures related to Theorem 4 . Conjecture 1 (Haj6s). I f all the n vertices have even degree, the edges can be covered by - edge-disjoint elementary cycles, or less.

[;I

Conjecture 2 (Gallai). The edges of a connected graph of order n can be n+l covered by 7edge-disjoint elementary chains.

[

]

Conjecture 3 (Nash-Williams). For n 2 15, ifthe number of edges is divisible 3n by 3 and if each vertex has even degree 2 - 3 then the edges can be covered 4 by edge-disjoint triangles. Theorem 5 (Erdos, Goodman, P6sa [1966]). Let G be a simple graph of order n; then the edges of G can be corered by [n2/4] edge-disjoint cliques, each of cardinality < 3. 1. The theorem is true for n = 2. Assume that the theorem is valid for all graphs with n - 1 vertices, we shall show that it is also valid for a graph G of order n > 2. First, we shall show that

[;I

=

['" ,'"] + [5] .

I f n = 2 k, then

If n

=

2k

+ 1, then

[;I

2. If G contains a vertex xo with degree dG(xo)< - , the subgraph Go

[

obtained by removing xo can be covered by (" ~

'I2] cliques of cardinality

< 3 that are pairwise disjoint. Thus, the minimum number of disjoint cliques that cover the edges of G is less than or equal to

238

GR-APHS

.

['"

;"'I

[;I.

+ [I] =

[;I

3. If each vertex x satisfies d&) > - , the number k = minxex dG(x) can be written as

k=[i]+r

,

r>O ,

where

Let x, be a vertex of degree k 2 2 r. We shall show that the subgraph generated by T,(xl) contains a matching of cardinality r. In fact, if this subgraph contains a matching of only r - 1 edges bl,yzl, Iy3, ~ 4 1 ,..., L V Z-~3, Yzr - 21, vertex YZ,- 1 is not adjacent to ~ G ( X I ) {yl,..- 3 Yzr - 21. Thus,

< (2 r - 2) + n - k = =2r -2

+n

-[;I-

r= n-

[;I+).-

2<

[;I

,< - + r - l = k - l , which contradicts that k equals the minimum degree.

4. Let TG(xl) = {yl,yz,. .., Y k } , and let G, be. the partial subgraph obtained by removing x1 and the r edges of the matching [yl ,yzl, . .., b z r - I, yzr] defined above. The edges of graph G can be covered by: - r triangles [x,,yl , y Z , xll,. .., [xl,Y Z r - 1, Y Z r , ~ 1 1 , - k - 2 r isolated edges [x,,~ 2 -,11, ..., [XI,Ykl, - the (n - 1)' disjoint cliques that cover G,.

[7

1

Thus the minimum number of disjoint cliques that cover G is less than or equal to

['"i')'] + r + (k- 2r) = Q.E.D. Remark. This theorem is the best possible, i.e., the edges of some graph

239

COVERMG EDGES WITH CHAINS

G, of order n cannot be covered by less than

[ g]cliques of cardinality < 3.

For an even n = 2 k, let G, be the complete bipartite graph minimum number of cliques needed to cover all the edges is n2 = n2 k2 = -

&k.

The

4

For an odd n = 2 k + 1, let G, be the complete bipartite graph K,,,,,. The minimum number of cliques needed to cover all the edges is

k(k

+ 1) =

4

3. Counting eulerian circuits

Let G = ( X , U) be a (directed) graph. A circuit of G is defined to be eulerian if it uses each arc exactly once. In this section, we shall study the number of distinct eulerian circuits in G. Graph G is defined to be pseudo-symmetric if the number of arcs entering vertex x equals the number of arcs leaving vertex x, for all vertices x, i.e. dG+(X)

=

&(X)

(x EX).

This terminology is consistent because each symmetric graph is also pseudosymmetric. Theorem 6. A graph possesses an eulerian circuit and only if, it is connected (except for isolated vertices) and is pseudo-symmetric. The proof is exactly the same as the proof of Theorem 1. EXAMPLE. What is the longest circular sequence that can be formed from the digits 0 and 1 so that no k-tuple of k consecutive digits occurs twice? Since 2k distinct k-tuples can be formed from the numbers 0 and 1, the sequence cannot have more than 2, entries. Using Theorem 6 , we shall show that there does exist such a sequence with 2k entries.' Consider a graph G whose vertices represent the different (k - 1)-tuples of 0 and 1 with arcs from vertex (a,,az, ..., ak- ,) to vertex (aZ,a3,..., a*-', 0) and t o vertex (az, a3,...,a,-1, 1). Since graph G is pseudo-symmetric, it possesses an eulerian circuit. If (al,a z , ..., ak-') is the first vertex of this circuit, (az,a3, ..., ak) the second vertex, (a3,ag,..., a,+') the third vertex, etc., the required sequence will be orl, a z , a3,a4, .... For k = 4, the graph in Fig. 11.7 provides several circular sequences of 2' = 16 entries. For example, (l) This problem occurs in telecommunications when it is necessary to determine the current position of a labelled cylinder by reading from the cylinder a sequence of only k of its labels.

240

abcdefghijklmnpq abcdkijefghlmnpq abcdkipghlmnjefq abcfghijedklmnpq abhijklmnpgcdefq abhijedklmnpgcfq abhijefgcdklmnpq abhipgcdklmnjefq

GRAPHS

oooO101001101111 oooO101101001111 OOOO101100111101 OOO0100110101111 = oooO110111100101 = 0OOO110101111001 = oooO110100101111 = oooO110010111101 = = = =

2 c

L-J 111'

11 11

Fig. 11.7

(Circular sequences obtained by permuting i and lmn have been omitted.) There are 16 different solutions. The problem of counting the eulerian circuits is closely related to the problem of counting the spanning arborescences, see Theorem (21, Ch. 3). Theorem 7. Let G be a connected pseudo-symmetric graph, and let x1 be a vertex of G. Then G has a partial subgraph which is an arborescence rooted at xl . Moreover, this arborescence H can be constructed by travelling through an eulerian circuit starting at x1 and placing in H the first arc used to enter each certex.

From Theorem (13, Ch. 3), H is an arborescence rooted at x1 since: 1. each vertex distinct from x1 is the terminal endpoint of exactly one arc of H , 2. x1 is not the terminal endpoint of any arc of H, 3. H does not contain any circuits, because if a path formed from the arcs of H goes from x to y, then vertex x has been visited before vertex y, and therefore no paths exist in H from y to x. Q.E.D. Theorem 8 (Aardenne-Ehrenfest, de Bruijn [1951]). In a connectedpseudosymmetric graph G, let A l denote the number of distinct arborescences rooted

241

COVERING EDGES WITH CHAINS

at x1 that are partial graphs, and let rk denote the outer demi-degree (or inner demi-degree) of vertex Xk. There are exactly n

distinct eulerian circuits. Note that two eulerian circuits are not considered as distinct if one can be obtained from the other by a circular permutation of the arcs. Consider an arborescence H rooted at x1 that is a partial graph of G. We shall show that there are exactly

n n

(rk - l)!

k= 1

distinct eulerian circuits that produce H as in Theorem 7. Number the arcs entering xk from 1 to rk and let O-(Xk)

= { Uk(l), uk(2), ..*> uk(rk)}

-

Suppose that &(rk) is the arc of H in O-(xk), for each k # 1. Suppose that arc u,(l) is fixed. Then there are exactly

n (rk - 1 ) n

!

k= 1

possible numberings. Each numbering corresponds to exactly one circuit as follows: Starting from x l , travel in reverse through all the arcs of the graph by choosing at vertex xk the unused arc of O-(xk) with the smallest possible number. This procedure defines a circuit because the route can terminate only at x1 (when any other vertex is encountered there is always an exit arc because the graph is pseudo-symmetric). We shall show that this circuit is eulerian. If it were not eulerian, then there is an arc & ( j ) that is not used by the circuit. Since rk 2 j , arc uk(rk)has not been used. Hence, arc up(r,), incident into the initial endpoint x , of arc Uk(rk)r has not been used, etc. Since each of these unused arcs belongs to H , the process eventually arrives at x l . But this is a contradiction, since the procedure that generated the circuit stopped at x l , and consequently, all arcs incident to x1 have been used. Thus each numbering corresponds uniquely to an eulerian circuit such that the first arc of the circuit to enter a vertex is present in the arborescence H. Since the number of distinct arborescences H is A ] , and is known by Theorem (21, Ch. 3), the proof is complete.

Q.E.D.

242

GRAPHS

Corollary 1. In a pseudo-symmetric graph, the number of arborescences rooted at x k that are partial graphs is independent of which vertex xk is chosen. Theorem 8 can also be stated for vertex xk instead of for vertex xl. Since the number of eulerian circuits does not change, A , = Ak. Q.E.D.

Corollary 2. In a graph G with m arcs and with order n that satisfies m > 2 n, the number of eulerian circuits is eren. If G is not connected and pseudo-symmetric, the result is obvious. If do+(x) < 2 for each vertex x , then

which is impossible. If d,'(x) 2 3 for x = xk, then (rk - l)! is even and the formula in Theorem 8 shows that the number of distinct eulerian circuits is even. Q.E.D. Note that this result is the best possible: each graph shown in Fig. 11.8 satisfies m = 2 n and contains exactly one eulerian circuit.

Fig. 11.8

Corollary 2 can be extended by using a fundamental theorem of algebra, due to Amitsur and Levitzki [1950], that was reformulated in more combinatorial terms by M. P. Schutzenberger [1958] as follows. Let G be a graph of order n whose arcs are numbered 1,2, ..., m. To each , beginning with arc 1, there corresponds a eulerian circuit ( l , . j 2 , j 3 ...,j,,,) permutation

243

COVERING EDGES WITH CHAINS

An eulerian circuit / I is defined to be odd (respectively, even) if the corresponding permutation is odd (respectively, even). We shall show that if m > 2 n , the number of even eulerian circuits equals the number of odd eulerian circuits. (Note that if the arcs are numbered differently, the equality is preserved.)

Lemma. Let

7c

= ( j , ,j,,

...,j,,,)be a permutation

of 1, 2, ..., rn and let

ak = ( j l , j z ,... j k - l , j k + ~ , j k +... 3 , j,)

be a permutation of degree m - 2 obtained from a by the elimination of two consecutive indices j,, j,, :then n and nk have the same parity if and only if one of the following conditions hoih:

,

(1) (2)

and and

jk <jk+1 >jk+l

jk

Jk + J k + l

jk + j k + l

odd* even.

Permutation x = ( j l ,j,, ...,j,) becomes (J,, j , , ...,jk-,,j,,, , ...,j,, j,) by a product of rn - k transpositions of two consecutive terms. Similarly, this new permutation becomes (jl, j , , ...,jk-l,jk+2,jk+3,. . . , j m 7 j k , , j k + l ) by a product of m - k transpositions. Thus 71 becomes (j,,,j , , . . . , j m , J k , j,,,) by a product of 2(m - k) transpositions. Two permutations are of the same parity if one can be obtained from the other by an even number of transpositions; thus, permutations a and ( j , ,j , ,' ...,j,,,,j , , j k + ,) have the same parity. By removing j , and j k + from this latter sequence, the number of inversions is decreased by

,

I { i / i # j k , j k +1 , i > j k 1 1 + 1 { i I i If j , < j,+

# j,+, ,i > jk+ 1 I

I

this number equals (rn

- j,

- 1) + (m - j k + J = 2 m - 1 - ( j k

If j , > j k + l , this number equals (m -jk) + (m - j k f l= )2m

+jk+l)-

- ( j k+ j k + J .

Since permutations n and have the same parity if, and only if, this number is even, the lemma follows. Q.E.D.

Theorem 9 (Schutzenberger [1958]). In a graph G = (X,U)of order n with m arcs that satisfies m > 2 n, the number of ecen eulerian circuits equals the number of odd eulerian circuits. 1. We may assume that m

=

2n

+ 1. Otherwise we have m - 2 n = k > 1.

244

GRAPHS

Then, construct a graph G ’ from G by adding k - 1 new vertices and by replacing one arc ( x , y ) E U by an elementary path from x to y that passes through each of these k - 1 new vertices. Graph G‘ satisfies m‘ - 2 n’ = (m k - 1) - 2(n k - 1) = m - 2 n - k + 1 = 1 ; hence, G ‘ has as many even eulerian circuits as it has odd eulerian cycles, and the same result follows for G.

+

+

2. Note that the theorem is true for a graph of order 1 or 2. We shall assume that it is valid for all graphs of order < n, and we shall show that it is also valid for a graph G of order n > 2, with m = 2 n 1.

+

3. We may assume that G is connected and satisfies (x EX) d,+(x) = dG(X)

.

Otherwise, from Theorem 6, the number of eulerian circuits equals 0, and the theorem is proved. 4. We may assume that G is a 1-graph. Otherwise, there are two arcs il and iz with the same initial endpoint and the same terminal endpoint.

With the eulerian circuit corresponding to a permutation I[, we can associate the eulerian circuit p’ corresponding to the permutation n’ obtained from n by transposing il and iz. Since p and p’ have different parities, the theorem follows.

5. If G contains a vertex x with &(x) = &(x) = 2 and with a loop at x, the theorem can be proved as follows: The three arcs incident to x are of the form (x,,x) = io, ( x , x) = i,, (x, XJ = i 2 . Consider the graph G‘ obtained from G by removing x and all arcs incident to x, and by adding an arc io = (xo,xz). Graph G’ has order n’ = n - 1 and satisfies m’ - 2 n’ = (m - 2) - 2(n - 1) = m - 2 n = 1. Thus, from part 2, G’ has as many even eulerian circuits as it has odd eulerian circuits. To each of the corresponding permutations, add the entries il, iz after the entry io. In this way, each eulerian circuit of G is obtained. If, for example, il < iz and il + iz is even, two circuits of the same parity in G‘ correspond to two circuits with the same parity in G. Thus, the theorem is valid for G. 6 . If G contains a vertex x with d,+(x) = &(x) = 1, the theorem can be proved as follows: By making the transformation shown in Fig. 11.9, a graph C, is obtained for each arc i E w - ( x o ) . From Part 5, each graph G,has the required property. To each eulerian circuit in G i , there corresponds an eulerian circuit in G .

245

COVERING EDGES WITH CHAINS

G

I

Fig. 11.9

To each eulerian in G there corresponds an eulerian circuit in only one of the G1. Hence G has the required property,

Fig. 11.10.

7. In all the other cases, we shall show that the outer demi-degree of each vertex in G equals 2, except for one vertex x , with d$ ( x , ) = 3. From Part 6 , we may assume that dC+ ( x ) 2 2 for all x, and consequently 2 n 1 = rn = C dG+(x) z 2 1 1 .

+

xax

246

GRAPHS I

I

Fig. 11.11.

Thus, d; (x) = 2 for all x # xl. Since G has order configuration shown in Fig. 11.10.

ti

> 2, it contains the

8. The configuration in Fig. 11.10 can be altered in four ways as shown in Fig. 1l.! 1. These alterations yield the graphs G , , Gz, G , , G 6 , each of order n a n d with rn - 2 n = 1. Let n(G), n,(G) and nl(G) respectively denote the number of eulerian circuits, the number of even eulerian circuits and the number of odd eulerian circuits in graph G. Then

n(G) = n(G1)

+ n(GJ - n(G6) - n(G,) .

From Part 6 ,

~ o ( G I= ) ndGl), Furthermore, from Part 5, n0(G6)

Thus,

=

n1(G6),

no(Gz) = n , ( G z ) . = n1(G7)

247

COVERING EDGES WITH CHAINS

Thus, G has the required property.

Q.E.D. EXERCISES 1. Let G = ( X , E ) be a simple connected graph regular of degree 4, and if (E', F') is a partition of the edges into 2 factors; show that there exists an eulerian cycle in G that alternately passes through edges of E' and F'. (Kotzig [1956]) 2. Use Theorem (15,Ch. 3), and Theorem (6, Ch. 1I), to give a necessary and sufficient condition that a semi-functional 1-graph possesses a hamiltonian circuit.

3. A simple graph L(G) is defined to be the line-graph of G = ( X , E) if each vertex of L(G) represents an edge of G and if two vertices are adjacent in L ( G ) if they represent adjacent edges of G. (These graphs are studied in Chapter 17, Section 4.) Show that if G possesses an eulerian cycle, then L ( G ) possesses also an eulerian cycle. Show that the converse is not true. Hint: Consider a graph G with 4 vertices a, b, c, d and 3 edges ab, bc, bd. (Chartrand [1964]) 4. Show that if G possesses an eulerian cycle, then L(G) possesses a hamiltonian cycle. Show that the converse is not true. Hint: Consider the graph G with 4 vertices a, 6 , c, d and 6 edges ab, ac, ad, bc, bd, cd. (Chartrand [1964])

5. Show that if G possesses a hamiltonian cycle, then L ( G ) possesses a hamiltonian cycle. Show that the converse is not true. Hinr: Consider the graph G with 5 vertices, a, b, c, d, e and 6 edges ab, ac, ad, eb, ec, ed. (Chartrand [1964]) 6. Show that if C is a regular multigraph of order n and odd degree h = 2k

h

>n

+ 1, then G has a factor.

Hint: Use the proof of Theorem 2, and Corollary 1 to Theorem ( 5 , Ch. 7).

+ 1 with

CHAPTER 12

Chromatic Index

1. Edge colourings The chromatic index q(C) of a graph G is defined to be the smallest number of colours needed to colour the edges of G so that no two adjacent edges have the same colour. Aq-colouring of the edges is defined to be a partition of the edge set into q subsets that are matchings. Clearly, q(G) 3 max dG(x)

a

X E X

In this chapter, we may assume without loss of generality that G is a multigraph without loops.

EXAMPLE 1. Scheduling an oral examination. At the end of the academic year each student must be examined orally by each of his professors. How can the examinations be scheduled so that they end as soon as possible? Let X be the set of students, and let Y be the set of professors. Form a bipartite multigraph G = ( X , Y, E ) in which a E X is joined to b E Y by k edges if, and only if, student a must be examined by professor b exactly k times. If the edges of this graph are coloured so that no two adjacent edges have the same colour, each colour can correspond with one examination period. Hence all examinations can be completed in q(G) time periods, and not less. EXAMPLE 2. The Lucas schoolgirls problem. Each day the 2 p schoolgirls of a boarding house take a walk in p rows of two; can they take 2 p - 1 walks consecutively without any two girls walking together more than once? This is possible, if and only if the chromatic index of the complete graph KZp with 2 p vertices is equal to 2 p - 1, each colour representing a walk. EXAMPLE 3. Latin square. A latin square is an n x n matrix with entries 1 , 2 , .,., n such that no entry appears twice in the same row and no entry appears twice in the same column. Let k < n ; in an n x n tableau, place only entries 1, 2, ..., k, so that no entry appears twice in the same row or in the same column. Is it possible to 248

249

CHROMATTC INDEX

+

+

place the entries k I , k 2, ..., n in the empty positions so that a latin square is formed? Represent the rows of the matrix by vertices a , , a z , ..., a,, and the columns by vertices b l , b z , ..., b,; join a, and 6, if and only if the position in row i, column .i is empty. The latin square can be completed if, and only if, the edges of this bipartite graph can be coloured with n - k colours k 1, k 2, ..., n, each edge [ai,bj] coloured with a corresponding to an entry a at the intersection of row i and column j .

+

+

Theorem 1. The chromatic index of a simple complete graph G of order n (and maxinium degree h = n - 1) is q(G)

{

= =h

if n is eoen f n is odd.

+1

CASE1: n is even. Number the vertices 0, 1, 2, vertices as shown in Fig. 12.1.

..., n

-

1 and place the

nn-3

-+I

2

-

2

Fig. 12.1

Let the first perfect matching (i.e., the edges of the first colour) be

n n 11. ~ 0 , 1 1 . ~ 2 , n - 1 1 . ~ 3 . n - ,...,[i-,~+ 21 These edges are the dark edges in Fig. 12.1. By adding 1 modulo n to the index of each vertex except vertex 0 in the above matching, we obtain another perfect matching, which corresponds to a rotation around 0 of the dark edges in the figure. This operation can be performed n - 1 times, and each time a new colour is assigned to the resulting matching. No edge is coloured twice, because each time the coloured edges form a different angle with the horizontal. However, each edge of G has been coloured, and therefore q(G) = n - 1.

250

GRAPHS

CASE2: n is odd. Consider the graph G‘formed from G by adding a vertex xo that is joined to each vertex of G. From Case 1 above, we have

q(G’) = (n

+ 1) - 1 = n.

Thus the edge of G can be coloured with n colours. The edges of G cannot be coloured with n - I colours, because then each colour would correspond to a perfect matching, and a graph with odd order has no perfect matching. Thus q(G)=n=h+l.

Q.E.D. Theorem 2. The chromatic index of a bipartite multigraph G = ( X , Y,E ) with maximum degree h is dG) = h .

From Corollary 4 to Theorem (5, Ch. 7), graph G contains a matching El that saturates every vertex of degree h. Colour El with the first colour. Consider the bipartite multigraph ( X , Y, E - E l ) )whose maximum degree is h - 1. This graph contains a matching E2 that saturates every vertex of degree h - 1. Colour E2 with the second colour. By repeating this operation, the edges of G will be coloured with h colours. Q.E.D. APPLICATION TO LATIN SQUARES. The theory of latin squares, which was founded by Euler, can use Theorem 2 in a very interesting way: a latin square of order n is formed when the entries 1,2, ..., n are distributed into the n2 positions of an n x n square so that no entry is repeated in any row and no entry is repeated in any column. We shall show:

Let T be a p x q rectangle whose positions contain the numbers I , 2, ... n such that no number is repeated in any row and no number is repeated in any column. Let m(k) denote the number of times that k appears in T. A necessary and suficient condition that n - p rows and n - q columns can be added to form a Latin square is that )

m(k)>p+q-n

( k = 1,2y..,yn).

Necessity. Let m’(k) denote the number of rows of Tin which k does not appear. If a latin square can be formed, then m’(k) < n - q. Hence m(k) = p

- m’(k) 2 p - n + q .

251

CHROMATIC INDEX

Sujiciency. Let N = { ly2,...,n}, P = { 1 , 2,... y p } , form the bipartite graph G = (P,N , E ) in which i E P and k E N are joined by an edge if number k does not appear in the i-th row of T.Then,

dG(i) = n - q (irz P ) dG(k)= m’(k) < n - q ( k EN> . From Theorem 2, the edges of G can be coloured with n - q colours. Complete T by adding n - q columns; the first additional column will be defined by the endpoints in N of the edges of the first colour, the second column by the endpoints in N of the edges of the second colour, etc. In this way, we obtain a p x n rectangle T, that satisfies also the hypothesis of the theorem because @(y) = p = ( p n) - n. Then n - p rows can now be added to complete T, and this yields a latin square. Q.E.D.

+

Consider the problem of colouring the edges of a bipartite graph with colours a l , a2, ..., a, such that there are exactly mi edges with colour ai. The following theorem gives conditions for the existence of such a colouring. Theorem 3 (Folkman, Fulkerson [1966]). Let G = (A’, Y, E ) be a bipartite multigraph with maximum degree < q. Consider a sequence

m1 3 m, 2

... 2 m q ,

with 4

xm,=m=IEl. ,=I

Form the conjugate sequence (my, m,*, ...) where mT denotes the number of mi that are 2 j (see Chapter 6). If there exists a colouring of the edges in q colours a l , a s , ,.., a, with exactly miedges with colour u i , for all i, then mdA, B) 3

jz

c

(X-AI+ IY-81

(A c X , B c Y ) .

mi*

This necessary condition is also sujicient ,for the case:

m , = m, =

= m,

3 mk+l = mk+2=

= m4.

Necessity. If there are m,edges with colour a,, then the number of edges with colour aithat join A and B is

m,(A,B) = m, - m , ( X - A , Y) - m,(X, Y - B)

+ m,(X-

A, Y

- B)

a m , - IX- A ( - IY

2

-BI.

252

GRAPHS

4

2

C max { 0, mi - I X - A I - I Y - B I 1 =

i= 1

j >

c

rn?

JX-AJ+IY-B(

.

(The last equality can be verified from the Ferrers diagram, see Chapter 6 , Section 1.) Suficiency (for the special case). Consider such a sequence (mi).See Fig. 12.2. Let k

h” = q - k ; h‘+h” = q . We shall show first that G is the union of two disjoint partial multigraphs G’= ( X , Y, E’) and G“ = ( X , Y,E”) with IE’( = m’, IE”I = m ” , h’= k ;

max &(Z) < h’ , max dG”(z) < h” . From the lemma to Theorem (7, Ch. 7), it suffices to show that for all A , B,

- h’l X - A I - h‘ I Y - B I , B) 2 m” - h” 1 X - A 1 - h” I Y - B I .

(1)

m,(A,B) 2 m’

(2)

m&4,

n1‘ -

h’

m” h”

m: -c

h‘ = k

-q Fig. 12.2

h“

=q -

k

253

CHROMATIC INDEX

Inequality (1) can be easily verified from the Ferrers diagram (Fig. 12.2). If m''

- < IX- A1 h"

+ JY-BI,

then

=

m'- h ' 1 X - A 1

- h'IY-BI.

If m"

- > l X - A I + ( Y - B ( ,

h"

then j

> IX-A

cI

- I Y-BI

rnf 2 m'

-

- A 1 - h' I Y - B I .

h' 1 x

In both cases the inequality of the hypothesis yields inequality (1). To verify inequality (2), note that if m" -->JX--AI+IY-Bl,

h"

then

= m"

- h" ( X- A I - h"l Y - B I ,

and if

m" h
+I

Y-BJ,

then

Thus inequality (2) is satisfied in both cases.

r

From Theorem (7, Ch. 7), multigraph G' (with m' edges and maximum m' = m, edges that saturates degree d h') contains a matching E; with h each vertex of maximum degree. Colour the edges of EI with the first coiour.

254

GRAPHS

The partial multigraph generated by E' - E; has maximum degree < h - 1. From Theorem (7, Ch. 7), this graph contains a matching Ei with m' - m 1 = ml edges. Colour the edges of Ea with the second colour, etc. h' - 1 Thus G ' can be decomposed into h' = k matchings of m, edges. Similarly, G" can be decomposed into h" = q - k matchings of m,,, edges. Thus G can be coloured with q colours so that mi edges are of colour at for all i. Q.E.D.

2. The Vizing theorem and related results This section presents some bounds for the chromatic index of a multigraph without loops. The following very simple theorem gives a lower bound. Theorem 4. Let G be a multigraph without loops, with m edges, with maximum degree h, and let t be the cardinality of a maximum matching. Then,

Consider a colouring of the edges with q = q(G) colours tll, az, ..., a,, and let E, denote the set of edges with colour at. We have

m m

Hence, q 2 t

= IEi

, and q 2

I + IE2I +

+ I Eg I < q t .

[ 51*. The theorem follows. Q.E.D.

Uncolonred Edge Lemma. Let G be a multigraph without loops and let q(G) = q 1. Suppose that a set C = { u l , az, ..., a, } of q colours has been used to colour all the edges of G except one edge [a, b],, and let C, denote the set of colours used for the edges incident to vertex x. Then

+

Icaucbl=q

1 c,,n c b I = &(a)

I ca- c b I I cb - c, I

+ &(b) - 4 - 2

- dG(b) + 1 = 4 - &(a) + 1 =

4

*

No colour of C can be missing from both C, and C b , since then [a, b], could be coloured with this missing colour. Thus C = C, u C,,, and = I CI =

I

cbl = I can c b l

+ Ic a -

c b l

+I

c b

- cal

*

255

CHROMATIC INDEX

Furthermore,

By elimination, these three equalities yield :

Theorem 5. If G consists of a cycle [xl , xz, ..., x,,, x,], with possibly more than one edge between two consecutive vertices, and i f G has m edges and maximum degree h, then

ih

i f n is even

If n is even, the proof is evident because then G is bipartite and Theorem 2 applies. Thus we may assume that n = 2 k 1 odd. Note from Theorem 4 that

+

( [y] * ]

4G) 3 maJr h ,

= max

( h , [f ] * 1.

It remains to show that

This is true for a multigraph with 2 k + 1 edges, because q(G) = 3 for a 1. cycle G without multiple edges. Therefore, we may assume that m > 2 k Suppose that the theorem is valid for all multigraphs with less than m edges, 1 edges. and consider a multigraph G with m > 2 k

+

+

1. If we remove from G an edge [a, b],, so that a and b remain adjacent, the resulting graph G' is of the same type but with m' = m - 1 edges. Thus, from the induction hypothesis, the edges of G' can be coloured with a set C of q colours, where

For rx E C, let E, denote the set of edges with colour a.

256

GRAPHS

2. We shall assume that q(G) = q + 1 and we shall show that this leads to a contradiction. There is in C a colour y with I Ey I c k since, otherwise,

and

which contradicts q 2

I;[

**

Two cases must be considered: y E Cb - C, (or y E C, - cb) and y E C, n c b . From the Uncoloured Edge Lemma, C = C, U C, and no other cases are possible. CASE1: y E Cb - C., From the Uncoloured Edge Lemma,

I c, -

cb

I = 4 - dG(b) -k 1 2 h - &(b) -k 1 2 1 .

Thus there exists a colour a E C, - cb,and clearly a # y. Let G(a, y ) denote the partial graph of G generated by the edges with colour a or y. The connected component of G(E, y) that contains 6 is an elementary chain with endpoint b (because a # c b ) and does not contain vertex a (because I EI I < k). By interchanging the colours y and a along this chain, colour y is no longer incident to vertices a and b and, therefore, [a,b], can be coloured with y, which is a contradiction. CASE2: y E C, n cb. From the Uncoloured Edge Lemma, there is a colour a E C, - Cb and a colour j? E Cb - C., Clearly, y # a, j?. Consider the connected component of graph G(B, y) that contains a. This component is an open chain paS[a,x ] that contains b if y joins vertices a and b, and does not contain b if y does not join vertices a and b. Since this chain begins with an edge with colour y, by interchanging y and j? along this chain a new colouring with I E, I < k is obtained. Furthermore, in this new colouring, y E cb - C,, which is Case 1. Q.E.D. We shall now give an upper bound for the chromatic index. The strongest theorem of this type is due to Vizing [1964], and was rediscovered by R. P. Gupta [1966]. It is given in a slightly stronger form below: Theorem 6. Let G be a multigraph without loops. Let [a, b], be an edge of G, and let G' = G - [a, b ] , . If the edges of G' can be coloured with q colours, where q 2 &(a), q 2 d,(b), and if

251

CHROMATlC INDEX

x

+

=-

rda)

< 4,

d d x ) mG,(a,x ) then the edges of G can be coloured with q colours. E

Suppose that G cannot be coloured with q colours. Then,

+

d G ' ) = 9, q(G) = q 1. 1. Let C, denote the set of colours used for the edges of G' incident to a vertex x E I',,(a). Then,

I C - C, I

=4

- &(X)

2

W I ~ ( UX) ,

.

Therefore, there exists a mapping g(e) of the edge set U,.(U) into the set C of colours such that (0

el = [a, 4

(ii)

ek = [a, x ] k ,

*

1

ej

s(eJ

= [a, x l j ,

# C, , k#j

g(ek) # g(e,).

=>

2. We shall define a sequence y o , yl, y,, ... of distinct colours in C, as follows : Let yo be a colour in C, - C,; such a colour yo exists because, from the Uncoloured Edge Lemma, IC,-C*l=q-d,(b)+l>I. Let e, = [a, xlll be the edge with colour yo that is incident to vertex a. Let y1 = g(el). Clearly, y1 # y o , since y1 4 C,, and yo E Cxl. We have y1 E C , because, otherwise, y1 $ C,, y1 = g(el) $ C,, , and we can recolour

-

b,X,ll

[a, b ] ,

+

with with

Y1

9

yo.

which contradicts q(G) = q 1 . Thus, let e, = [a, xZlzbe the edge of colour y1 incident to a. Let y2 = g(ez) . If y z = yl, or y z = y o , terminate the sequence. Otherwise, consider an edge e3 of colour ya incident to a, etc. In general, if an edge ek = [a, Xk]k is found, and if y r = g(ek) belongs to { yo, yl, ..., y k - l } , terminate the sequence. If Yk

we have recolour

Yk

E

f YO, Y1,

C, because, otherwise,

Yk

Yk-1

$ C,,

9

yk

= g(ek) 4 C,,, and we can

[a,xklk [a, x k - i l k - 1

with with

Yk,

[a,bl,

with

yo,

.....

which contradicts q(G)

=

q

+ 1.

Yk-i

3

258

GRAPHS

Since graph G is finite, this process will terminate with an edge t?k = [a, x k ] k such that dek) = Y j j < k . Consider the sequence [a,b l O ~ X 1 l l ~[a, XZ12, [a, X k l k of distinct edges. Vertices b, xl,x2, ..., x k are not necessarily all distinct; however, it follows from (ii) that X k # x,, since ek # e, and 3

g ( b , Xklk) = g(b, Xjl,). 3. Let B be a colour of Cb - C, (since q 2 dG(a),this colour B exists by the Uncoloured Edge Lemma). Consider the connected component of graph G(B, y,) that contain vertex a. This component is an alternating chain of colours y, and B, and one of its endpoints is vertex a (because Ca),Denote this chain by psr, [a,21. At most one of xi, x, can be on this chain because y, $ C,, and yr $ Cxk;besides, if xi (or xk)lies on this chain, it is an endpoint.

a$

Fig. 12.3

CASE1. y, # yo and z # x,. Thus, X, $ PgYJaY 21. So, colours y, and B can be interchanged along chain pgy,[a, z ] to form a new colouring with y, $ C, and y, $ Cx,.We can then recolour

[a Xjlj [a, +lIj.-l 9

This contradicts q(G) = q

+ 1.

with with

yj, Yj-1

9

259

CHROMATIC INDEX

CASE2. y j # yo and z = x j . Thus X, I P B ~[a, , zIColours y j and fl can be interchanged along this chain to form a new colouring with y j .$ C, and y j 6 Cxk.We can then recolour with with with

[ a ,x k l k [a, X k - l l k - 1 [ a ,X k - 2 1 k - 2

yj yk-1

9

Yk-2

3

..........................

[a, x j + ~ l j + ~with [ a ,X j + l l j + l with to 3 xjl j with [ a , xj-11,-1 with

yj+2

,

Yj+l

Y

p, ~ j - 7i

.......................... with with

[a,XI11 [ a ,4 0

This contradicts q(G) = q

yl, yo.

+ 1.

CASE3. y j = yo and z # b. Thus, b I PBy0[a, 21. Colours yo and fl can be interchanged along chain pBy0[a, z ] . We can then colour [a, blOwith yo. This contradicts q(G) = q + 1. CASE4. y j = yo and z = 6. Colours yo and j? can be interchanged along chain ,uBro[ay b]. We can then recolour [a xklk with yo, [a, ~ k - l l ~ - ~with Yk-1 , Y

.......................... with with

[a, X l l l [a, bl,

y1

/?.

+

This contradicts q(G) = q 1 . Since we find a contradiction in each case, the theorem follows. Q.E.D. Corollary 1. Let G

=

( X , E ) be a multigraph without loops, and let

d&) = d&)

+ max m&, vex

Then q(G)

< max d&). xex

y)

.

260

GRAPHS

Let G’be a partial graph of G with q(G’)

< q = max d&) xcx

and with the maximum possible number of edges. If G’ # G, then there exists an edge [a,bl0 in graph G - G’. Let G = G‘

+ [a, b I 0 .

For each x E X , we have dG,(x)

+ mG,(a, X ) Q

<4 <4 . 9

dG4x) < dG(x) < &(x)

Hence, from Theorem 6, the edges of G can be coloured with q colours. This contradicts the maximality of G‘. Q.E.D. Corollary 2 (Vizing’s theorem [1964]). Let G be a multigraph without loops of multiplicity max,,y mG(x,y ) = p , and with maximum degree h. Then (I

dG) G h

+P

*

Clearly,

d&) = dG(x)

+ max mG(x,y ) < h + p . YSX

Thus, from Corollary 1, q(G)

< max d&) < h +- p . Q.E.D.

Corollary 3. Let G be a simple graph with maximum degree h; then q(G) equals h or h 1.

+

The proof follows by setting p = 1 in Corollary 2. Corollary 4. Let G be a multigraph without loops and with maximum degree h = 3 ; then q(G) equals 3 or 4. If G contains two vertices x and y such that rn,(x, y ) = 3, then a connected component of G contains only these two vertices, and its edges can be coloured with only three colours. Therefore, we may assume that rnG(x,y) < 2 for all x, y E X. By removing an edge between each pair x, y with mG(x,y) = 2, a simple graph G‘ with maximum degree < 3 is formed, and, from Corollary 3, q(G’) < 4. Since each edge of G - G’is adjacent to at most three colours, it can be coloured with the missing colour. Q.E.D.

261

CHROMATIC INDEX

Theorem 7. (Ore [ 1 9681). ZfG is a multigraph without loops and with maximum degree h, then

where the inner maximization is taken over all elementary chains [x, , x 2 , x 3 ]of length 2.

Clearly, the theorem is valid for graphs with 1,2, or 3 edges. If the theorem is true for all graphs with m - 1 edges; we shall show that it is also true for a graph G with m edges. Let = max

[ h, max [z1 ( d G ( x l ) + dG(x2) +

dG(x3))]

].

If edge [a, blOis removed from G, then the remaining multigraph G' can be coloured with q colours because

We shall assume that q ( G ) = q contradiction.

+ 1 and we shall show that this leads to a

1. Since q 2 h, the Uncoloured Edge Lemma yields

I c a - c b I = - dG(b) + 1 3 h - &(b) + 1 > 0 1 C, - C, I = q - &(a) + 1 2 h - dc(a) + 1 > 0. Thus, let a E C, - cb,/3 [a, allo with colour u.

E

Cb - C,. Let a, be vertex joined to a by an edge

2. We shall show that C,, 2 C,, - C., The connected components of the partial graph G(a, B) generated by colours a and p consist of either isolated vertices or alternating even cycles coloured u and 8, or alternating open chains coloured a and B. Since a E C, - cb, p E cb - C,, the connected component of G(a, p) that contains vertex a is neither an isolated vertex nor an alternating cycle. Therefore, it is an open chain with endpoint a that contains a l . Furthermore, b is the last vertex of this alternating chain because, otherwise, by interchanging colours a and B along this chain, colour a could be removed from set C, and [a, blO could be recoloured a. Thus p E C, for each E c b - C,. Hence

ca,I3 Cb - c,.

262

GRAPHS

3. We shall show that C,, 3 C, - C,.Suppose that this is not so. Let tl' E C, - C, and a' $ Gal. Let a2 be the vertex joined to a by an edge with colour a'. We see, as in Part 2, that the connected component of the partial graph

0

Fig. 12.4

G(a', j?) that contains a is an alternating chain with endpoints a and b. Thus the connected component of G(a', j?) that contains a, is an alternating open chain p [ a l ,z ] that contains neither a nor 6. Clearly, z # a , , because j? E C,, . Colours a' and j? can be interchanged along chain p [ a l , z]. Then [a, allo can be recoloured with j?, and [a, bI0 can be recoloured with a, which contradicts q(G) = q 1. Hence

+

c,,

3 Cb

- c,.

4. From Parts 2 and 3, cu,

= (cu

-

cb)

(cb

-

c4)

*

Hence dG(al)

=

Ico, I > I

c4

- cb I + I cb

2 (4 - dG(b) + 1)

- ca I 2

+ (4 - dG(a) + 1)

Thus

4

1

+ < 2 ( d G ( a l ) + dG(a) + dG(b))

Y

which contradicts the definition of q. Q.E.D. Corollary (Shannon [1949]). Let G be a multigraph without loops and wirh maximum degree h. Then

The proof follows immediately from Theorem 7.

263

CHROMATIC INDEX

Remark I . The above corollary can also be deduced from Vizing's theorem. Let h > 0, and suppose there exists a multigraph G with maximum degree < h such that q(G) = 32h 1 . By removing some edges if necessary, we may assume that G is critical with respect to this property, i.e. if we remove

[-1 +

from G any edge, the resulting graph has chromatic index

G contains two distinct vertices a and b with

because, otherwise, m,(x, y )

<

for each pair x, y , and the Vizing theorem

would imply

dG) Q h

+ [;ih]

=

3h

Let [a, 61, be an edge joining a and b. The multigraph G = G can be coloured with q =

- [a, b],

colours (because G is critical); for one such

colouring,

Q 2h

Q

-[

g] -

1- 1

[Y]

-1.

Thus edge [a, b], can be coloured with one of the q colours, which contradicts q(G) = q + 1. Remark 2. It is easy to show that the bound

[y ]given by the corollary

can be attained for each value of h. Consider the multigraph Gh consisting of three vertices a, 6, c,

El *

c and between b and c.

edges between a and b and

edges between a and

264

GRAPHS

a

N

Fig. 12.5

(See Fig. 12.5). Each edge of Gh requires a different colour, and the total number of edges is

i

i

3h 3 k = [ 7 ]

if

h=2k

3h 3 k + l = [ l ]

if

h=2k+l.

Thus Gh is a multigraph with maximum degree h and chromatic index

[YIVizing [1965] has also shown that if a multigraph G with maximum degree h does not contain a subgraph G,, as defined above, then

If h is even, this result can be improved by the following theorem:

Theorem 8. Let G be a multigraph without loops and with etlen maximum degree h = 2 k. Let 2 < p < k. I f G does not contain G,, as a subgraph, then

Two cases must be considered:

CASE1 :p = k. Let G be a multigraph with maximum degree < 2 k such that q(G) = 3 k. We may assume that G is critical with respect to this property. It remains to show that G is identical to the graph G Z kshown in Fig. 12.5. G contains two distinct vertices u and b with %-(a,

b) 2 k

Y

265

CHROMATIC INDEX

because, otherwise, the Vizing theorem would imply

which contradicts q(G) = 3 k . Let [a, bIo be an edge joining these two vertices. Let G' = G - [a, b],

.

Multigraph G' can be coloured with 3 k - I .colours, and for such a colouring, 3k-1=IC,,uCbI< < (&(a) m d a , 4 )

-

+ (dG(b) - m ~ ( ab)) , + (mG(a,b) - 1 ) =

+

-

d,(b) mG(a, b) - 1 < <2k+2k-k-1 =3k-1.

= dG(a)

Thus

Similarly, graph G' contains a vertex b, adjacent to b with

because, otherwise,

max (dG,(bl)

+ mG.(b, b,))

G 2k

+ (k - 1) = 3 k - 1 ,

b1

which by Theorem 6 implies that q(G) < 3 k - 1, which contradicts q(G) = 3k. Since mG,(b,a) = mG(a,b) - 1 = k - 1 , then b, # a. Furthermore, since dG(b) < 2 k and mG(b,a) = k , we have mc(b, b,) = k

.

By applying the same argument to the pair b, b,, we see that G contains a vertex b, Z b, with mG(b,, b,) = k, etc. Vertices a, b, b,, b,, ... form a chain of k-tuple edges, that can terminate only at vertex a. Therefore, G has a cycle p such that each pair of two consecutive vertices is joined by k edges. Since G is critical, G is connected and, therefore, G can have no other edges. If cycle p has even length, then q(G) = 2 k (from Theorem 5), which contradicts q(G) = 3 k.

266

GRAPHS

If cycle p has odd length no 2 5, then, from Theorem 5,

2m = 2 k + 2[ kx ] *

<2k+ which contradicts q(G) = 3 k . Thus, cycle p has length 3, and G

[?I* =

<

<3k,

Gzk.

CASE2 : p c k. Let G be a multigraph with maximum degree 2 k that does not contain a

[;I

subgraph G2p.We shall show that G can be coloured with 3k - - colours. By adding vertices, G can be imbedded in a regular multigraph of degree 2 k that contains no subgraph G a p ;thus, we may assume that G is regular. From the Petersen theorem (Theorem 2, Ch. ll), the edge set E of G can be decomposed into k factors E l , E,, ..., Ek.The graphs Hi = ( X , Ei) are regular of degree 2; thus q(H,) < 3, and, consequently,

dG) =

c4w3

Q

I

3k.

Note that, if h is even, this is Shannon’s result. Let F = El u E2u-u E p . Multigraph ( X , F ) has maximum degree 2 p and contains no subgraph G z p . From Case 1, ( X , F ) has chromatic index < 3 p - 1. Thus

Q.E.D. Corollary (FiamEik, JucoviE [1970]). Let G be a multigraph without loops and with maximum degree h. I f G does not contain a partial subgraph G4 (see Fig. 12.5), then

If h

=

2 k is even, then. from Theorem 8 with p q(G) < 3 k -

k

h + l

=3[

-1 2

=

2,

h + l -[ TI

261

CHROMATIC INDEX

+

If h = 2 k 1 is odd, then G can be embedded in a graph G‘ with maximum degree h’ = 2 k + 2 that contains no subgraph G4. Thus k + l h+l =3 dG) < d G ’ ) < 3(k 1) . Q.E.D. The table below gives upper bounds for the chromatic index of a multigraph G of maximum degree h. These bounds are calculated from the corollary to Theorem 7 and from Theorem 8. h = 3 q(G)< 4 h = 4 q(G) < 6 ; if G G4, q(G) < 5 h = 5 q(G)< 7 ; if G $ G 5 , q(G)< 6 h = 6 q(G)< 9 ; if G + G 6 , q(G) < 8 h = 7 q(G)< 1 0 ; if G + G7, q(G)< 9 h = 8 q(G)<12; if G $ G g , q(G)<11 a n d i f G G 4 , q(G) < 1 0 . We conjecture : Conjecture. Let G be a multigraph without loops and with maximum degree h. If G $ G,, where 4 6 s 6 h, then 3h

[F]

+

+

+

If both h and s are even, the conjecture becomes Theorem 8. If h = s is odd, the conjecture can be proved as in Case 1 of Theorem 8. Another conjecture is the following: Conjecture (Vizing [1965]). Let G be a simple graph with maximum degree h. Then the vertices and edges of G can be coloured with h + 2 colours so that no two adjacent oertices have the same colour, no two adjacent edges have the same colour, and no vertex has the same colour as an edge incident to it. 3. Edge colourings of planar graphs (abstract) From Corollary 3 to Theorem 6, we know that if G is regular of degree 3, then q(G) < 4.It has been conjectured that the chromatic index of a regular planar graph of degree 3 without isthmi is exactly 3. This conjecture has never been proved. However, by studying colourings of such graphs, many results related to the famous “four colour conjecture” have been discovered : Theorem 9. For a regular planar multigraph G of degree 3 without isthmi, the following conditions are equivalent: (1) the faces of G can be coloured with 4 colours so that no two adjacent faces have the same colour,

268

GRAPHS

(2) the edges of G can be coloured with 3 colours so that no two adjacent edges have the same colour, ( 3 ) each tlertex can be assigned a coeficient p ( x ) , where p ( x ) equals 1 or - 1, such that each face p satisfies

+

C

p(x)

I0

(mod 3 ) .

EEll

(1) 3 (2) If the faces of G are coloured with 4 colours a, fl, y, 6, label with a “0” each edge separating an a-face from a b-face or separating a y-face from a &face. Label with a 1 ” each edge separating an a-face from a y-face or separating a 8-face from a &face. Finally, label with a “2” each edge separating an a-face from a &face or separating a fi-face from a y-face. Two adjacent edges cannot be labelled with the same symbol (since this would imply that two adjacent faces have the same colour). Hence condition (2) is satisfied, “

(2) 3 (1) Suppose that the edges of G can be coloured with three colours 0, 1, 2. The edges with colours 0 and 1 form a regular graph of degree 2, and, therefore, its faces can be coloured with only two colours p and q. The edges with colours 0 and 2 also form a regular graph whose faces can be coloured with two colours r and s. Thus, each face of G can be coloured with one of the four combinations pr, ps, qr, qs. I f two faces are separated by an edge of colour 0, they are coloured with distinct combinations. The same is true for two faces separated by an edge of colour 1 or of colour 2. Thus, the faces can be coloured with 4 colours so that no two adjacent faces have the same colour. (2) 3 (3) Consider a 3-colouring of the edges with colours 0, 1, 2. If the three edges incident to vertex x are 0, 1 , 2 in clockwise order, let p(x) = 1 ; otherwise, let p ( x ) = - 1. Now follow the contour of a face p in the clockwise direction. Each time a vertex x with p ( x ) = 1 is crossed, the index of the colour decreases by one, and each time a vertex x with p ( x ) = - 1 is crossed, the index increases by one (modulo 3). Therefore, the algebraic sum of all the coefficients of the vertices on contour ,u is necessarily a multiple of 3.

+ +

(3) a (2) Suppose that the coefficients p(x) satisfy condition (3). Starting from some arbitrary edge to which we assign the label 0, we shall now give each edge a label, either 0 or 1 or 2; this is done

269

CHROMATIC INDEX

step by step with the help of the coefficients p(x) in such a way that the three edges incident to a vertex x are 0, 1, 2 counterclockwise if p(x) = 1, and in the reverse order if p(x) = - 1. If G is connected, each edge [a, b] will be given a label g(a, b). It remains to show that this labelling is consistent. Let v be an elementary cycle which contains a set F of faces. Let x E X , if vertex x is incident to exactly i faces in F. Let E ( X , v ) = - 1 for x E XI, 1 for x E X,, 0 for x E X,.If [a, b] and [b, c] are two consecutive edges of v in a counter-clockwise tour, then

+

+

g(b, c) = g(a, b)

+

E(X,

v)p(x) (mod. 3 )

.

Moreover, we have (modulo 3),

2

E(X, V ) P ( X )

XEV

2

= 1=0

(i

2 dx)) = 2 2

ZEXj

~ l e FX

p(4 = 0.

E V

The consistency of the labelling on v follows.

Q.E.D. Corollary 1. If the number of edges bounding each face is a multiple of 3, then the edges of the graph can be coloured with only three colours. Let p(x) = 1 for each vertex x:condition (3) of the theorem is satisfied. Q.E.D.

+

Corollary 2. If the number of edges bounding each face is a multiple of 2 then the edges of the graph can be coloured with only three colours. Let v be an elementary cycle, that contains faces p l , p2, ... .The length Z(v) of v satisfies

l(v)

l(pl)

+ 1(p2) +

(mod. 2) .

This implies that v has even length; therefore each cycle has even length. Thus, from Theorem (4,Ch. 7), each vertex x can be assigned a coefficient p(x) = 1 or - 1 so that no two adjacent vertices have the same coefficient. Clearly, these coefficients satisfy condition (3) of Theorem 8. Q.E.D.

+

Conjecture. For a simple planar graph with maximum degree h 2 6, the chromatic index equals h.

=

For h = 2, 3, 4, 5, it is easy to construct a simple planar graph G with q(G) h + 1. For h 2 8, the conjecture was proved by Vizing [1965].

270

GRAPHS

EXERCISES 1. Consider the complete graph K,,,where n is even. Denote its vertices by 0, 1 , and consider the function f ( a , b ) = n + b m o d ( n - 1 ) if a , b # n - 1 , =2a mod@-1) if a # n - l , b = n - 1 , =26 mod(n-1) if a = n - l , b # n - 1 . Show that fdefines a (n - l)-colouring of the edges, i.e., (1) 0 c f(a,b) c n - 2, (2) f ( a , b) is an integer for a # b, (3) f(u, b) = f ( b , 4, (4) f ( a , b) # f ( a , c) if b Z c.

..., n - 1,

2. In a bridge tournament, 4 p players simultaneously play on p tables, and no player wants to have the same partner twice. Show that it is possible to organize 4p - 1 games so that, for a permutation f on the set of chairs, each player seated in chair x will play his next game in chair f ( x ) .

3. Let G be a simple connected graph that decomposes into two connected components C1and Ca when vertex a is removed. If G1 is the subgraph of G generated by ClU { a }, and if Ga is the subgraph of G generated by Czu { a }, show that q(G) = max

{ ~ ( G I )d, G 3 , d&)

.

4. Let ml 2 m2 3 - * * > m, be a non-increasing q-tuple of integers. Consider the relation (mi) i (mi)defined by k

k

i= 1

I= 1

C mi s C mr mi= i= 1

If ml 2 mj

(k = 1,2, ..., q - 1)

mt. i=1

+ 2, the q-tuple (m;)defined by

mi = mt- 1 mi = mj 1 mi = mk (k # i ,j ) is called the transfer af(m,) by (i,.i). Show that a transfer of m defines a q-tuple rn i m, and that each q-tuple m" with ma < m can be obtained from m by a finite number of transfers.

+

5. Let ml > mz & * - > m, be a non-increasing sequence of integers, and let G be a multigraph with

Zmi edges coloured al, aa, ..., a,, where the number of edges coloured a, equals ml, for all i. Using Exercise 4, show that if (mi)is a sequence with (mi)i (ml), then G can be coloured in q colours with mi edges coloured a,, for all i. Hint: Note that in the partial graph G(al, a,) generated by the edges with colour at and a!, a connected component is an alternating open chain whose extremities have m

=

different colours. The interchange of colours al and by ( i , j ) .

a, along

this chain defines a transfer

(Folkman, Fulkerson [I96911

27 1

CHROMATIC INDEX

6. Show that a simple regular graph of degree /I with order has chromatic index equal t o h + 1.

7. Let G be a simple graph with maximum degree h such that q(G) 7 h + 1, and such that q(G - e ) = h for every edge e. Show that: 1. Each vertex adjacent to a vertex of degree k is adjacent to at least h - k + 1 vertices of degree h. 2. G possesses an elementary cycle of length 3 h + 1. (Vizing [1965]) 8. Let G be a connected simple regular graph of degree 3 with m edges and with q(G) = 3. Colour the edges a, 8, y or 0 (“no colour”) so that the three edges incident to the same vertex are either all uncoloured or each has a different colour or one is uncoloured and the other two have the same colour. If edge el is coloured et, then the colouring defines a vector e = (el, e2, ..., em). Any colouring that satisfies the above conditions is called a good colouring. Define an operation (called “sum”)as follows:

+

O+O=a+a=@+B=y+y=O, 0 + a = a + O = a, 0 B = + O = /I,0 y = y 0 = y , B+y=y+B=a, y+a=a+y=B, a+B =8+a=y.

+

+

+

Show that: (1) a good colouring

e

is characterized by the property

(2) the vector sum of two good colourings is also a good colouring, (3) the set of all good colourings together with the operation +, is a n abelian group, (4) if G is planar and without isthmi, a set of generators of this abelian group consists of the union of a fundamental basis of cycles coloured with a, and of a fundamental basis of cycles coloured with 8. (Vigneron [19461, [19591, [I96111

CHAPTER 13

Stability Number

1. Maximum stable sets

Consider a simple graph G = ( X , E). A set S c X is defined to be stable if no two distinct vertices of S are adjacent. In other words, S is stable if, and only if, F,(S) n S = 0 . Let 9 'denote the family of all stable sets of G. Then,

0QIEy

SeY,AcS =. A € . % @ . The stability number a(G) of G is defined to be the maximum cardinality of a stable set, a(G) = max I S I

.

S€Y

EXAMPLE1 (Gauss). Problem of the eight queens. Can eight queens be placed on a chessboard so that no queen can capture another queen? This famous problem is equivalent to finding a maximum stable set of a simple graph G with 64 vertices and with y E T,(x) if squares x and y are in the same row or in the same column or in the same diagonal. 8

8

7 6 5 4

7 6 5 4 3 2

3

2 1

1 (7263 1485)

(61528374) Fig. 13.1 212

213

STABILITY NUMBER

This problem is more difficult that it would appear at first sight, and Gauss initially believed .that it had only 76 solutions; in 1854, Schachzeirung, a Berlin chess journal, gave only 40 solutions. In fact, there are exactly 92 solutions, which can be obtained from the following permutations : (72631485) (3 584 1726) (16837425) (5 1468273)

(6 1528374) (46152837) (57263184) (4275 1863)

(584 17263) (57263148) (48 157263) (3 528 1746)

Each of these permutations gives a possible diagram (see Fig. 13.1) and from each of the diagrams we can obtain eight solutions by rotating the chessboard and by reflecting the chessboard with respect to the principal diagonal. The last permutation (35281746) gives only four distinct solutions because it yields the same diagram after rotation of the chessboard. Recently, it has been shown that it is always possible to place k queens on a k x k chessboard for all k 2 4 (Hoffman, Loessi, Moore [1969].)

EXAMPLE 2. Cotlering a chessboard with tetraminos. Consider the problem of covering the 64 squares of a chessboard with 16 tetraminos, of various shapes, each covering exactly 4 squares. For example, if there are 15 L-shaped tetraminos and 1 square tetramino (Fig. 13.2), we can see that no covering is possible by the following argument: if the squares of the chessboard are coloured black and white as in Fig. 13.2, then the number n, of white squares and the number n, of black squares covered by an L-shaped tetramino always satisfies n, - n2 = 2 modulo 4.

L-shaped tetramino

square tetramino

Fig. 13.2

274

GRAPHS

In general, the problem reduces to verifying that the stability number of a certain graph G equals 16. To each tetramino i together with the position A occupied by this tetramino, there corresponds a vertex x(i, A). Join vertices x(i, A) and x(k, v) either if i = k , or if i # k and the set of squares occupied by tetramino i placed in position A overlaps with the squares occupied by tetramino k placed in position v. Each stable set with 16 vertices of the resulting graph corresponds to a covering of the chessboard with 16 tetraminos, and vice versa. Numerous analogous problems have been proposed by Golomb [19651. The construction of a maximum stable set is a particular case of the problem of finding a minimum transversal set of a hypergraph, which is discussed in Chapter 18. Consider a simple graph G = ( X , E). A clique is defined to be a set C c X such that each pair of distinct vertices in C are adjacent. Let %' = (Cl, C,, ..., ck)be a partition of X into cliques. Let B(G) = min I Q I 0

denote the smallest possible number of cliques that partition X

Theorem 1. Let G be a simple graph then a(G) < t?(G). Furthermore, [f S is a stable set and if %? is a partition into cliques such that I S I = I 9? 1, then S is a maximum stable set and V is a minimum partition. For each stable set S and each partition %? = (C, , C , , . .., ck), we have ISnC,I

Hence, I S

I < I V? 1.

<1

(i = 1, 2,

... , k).

Therefore

a(G)=maxISI~minI~l=e(G). Furthermore, if I So I

=

I so 1

I %?, I, the above inequality implies that = a(@,

I vo I = e(G) -

Q.E.D. There are many classes of graph G with a(G) = O(G). (See for instance the graphs considered in Examples 1 and 2.) Other classes of graphs with this property are studied in Chapter 16.

Remark. Given a minimum partition %? = (Cl, C,, ..., C), for a graph G with a(G) = 8(G), it is easy to construct a maximum stable set by a backtracing procedure : Select any vertex x1 E C1 Choose a vertex x, E C, that is non-adjacent to x l . Choose a vertex xa E C, that is non-adjacent to x , , x,, etc. When this procedure can no longer continue, backtrack (as in the

.

275

STABILITY NUMBER

exploration of an arborescence described in Chapter 3). Proceed until a maximum stable set has been found. Given a simple graph G = ( X , E ) and a stable set S sequence relative ro S is defined to be a sequence

= ,'A

an alrernaring

f J =( Q l , b l , a 2 , b Z , a 3 , . . . )

of distinct vertices alternately belonging to A satisfies the following conditions : (I) 0 1 € A , (2) b, E B - { b l , b z , ..., b,-, 1 and

{ a1

rG(bi)

9

a2,

=

X - S and to B

-.., al } # @

=

S that

3

(3) a l r l ~ A- { a l , a z ,...,a, ) a n d rG(ai+l)

rG(at+l)

{ 61, 6 2 , { a19 a t ,

6, 1 # 0 ,

1= @

.**$

*

An alternating sequence is said to be niaxirnal if no more vertices can be added to it without violating (2) or (3). We shall see that for the maximum stable sets, alternating sequences play a r6le similar to that of alternating chains for the maximum matchings. Recall from Theorem (4, Ch. 7) that a graph without odd cycles possesses a partition ( A , B ) of its vertices into two stable sets A and B ; such a partition is called a bicolouring of the vertices.

Lemma 1. Let G be a tree, and let ( A , B ) be a bicolouring of its rertices with 2 1 A 1. Then, graph G has at least one pendant certex in B. Suppose that all pendant vertices of G are in A . Let A, c A be the set of pendant vertices. Let B, = B be the set of pendant vertices of the tree GX--AI.Let A , c A be the set of pendant vertices of the tree GX-Al-Bl, etc. We have I A , I 2 1 B, I, because each pendant vertex of G x - a l is not pendant in G, and can be mapped into one of its neighbours in A , by an injection. Similarly, 1 B1 I 2 I A , 1, etc.... Thus, we can write:

1B]

I A 1 I >, ID1 I If A , , , #

I A2 I 2 I B2 I 2

*..

2 I Bq I 2 I A q + l

Bq #

a ; Bq+l= 0 .

a

a

121, we have IAI'

which contradicts that 1 A

I

IBI.

I 9

276

GRAPHS

If A , , , = 0, then set B, reduces to a single vertex (that is not pendant in G,i,"BJ. Thus, I A , I > I B, I, and

c 1

c a

I A I = i= 1 141> i=1 I 4 1 = I B I In both cases, a contradiction results. Q.E.D.

Lemma 2. Let G be a tree of order n, and let ( A , B ) be a bicolouring of its vertices such that I A I = I B I or I A I = I B I + 1. Then, there is an alternating sequence (a,, 6 , ,a,, b 2 , ...,) that uses each vertex of G exactly once. 1. Clearly, the lemma is true for n = 1 and n = 2. 2. If the lemma is valid for n = 2 k, we shall show that it is also valid for n = 2 k + 1. Since 1 A I = I B I + 1 > 1 B I, there exists by Lemma 1 a , A . For G x + ), there is an alternating sequence pendant vertex Q ~ + in ( a , , b,, .. ., bk) that uses all vertices, by the induction hypothesis. Hence ( a , , b l , ..., b k ,a k + , )is the required alternating sequence. 3. A similar argument shows that if the lemma is valid for n = 2 k 1, then it is valid for n = 2 k + 2. Q.E.D. Theorem 2. A stable set B is maximum i f , and only if, there is no maximal alternating sequence of odd length. 1. If such an alternating sequence a exists, then B is not a maximum stable set, since B' = ( B - a) U (a - B ) is a stable set with greater cardinality. 2. Let A be a maximum stable set, and let B be a stable set with I B I < 1 A I. We shall show that there exists a maximal alternating sequence of odd length relative to B. Let Bo = B - A , and let A . = A - B. Hence, I BO I < I A . 1. Let A , V B,, A 2 U B,, ,.., Ak U Bk be the connected components of subgraph GAOUBO, where A , c A o , B, c Bo for i = 1, 2, ..., k. Thus,

+

Hence, there exists an index i with I Bi I < I A, I. Let i = 1 be this index. 1 = I A , 1, a spanning tree of the connected subgraph G , i l V B l If I B1 I has a bicolouring ( A l , B,). By Lemma 2, its vertices constitute a maximal odd alternating sequence a relative to B,. Clearly, sequence CT is also for G an alternating sequence relative to B ; furthermore, sequence 0 is a maximal alternating sequence because if b E B - 6,either b E B - A = Bo and b is not adjacent to any a, € 6 ,or b E B n A and b is not adjacent to any a, E CT (since A is a stable set).

+

277

STABILITY NUMBER

+ +

If I B1 I 1 c I A , I, then pendant vertices can be removed from the spanning tree of G,,,,, until the remaining tree has a bicolouring ( A ; , B,) with I B1 I 1 = I A ; I, by Lemma 1. Then, as before, its vertices constitute the required sequence u. Q.E.D. meorem 3. A vertex of G is deJined to be "free" if it belongs to at least one maximum stable set but not to all maximum stable sets. Let B be a stable set. A vertex x is free and only is, there is an alternating sequence u = (a,, b, , a2, ..., aq, b,) relative to B that contains x and such that * I ' , ( b ) n { a , , a , ,..., a,] = bEB-a

a.

1. If such a sequence u exists and if x E 0 , let

A = X-B, B' = ( B - 0) u (0 - B ) , A' = ( A - a) u (0 - A ) .

B' is also a stable set, and since 1 B' 1 = I B I, B' is a maximum stable set. If x E B, then x $ B'; if x $ B, then x E B'. Thus, vertex x is free.

2. Let B be a maximum stable set that contains x , and let A be a maximum stable set that does not contain x. Let Bo = B - A. Let A . = A - B. Clearly, x E Bo. In the subgraph generated by A , U Bo, let A, u B1 denote the connected component with A1 c A0 , B , c BO, X E B ~ . If I B, I < I A, I, it is easy to see that there exists in G a maximal alternating sequence of odd length relative to B (as in the preceding theorem). This contradicts that B is a maximum stable set. Similarly, 1 B1 1 > I A , I contradicts that A is a maximum stable set. Thus

141= 1 . 4 1 1 . Consider a spanning tree of the connected graph generated by A, u B,. From Lemma 2, its vertices constitute an alternating sequence (T = (a,, b l , a2, ..., b,) of even length in graph G,,,,,. Sequence u is also an alternating sequence of even length in graph G. Furthermore, if b E B - a, either b E B - A = B,, which implies that vertex b is non-adjacent to { a,, a2, ..., u q } , or b E B n A , and b is nonadjacent to { a,, as, ..., aq} since A is a stable set. Thus, sequence o is the required alternating sequence. Q.E.D.

278

GRAPHS

2. The Turh theorem and related results No good algorithm is known for determining the stability number of a graph; however, there are several known bounds for a(G) in terms of various other invariants. Theorem 4 (J. C. Meyer [1972]). Let G = ( X , E ) be a simple graph of order n with vertices X I , x,, ..., x , such that 1 Q d~(x1Q ) ~ G ( x , )Q Jf for an integer p, 2 Q p Q n,

< do(&) .

dc(xn) + + ~ G ( X , - , + 2) Q n - P , then each stable set with fewer than p oertices is contained in a stable set with p vertices. The theorem will be proved by an induction on p. If p = 2, and if dG(xn)Q n - 2, then dG(x)< n - 1 for all x E X, and each vertex belongs to a stable set with two vertices. Thus the result is true. Let p 2 2. We shall assume that the result is true for p, and we shall show that it is also true for p + 1. Suppose that dG(xn) ... d G ( ~ n - P < n - p - 1 . Then, a fortiori, dG(x,) .-. dG(xn-,+,) Q n - p - 1 < n - P . Hence, from the induction hypothesis, a stable set S with fewer than p vertices is contained in a stable set So with p vertices. Let So = { y l , ..., y,). Then, dG(y,) -.. dG(yp)Q dG(xn) -.. d O ( ~ n - P +Ql )n - p - 1 . Therefore, l~,(So)I < n - p = I X - S o l . Consequently, there exists in X - Soat least one vertex that is not adjacent 1 vertices. to So.Therefore, S is also contained in some stable set with p Q.E.D.

+ +

+ +

+ +

+ +

+

Corollary (Berge [1960]). In a simple graph G with order n and maximum

degree h, each,maximal stable set has cardinality Let p =

n

-

at least.

[ h : ll*

*. Clearly

+

+ +

dG(xn) dG(xn - 1 ) ... d ~ ( x n - p+2) Q h(p The definition of p implies that h(p - 1) < n - p.

- 1)

219

STABILITY NUMBER

Thus 4AXn)

+ ddxn - 1) + ... + &(xn

- p + 2)

6 n -P

and the corollary follows from Theorem 4.

Q.E.D. EXAMPLE. There are 99 bridge players. No player will play with someone whom he does not know. Let h be the maximum number of players unknown to the same player. For what values of h can a bridge table (four players) be formed ? Let the vertices of G represent the players, and let the edges of G join pairs of players who do not know one another. If h = 32, then

and no table of four is possible. On the contrary, if It = 31, then a table can be formed. In fact, given any three players who know one another, a fourth can be found to play with them. Remark 1. Clearly, the above corollary implies that

For each n and for each h < n, we shall show that this inequality is the best possible. 1. We shall show first that if a and b are two integers, then

*

a-1

We can write:

a=qb+r, If r

=

0, then a - 1

=

~ < r < b ,q = [ t ] .

(q - l)b

[;I*

=

If r > 0, r < b, then a - 1 = qb

[;I

*=

+ (b - l), 0 < b - 1 < 6. Therefore, q =

I-[

a - 1

+1

+ ( r - l), 0 < r - 1 < 6. Therefore,

q + l = [ a +- 1J + l .

2. Let

Since n - 1 = ( p - l)(h -t- 1) + r, 0 tices can be formed from a clique with r



+ 1,

a graph with n ver-

+ 1 > 0 vertices a n d p - 1 disjoint

280

GRAPHS

(h + 1)-cliques. Clearly this graph has maximum degree h since, for all p , (even for p = l), it contains a (h + 1)-clique. Furthermore, its stablity number equals p . Remark 2. Related to Theorem 4 is a conjecture, due to P. Erdos 119671, that was recently proved by Hajnal and SzemerCdi [1970]: The vertices of a graph of order n and maximum degree h can be partitioned into h

+ 1 disjoint stable sets each with [-j *or [+I

vertices.

Remark 3. For planar graphs, the lower bound on a(G) given by Remark 1 seems weak. The following conjecture suggests an improvement:

Conjecture. (P. Erdos). Every planar graph G of order n has stability number n

a(G) Z

5.

This conjecture is weaker than the famous "four colour conjecture", which states that the vertices of a planar graph can be partitioned into 4 stable sets S , , S,, S,, S,, since this implies: n=ISl[+IS,I+ISJI+IS,J<4a(G).

It can be shown that a(G) 2

5

for each planar graph G of order n. See the

corollary to Theorem (12, Ch. 14).

Theorem 5 (Turdn [1941]). I f n and k are two integers, n 2 k > 0, puz

let r be an integer such that: n=k(q-1)+r,

O
Let Gnakbe the simple graph that consists of k disjoint cliques, of which r have q vertices and k - r have c - 1 vertices. Then, every graph G with n rertices and stability number < k that has the minimum possible number of edges is isomorphic to Gn,k. 1. Consider the various values for n given below:

; 1:

fk";:... 3k

3k+l 3k+2

...

4k

. ..

... ... ...

4k+1 4k+2

...

(4

... ... ...

+ l ) k ...

281

STABILITY NUMBER

The proof is immediate for the values of n in the first column; we shall show by induction that if the theorem is true for the q-th column, then it is true for the ( q + 1)-st column, i.e. for graphs with order

n=k(q+l)+r, O k. Subgraph G x - s has order n - k and stability number d k ; hence, by the induction hypothesis, m(GX-S)

Since Gn,kcan be formed from disjoint cliques in G, - k,L , m(Gn,k)

2

m(Gn-k,k)

Gn-k,k

by adding a vertex to each of the

-

- m(G,-k,k) =

*

Furthermore, since m(G) < m(Gn.k)by the definition of G, it follows that

n

- k = I X - S I Q mG(X-

S , S) =

= m(G) - m(Gx-s) d m(Gn,k)

- m(Gn-,J = n - k .

Hence,

(

m(G) = m(Gn,k) m(Gx-s) = m ( G n - k , k > 9

9

and, by the induction hypothesis, GX-S

=

Gn-k,k

Thus,G is formed from k disjoint cliques C1,C,, ..., c k and one stable set S.

3. The above inequalities show also that I X - S l =mG(X-S,S). Thus, each vertex x E X - S is joined to S by exactly one edge. Let s(x) denote the neighbour of x in S. If x E C,, y E C,, i # j , then s(x) # s(y) since, otherwise, the set would be a stable set with k + 1 vertices. If x E C, y E C,, then s(x) = s(y) since, otherwise, the number of cliques C,would be > I S 1 = k. Thus, graph G is isomorphic to graph G,, k . Q.E.D.

282

GRAPHS

Corollary 1. If G is a simple graph with n vertices and m edges, and with a(G) = k , then

where q

=

[:I-*

. Equality holds if, and only if, G is isomorphic to graph G,,,

Graph GnSkconsists of r cliques of q vertices and (k vertices. Thus the number of edges equals

- '(rq

=-

2

k.

- r ) cliques o f q - 1

+ kq - rq - 2 k + 2 r ) =

1

= z ( q - 1) (n

+ n - k(4 - 1 ) ) =

-k

Q.E.D. Corollary 2. If G is a simple graph with n vertices and m edges, then I1

a(G) 2

2m+n Equality holds i f , and only i f , the connected components of G are cliques with the same cardinality. 1. If a(G) = k , n = k(q - 1) + r, then from Corollary 1, 1 1 ma2(4- 1)(n-k+r)=-(n-r)(n-k+r). 2k It is easy to see that (n - x)(n - k x ) has its minimum value in [I, k] only for x = k . Hence 2 k m 2 ( n - k ) n. or n2 k > 2m+n' 2. If G consists o f p cliques with cardinality no, then a(G) = p , and

+

n2 2m

+n

-

P2 nf pno(no - 1 )

+ pno = p

= a(G) .

283

STABILITY NUMBER

Therefore, equality holds in the inequality of the corollary. Conversely, if equality holds, then G = G,,k, and r = k. Thus, G has the required form. Q.E.D. Corollary 3. I f G is a simple graph with n vertices and m edges, then 2n -m a(G) 2 . Equality holds if, and only & each connected component of G is either a 2clique or a 3-clique. We may assume that G is connected (otherwise, the result could be demonstrated for each connected component). 1. If G is a 2-clique, then 1 = a(G)

4-1 >- 1, 3

and equality holds throughout. If G is a 3-clique, then 1 = a(G)

6-3 >= 1, 3

and equality holds throughout. 2. Let G be a simple connected graph with n vertices and m edges. First suppose that m 2 n. From Corollary 2,

Note that the inequality n2

2m+n

> 2 n 3- m

is equivalent to n2 - 3 m n

f

2m2 2 0

or (n - m ) ( n - 2 m ) 2 0 .

Thus, if m 2- n, we have

Equality can hold only if m

(because m

It

is not excluded since 2 m 2 n) and if G is a clique. Thus, equality holds only if G is a 3-clique. = it

= -

284

GRAPHS

3. Now, suppose that 1 < m < n ; then m = n - I because G is connected. Thus, G is a tree. Since G is bipartite and n > 2, 2n-(n-1) 2n-m n n + l a(G) 2 -- 2 -2 3 3 3 ' Equality holds only if n n + l o$G)=-e2 3 ' i.e. if n = 2, i.e. if G is a 2-clique. 4. Finally, suppose that m = 0 and n = 1 ; then, clearly,

-

Q.E.D. Corollary 4 (Zarankiewicz [1947]). Let G be a simple graph with n vertices and with maximum degree h, and let k = [h- Then, a(G) 3 k . Besides, if G does not consist of k disjoint cliques each with cardinality

2, we

k

have

a(G) > k.

The number m of edges in G satisfies

Hence, from Corollary 2,

If G does not consist of k disjoint cliques with the same cardinality, then, from Corollary 2,

Q.E.D. Remark. We shall show that Corollary 4 implies:

If n is a multiple of h -t 1, let k

n 1' Then,

=h

a(G) 2

+

n h + l '

285

STABILITY NUMBER

If n is not a multiple of h

+ 1, let k = [ -h1. +n l

Then,

In both cases, Q.E.D.

3. a-Critical graphs Graph G is said to be a-critical if each partial graph G’ obtained from G by the removal of an edge satisfies a(G‘) = a(G) + 1. See Zykov [1949]. An a-critical graph is necessarily a simple graph. Property 1. A graph G with a(G) = k has an a-critical partial graph H with a(H) = k. If a(G) = k, any graph H obtained from G by the removal of an edge satisfies a(H) = k or k + 1. Successively eliminate the edges of G whose removal does not change the stability number until no more such edges exist. The remaining graph is a-critical and is a partial graph of G. Q.E.D. Property 2. In an a-critical graph G with a(G) = k , for each vertex x , there is a stable set S, with I S, I = k - 1 whose union with x is a maximum stable set. Set S, is called a “cell” of x. 1. Let x be a vertex of G. If there exists a vertex a adjacent to x, then the removal of edge [a, x ] creates a stable set S with k + 1 vertices. Thus, a, x E S, and the set S - { a, x } is a cell of x. 2. If x has no neighbours, each maximum stable set S contains x, and therefore, S - { x } is a cell of vertex x. Q.E.D. Property 3. In an a-critical graph, two vertices a and b have a common cell [L and only i f , they are adjacent. Let G be an a-critical graph with a(G) = k. 1. If a and b are adjacent, the removal of edge [a, b] creates a stable set S with k + 1 vertices. Thus, S - { a , b } is a cell of both a and b. 2. If a and b are non-adjacent, they cannot have a common cell S because then S u { a, b } would be a stable set with k + 1 vertices. Q.E.D. Property 4. Let G be an a-critical graph with a(G) = k. I f S , isa stable set with k - I vertices, then the set C, of vertices that hatie Soas a cell is a clique; furthermore, C, is disjoint .from So and no edge goes from C, to S o .

286

GRAPHS

The proof follows from Property 3.

Property 5. In a connected, a-critical graph G of order 2 3, each vertex has a degree 2 2. Suppose that there exists a vertex a with dG(a)< 2. Vertex a cannot be isolated since G is connected. Therefore, a is pendant and there is a vertex x with d,(a) = 1 . [ a , x] E E , Since n 2 3 and G is connected, edge [a, x ] is not an isolated edge, and therefore x has a neighbour b # a. Let s b , be a cell of both b and x. Then, 1 SbwI = k - 1, and a $ s b x (since a is a neighbour of x). Vertex a is not adjacent to S b x , since dG(a)= 1. But, then, s b x U { a, b } is a stable set with cardinality k 1, which contradicts a(G) = k. Q.E.D.

+

Property 6. Gigen two adjacent edges [a,b] and [b, x ] in an a-critical graph, there is a maximum stable set Sosuch that (1) a , b $ S o , X E S O , (2) only edge [b, x ]joins b to So. Let

s b x

be a cell of both b and x (which exists by Property 3), and let

so = Sb, " { x 1 Since a $ So (because a is a neighbour of b and is distinct of x), and since b 4 So, Property (1) follows. (2) is immediate. Q.E.D. The structure of a-critical graphs has been extensively studied. First, it has been proved by Beineke, Harary and Plummer [I9671 that in an a-critical graph, any two adjacent edges lie on a common odd cycle; Andrasfai [1967] has proved that in an a-critical graph each non-isolated edge lies on an odd cycle without chords. The following theorem generalizes both of these results :

Theorem 6 (Berge [1970]). In an a-critic'al graph G with a(G) = k, any two adjacent edges [a, h], [b, x] lie on a comnion odd eletnentary cycle without chords. 1. If edge [b, x ] is removed from G, a stable set S b , with cardinality k 1 is formed. Let B = S,, - { b } . Clearly, B is a maximum stable set in G, and a, b $ B, x E B. Only edge [b, x ]joins b to B.

+

STABILITY NUMBER

281

2. Clearly, stable set B is not maximum in the partial graph G - [a, b ] ; therefore, by Theorem 2, there exists in G - [a, b] a maximal alternating sequence rJ = (a1,b,,a2,b2, ...,a,), with a, E X - B and b, E B for all i. Since set T = ( B - 0 ) u (CT - B ) is a maximum stable set in G - [a, b ] , we have a, b E T. Hence, a, b E Q - B. The subgraph of G - [a, b ] generated by r~ is connected and has bicolouring (Q n B, r~ - B). Let p be the shortest chain connecting a and b. Since a, b E r~ - B, p together with edge [a, b] form an odd elementary cycle without chords in G. This cycle contains [a, b ] and [b, x ] (since only [b,x ] joins b to r~ n B). Q.E.D.

Corollary 1. A connected a-critical graph has no articuIation vertices. Suppose that a is an articulation vertex. Let B and C denote the connected components obtained by removing a, and let [a, b] and [a, c] be two edges with b E B and c E C. No elementary cycle can pass through these two edges. This contradicts Theorem 6 . Q.E.D.

Corollary 2. No clique of a connected a-critical graph is an articulation set. Suppose that A,, is a clique that is an articulation set. We shall show that the graph is not a-critical. Let A c A . be a minimal articulation set. Then, I A > 1 since, otherwise, the graph is not a-critical by Corollary 1. Let B and C be the two connected components resulting from the removal of A . Each vertex a E A is adjacent to both B and C (otherwise, A - { a } would be an articulation set, which contradicts the minimality of A). Let [a, b] and [a, c] be two edges with b E B and c E C. Each elementary cycle that contains these two edges contains a vertex a’ E A distinct from a. Since A is a clique, this cycle has a chord [a, a’]. From Theorem 6 , this contradicts the hypothesis that the graph is acritical. Q.E.D.

Corollary 3. A connected a-critical graph G is either a clique, or contains an odd cycle of Iengrh 2 5 wirhout chords.

If G is not a clique, then there exist two non-adjacent vertices a and 6 . Let [a, x1 ,x2, ..., b] be a shortest chain connecting a and b. Clearly, x 1 # a, b.

288

GRAPHS

No triangle contains both [a, x,] and [xl ,x,]. Thus, from Theorem 6, there is an odd cycle of length 2 5 without chords, Q.E.D. The following theorem characterizes a-critical graphs G with a(@ =

0. Theorem 7 (Berge [1960]). For an a-critical graph G = ( X , E), the following conditions are equivalent: (1) the smallest number B(G) of cliques that partition X satisfies WG) = 4G),

(2) graph G consists of a(G) disjoint non-adiacent cliques; ( 3 ) each cycle of length 5 in the complementary graph G = ( X ,B,( X ) - E ) has at least two chords. (1) * (2) Let B(G) = k, and let C , , C,, ..., C, be k cliques that partition X . Suppose cliques C1and C, are adjacent and let a, E C,, a, E C, with [a,, a,] E E. The graph H obtained from G by removing edge [ a l , a,], satisfies a(H) = k

Since (C, ,C,,

+ 1.

..., C,) is also a partition into cliques, it follows that B(H) Q k

Thus,

k

+ 1=a

(

G ~e(H) G k ,

which is a contradiction. Thus, condition (2) holds. (2) =. (3) Let ji = [xl ,x,, x 3 , x4, x 5 , x,] be a cycle in G. We have: [ x i , x ~ I [, ~ z , X ~[ xI3,, X 4 I , [x4,x5], [ X ~ , ~ , ] E ~ Z-( EX .) Let C1,C 2 ,..., C, be k disjoint non-adjacent cliques that form graph G. We shall show that ji has at least two chords in G. If k = 1 or 2, cycle ji cannot exist; therefore, we may assume that k 2 3. Cycle ji encounters at least three different C, successively; for example, suppose x1 E

c,,

xz E

cz,

x3 E c3.

STABILITY NUMBER

289

We may assume that either vertex x4 or vertex x, belongs to C, Otherwise, j i would have two chords in G and (3) would hold.

Fig. 13.3

For example, suppose x4 E C, (see Fig. 13.3). Then [x,, x4] and [xz, x,] are two chords of ji in c, and (3) holds. (3)

3

(I) First, we shall show that if a, x, y are three disrinct vertices of G with b,XIEE, b,YlEE, then [x, y ] E E. Suppose that [x, y ] q! E, and consider a common cell Sax of a and x and a common cell S,, of a and y. (From Property 3, these cells exist.) Since [a,y ] E E, y $ Sax. Similarly, x 4 Sag. Cell Saxis not a cell of y, because [ x , y ] g E and Property 3. Let b be a vertex of S,, that is adjacent to y. Similarly, let c be a vertex of S,, that is adjacent to x (see Fig. 13.4).

290

GRAPHS

We have b # c, because one of these two vertices is a neighbour of x and the other is not. Therefore, [x, y, c, a, b, x] is an elementary cycle of (7 of length 5, and has at most one chord in c,which contradicts (3). Thus, the binary relation defined by ‘‘x is adjacent to or identical to y ” is an equivalence relation. The classes of this equivalence relation are disjoint, non-adjacent cliques of G . Hence, .(G) = O(G). Q.E.D.

Theorem 8 (Hajnal [1965]). In an a-critical graph G wifhouf isolated uertires, each stable set S satisjies I T,(S) 2 I S I.

I

If I S I = 1, the result is evident, since G has no isolated vertices. If the result is valid for all stable sets with fewer than q vertices, we shall show that it is valid for a stable set S with q 2 2 vertices. Let

s = { XI,

x2,

..., xq } .

Let So = {XI, x2,

Suppose I T d S ) I < I S 4

I.

- 1 = I so I

* * a ,

Xq-1

Then, by the induction hypothesis, we have 4

I rG(S0) 1 < I T d S ) 1 d q - 1 .

.Yq

S

Fig. 13.5

Hence, T,(S)

=

1.

T,(So). Furthermore, we have

291

STABILITY NUMBER

Thus, from the Konig-Hall theorem (Theorem 5, Ch. 7), Socan be matched and the q - 1 vertices of T,(So), say a , , a,, ..., a , - , , satisfy into f,(s,),

(i = 1, 2,

[Xi, U J E E

...

)

q - 1).

Since f&q) # 0 (because G has no isolated vertices), there exists a vertex of T,(So) = I',(S) that is adjacent to x,. Without loss of generality, let a , be this vertex; let D be a common cell of a, and xl. Then, a,, x1 4 D and x, D (because x, is a neighbour of a,). Thus, set D can intersect set S u f , ( S ) only at the following vertices: f$

x,ora,;

x 3 0 r a 3 ; x 4 0 r a 4 ;... ; ~ ~ - ~ o r a , - ~ .

Therefore,

I (su ~ G ( S n) ) D 1 < 4 - 2 . If a(G)

=

k, the stable set T = D - (S u I',(S)) satisfies

I TI = I D I - I (SuTG(S))nDI

(k - 1)

- (4 - 2) = k

-

q

+I

and, hence,

I S u T I = I TI

+ I S 1 2 (k - q + 1) + q = k + I .

Furthermore, set S u T is stable (because T n T,(S) dicts a(G) = k.

=

0). This

contra-

Q.E.D. Lemma. Consider a graph G with no stable set of cardinality k S , , S, ..., S, are stable sets with k elements, then

+ 1. r f

-

I f p = 1, the result is evident. If the result is valid for p 1 stable sets, we shall show that it is also valid for p stable sets S1, S, ..., S., For k = 1, 2, ..., p , let k

k

Bk

=

n si.

i= 1

Clearly, set B,-l is non-adjacent to set A D - , ; thus, no vertex of set

292

GRAPHS

(Ap-, n Sp) u B p - , is adjacent to another vertex of the set and, consequently, the set is stable. Hence (1)

~ ( A , - , ~ s , ) ~ B , -<~u~( G ) = k . /

Furthermore,

By comparing equations (1) and (2), we have

1 + 1 B,-, I 2

Since the induction hypothesis yields that 1 A,-,

2 k,

IB,I=IB,-1nS,I=IB,-,I-IB,-1-~,I~

2 lBp-11

- JB,-1 - ( A , - l n S p ) I

2

~(2k-~Ap-~l)-(lApI-1Ap-~l)=2k-IApl~

Q.E.D. Theorem 9 (Hajnal [1965]). Let G be an u-critical graph without isolated vertices with u(G) = k and I X 1 = n. Then

do@)


(x E X) .

Suppose there is a vertex a with 1 TG(a)I > n - 2 k + 1. We shall show that there exists a stable set Sowith I Tc(So) I < I SOI. Let S1,S,, ..., Spbe the stable sets with cardinality k that contain vertex a. If we let A = U S,, B = S,, then, from the lemma,

n

+

(A( (B(22k. Furthermore, each vertex x E A is non-adjacent to set B - { a ) . Besides, if y e TG(a), there exists a common cell S,,, of a and y (from Property 3). From the definition of B, Thus, Sagu { y ) is a stable set that contains B - { a }, and again, y is nonadjacent to set B - ( a ). Finally, by letting So = B - { a ),

293

STABILITY NUMBER

and so

rG(&) c X - A - r,(a). - ( 2 k - ( B I ) - ( n - 2 k + 1) = IBI

=

- 1 = I&l.

Thus, set So is stable and satisfies To(&) I < I So I, which contradicts Theorem 8. Q.E.D. Corollary 1. Let G be an u-critical graph without isolated vertices. Then n u(G) < 5 . Equality holds i f , and only i f , each connected component of G is a 2-clique. Let k = u(G); then each vertex x satisfies 1 < d&) < n - 2 k 1. Hence, n-2k20. Equality can hold only if each vertex has degree 1, i.e., if all edges of G are isolated edges. Q.E.D.

+

Corollary 2 (Erdos, Gallai [1961]). An u-critical graph G with u(G) and I X I = n contains at least 2 k - n isolated vertices.

=

k

If G has n isolated vertices, then the result follows because k = u(G) = n. If G has p > n isolated vertices, the graph G’ obtained from G by removing the isolated vertices satisfies n’=n-p, a(G’) = k‘ = k p .

-

From Corollary 1, graph G’, which is u-critical and without isolated vertices, satisfies 0 < n’ 2 k‘ = (n - p ) - 2(k - p ) .

-

Q.E.D.

I. Consider a graph G consisting of an odd cycle with n = 2 k + EXAMPLE 1 vertices. Then a(@ = k , and G is a-critical. This graph is the only connected u-critical graph with n - 2 k = 1 (because, from the Hajnal Theorem, d&) d 2 for all x , and neither an open chain nor an even cycle is u-critical).

294

GRAPHS

EXAMPLE 2. Consider the graph G with four vertices a, 6, c, d and 6 chains p&, 61, p&, cl, pda, dl, p&" cl, p5[b,dl, pdc, 4 of odd length 2 3 that are vertex-disjoint (except at their endpoints), Let u(C) = k ; then n-2k=2. As seen from Fig. 13.6, the removal of any edge [x, y ] increases the stability number, and, therefore, G is a-critical. Andrgsfai [1967] has shown that

k=l n=16

Fig. 13.6

the graphs of this type are the only connected a-criticalgraphs with n - 2 k = 2. The structure of connected a-critical graphs with n - 2 k = 3 is not known. So far, only the a-critical graphs with connectivity 2 have been characterized (Wessel [1970]). Several methods for constructing new classes of u-critical graphs have been described by Plummer [1967],Gallai and others (see George [1971]). The following result gives the maximum possible number of edges in a-critical graphs.

Theorem 10 (Erdos, Hajnal, Moon [1964]). I f G is an a-critical graph with n t?erticesand m edges, and with u(G) = k , then

G consists of a stable set with k - 1 i>ertices Equality holds f , and only and a clique with I I - k + 1 rertices. Let .f,(n) denote the maximum possible number of edges in a graph G with the properties: (1) G has at least one stable set with cardinality k , (2) the removal of any edge creates a new stable set with cardinality k + 1, (3) G has order n.

295

STABILITY NUMBER

1. First, we shall show that

Note that

h ( k + 1) = 1

because each graph satisfying (l), (2) and (3) with k + 1 vertices and with a maximum number of edges consists of a 2-clique and a stable set of k - 1 vertices. Consider a graph G that satisfies (I), (2) with order ii > k 1 and with &(n) edges. Let a and b be two adjacent vertices. From (2), the removal of edge [a, b] creates a stable set S with cardinality k + 1, and a E S, b E S. Consider the graph G' obtained from G by removing vertex b and each edge [a, z ] with z 6 T,(b). For G', set S - { b is stable with cardinality k. The removal of an edge of G' creates a stable set with cardinality k + 1, since this creates in G a stable set with cardinality k + 1 that either contains a and not b or contains neither a nor b. Hence the number of edges in G' is

+

nt(G')

< fk(n - 1).

Furthermore, the number of edges removed from G to form G' is m(G) - m(G) < I Hence

+(X-

Sl = 1

+ n - (k + 1)

ji(#)= m(G) < m(G') + n - k 6 fk(n - 1)

=n

-k

.

+ ( i ~- k ) .

Thus,

)fk(n

... fk(k

fk(n) Gfk(n - 1)
- l) + (n - k) - 2 ) + (n - 1 - k)

+ 1) = 1 .

Hence,

fk(n) 6 1

+ 2 + ... + (n - k) = (n - k ) ( n2- k + 1)

2. Denote by G,(n) a graph consisting of the union of a (n - k + 1)clique and a stable set of k - 1 vertices. Clearly, graph G,(n) satisfies (I), (2) and (3), and has

(" -

-k ')edges.

296

GRAPHS

Thus

and G,(n) is a graph with the maximum possible number of edges. 3. We shall show that each graph of order n withf,(n) edges that satisfies (1) and (2) is isomorphic to G,(n). Clearly, this is true for n = k 1. By induction, we shall show that it is also true for a graph G of order n > k 1. Consider an edge [a, b] of G, and construct a graph G' by removing vertex b and each edge [a, z] with z q? T,(b). As before, m(G) - m(G') ,< n - k. Hence

+

+

m(G') >/fk(n) - (n - k ) =fk(n - 1). From Part 1, G' satisfies conditions (1) and (2) since G has order n - I and has the maximum possible number of edges, it follows from the induction hypothesis that

G'

3

Gk(n - 1)

Kn-k L' s k - 1 ,

where Kn-, is a clique with n - k vertices and Sk-lis a stable set with cardinality k - 1. We shall consider two cases. 1 : U E S k - 1 (See Fig. 13.7). Since the removal of edge [a, b] creates in G a stable set with cardinality k 1, there exists a vertex y with CASE

+

YEKn-k,

Y # r ~ ( b ) ,r d Y ) n s k - t

=

@*

Furthermore, if c E Kn-%and c # y , then c is not adjacent to a because, otherwise, a stable set with cardinality k + 1 could not be created in G by removing edge [a, c]. Furthermore, c is adjacent to b, since the number of edges removed from G to form G' is equal to

fk(n) -&(n

- 1)

Fig. 13.7

=n

-k .

297

STABILITY NUMBER

+

Then, if edge [b, cl is removed from G , a stable set with cardinality k 1 cannot be created, which contradicts the hypothesis. Thus, Case 1 cannot occur. CASE 2: a E Kn-k. Vertex b cannot be adjacent to set in G (because the removal of [a, b] would create a stable set with'cardinality k 1). On the other hand, b is adjacent to each vertex of Kn-k (because the number of edges removed from G to form G' equals n - k). Thus, G is the union of an (n - k + I)clique, Kn-ku {b}, and a stable set & - l . Q.E.D.

+

4. Critical vertices and critical edges

This section generalizes some of the results of Section 3. A vertex x of G [a, b ] should create a stable set with cardinality k 1). Also, b is adjacent to each vertex of K n - k (because the number of edges removed from G to I)-clique form G' equals n - k). Thus G is the union of an (n - k K n T ku { b } and a stable set & - I . Theorem 11. The endpoints of a critical edge are non-critical vertices.

+

+

Let G' be the graph obtained from G by removing a critical edge [ x , y ] . Then, a(G') = a(G) 1 Hence,

+ .

a(G)

2 C ~ ( G ~ -={ ~a(Glu-{x,) )) 2 a(G') - 1 = a(G).

Therefore, a(Gx-(x)) = a(G) 9

and vertex x is not critical.

Q.E.D. Theorem 12. The edges incident to a critical vertex are non-critical edges. Let x be a critical vertex of G, and let [ x , y ] be an edge incident to x . Then, a(Gx-(x)) = a(?)

-1

*

Let Ex denote the set of edges incident to x. Then, a(G)

< a(G - [x, y ] ) Q a(G - EJ

=1

+ a(Gx-

= a(G).

Thus a(G - [x, y ] ) = a(G), and edge [x, y ] is not critical.

Q.E.D.

298

GRAPHS

Theorem 11 shows that the only critical vertices in an a-critical graph are the isolated certices. The following theorem applies this result. n Theorem 13. I f a graph G with order n has no critical certices, then a(G) < - . 2 Furthermore, if equality holds, then

n

u(G) = B(G) = - . 2 n n Suppose a(G) 2 - ; we shall show that a(G) = B(G) = 2 2’ Since G has no critical vertices, each vertex of G belongs to the complement of a maximum stable set. After removing an edge without changing the stability number, this remains true a fortiori. Suppose that enough edges are removed to form an a-critical graph H with a ( H ) = a(G). Then, graph H has no critical vertices (and no isolated vertices) and satisfies or(H) 2 !. 2 11 n From Corollary 1 to Theorem 9, a ( H ) = - , and H consists of - pairwise 2 2 disjoint edges. Hence, n n - = a ( H ) = a(G) < B(G) < B(H) = 2 2 Therefore

n

a(G) = B(G) = 2

Corollary.

. Q.E.D.

u,for each rertex x, the subgraph Gx-{,,contains a stable set

S with I S I > f , then G contains a stable set So with 2 n IS,/ > - + 1. 2 From Theorem 13, G has at least one critical vertex a. Hence

a(G) = a(Gx-

+ 1 > 2--n + 1 . Q.E.D.

5. Stability number and vertex-coverings by paths In this section, we shall assume that G is a I-graph. Theorem 14 (Gallai, Milgram [1960]). In a 1-graph G a(C) elementary paths that partition X .

=

( X , U ) , there exist

STABILITY NUMBER

299

Let M = { p l , p2, ..., p k } be a family of elementary pairwise disjoint paths that cover X . Let A ( M ) = { a,, a2, ..., a k } be the set of initial vertices of the paths in M. A family M always exists since all paths of length 0 can be chosen. We shall show that there is a family M' of elementary paths that partition X such that I M' I 6 a(G) and A ( M ' ) c A ( M ) . (This proposition is in fact stronger than the theorem.) 1. Clearly, this proposition is true for a graph with one vertex. If it is valid for all graphs with less than n vertices, we shall show that it is also valid for a graph G with n vertices. 2. If I M I > a(G) 1, we shall show that M can be replaced by a family with I R I < a(G) 1 and A ( @ ) c A ( M ) . Consider the subgraph generated by Xl = X - p l . The stability number of this subgraph is

+ +

a

.W,J G dc>Y and M1= { p 2 , ..., p k } partitions the vertex set of G x l . Hence, from the induction hypothesis, there exists a family R1in GX, such that I Gi I G ~ ( G x , < ) a(G) , 4 M i ) c A(Mi). Thus, family = U { p l } is the required family of G. 3. Clearly, if I @ I 6 cc(G), the proposition holds. If I R I = a(G) 1, set A ( @ ) = { a,, a 2 , ... } is not stable, because I A(H) I > a(G). Thus, there is an arc joining two of its vertices; let this arc be (a,, a2). If the path ,iil E @ which begins at a, has zero length, then the proposition holds. Otherwise, let

al +

...

Cll = Ca1, b l , c1, 1. The subgraph generated by XI = X - { a } possesses a family M , with

I

I ,< a(G) A ( M , ) c { 61, a2, a3 ... ] 3

I' 1

I

I I

Fig. 13.8

300

GRAPHS

If b, E A(A4,) or if a, E A ( M , ) , the proposition follows by adding vertex a, to one of the paths in M , . If b, $ A ( M , ) , a, 4 A ( M , ) , then I M I I < a(G), and the proposition follows with

M‘ = M , u [ a , ] . Q.E.D. Remark. This result is the best possible since for each simple graph it is possibl‘e, by assigning a direction to its edges, to form a I-graph G with min I M I = a(@. M

To find this direction, simply choose a ‘maximum stable set So and direct each edge incident to So towards So. Then, I M 1 2 a(G) for all families M .

Corollary 1 (RCdei). A complete I-graph contains a hamiltonian path. Since a(G) = 1 for a complete graph G, the existence of a path that covers all the vertices of G follows from Theorem 14. Q.E.D. Corollary 2 (“Dilworth’s theorem”). I f G is a transitire 1-graph and i f M is a family of paths that partitions its rertex set, then min I M I = a(C) . M

Theorem 14 implies that min I M I 6 a(G). Since G is the graph of a partial order, each path generates a clique. Therefore, m(G) 6 O(G) 6 min I M 1, and equality holds. Q.E.D. APPLICATION (“Sperner’s theorem”). If X is a set of n elements, and if F is a family of distinct subsets of X such that no member of 9 is contained then in another member of F,

This famous result has recently been extended in several directions (Erdos and Katona, Kleitman, Meshalkin, etc). We shall show that the Spencer theorem is an easy consequence of Corollary 2. Consider the graph G whose vertices represent the different subsets of X . Place the vertices of G into n + 1 rows 0, 1, 2, ..., n where the h-th row contains the vertices corresponding to the subsets with n - h elements. There are

(3

such subsets. Join vertex a to vertex b by an arc (a, b) if the corres-

ponding sets A and B satisfy A

3

B.

301

STABILITY NUMBER

The maximum cardinality of a family F with the required property equals the stability number a(G). Since the set of elements in a row is stable, there exists a stable set with

([ij)

members in row h =

From the Dilworth theorem, this stable set is maximum if and only if

(,,)

the vertex set of G can be partitioned into

paths.

To show that this is possible, notice that, given two consecutive rows, the smaller can always be matched into the larger: Since two consecutive rows form a bipartite graph in which all vertices of the same class have the same degree, this follows from Corollary 4 to Theorem (5, Ch. 7). For all h, let Eh be a maximum matching between the rows h and h 1. Then the union of the ? ,?h define the edges of the required paths. Q.E.D.

+

EXERCISES 1. Using the Turfin theorem, show that the minimum number of edges in a graph G with n vertices and stability number k equals n-

1

c [f].

i=o

2. If n 2 4 ( p

+ l), and

if each subgraph G, with

a(GA) < p, then show that m

<

I A I = 2p + 1

[TI.

(Las Vergnas) satisfies I A I -

(Erdos, Galhi [1961])

3. Consider a simple graph G = (X, E) and a partition (XI, X,, ..., X,) of X such that P

dGxj).

=

i= 1

Show that a(C) = O ( c ) if ~ ( G x ,= ) Q(Gx4) 4. If the partition (XI, X,,

(i

=

1 , 2,

...,p )

I

..., X,) above has more than one class, then

G is said to be decomposable. Show that a connected a-critical graph is not decomposable. (Plummer) 5. For a non-decomposable graph G that is not a clique, show that a(C) c O ( c ) . 6. Let G = (X,E ) be a connected graph, and let Eo be a maximum matching. Suppose

that Eo is not a perfect matching. Let XI denote the set of vertices that can be reached

302

GRAPHS

from an unsaturated vertex by an even alternating chain (and by no odd alternating chain). Let Xzdenote the set of vertices that can be reached from an unsaturated vertex by an odd alternating chain (and by no even alternating chain). Show that if XI U X, = (Berge [1957]) X, then XI is a maximum stable set of graph G. 7. If A c X is a set of vertices in G, let S(A) denote a maximum stable set of the subgraph generated by A. Given a maximum matching Eo, let MI, M2, ..., denote the con-

nected components of the subgraph generated by the “mixed” vertices. Let 4 , I,, ..., denote the connected components of the subgraph generated by the “inaccessible” vertices. Let

S =X1u S(M1) u S(M2) u ...u S(1l) u S(I2) u ... If there is at most one connected component of mixed vertices, show that S is a (Berge [1957]) maximum stable set of G. 8. Show that in a graph G without critical vertices,

1 r,(S)I 2 I S I for every stable set S.

9. Recall that if G is an a-critical graph without isolated vertices then G has no critical

vertices. Show that the converse is not true by considering the following graphs: (1) the complementary graph of a cycle of length 7 without chords, (2) C, augmented with a vertex that is joined to each vertex of C,.

c,,

I:[

10. If G is a graph with n vertices and m edges with rn >

triangle. Show that there exists a graph of order n with

, show that G contains a

[TI

edges that contains no triangles.

11. Show that the number m(G) of edges in a graph G with a(G) = k is greater than or equal to 2k 1. Show that m ( G ) = 2 k 1 if, and only if, G is an elementary cycle of length 2 k 1. (Bandy)

+

+

+

12. Let A = { ao,al, ..., a,} be an articulation set of graph G. If, for an integer k with 0 G k < p , each vertex a,, al, ..., ak is adjacent to vertices x1 and x2, belonging to different components of G x --A, and if each vertex ak + a k , 2r ..., a, is adjacent to a0, then G

is not a-critical. (Wessel [19701) 13. Show that the complementary graph of the Petersen graph (Fig. 10.11) is a-critical. 14. Let G be a 1-graph. Show that there exists a stable set S such that each vertex of G is the initial endpoint of a path of length < 2 with a terminal endpoint in S. (Lbvasz, Chvhtal)

CHAPTER 14

Kernels and Grundy Functions

1. Absorption number

For a 1-graph G each x $ A ,

=

(X,r),a set A

c X is defined to be absorbant if for

T(x)n A #

125.

Let &’ denote the family of all absorbant sets of graph G . Then XEd, A E ~ A, ’ I A *

A‘€&.

The absorption number of graph G is defined to be B(G) = min I A

I.

A € ad

This section studies the absorbant sets with minimal cardinality.

EXAMPLE 1. Radar stations. A number of strategic locations x l , x2, ... (called cells) are kept under the surveillance of radar. Radar in cell x4 can survey x1 or x2 or x3, as shown in Fig. 14.1. Similarly, cell x2 can be surveyed by radar located in x3 or x5, etc. What is the minimum number of radar stations needed to survey all the cells? It is the absorption number of the graph in Fig. 14.1, which is equal to 2, since radar stations placed at x2 and x4 are sufficient.

s2 A={

s-2. S S

}

Fig. 14.1

303

304

GRAPHS

EXAMPLE 2. Chessboard controlled byfive queens. What is the minimum number of queens that can be placed on a standard chessboard so that each square is controlled by at least one queen? Five queens are sufficient because the absorption number /(G) of the graph G, defined by the moves of a queen on a chessboard, is 5 (see Fig. 14.2). Note that the same placement of the 5 queens also controls a 9 x 9 chessboard. The placement at the right in Fig. 14.2 shows how five queens can control an 11 x 11 chessboard. No general results are available for the maximum size n(k) of a chessboard that can be controlled by k queens.

Fig. 14.2

Finding a minimum absorbant set is a special case of the minimum transversal problem, for which constructive algorithms are given in Chapter 18. Here, we shall give bounds on the absorption number. Proposition 1.

If G is a

1 -graph with n vertices and m arcs, then

B(G) 2 n - m . Let A be a minimum absorbant set. Each vertex of X - A is the initial endpoint of an arc going into A . Thus, n-IAl=IX-Al<m. Hence,

p(G) = I A

I2n

-

m

.

Q.E.D. Proposition 2. I f G is a I-graph without loops with n uertices, then B(G)


- max d F ( x ) . xsx

305

KERNELS AND GRUNDY FUNCTIONS

Let xo be a vertex with d;(xo)

=

d;(x). Since the set

max,.,

A =x

- z-,-(XO)

is absorbant, we have B(G)

< IA 1 = n

- max d G ( x ) . xsx

Q.E.D. If G is a simple graph, its dominance number P*(G) is defined to be the absorption number P(G*) of the symmetric graph G* obtained by replacing each edge in G by two oppositely directed arcs. As in the.Tura’n theorem, we shall calculate the maximum number of edges possible in a simple graph with a given order and dominance number. Theorem 1 (Vizing [ 19651). I f G is a simple graph with n uertices, m edges and B*(G) = k 2 2, then

< [ 15 ( n - k ) ( n - k + 2 ) ] . For each n and each k, equality holds if and only if G is isomorphic to a graph m

G , , k obtained by remocing from a (n - k)-clique the edges of a minimum covering and by joining its vertices to each rertex in a stable set of cardinality k.

1. First, we shall show that 1

rn<-(n-k)(n-k+2).

2

Clearly, this inequality is satisfied for n = 2. We shall assume that it is satisfied for all graphs with order < n, and we shall show that it is satisfied for a graph G with order n > 2 and with P*(G) = k > 1. Let G* be the symmetric graph obtained by replacing each edge of G by two oppositely directed arcs. Let xo be a vertex with maximum degree. From Proposition 2, we have

I TG(xo)I = d,(x,,)

= max d , ( x )


-k

,

X

Thus, we may write:

I TG(x,)I = n - k - r ; Let S

=

X

0

< r < n - k.

- { x o } - f G ( x o )then IS1 = k + r -

1.

I f y E TG(xo),the set (S - TG(y))U { x o , y } is absorbant, and, therefore,

306

GRAPHS

Thus,

k'+r-1 -)fG(y)nSI+2>k or

I r G ( y )n S I < r + I . Furthermore, if G, is the subgraph generated by set Sand if A is a minimum absorbant set of graph G,, then

IAu{xo)I~k (because A

U { xo}

is absorbant in G). Hence, P*(G,) 3 k

-1.

By the induction hypothesis, the number of edges in G , is

nt(G,)

< 1 (k + r =

- 1-

(k - 1)) ( k

+ r - 1 - ( k - 1) + 2 ) =

1

- r(r + 2 ) . 2

Therefore, the number m ( G ) of edges in G satisfies

+ I fG(xo) I + rcixo) 1 (I fJy) n S 1 + I f',(y> I) < < r(r + 2) + ( n - k - r ) + ( n - k - r ) ( r + 1) + ( n - k - rI2 =

2 m(G) = 2 m(G,)

YE

=

(n

< (n

- k ) ( n - k + 2) - r(n - k - k)(n - k + 2).

-

r ) ,<

The required inequality follows.

2. We shall show that there exists a graph G,,k with order n and with P*(Gn,J = k , such that nt(G,,,) =

[21 ( n - k) ( n - k + 2 ) ] .

For k = 2, consider the graph G,,, defined by the union of a set A , = Bn-,= { xg. ..., x, } where each vertex of A 2 is joined with each vertex of B n W 2each , vertex of Bn-2is joined with every other vertex of B n - 2 ,except for some pairs of vertices that constitute a minimum covering of B,-z(see Ch. 7, Q 2). { xl, x2 } and a set

307

KERNELS AND GRUNDY FUNCTIONS

=

[?(" 1

-

214.

Thus, for this graph G,,. the required equality holds. If k > 2, define graph G,,.k by adding to graph G,,-k+z,2a set of k isolated vertices. Then,

[2

1 m(Gn,,) = n i ( G , , - k + 2 , z = ) -(n - k

-2

+ 2 - 2 ) ( n - k + 2)]

Again, equality holds. The inequalities in Part 1 can be used to show the uniqueness of graph G,,.k . Q.E.D.

Corollary.

If G is a simple graph with n ilertices and m edges, then P*(G) < I I + 1 - ,, 1 + 2 ni

From the inequality of Theorem 1, we have (n - k)2 + 2(n - k ) - 2 111 2 0 . Since n - k 2 0, then

or

< ?I + 1 -,1 + 2 m . ~-

k

Q.E.D. 2. Kernels

For a 1-graph G = ( X , r),a set S = Xis defined to be a kernel if it is both stable and absorbant, i.e. if (1) (2)

S

+

x $S

+

XE

T(x)n S = T(x) n S #

0 0

(stable) (absorbant).

308

GRAPHS

Not every graph has a kernel (see graph in Fig. 14.3), and if a graph possesses a kernel, this kernel is not necessarily unique (see graph in Fig. 14.4). In this section we shall present existence and uniqueness theorems.

Fig. 14.3. Graph without kernels

Fig. 14.4. Graph with two kernels

EXAMPLE 1 (Von Neumann, Morgenstern [1944]). The concept of a kernel was first presented (under the name solution) in game theory. Suppose that n players, denoted by (l), (2), ..., (n), can discuss together to select a point x from a set X (the “situations”). If player (i) prefers situation a to situation 6, we shall write a > *b. The individual preferences might not be compatible, and, consequently, it is necessary to introduce the concept of effecticepreference. The situation a is said to be effectirelypreferred to b, or a > b, if there is a set of players who prefer a to b and who are a11 together capable of enforcing their preference for a. However, effective preference is not transitive, i.e., a > b and h > c do not necessarily imply that a > c. Consider the 1-graph (A’, r),where T(x) denotes the set of situations effectively preferred to x. Let S be a kernel of the graph (if one exists). Von Neumann and Morgenstern suggested that the selection be confined to the elements of S. Since S is stable, no situation of S is effectively preferred to another situation of S. Since S is absorbant, for every situation x q! S, there is a situation in S that is effectively preferred to x , so that x can be immediately discarded. EXAMPLE 2. Basis of axioms. Consider a “theory”, i.e., a set of propositions a, 6, c, ..., that we shall represent by vertices; add arc (a, b) if proposition b implies proposition a. The resulting graph G = ( X , U ) is transitive, i.e.: ( a , C)E u . (U,b)E u, ( 6 , C)E u A basis ofaxioms of the theory is a set B of propositions (called “axioms”) such that: (1) each proposition not i n B follows from one of the axioms, (2) no axiom follows from another axiom. The problem of finding a basis of axioms reduces to finding a kernel of G. It will be shown later that each transitive graph has a kernel.

309

KERNELS AND GRUNDY FUNCTIONS

Proposition 1. A necessary and suficient condition for a set S c X to be a kernel of a 1-graph G = ( X , r)is that its characteristicfunction cps(x)satisfies cp,(x) = 1

- max cp,(Y)

*

YE r(x)

Recall that the characteristic function cps(x)of set S is defined by cp,(x>

( == O

if if

XES, x$S.

If T ( x ) = 0,we define max cp,(y) = 0 . rw

YE

1 . Let S be a kernel. Since S is stable. we have cp,(x) = 1

=s

XE

Since S is absorbant, we have cp,(x) = 0

e-

max cp,(y) = 0 .

S

x$ S

-

Y E

rw

max cp,(y) = 1 .

IE r w

Combining these, the required formula follows.

2. Let cp,(x) be the characteristic function of a set S which satisfies the formula; then we have XE S =s cp,(x) = 1 =$ max cp,(y) = 0 =. T(x)n S = @ Y Er w x$S

*

cp,(x) = 0 Y

max cp,(y) = 1 rw

=$

T(x)n S #

0.

Thus, S is a kernel.

Q.E.D. Proposition 2. If S is a kernel, then S is a maximal stable set and a minimal absorbant set. Let S be the kernel of a I-graph G = (A', r).I f a 6 S, the set S u { a } cannot be stable because r ( a ) n S # fa. I f b E s, the set T = S - { b } cannot be absorbant because b 6 T and T(b) n T = 0. Q.E.D. Theorem 2. If G = ( X , r) is a symmetric I-graph, then G has a kernel. Furthermore, a set S c X is a kernel 5f, and only S is a maximal stable set. I . Clearly, a maximal stable set S of G is absorbant (otherwise, there would exist a vertex x 6 S non-adjacent to S , and S could not be a maximal stable set). Thus, S is a kernel.

310

GRAPHS

2. Conversely, if S is a kernel of a symmetric graph G, then S is a maximal stable set (because, otherwise, S would not be absorbant). Q.E.D.

Algorithms to construct all kernels of a graph have been presented by Roy [1970] and Rudeanu [1966]. Theorem 3. r f G = (X,r) is a transitiue I-graph, each minimal absorbant set has cardinality /I(G). Furthermore, a set S c X i s a kernel if, and only i f , S is a minimal absorbant set Let G

=

( X , r) be a transitive 1-graph, i.e.: yEr(X),

.Erg)

9

ZE T(X).

Consider the strongly connected components of G . If C is a component with m:(C, X - C ) = 0 (i.e., no arcs leave C ) , C will be called a terminal component. Since the graph obtained from G by contracting each strongly connected component contains no circuits, G possesses terminal components. Let C1, C 2 ,..., Cq be the terminal components of G. Each of these is a complete symmetric subgraph, because G is transitive. 1. If A is a minimal absorbant set, then A contains at least one vertex from and each x E C, satisfies each terminal component. Otherwise, A n C, = 0, x#A,

T(x)nA=

0,

which contradicts that A is absorbant. Let a, E A n Cifor each i. Consider the set A’

=

a , , a 2 , ..., a4 1 .

A’ is also an absorbant set because each x $ A’ is the initial endpoint of an arc leading into A‘ (because of transitivity). Hence A’ = A , and each minimal absorbant set is also a minimum absorbant set. 2. If S is a kernel, then S is a minimal absorbant set by Proposition 2. Conversely, if A = { a,, a2,. . ., aq } is a minimal absorbant set, then A is stable because no arc leaves a terminal component; therefore, A is a kernel. Q.E.D. Corollary 1. A transitice I-graph has a kernel, and ull of its kernels have the same cardinality.

Consequently, in Example 2, all the axiom bases have the same cardinality.

KERNELS AND GRUNDY FUNCTIONS

Corollary 2. In a graph G

=

311

( X , U ) , there is a set B c X such that

(1) no paths join trvo distinct t'ertices of B, (2) each vertex x 6B is the initial endpoint of a path leading into B.

The corollary follows by applying Corollary 1 to the transitive I-graph

( X , P ) obtained by creating an arc (x, y ) if there is a path from x to y in G. Q.E.D. Theorem 4. A 1-graph M'ithout circuits possesses a kernel, and this kernel is unique. Given a 1-graph G

=

( X , f) without circuits, consider the sets

X ( 0 ) = ( X / X E X , f ( x ) = fzr} X ( l ) = ( x / x # X ( O ) , m)C X < O ) > X(2) = { x / x $ X ( O ) u X(t), f ( x ) c X ( 0 ) u X ( 1 ) } , etc... These sets are pairwise disjoint, and x E X ( k ) if, and only if, the longest path from x has length k. Since G contains no circuits, the sets X ( k ) form a partition of X . A characteristic function qs of a kernel Sc a n be successively defined on the sets X(O), X(1), X ( 2 ) , etc. by the equality of Proposition 1. Furthermore, cpS(x)is uniquely defined for each x. Q.E.D. The following theorem due to Richardson shows that a graph without odd circuits possesses a kernel. The original proof of this result was long and involved. The less complicated proof presented here is due to Victor Neumann. A semi-kernel of a 1-graph G = ( X , r) is defined to be a non-empty stable set S c X such that each x adjacent to S has at least one successor in S. Lemma. if for each non-empty subset A c X , the subgraph GAhas a semikernel, then G has a kernel. Let S be a maximal semi-kernel of C , and let A = X - S - fG(S). If A = fzr, then S is a kernel. If A # fzr, then GAhas a semi-kernel T. Sets S and T are nonadjacent; thus, S u T is stable. Each x $ S v T that is adjacent to S u T has a successor in S u T. Thus, S u T is a semi-kernel of G, which contradicts the maximality of S. Q.E.D. Theorem 5 (Richardson [1953]). If C circuits, then G has at least one kernel.

=

( X , r) is a I-graph without odd

312

GRAPHS

From the lemma, it suffices to show that G possesses a semi-kernel. We may assume that G is connected. Let XI be a strongly connected component of G with I'(X,) XI.If 1 Xl 1 = 1, then X1is a semi-kernel. If 1 X , I > 1, and if xo E X , , let x be a vertex in XI distinct from x o . Then, all paths p [ x o , x] remain in the interior of XI ; and they have the same parity (since, otherwise, an odd circuit could be formed with a path p[x, x,,]). Let S denote the set of all x E XI such that all paths p[xo, x ] are even. Set S is stable. If z E Xis adjacent to S, then each successor of z belongs to S. Thus, S is a semi-kernel. Q.E.D. Note that the graph in Fig. 14.3 has no kernel: therefore it contains an odd circuit. Similarly, the graph in Fig. 14.4 does not have a unique kernel, therefore it contains a circuit.

3. Grundy functions Consider a 1-graph G = ( X , r)without loops. A non-negative integer function g(x) is called a Grundyfunction on G if for every vertex x, g(x) is the smallest non-negative integer which does not belong to the set { g(y) / y E f ( x ) }. This concept originated with P. M. Grundy El9391 for graphs without circuits. It was extended to 1-graphs by Berge and Schutzenberger [1956]. The Grundy function can also be defined as a function g(x) such that (1) g(x) = k > 0 implies that for each j < k , there is a y E T ( x ) u+th g(y>= J , (2) g(x) = k implies rhat each y E T(2) satisfies g(y) # k. A pseudo-Grundyfunction is defined to be a function that satisfies property (1) above and (2') g(x) = 0 impIies that each y E T ( x ) satisfies g ( y ) # 0. We shall see in the following development that a Grundy function (or a pseudo-Grundy function) determines a kernel of a graph. Remark 1 . Some graphs have no Grundy function. Some graphs have more than one Grundy function. (See Fig. 14.5 where the value of a Grundy function is written next to each vertex.) 0

--1 --

1

0

0

0

07-T 1

Fig. 14.5

313

KERNELS AND G R U N D Y FUNCTIONS

Remark 2. If G has a Grundy function g(x), then G has a kernel, since the set S = ( x / x E X , g(x)=O} satisfies simultaneously : (1)

XE

S

=-

g(x) = 0

*

T(x)n S = 0 ,

min g(y) > 0 rw

YE

(2)

x$ S

*

g(x)

>0

=>

min g(y) = 0 YE

3

T(x)nS #

0.

r(x)

The converse is not true. It is left to the reader to verify that the graph G in Fig. 14.6 has a kernel { d } but has no Grundy function. Theorem 6. I f G is a 1-graph such that each subgraph has a kernel, then G possesses a Grundy .function. Let G be such a graph, and let So be a kernel of G. Let S, be a kernel of GI = Gx-so ; let S, be a kernel of G, = Gx-csovsl,,etc. The sets S, form a partition of X . Let g ( x ) be an integer defined by: g(x) = k

X E & .

0

We shall show that g(x) is a Grundy function of G. 1. Let g(x) = k ; we shall show that for each j < k, there exists a vertex T ( x ) such that g(y) = j . Since x E S,, and k > j , vertex x is present in graph G,. Since x $ S,, there is a vertex y E S, such that y E T ( x ) ; thus, there is in T ( x ) a vertex y with g(y) = j .

y

E

2. Let g(x) = k ; there is no vertex y then S, would not be stable.

E

T ( x ) such that g ( y ) = k , because

Thus g(x) is a Grundy function.

Q.E.D. Corollary 1. A symmetric graph possesses a Grundy function. This follows from Theorem 2. Corollary 2. A transititle graph possesses a Grundy function. This follows from Theorem 3. Corollary 3. A graph without odd circuits possesses a Grundy function. This follows from Theorem 5.

314

GRAPHS

Theorem 7 (Grundy [1939]). A graph G without circuits possesses a unique Grundy function g(x). Moreoter, .for each rertex x , g(x) does not exceed the length of ?lie longes? path from x. As in Theorem 4, consider the sets:

a}

X(0) = { X / X € X , T ( X ) = X(1) = ( X / X # X ( O ) , T(x) = X(O)> X(2) = ( x / x $ X ( O )u X(1),T(x) c X(0)u X(l)} etc. Clearly, these sets partition X,and x E X(k) if, and only if, the longest path from x has length k. The values g ( x ) can be successively defined on the sets X(O), X(l), etc. If x E X(O), let g(x) = 0. If x E X(1), let g(x) = 1. If for each y E X ( k ) , the value g ( y ) is uniquely defined and satisfies g ( y ) < k, then for x E X ( k l), the value g ( x ) is uniquely defined and g(x) < k I. Q.E.D.

+

+

Before demonstrating the fundamental properties of Grundy functions, it is necessary to define the Cartesian sum of I-graphs: r,),..., G, = (X,,r,) be 1-graphs, and let Let G1 = (X,,rl),G, = (X2, P = { 1, 2, .. ., p }. The normal product of these I-graphs is defined to the 1-graph G = G1.G, ... G, with vertex set

. . x = x, x x2 x .'. x x, = I7 xi ieP

and with correspondence

The Cartesian sum of these graphs is defined to be the 1-graph G = + G, with vertex set X = Xiand with correspondence G, G,

+

n,,

+

r(x1,x2,*.-,xp)=

u

({xl>x

***

x { x i - ~ ~ x G ( x J x { x i + l > xx { x p > ) *

ieP

Finally, the Cartesian product of these graphs is defined to be the 1-graph

G = G, x G, x with vertex set X

=

x

G,

niEP Xi and with correspondence

315

KERNELS AND GRUNDY FUNCTIONS

mG

H

h

II

I

QY

d

G+H ox

Gx H

b.v

nx

hx

Fig. 14.6

EXAMPLE. Consider two machines, and let X , denote the set of possible states for the first machine. For xl, x l E X , , let x i E f , ( x , ) if state x i can follow state x,. Thus, the first machine defines a graph G, = (XI, rl). Similarly, the second machine defines a graph G, = ( X , , f2). Let ( x l , x,) describe the state x1 of the first machine and the state x, of the second machine. If the operator works both machines simultaneously, graph G1 x G, represents the possible changes of situations. If the operator works only one machine at a time, graph G, + G, represents the possible changes of situations. If the operator can work either one machine or two machines simultaneously, graph G, .G, represents the possible changes of situations. Remark. The connectivity of a Cartesian product of I-graphs has been studied by McAndrews [1963], who has shown the following result: If G and H are two strongly connected graphs respectirely with the sets M ( G ) and M ( H ) of circuits, then the Cartesian product G x H is strongly

316

GRAPHS

connected g, and only i f , the lengths of the circuits in M ( G ) V M ( H ) are relatiilely prime. Similar properties for the Cartesian sum have been studied by Aberth [1964],who has shown that: Graph G H is strongly connected i f , and only i f , both G and Hare strongly connected. Other properties of G H a n d G x H have been studied by Picard [I9701 and Pultr [1970].

+

+

Proposition 1. If 1-graph G = ( X , r)possesses a kernel S,and i f I-graph H = ( Y, A ) possesses a kernel T, then the normal product G. H has the Cartesian product S x T CIS a kernel. It is left to the reader to verify the stability and absorption of set S x T.

Does the Cartesian product of two graphs, each with a kernel, also have a kernel? The example in Fig. 14.6, due to C. Y . Chao [1963],provides a counter example: graphs G and H respectively have kernels { d } and { y }, but graph G + H does not possess a kernel, because such a kernel would necessarily contain vertex dy, (since r ( d , y ) = 0 )and two of the three vertices ax, bx, or cx, which contradicts stability. Later, it will be shown that if both graphs G and H possess a Grundy function, then graph G H has a kernel. Before demonstrating these results some preliminary developments are required. The binary expansion of an integer p is a sequence (p', p2,...,p k ) o f digits such that

+

p =pl

+ 2p2 + 4p3 + + 2 ***

k - y

Its binary form is k

P=P P

k-1

2

*'*PP

1 '

For example:

Decimal Binary form form

-

-

0 1

2 3 4 5

= = = = = =

0 = 1 = 10 = 11 = 100 = 101 =

Binary expansion

Decimal form

-

-

(0) (1) (0,1) (1,l) ( O , O , 1) (1,0, 1)

6 7 8

Binary form

Binary expansion

-

=

= =

9

=

10 11

= =

110 111 1000 1001 1010 1011

(0, 1, 1) (1,1,1) ( O , O , O , 1) ( l , O , O , 1) = (0,1 , 0 , 1) = (1,1,0,1)

= = = =

317

KERNELS AND GRUNDY FUNCTIONS

Let [nIc2,denote the integer n modulo 2 (which is the remainder of n divided by 2). The
+

+

P = ( [ t1=1 P:]

(2)

I'&

*[%I

i=l

(2)

i=1

(2)

,...).

For example, we obtain :

i- 7 = (1,l) i(1,1,1) = (O,O, 1 ) = 4 1 -3l-F 11 = (1) i(1,1) i(1,1,0,1) = ( I , o , o , 1) = 9 , 3

etc.

Proposition 2. The digital sum has the following properties:

iq) $ r = p i- ( q ir ) , ( 2 ) Existence of an element 0 satisfying: p + 0 = p , (1) Associativity: ( p

(3) For each p , existence of an element (- p ) satisfying: p (4) Commutativity: p

+. (- . p ) = 0,

+ q = q +p.

The proof follows immediately. Thus, the first three properties imply that the integers under the binary sum operation form a group. The last property shows that this group is abelian.

Corollary. Given two integersp andq, there exists a unique integer r such that p+r=q. This is a well known property of groups. Theorem 8. If the 1-graphs G, = ( X I ,r),..., G, = ( X , , r,) respectively possess Grundy functions gl(x), ...,g,(x), then the function x2,

mn.3

X")

=g,(d

-r- gz(x2) ... ?- g"(x")

is a Grundy function for the Cartesian sum G = GI Consider a vertex ( x l , x,, ..., x,,) of G, and let

*.*I . . p1 < .-.+ p"

gi(xJ = pi = (P:, P:, g(x1,

..., x.)

=

p2

4-

where

P k = [ i=1 id

]( 2 )

+ G2 + . + G, . (i = 1 , 2 , ..., n)

= p = (pl,pz,

...)

318

GRAPHS

1. We shall show that for each q < p , there is a vertex 0 1 ,

Y,,

.*.)

Y n ) E T(x1, ~

2 ee.3,

xn)

such thatg(y,,yz, * . . , Y n ) = q Let k, be the largest k for which pk # qk.Then, pko = 1, q k o otherwise, we would have

0

+

c 2k-' pk > 2ko-1 + kalc 2k-1

k31

=

0, because,

qk

k
kcko

and

2ko-1 <

c 2k-'(pk - qk),

kal

kCko

which is impossible. Thus,

Without loss of generality, suppose that pto = 1. From the preceding corollary, there is an integer r such that p $ r = q, and, consequently,

PI i r < P l . Hence, there is a vertex y,

E

f , ( x , ) in graph G, with g,(y,)

g01,x2,x3, .-,x,) = C p r

=

p,

+ r. Hence

1- r = 4 .

2. We shall show that if (y,, x z , ..., x,) E T(xl,x 2 , ..., xn), then S(Yl3 ~2 s xn) # P * If y1 E T l ( x , ) , then g l ( y l ) # g,(xl). Therefore, if we had * a +

gl(Y1) -F gZ(x2) i*.*

4gn(xn) = gl(xl) 1.g,(xz) t ign(Xn) e.0

then, from the preceding corollary, gl(yl) = g l ( x l ) ,which is a contradiction. Therefore, g is a Grundy function for G. Q.E.D. 4. Nim games

Given two players A and B and a 1-graph (X,r),the following game can be defined: Starting from some initial vertex x,, player A selects a vertex x1 from f(x,,). Player B selects any vertex x2 from T(x,). Next, player A selects any vertex xg E T(x,), etc. If a player selects a vertex xi with T(xJ = 0,then

KERNELS AND GRUNDY FUNCTIONS

319

that player wins, and his opponent loses. Clearly, if there are circuits in the graph, the game need not terminate. This game is called a Nim-type game. We shall study characterizations of its winning positions, i.e., those vertices which must be chosen by a player in order to win no matter how his opponent plays. EXAMPLE 1. Fan Tan. There are p piles of matches. Each of two players alternately select a pile and remove one or more matches from it. The player who removes the last match wins.

EXAMPLE 2. Two players play with the same rules, but the last player to remove matches is the loser. This game was popularized by the film “Last Year at Marienbad” by Alain Resnais; it reduces to a Nim game by adding one vertex to the graph. EXAMPLE 3. Each of two players alternately place a tile that covers three squares in a straight line on an n x n chessboard. No square can be covered twice. The last player to place a tile wins. If n is odd, the first player can win by placing his first tile in the centre of the chessboard. Then, he should place his second tile in a position that is symmetric with respect to the centre of the chessboard to the first tile placed by his opponent, etc. (This is always possible since the centre of the chessboard has been covered.) Thus, his opponent must lose. EXAMPLE 4 (Withoff [1907]). Let X be the set of non-negative integral points in the plane. From an integral point ( p , q ) in X,a player can select any

Fig. 14.7

320

GRAPHS

point such that the value of one of the co-ordinates decreases and the value of the other co-ordinate remains unchanged, or such that the value of both coordinates decrease by an equal amount. The players take alternate turns. The first player to select the origin (0,O)wins. For example, from point (4, l), the following choices are available: (4, 0), (3, 11, (2, 11, (1, 11, (0,11, (3,O). The graph ( X , r) defined by this game contains no circuits, and, consequently, it possesses a Grundy function, indicated in Fig. 14.7. The winning positions are circled. Note that these positions are distributed symmetrically around the main diagonal. The above game is due to R. Isaacs [1958], but it was noticed by J. Kenyon [I9671 that it is equivalent to a game invented by Withoff in 1907. Theorem 9. If the graph ( X , r)possesses a kernel S, and $ a player chooses a vertex in S, this choice assures him o f a win or a draw.

If player A chooses a vertex x1 in S, either f ( x , ) = /21 (victory), or his opponent B will be forced to choose a vertex x2 in X - S. Then, player A can choose again a vertex x s in S, etc. The game ends when one of the players chooses a vertex xk with f ( x k ) = 0.Clearly, xk E S, and the winning player cannot be B. Q.E.D. To win games of this form, a player could calculate a Grundy function g(x) (if there exists one) and then play on the kernel

s = { x I g(x) = 0 ) . If the initial vertex xo satisfies g(xo) = 0, then player A is in a dangerous position because his opponent can get a win or a draw. If g(xo) # 0, then player A can assure himself of a win or a draw by choosing any successor x1 of xo with g(x,) = 0. Consider n Nim games (XI, r,),(X,,f,), ..., (X,,, r,). Suppose now that each player can play only one of the Nim games during his turn; the first player who cannot play a t all is the loser. This situation is in fact a Nim game on the Cartesian sum of the graphs ( X i , ri).The following theorem gives a winning strategy. Theorem 10. Consider n Nim-games G, = (X,,fl),G2 = (X2, fd,..., G, = ( X , , r,) with Grundyfunctions g , , g,, ..,,g , . A player will not lose the game on the Cartesian sum G = 2 Gi if he chooses a position x = (XI,X2,...,~,)

321

KERNELS AND GRUNDY FUNCTIONS

such that g,(xd

ig,(x,) i.*.ig,(x,)

= 0.

The proof follows immediately from Theorem 8 and Theorem 9.

EXAMPLE. Fan Tan. Each of two players alternately selects one of p piles of matches and removes at least one match from it. The player who removes the last match wins. rl),( X 2 ,r2),..., (A',,,r,,), This game is the Cartesian sum of the games (XI, where x k E X, represents the state of the k-th pile. Clearly g k ( x k ) equals the number of matches in the k-th pile. From Theorem 10, a position is winning if, and only if, the digital sum of the number of matches in the piles is zero. Consider a Nim game G = (X,r) without circuits and with a Grundy function g(x). Let $ be any operation that associates a vertex z to every ordered pair of vertices ( x , y). This is written as z = x y. For S c X and T c X,let

<

s iT =

{s

-f

t/S€S,

t€T}.

The following result produces a winning strategy: Theorem 11 (Grundy [1939]). Given a l-graph G and an operation $ such that

r(x

=

(X,r) without circuits

iY ) = (miy ) u (x ir) 7

then the Grundy function g of G satisfies for all x, y E,'A

4Y ) = g(x> I.gcY)

(1)

*

We shall prove the theorem by induction. Consider the sets

n o ) = { x / m )= @> W )= { x / W )= m>>

~ ( 2= ) { x / ~ ( x c) ~ ( l ) ) etc , ...

Since the graph contains no circuits, X thermore, we have : x

iy EX(0) *

(r(x)

=

X ( p ) for some value of p . Fur-

iy ) u (x ir ( y ) ) = 0 =-

Thus, x $ y E X(O), implies that: g(x

iy ) = 0 = 0 i0 = g(x) ig(y) .

x , y EX(0).

322

GRAPHS

Therefore, equation (1) is satisfied on X(0). Suppose that equation (1) is satisfied on X ( k - 1); we shall show that it is satisfied for a vertex z = x $ y in X ( k ) . If this were not true, then either g ( 4 > g(x)

i g(y)

or g(4

< g(x> i- d Y )

CASE1. Suppose g(z) > g(x) -ig ( y ) . There is a vertex z1E T ( z) such that = g(x)

id Y )

*

Without loss of generality, we may write z1 = x1 zIE r ( z ) c X(k - l), we have g ( z d = g(x1

+ y, x1 E T(x). Since,

iY ) = i d x i ) iA Y )

Hence, g(4

id Y ) = g ( x J ig ( y ) -

Therefore, g(x) = g ( x l ) , which contradicts that x 1 E T(x).

CASE2. Suppose g ( z ) < g ( x ) integer r such that g(z> = r

4 g(y). Then, as in Theorem 8, there is an

i- g(y)

.c g ( 4

9

Since there is a vertex x1 E T(x) with g(xl) = r, then

Since x1


d.4 = E

id Y ) -

T(z) c X ( k .- l), we have

iY ) = g ( x J 4g(Y) = g ( z ) which contradicts that x1 iy E T(z). g(x1

9

Thus, in both cases, we have a contradiction.

Q.E.D. Corollary. Let G = ( X , r) be a I-graph without circuits, and let be an operation on X such that (1) x i - y = y i x ,

(2) x

i( y i2) = ( x iy ) -f 2 ,

323

KERNELS AND GRUNDY FUNCTIONS

(3)xI.z=y+z

=-

x = y ,

(4) there exists a vertex e E X such that x

$

e = x for

all x E X and such that

r ( e ) = 0,

(5) r ( x

iy ) = ( f ( X ) iv) u ( x im ) ) .

A vertex z is deJined to be “irreducible” if there exist no vertices x and y distinct from e such that z = x + y .

Each vertex x E X can be uniquely uritten as a digital sum of irreducible vertices. If the calue of the Grundy function g(z) is known for all the irreducible vertices, then g(x) is defined for x by g(x) = g(z1

i

z2

i

-) = g(z1)

4g(z2) 4 . *’.

The uniqueness of this decomposition is a well known result from the theory of semi-groups.

EXAMPLE. Two players alternately choose one of n piles of matches and divide the chosen pile into two unequal piles. The last player who can make such a division is the winner. A position x is represented by the numbers XI,X2, ..., Xk that denote the numbers of matches in each pile. Let - X t Y = ( X 1 , x 2 ,...Y~k)/(Yl,YZ,...,~~) =(xl,X1,...,xk,Yl,...,y1). The operation $ satisfies: (l)x+y=y+x,

(2) x

i( y -F z ) = ( x -f

.

(3)x+z=y+z

y)

*

-F z , x = y ,

(4) There exists a position e (no matches) such that x $ e = x for all x and such that r ( e ) = 0,

(5) r ( x

iy ) = ( r ( x ) -F y ) u (x iT ( Y ) ) .

Thus, the value of the Grundy function g for the irreducible positions permits the calculation of the other values by means of a simple digital sum. An irreducible position corresponds to a single pile with XI matches, and some values of g(fJ are given below

324

GRAPHS r

TI = 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 g ( ~ , ) = 0 0 1 0 2 1 0 2 1 0

2

1

3

2

1

3

2

4

3

0

EXERCISES 1. Let X be an integer such that max { n - rn, 1 } Q

A < [n + 1 - , / i T G i l .

Show that there exists a simple graph G with n vertices and m edges such that the corresponding directed symmetric graph G* satisfies p(G*) = A. 2. Show that in the Nim game in Example 4, the abscissa of the n-th winning position below the diagonal is

[-“I.

3. Show that for the Marienbad game (Example 2) there is a Grundy function g’(x) that can be constructed from the Grundy function for Fan Tan (Example 1) as follows: g’(x) = g(x) if in position x , there is at least one pile with more than one match g’(x) = 1 - g(x) if no pile has more than one match. 4. Two players are presented with n piles of matches. Let q < n. Each player must alternately choose 9 piles and remove at least one match from each of the chosen piles. The last player to remove matches wins. In order to determine if a position is winning, it is necessary to calculate in binary the number ( p : , p % ,...) of matches in the k-th pile and to show that for each integer r,

+

where [ h ] o + 1 7 )denotes the integer remainder resulting from the division of X by q 1. (E. H. Moore [1909]) Does this rule define a Grundy function? 5. Show that if GI, G,, _..,G , are 1-graphs respectively with pseudo-Grundy functions g,, g,, ..., g,, their normal product GI . Gz . ... . G , possesses a pseudo-Grundy function g ( m , x z , ..., x d = maxgdx0 icp

(Butsan, Varvak [1966]) 6. Let G and H be 1-graphs. Show that their Cartesian sum G

circuit if, and only if, both G and H contain eulerian circuits. 7. Let

+ H contains an eulerian

(Aberth [1964])

G be a graph that is not strongly connected. Let p be the cardinality of a kernel

of graph 6, the transitive closure of G. Let graph H be obtained from G by reversing the direction of each arc. Let q be the cardinality of a kernel of graph fi. Show that the minimum number of arcs that must be added to G to make a strongly connected graph equals max { p, q 1. Hint: Use Exercise 14, Chapter 7. (B. Roy [1962])

CHAPTER 15

Chromatic Number

1. Vertex colourings

The chromatic number of a graph G is defined to be the smallest number of colours needed to colour the vertices of the graph so that no two adjacent vertices have the same colour. The chromatic number of graph G is denoted by y(G). A graph G with y(G) < k is called k-colourable. A k-colouring of the vertices S,, ..., S,) of the vertex set into k stable sets, each repreis a partition (Sl, senting the vertices of one of the k different colours. In this chapter, we shall assume that graph G is simple.

EXAMPLE 1. The chromatic index (see Chapter 12) of G is equal to the chromatic number of a graph L(G) formed in the following way: The vertices of L(G) represent the edges of G, and two vertices are adjacent in L(G) if, and only if, the corresponding edges in G are adjacent. Thus, results for the chromatic number yield results for the chromatic index. EXAMPLE 2. Is it possible to colour all the countries on a geographic map with four colours so that no two countries with a common border have the Same colour? This famous problem, called the “four colour problem” has not yet been solved. A geographic map corresponds to a planar multigraph G whose edges are the borders between countries. Colouring the countries corresponds to colouring the vertices of the topological dual G* (see Ch. 2, 42). Thus, the map colouring problem is equivalent to showing that all planar graphs are 4-colourable. EXAMPLE 3. Let X represent the inhabitants of an aquarium. Consider the 1-graph G = ( X , f)with & ( x ) < 2 for all x E X . Is it possible that this graph represents a family tree, i.e., y E f(x) implies that y is the child of x? Clearly, a necessary condition is that the graph contains no circuits and its vertices can be coloured with two colours ctl (male) and a2 (female) such that r; x ) contains at most one element of each colour, for each x . 325

326

GRAPHS

Fig. 15.1. Geographic map corresponding to a 4colourable graph G. The coloun 0, 1, 2, 3, 4 correspond to the values of a Grundy function for the symmetric (directed) graph G*

It is also necessary that the simple graph H and only if,

=

( X , E ) , where [x,y ] E E if,

r x n r y # %, is bicolourable. EXAMPLE 4. Examination schedule. At the end of the term, each student must take an examination in each of his courses. Let X be the set of different courses and let Y be the set of students. Since the examination is written, it is convenient that all students in a course take the examination at the same time. What is the minimum number of examination periods needed? For each examination x , let S(x) c Y denote the set of students who must E ) where [x, x’] E E if S(x) n write examination x. Form graph G = (X, S(x’) # i.e., if examinations x and x’ cannot be given simultaneously. Each vertex colouring of this graph corresponds to a possible examination schedule, and vice versa. Thus, the examination scheduling problem is solved by finding y(G).

a,

EXAMPLE -5. The schoolgirl problem (Kirkman). This famous problem can be stated as follows: Each day the fifteen schoolgirls a, b, c, ..., m, n, o from a

327

CHROMATIC NUMBER

boarding school take a walk in 5 rows of 3. Is it possible for them to take 7 walks such that no two girls walk together more than once? By considering symmetry, Cayley found the following solution : ~

Sunday

l

Monday

a ! bd chm din ejo

l

Tuesday Wednesday Thursday

c

abe

g

ghk ijm

Friday

Saturday

ahj bmn cdg efi klo

aci bho dkm eln

fW

The schoolgirl problem is related to the well known Steiner problem: Is it possible for the 15 schoolgirls to form successively 35 distinct triples such that no two girls appear together twice in a triple? To solve this problem, form a graph G whose vertices are the

(y)

=

455 possible triangles. Two

vertices are joined together if the corresponding triangles have two girls in common. Thus, the problem reduces to finding a maximum stable set S, Clearly, 1 S I < 35, because no girl can appear in more than 7 distinct triples of S, and therefore there are at most 15 x 7 x 3 = 35 triangles), If a stable set S has 35 vertices, then S is a maximum stable set. To check if a solution S of the Steiner problem yields a solution of the schoolgirl problem, construct a graph G’ whose vertices are the 35 selected triples, and with two vertices of G‘ triples joined by an edge if the corresponding triples have one girl in common. If the chromatic number y(G’) = 7, the schoolgirl problem is solved. If ?(GI) > 7, it is necessary to select another set S. A set of triples which meets the conditions of the Steiner problem is also called a Steiner triple system of order 15, or a (15, 3, 1)-block design. It was proved by the Reverend T. J. Kirkman, in 1847, that a Steiner triple system of order n exists if and only if n = 1 or 3 (mod 6). Now, if n = 6 k 3 girls take a walk each day in 2 k + 1 rows of three, can an arrangement be made for 3 k + 1 different days such that any pair of girls belong to the same row on exactly one day of the 3 k + 1 days? The existence of such an arrangement for all k 2 0 was proved by Ray-Chaudhuri and Wilson, [1971].

+

328

GRAPHS

Algorithms to determine chromatic number. Several algorithms exist for the determination of a minimum vertex colouring of a graph. The most efficient of these algorithms use the following two principles :

I. Principle of contraction-connection. Consider a simple graph G with two non-adjacent vertices a and b. The connection G 1 ab of graph G is obtained by joining two non-adjacent vertices a and b with an edge. The contraction G : ab of graph G is obtained by shrinking set { a , b } into a single vertex c(a, b) and by joining it to each neighbour in G of vertex a and of vertex b.

P

Fig. 15.2

C

CHROMATIC NUMBER

329

A colouring of G in which a and b have the same colour yields a colouring of G : ab. A colouring of G in which a and b have different colours yields a colouring of G 1 ab. Repeat the operations of connection and contraction, until the resulting graphs are all cliques. If the smallest resulting clique is a k-clique, then y(G) = k. For example, the graph G in Fig. 15.2 yields via the connection-contraction process a 3-clique. A 3-colouring.of the original graph is obtained by recovering the original from the 3-clique. Vertices a and dwill be coloured a ; vertices e and b will be coloured /?; vertex c will be coloured y. 11. Principle of separation into pieces Suppose that during the above contraction-co’nnection process, we encounter a graph with an articulation set A that is a clique; the removal of A creates connected components C1,C, C, ..., and, clearly, the subgraphs GAuc, = H 1 2 ,GAuc, = H,, GA,,, = H 3 , ... can be coloured separately so that the colours of the vertices of A do not conflict.

Fig. 15.3

For example, graph G in Fig. 15.3 is 3-colourable because each subgraph Hi is 3-colourable. A p i e c e of graph G relative to the articulation set A is defined to be a sub-

graph of G of the form H, be coloured separately.

=

GAucl. Clearly, the pieces of a graph can

330

GRAPHS

Theorem 1. I f G is a graph of order n, then

1

y(G)a(G) 2 n 9 y(G) a(@ < n

+

+ 1.

Let us colour the vertices of G with q = y(G) colours a l , a z , ..., a p .Let S, denote the set of vertices with colour a,. Since S, is a stable set, we have

Furthermore, if a maximum stable set S is coloured with the first colour, then n - a(C) colours can be used to colour X - S. Hence, y(G)

< (n - a'(G)) + 1 . Q.E.D.

Theorem 2 (Gaddum, Nordhaus [1960]). Let G b e the complementary graph of a simple graph G. Then y(G)

+ y(G) < n + 1.

Recall that the complementary graph of G = (X,E ) is defined to be G= (X,Pz(X)- E). Clearly, the theorem is true for n = 1 and n = 2. We shall suppose that the theorem is valid for every graph with n - 1 vertices, and we shall show that the theorem holds for a graph G with n vertices, n > 2. Let Go be the subgraph obtained by the removal of vertex xo. Clearly, we have

(1)

Y(G) < Y(G0) + 1 *

(2)

y(a

< y(G,) + 1

*

1. If in both (1) and (2) equality holds, then dG(xO)

2

ddxo) 2 y(G0).

Hence, y(Go)

Hence y(G)

+ y(Go) < &(xo) + &(xo)

= n - 1.

+ y(G) < (n - 1) + 2 = n + 1 .

2. If either (1) or (2) is not satisfied with equality, then, by the induction hypothesis, y(G)

+ y(G) < y(Go) + y(co)+ 1 < n + 1 .

331

CHROMATIC NUMBER

Hence, the theorem is valid in both cases.") Q.E.D. Remark. This bound is the best possible : A graph of type T,(n,p ) is defined to be a graph with n vertices firmed from a stable set S, with p vertices and a with n - p + 1 vertices such that clique Kn-p+l

I K"--,+, n s, I = 1 -

Fig. 15.4

If graph G is of type TI, then y(G) = n - p

+ 1,

Y G ) = P. Similarly, a graph of type T,(n,p) is defined to be a graph with n vertices formed from a cycle C, of length 5 without chords, a stable set S, with p < n - 5 vertices, and a (n - p - 5)-clique K,,-,- 5 , such that these three sets are disjoint and (1) each vertex C, is adjacent to each vertex of Kn-,,-,, (2) no vertex of C, is adjacent to a vertex of S,. If G is a graph of type T,, then (') A stronger inequality has been obtained by R. P. Gupta [1968]. A quasi-colouring wirh k colours is defined to be a mappingf : X+ (al,aa , ak) such that i # j implies the existence of two adjacent vertices x and y with f ( x ) = u, and f ( y ) = aI.A colouring with a minimum number of colours is a quasi-colouring because, otherwise, two colours a ' and a, are not adjacent and the vertices with any one of these two colours could be coloured by using only one colour. Let #(F) denote the maximum number of colours needed for a quasi-colouring of G. Thus, y(C) < I)(@. Gupta has shown that * a * ,

+

Y(G) v(G) < n Clearly, this result is stronger than Theorem 2.

+ 1.

3 32

GRAPHS

y(G) = ( n

- 5) + 3 = n - p - 2 ,

-p

y(E) = p + 3 y(5) = n + 1 . 0

Hence

y(G)+

K2

G E T2(11, 4)

Fig. 15.5

The bound given in Theorem 2 is attained by graphs of type TIand. T2. Furthermore, H. J. Finck [I9661 has shown that only the graphs of types TI and T, attain this bound.

Corollary. I f G is the complementary graph of a simple graph G of order n, then

By Theorem 2, we have 4 Y(G)y(G) G (Y(G) -

y(G)’ + 4 Y(G) y(G) G (w) + Y(G)’G (n + I)* .

Q.E.D. Remark. This bound is the best possible. If n is odd, then each graph G of type

satisfies

If n is even, then each G of type

333

CHROMATIC NUMBER

satisfies

Theorem 3. If G is a simple graph with n vertices and m edges, then y(G)

n2 >7 n -2m

Let y(G) = q. Let S1, S2, ..., S, denbte the sets of vertices of colour a l , respectively. The adjacency matrix of graph G has the form:

a z , ..., a,,,

s,

-I

I

s 2

...

lo

___-

0 Let n, = I Si I. Let No denote the number of 0 entries in this matrix, and let Nl denote the number of 1 entries in this matrix. Then,

Nl = 2 m ,

This follows from the Cauchy-Schwartz inequality I x I I y with x = ( n l , n2, ..., n,) and y = (1, 1, ..., 1). Thus, the total number of entries in the matrix satisfies:

n2 = N ,

n” + N1 2 2 m + -

Hence,

(n2 - 2 m) q 2 n2 .

4

1

2

I ( x, y ) 1,

334

GRAPHS

Hence the required formula.

Q.E.D. n2 Remark. In order to have equality, it is necessary that No = - and that 4 I

IXl.IYI = I(X,Y>l,

which implies that ( n l , n 2 , ..., n,) is proportional to (1, 1, ..., 1). Thus, equality holds only if

n , = n2 =

= n,

’

and if the matrix has only 1 entries outside of the blocks S, x S , , i.e., the graph is formed from q stable sets S , , S2,..., S, of the same cardinality with two vertices joined together if, anh only if, they belorg to distinct S,.

Theorem 4. (Roy [1967], Gallai [1968]). If G = ( X , E ) is a simple graph with y(G) = q, then, for each orientation of its edges, there exists an elementary path with q uertices. Furthermore, there exists an orientationfor which there is no path with more than q uertices. 1. Suppose that G = (X, r) is a 1-graph. Let t ( p ) denote the number of vertices encountered by an elementary path p (i.e., the length of path p plus one). We shall show that max, t ( p ) 2 y(G) where the maximum is taken over all elementary paths p of G. More precisely, we shall show that the vertices of G can be coloured with max, t(p) colours. By Corollary 2 to Theorem (3, Ch. 14), there exists a set Sfsuch that each vertex of G can be reached by a path starting from S:, and no two vertices of Sy are connected by a path. Let

s,”= r(s:)- s: ; s: = r(s;)- (s:u S,”) . ...................... Clearly, these sets partition X. Let (X, A ) denote the partial graph generated by the arcs ( x , y ) of G such that X€S,O, y E S : + l .

If, for some index i, there exist two vertices a E S, and b E S, with b E r ( a ) , then define a partition (Si,S:, S ; , ...) of X, where

335

CHROMATIC NUMBER

s; s:

=

q

=

sp - { b } ,

f . j = 1 , 2,..., i -

1)

s:+1= s;+lu{bL s;+2

=

sP+z ”

u d(&)) .....................

s;+3= $+3

-

Repeat the process until all sets in the partition are stable. The initial partition (S:, S,”,S,”, ...) has the property that for each p and for each x E S,O, there exists an elementary path p x terminating at x with t(p,) = p . All the successive partitions have this property because if (Sf, Sg, S,”, .. .) satisfies the property, then (Sf+ l, S,k S,”+l, ...) also maintains this property for all vertices that do not change class (and also for the other vertices, obviously). Thus, a partition into h stable sets with the desired property can be obtained. Hence, +

y(G)

Gh

< max W . P

2. This result is the best possible: let the vertices of a simple graph G be coloured with q = y(G) colours, a l , aZ, ..., a,, and let each edge [x, y ] be directed from x to y if x is coloured a, and y is coloured aj with i < j ; then it is evident that no path of the resulting graph contains more than q vertices. Q.E.D.

Corollary. Let G be a simple graph coloured with q = y ( G ) colours al,az , Then there exists an elementary chain that encounters consecutively the q colours at in this order.

..., a,.

Direct edge [x,y ] from x to y if x is coloured a t , y is coloured with a j , and i < j , From Theorem 4, there exists an elementary path containing exactly y(G) vertices. Clearly, this path has the required property. Q.E.D. It follows from Theorem 4 that a complete graph contains a hamiltonian path.

Theorem 5. Let ( S , ,Sz, .. , S,) be a q-colouring (not necessarily minimum) of a simple graph G, and let d , = max d G ( x ) . XESk

336

GRAPHS

Then, y(G) B max min { k

, dk + 1 } .

k
1. If S , is not a maximal stable set, add vertices to S to obtain a maximal stable set S ; . If S, - S; is not a maxima1 stable set in X - Sl, then add vertices to it to form a maximal stable set SL in X - S;, etc. This process defines a new colouring

( G , s;, *.*, s:) , with min ( k . r

2. Let i(x) denote the index i such that x E S;. Let x,, be a vertex with i(xo) = k ; since x,, is adjacent to each S; withj < k - 1 (by the maximality of Sj), then d,(x,) 2

Thus, for all vertices x, i(x)

k - 1.

< &(x) + 1 .

3. Let xoE S,. From Part 1, it follows that i(xo) < k and, consequently, i ( x o ) 6 max i(x) Q max (dG(x) xosr

+ 1) = dk + 1 ,

XSsk

Hence, y(G)

< max i(xo) < max min { k, dk + 1 } . xo

kbq

Q.E.D. Corollary 1. Let G be a simple graph. Iffor some integer q, the number of certices with degree 2 q is < q, then G is q-colourable. Assume that the vertices xi are indexed with decreasing degrees. Consider the n-colouring ({ x1 }, { x,}, ..., { x , } , ..., { xn}). Thus, d, = d,(x,). If k < q, we have

min{d,+ l , k } < q. If k 2 q

+ 1, we have

min{4+ 1,kl<min{d,+,+ l , k } ~ m i n ( q , k } ~ q . Thus, from Theorem 5, y(G)

< q. Q.E.D.

331

CHROMATIC NUMBER

Corollary 2. A simple graph with maximum degree h, is (h + 1)-colourable. The proof follows by letting q = h 1 in Corollary 1.

+

Corollary 2 can be improved by the following result, known as the “Brooks Theorem”; the proof presented here is due to Melnikov and Vizing [1969].

Theorem 6 (Brooks [1941]).Let G be a connected simple graph with maximum degree h. Then G is h-colourable, unless (1) h # 2, and G is a (h + 1)-clique, or (2) h = 2, and G is an odd cycle. Clearly, the theorem is true for h = 0, 1 , 2. Let G = (X,E ) be a simple graph with maximum degree h >, 3, with y(G) = h + 1, and that contains no (h + I)-clique. We shall show that this leads to a contradiction.

1. Since vertices that are not essential to the above properties can be removed, we may assume that G is minimal with respect to these properties. Let xo E X . The subgraph Go generated by X - { xo ) contains no (h 1)clique, and therefore y(Go) < h 1. Hence, y(Go) = h. This implies that dG(xo)B h because, otherwise, one of the h colours used to colour Gocould 1. be used to colour x o , which contradicts y(G) = h Thus, dG(xo)= h. We may assume that, if y,,y,, ...,y , denote the h vertices adjacent to xo,they are coloured with the colours ul, a2, . ., a,, respectively in a giuen h-colouring of Go.

+

+

+

.

2. Let C(a,, aj) be the subgraph of Go generated by the vertices with colours at or a, in the h-colouring of Go.Vertices y , and y , belong to the same connected componenr of G(a,, a,) because, otherwise, after interchanging colours ai: and a, in the connected component containing y,, xo could be coloured a [ , which contradicts y(G) = h + 1. 3. We shall show that the connected component of G(ui,a,) that contains y, and y,,is an elementary chain p[y,, y,] going from y, to y,. Vertex y r is adjacent to only one vertex with colour a, (otherwise, since dG(yi)Q h, vertex y, could be recoloured with a colour c ( k , k # i,j, and vertex xo could be coloured a!). Consider a chain in G ( q , a,) from y , to y,. We shall show that this chain is

unique. Let x be the first vertex of this chain with dGcar, a,)(x)> 2. I f x has colour a,, for example, then there are three vertices with colour a, adjacent to x ; since d,(x) ,< h, vertex x can be recoloured a*, k # i,j. This would disconnect y , and y, and contradict Part 2. 4. We shall show that the two chains pb,, y,] and p [ y l ,yk] that constitute components G(a,, aJ) and G(at, a,) cannot contain a common vertex z # y l .

338

GRAPHS

If such a vertex z existed, then it would have colour at and would be adjacent to two vertices with colour a, and to two vertices with colour aka Thus, h 2 dc(z) 2 4, and there is a fourth colour aI# a*, a,, a,, to recolour z. This would disconnect yi and y , and contradict Part 2.

+

5. Since G does not contain any (h 1)-cliques, there exist in Tc(xo)two non-adjacent vertices, say y , and y,. Consider the connected component p b l , y z ] of G(al, a,) and the connected component p ' [ y l , y 3 ] of G(a,, a&. Let x be the first vertex after y , in chain p [ y , , y,]. Since y, and y, are nonadjacent, x # y,. If colours a, and a3 are interchanged in the component of G(a,, as) that contains y, and y,, then vertex y1 is recoloured a3 and vertexy3 is recoloured a l . Furthermore, the new component H,, with colours a1 and a, that contains y, satisfies HI2

= P [ X , Y2l

Y

, and p ' [ y , , y3] have no common vertex since from Part 4, chains ~ [ y ,y,] (except rl). On the other hand, the component H23 with colours a, and a3 that contains y , satisfies H23

= &,

XI

9

since x has colour a2 and is adjacent to y,. Hence, x is a vertex common to the connected components H12and H Z 3 .This contradicts Part 4 and completes the proof. Q.E.D. The next section will present extensions of the Brooks theorem. 2. y-Critical graphs

A simple graph G is defined to be y-critical if for each vertex xo the subgraph Go generated by X - { xo } has chromatic number y(Go) < y(G). Note that in this case, y(G,) = y(G) - 1 because if y(G) = q 1 and y(Go) c q, then Gocan be coloured with q - 1 colours and vertex xo can be coloured with a q-th colour, which contradicts that y(G) > q. We shall now study the properties of y-critical graphs.

+

Property 1. A graph G with y(G) = q 1.

y(G) = q

+

+ 1 has a y-critical subgraph

with

If G is not y-critical, there is a vertex xo whose removal does not decrease the chromatic number. If the subgraph Go generated by X - { xo 1 is not y-critical, then again there is a vertex xl whose removal from Go does not

339

CHROMAnC NUMBER

decrease the chromatic number, etc. Eventually, this process locates a y-critical subgraph. Q.E.D.

+

Property 2. If G is a simple y-critical graph with y(G') = q 1, then d,(x) 2 q.for all x . If dG(xo)< q, the subgraph Go generated by X - { x o } can be coloured with q colours. At least one of these q colours is not adjacent to vertex x ; thus, x can be coloured with this colour, which contradicts y(G) = q 1. Q.E.D.

+

Pmperty 3. A y-critical graph is connected. The proof is obvious. Property 4. A y-critical graph contains no articulation set that is a clique. Let G be a y-critical graph with y(G) = q + I , and let A be a clique that is an articulation set of G. Since G is y-critical,

-

I A I = ?(GA) < 4 Let B, , B2,B3, . .. be the connected components of the subgraph generated by X - A , and let B; = B, V A , B2 = B; v A , .., be the corresponding pieces. Since G is y-critical, each piece can be coloured with q colours. Thus, G can be coloured with q colours, which contradicts y(G) = q 1. Q.E.D.

+

Property 5. A y-critical graph has no articulation points, This follows from property 4 since an articulation point is a 1-clique.

+

Property 6. If G is a y-critical graph with y(G) = q 1, and if A = { a, b } is an articulation set of G, then there are exactly two pieces B; and B, relatice to this articulation set and =

Y(W

= 4.

Clearly, a piece B' relative to A can be coloured with q colours. Three cases must be considered : (1) B' cannot be coloured with q colours so that a and b have the same colour, (2) B' cannot be coloured with q colours so that a and b have different colours, (3) There is a colouring of B' in which a and b have the same colour, and there is also a colouring of B' in which a and b have different colours. Case (3) cannot apply to a piece B' because the corresponding component B can be removed without changing the chromatic number of G. If Case (1)

340

GRAPHS

applies to B ‘ , then y(B‘) = q because if y(B’) = q - 1, vertices a and b can be coloured with a q-th colour, which contradicts that Case (1) applies to B’. Similarly, if Case (2) applies to B’, then y(B’) = q. Since y(G) = q 1, there is a piece satisfying Case (1) and a piece satiifying Case (2). Since G is y-critical, there is only one piece of each type.

+

Q.E.D.

(2 pieces)

(5 pieces or less)

(3 pieces or less)

@ (3 pieces or less)

@ (2 pieces)

@ (5 pieces or less)

Fig. 15.6. Various types of articulation sets in a y-critical graph

Property 7. I f G is y-critical with y(G) = q + 1 2 4, and if A = { a, b, c } is an articulation set, then one of the following cases occurs: I . G, contains only one edge and A has at tnost three pieces, each with chromatic number q. 11. G, contains exactly two edges, and A has at most two pieces, each with chromatic number q. 111. G, contains no edges, and A has at mostfire pieces, each with chromatic number q or q - 1.

We shall show only Case I. For example, suppose that a and b are the adjacent vertices in A . If piece B‘ is coloured with colours 1,2, ...,q, then vertices a, b, c may be coloured 121, or 122, or 123. However, not all of these

341

CHROMATIC NUMBER

three colourings are possible for B' because G is y-critical; it follows that y(B') = q (because if y(B') < q - 1, then using a q-th colour, all three of these colourings would be possible). If two of these colourings are possible for piece B;, then there is a piece B; for which only the third colouring is possible (and there are only two pieces since G is y-critical). If no piece admits two of the three colourings, then there exist three pieces BI, B;, Bj that respectively admit one of each of the three colourings. Q.E.D. Property 8. A y-critical graph G = ( X , E ) with y(G) disconnected by the removal of q - 1 edges, i.e.,

=

q

+I

cannot be

mG(A,X-A)aq ( A c X ,A # 5 2 / , X ) . We shall show that a contradiction results if there are two nonempty sets A and B of vertices G with A U B = X , A A B = 0 , mG(A,B) Q - 1 . Since G is y-critical, the subgraph GAis q-colourable; consider a q-colouring function for G A ,i.e., a mappingf(a) from A into { 1, 2, ...,q 1 such that

X,YEA, b , Y l E E =. f ( x ) Z S b ) . Let a,, a,, ...,a, be the vertices of A adjacent to B; choose function f so that (i = 1 , 2 , ...,r ) . f(aJ 6 i Let e,, e,, ., ,, ep denote the edges that join A and B, with indices chosen so that e, = [ a i , x ] , e, = [ a j , y ] , i < j =. s < t . Let g' be a q-colouring function for the subgraph G B .We shall show that by a sequence of transformations, this q-colouring of GB can become com1. If edge patible with the q-colouring of G A ,which contradicts y(G) = q e, is of the form [a,, b,], define a q-colouring g2(y) of G, by

+

= g'(y)

= g'(bJ

if

g'(v> f 2, g'(b3

if

g'(y) = g'@J

if

g'(y) = 2

.

Thus Now consider e,

=

[a,, b,], and similarly, define a q-colouring g3((y)by

342

GRAPHS

Thus,

This process can be continued, and since p edge ep gives a q-colouring gP+ with

=

rnc(A, B)

< q - 1, the last

gPf1(b,)> f ( a , ) *

ram btl

+

Thus G can be coloured with q colours, which contradicts y(G) = q 1. Q.E.D. Consider a y-critical graph G with y(G) = q G satisfies

min &(x)

q

+ 1; from Property 2, graph

.

xsx

To study the structure of G, we shall consider the set M =

{X

/ X € X ,

d&) = q ) ,

and study the structure of the subgraph G , generated by M.

Lemma 1. Let p = [a,, a , , a 2 , ..., ak-,, a,] be a cycle of G , and let f be a (q 1)-colouring of G with only certex a, having the (q + I)-st colour; then there exists a (q + 1)-colouringg of G ivitli only vertex a, having the (q + I)-st colour such that:

+

g(x) =fW d u o ) =f(a,) da1) = f ( a z >

( X 4 d

=f(a3)

..............

Let C Yctz, ~ , ..., a q + , be the colours in colouringf, where f(ao) = a g + l . Since dG(ai)= q, each colour a l , x Z , ..., Q, appears exactly Once in ~ G ( U , ) . Hence, interchanging the colours of a, and of a, gives a (q + 1)-colouring with only vertex a, having the (q 1)-st colour. This procedure can be re-

+

CHROMATIC NUMBER

343

peated around cycle p until vertex a, has colour a,,,. This produces the required colouring g . Q.E.D. Lemma 2. Let p = [a,, a,, ..., a,-,, a,] be a cycle of GM with no chord incident to certex a,; then p is an odd cycle.

+

1. Since graph G is ( q 1)-critical, there exists a (q + 1)-colouringfof G with only vertex a, having the (q + I)-st colour a,,,. Since dc(ao) = q, the set Tc(a,) - { a,, a k - , } contains exactly q - 2 vertices, and none of these vertices lies on p . Each of these vertices take a different colour in colouring f; let ul,a 2 , ...,a, - be these colours. 2. We shall show that only the colours and a, are assigned to vertices a,, a2, ..., a k - 1 in colouringf. If this is not true, then there exists in cycle p a vertex ai, i > 0, with colour u , , j # q - 1, q. After repeating i - 1 times the operation of Lemma and with 1, a colouring g is obtained with only vertex a. having colour g(x)

=f(4

(X#P)

d u o ) = f (a01 d a l ) = f (a3 = aj g(a2) = f (ai+ 1)

..............

d a k - 1)

= f ( a i - 1)

*

But, then, the set rc(ao) has two vertices with colour a, in colouring g. Thus, a, can be recoloured with a colour other than which contradicts y(G) = 1.

+

3. In colouring f, all vertices of cycle p, except a,, have either colour or u,. Since vertices a, and a k - l belong to T,(a,) and cannot have the same colour, cycle p is necessarily odd. Q.E.D.

aq-,

Theorem 7 . Let G be a y-critical graph with y(C) = q + 1; then dc(x) 2 q for all X , and each block of the subgraph GMgenerated by set M = { x x E X, dc(x) = q } is either a clique or an odd cycle without chords.

From Lemma 2, each even cycle of graph G M has at least two chords. Thus, from Theorem (7, Ch. 9), G M has the required form. Q.E.D. Remark. If q equals

h = max d,(x) , xsx

344

GRAPHS

the Brooks theorem follows from Theorem 7: Let G be a graph with max dC(x) = h

y(G) = h

and

+ 1.

Consider the y-critical subgraph G‘ of G with y(G’) = h + 1 (which exists by Property 1). Since dG.(x) 2 h (Property 2), and since dG.(x) < dG(x)< h, graph G’ is regular of degree It. In other words, the vertex set X’ of graph G’ is equal to the set M = { x / x E X’; d&) = h } . Furthermore, G’ is connected and has no articulation points (Properties 3 and 5). Since GL = G’ is a block, Theorem 7 shows that G’ is either a (h 1)clique or an odd cycle without chords. Hence graph G has either a connected component that is a (h I)-clique, or, if h = 2, an odd cycle without chords. The Brooks theorem follows.

+

+

The next theorem is also an extension of the Brooks theorem.

+ 1 2 4.

Lemma. Let G be a y-critical graph with y(G) = q clique C = { c l , c 2 , ..., cq } nith q tiertices such that (dG(ci)

- 4) G 4 -

G contains a

9

i=l

then in a q-colouring of X - C, the sets Ai = r G ( C i )

-c

contain a common colour. Suppose that d,(c,)

< dG(C2) < ..* G d G ( C q ) .

Since G is y-critical, then by Property 2, for at most q - 3 of the c i , then dG(cl)

=

dGG(c.2)

=

dG(c1)

dG(c3)

=

Z q. Since d,(c,)

-q 2 1

4.

Thus I A l I = 1. Let a1 be the colour of the unique vertex of A l in the q-colouring g of X - C. Suppose that there is a set A,, that does not contain colour al. (If several such sets exist, take index io to be as small as possible.) We shall now show that this is impossible, i.e., clique C can be coloured 1. with the q colours used for X - C, which will contradict y(G) = q Let a l , a z , ..., aq be the q colours of g ( X - C). Define the q-colouring g(x) of C successively in the following way: - let g(ci,) be colour a,;

+

345

CHROMATIC NUMBER

q # io. let g(c,) be any colour different from the colours of A , u

-if

{ Go 1;

-if

q - 1 # i,,, let g(c,-l) be a n y colour different from the colours of Aq-1 U { ci,, ~9 1; -etc. ... Note that, f o r j < q, a

a

Thus, (q-j+1) (lAjl-l)
or

Let g(c,) take a value in the set { ul, u z , ..., C A j " { cia, cq, cq-1,

***7

if the set

I ~ }

cj+l

1

has less than q vertices. This is possible if

-

+ 1 +(q - j +

q - j + l

- 1,

1) < q

i.e., if

-j+3+

4 - 3 q - j + l

< 0,

i.e., j2

- (q + 4 ) j

+ 4 q < 0.

Since the roots of this quadratic equation are j' c, can be coloured for:

=4

and j "

= q,

vertex

4<j
wemaytake:

=W

Z ) = dG(C3) =

Al = { a l } , Az = {az},

A3

=

4

7

{%).

If io # 2, 3, then g(al) = g(az) = g(a3) = ul. Moreover, C - { cl,c2, c 3 } has been coloured with q - 3 colours, including a l . Thus, clr c2, c3 can be coloured with 3 colours different from ul. If io = 3, then g(al) = g(a2) = al. Moreover, q - 3 colours have been

346

GRAPHS

used (including colour ctl for c3). Thus, c1 and c2 can be coloured with 2 colours different from ul. If io = 2, a similar result follows. In all cases, C can be coloured with the q colours that have already been used for X - C. Q.E.D.

Theorem 8 (Dirac [1952]). Let G be a graph with chromatic number + 1 and without any (q + I)-cliques. Let S = { x / d,(x) > q }. Then,

y(G) = q

If q = 1, the theorem is trivial because G cannot have both chromatic number 2 and no 2-cliques. If q = 2, then

1 (d,(x)

- 4) 2 0 = 4 - 2

XES

.

Suppose that the theorem is true for all graphs with chromatic number

< q ; we shall show that it is also true for a graph G with chromatic number y(G) = q that S #

+ 1, where q 2 3. (From the Brooks Theorem, we already know

0.) Suppose that G contains no (q +

I)-cliques and that

We shall show that this leads to a contradiction. We may assume that G is y-critical. (Otherwise, replace G by a y-critical subgraph.) Since d&) 2 q for all x E X , then

We shall now produce a contradiction for each of the following two possible cases :

CASE1. G contains no q-cliques. Let xo be a vertex of S. (By the Brooks theorem, S contains at least one vertex.) Let X - T be a maximal stable set that contains xo, and let GTbe the subgraph generated by T. Clearly, y(GT) < q 1 (because T # X and G is y-critical). If y(GT) < q, then G could be coloured with only q colours. Thus

+

y(G?-) = 4 ’

Let S’ =

( X / X € T ,

d,,(x)

>q

- 11.

347

CHROMATIC NUMBER

We have d,,(x)

+ 1 < d,(x)

(x

ET) ,

(because each vertex x E T is adjacent to X - T, since X - T is a maximal stable set.) Hence S' c S. In fact, S' is a proper subset of S since xoE S and xo$9.Hence,

C

xos'

[&,(x)

- (4 - I)] G

C

(dc(x)

- 4)

(dG(X)

- q) - 1 <

XES'

< XCE S

-3.

But, from the induction hypothesis, graph G, satisfies

c

X.S'

[dGT(X) -

(4 -

111 2 (4 - 1)

-2

9

which is the required contradiction. CASE2. G contains a q-clique. Let C = { cl, c2, ..., c,} be this clique. Set At = rc(ci) - C is not empty (because dG(cl) 2 q since G is y-critical). Set A = Tc(C) - C contains at least two vertices (because G contains no (q + I)-cliques). Renumbering the vertices if necessary, we may assume that d&,) < d&) < *.* < d,(cq). Since do(c,) - q 2 1 for at most q - 3 of the ci, from inequality (l), it follows that d,(c,) = 4. Thus, let al be the unique vertex of A ,

=

Fig. 15.7.

Tc(c,) - C. Since G contains no

348

GRAPHS

(q + 1)-cliques, there is a vertex c, that is non-adjacent to a,. (If several such vertices exist, let j be the smallest possible index.) Let

A, = { b l , b , , ... } . Let G' be the graph obtained from Gx-c by joining vertex a, to vertices b,, b 2 , .... Since Gx-c is y-colourable the graph G' - { a , } is q-colourable; hence y(G') < q + 1. Furthermore, from the lemma, graph G' cannot be coloured with q colours. Hence, y(G') = q 1.

+

+

Let G" be the y-critical subgraph of G' with y(G") = 9 1. Let graph H be obtained from G" by removing the edges of the form [a,, b,]. Graph H contains vertex a, and one of the vertices of A,, because if G" did not contain an edge of the form [a,, b,], then y(G") ,< y(GX-,) < 9. Note that dc(a1) 2

If I A,

I

dGrr(U1)

- I Aj 1

+ ( j - 1) 2

- I Aj

I - 1+j

I

2 2, we would have

1 [d,(x)

xss

4

- 41 2

1( d G ( C l ) - 9 ) + & ( a , ) i= j

2 (4 - j

- 4) 2

+ 1) (I A j I - 1 ) + (- I A, I - 1 + j ) =

+ 1) (I A, I - 1) - (1 A, 1 - 1) + j - 2 = = (9 - j ) (I I - 1) + j - 2 2

= (q

-j

2 (q - j ) + j

-

2 =q - 2.

This contradicts inequality (1). Hence, I A , I = 1. Let b be the unique ,vertex of A,. Graph H = ( Y , F ) can be formed from the (9 1)-critical graph G" by the removal of the only edge [a,, b]. Thus,

+

dH(a1) = d G v ( a 1 ) - 1 > ,9-l

d,(b) = dG,.(b) - 1 2 dH(x) = 4

(x

E

Combining this with Property 8 gives

y, x #

-1 01,

b) .

349

CHROMATIC NUMBER

c

(dGM

xss

c

- 4) 3 x s Y (dG(4

- 4) =

c

- Y)+

= m,(Y,X

(dH(X)

XEY

- q) 2 q

- 2

.

This contradicts inequality ( 1 ) . Corollary. Let G be a y-criticalgraph with y(G) cliques, and uith n tlertices and m edges; then 2ma(n+l)q-2.

=

q

+

Q.E.D. 1 , without ( q 1)-

+

Since dG(x)2 q for each vertex x, we have 2 rn =

c

&(x)

+ ( n - I S I) 4 2 4 - 2 + I S I q + nq - J S 1 4 =

XES

= (n

+ 1) 4 - 2 . Q.E.D.

Remark. The inequalities of Theorem 8 and its corollary are the best possible for q 2 3. To construct a graph G with

Y(G) = 4 + 1 9 such that the inequality of Theorem 8 holds with equality,

c = { c1, c2,

a*.,

Cq+1

}

and

D

{ c19

=

4

9

* * a 9

dq+l >,

and remove edges [Cl,

cz19

[Cl, 4 , 1 1 9

and add edge [ c z ,d q + J See Fig. 15.8. rl,

Fig. 15.8.

rn=ll,

n=7,

q=3

350

GRAPHS

+

Clearly, y(G) = q 1, and G contains no ( q + 1)-clique. Furthermore, c1 is the only vertex of S, and

d&)

=2q

-2

.

Thus,

Q.E.D. 3. The Haj6s theorem Let G be a simple graph. An elementary contraction on G is defined to be any operation that replaces two adjacent vertices a and b of G by a single vertex c and joins c to each vertex of f,(a) u T d b ) *

If the contraction operation is repeated enough times, a clique will be obtained. See Fig.. 15.9.

One of the most interesting conjectures concerning the chromatic number of a connected graph is due to Hadwiger:

Conjecture (Hadwiger [1943]). Ecery connectedgraph G with y(G) = q can be transformed into a q-clique by a sequence of elementary contractions. For q 6 4, Dirac [1952] and Halin have validated the conjecture. For q = 5, Wagner [1964] has shown that Hadwiger’s conjecture is equivalent to the four colour conjecture. Let gq be the class of all graphs with y(G) > q. Consider the following three operations in gQ: (I) Add vertices and edges. (11) Let GI and G, be two disjoint graphs. Let a, and b, be two adjacent vertices in GI, and let a, and b2 be two adjacent vertices in G2. Remove

CHROMATIC NUMBER

351

edges [ a , , b,] and [az,bz]. Add an edge [b,, b,], and contract set { a , , a, } into a single vertex. See Fig. 15.10. (111) Contract a set of two non-adjacent vertices into a single vertex. the resulting graph Clearly, if we perform these operations on graphs in ?Iq, belongs also to 9q.

Fig. 15.10

Theorem 9 (Haj6s [1961]). Euery graph G with y(G) > q can be obtainedfrom (q + l)-cliques by means of operafions (I), (11) and (111). Suppose that C = (X,E ) is a graph of order n which contradicts the theorem, and assume that G has a maximal number of edges with respect to this property. 1. We shall show that the relation "vertices x and y are identical or nonadjacent" is transitive. Suppose that this is not true: there exist three vertices x, y , z such that [x, y ] $ E, [y,z ] $ E and [x, 21 E E. From the maximality of G,the graphs G + [x, y ] and G 4- [ y , 21 respectively contain graphs G, and G, that can be constructed from K,, by the operations. We may assume that GI and G2 are disjoint (by duplicating). Let [ x l ,y,] be the edge in GI that corresponds to [x, y ] , and let [ y , , 2 4 be the edge in G2that corresponds to [ y , 21 (since [x, y ] appears in GI and [ y , z] appears in C,). Remove edge [x,, y,] from G, and edge [y,, z p ]from G2, contract { y L ,y, } into a single vertex and join vertices x1 and zz (i.e., perform operation (11) on graphs G, and G,). In the resulting graph G', identify all pairs of vertices that correspond to the same vertex in G, which are necessarily non-adjacent in G' (operation (111)). The resulting graph G" is a partial subgraph of G. Hence graph G can be constructed from graph G" by operation (I). This contradicts the definition of G.

2. The above equivalence relation divides X into classes S , , S,, ..., S, such that any two vertices belonging to distinct classes are adjacent and

352

GRAPHS

such that any two vertices belonging to the same class are non-adjacent. Since G cannot be constructed from K,,, by operation (I), we have G $ K4+ Thus k < q and, consequently, y(G) G 4

-

which contradicts y(G) > q. Q.E.D.

Using the Haj6s theorem, the four colour conjecture can be restated as follows : Every graph obtained from 5-cliques by operations ( I ) , (ZZ) and (ZZZ) is nonplanar. Clearly, this is true if the construction is limited to operations (I) and (11). Using the Haj6s theorem, the Hadwiger conjecture can be restated as follows: EiTery connected graph obtained from (q + 1)-cliques by the three operations can be transformed into K4 by elementary contractions. Heuchenne [19681 has presented more restrictive operations to construct from K , , , a family of graphs G with y(G) > q that can be transformed into K , , , by elementary contractions. He conjectures that every graph G with y(G) > q belongs to this family. +

4. Chromatic polynomials Let G = ( X , E ) be a graph with vertices xl, x 2 , ..., x,. In this section we shall enumerate the distinct A-colourings possible for graph G, i.e. the mappings f ( x ) from X into { 1, 2, ..., ,?)such that b,YIEE

*

f(x)+fCv).

The chromatic polynomial of G is defined to be a function P(G; ,?) that expresses for each integer ,? the number of distinct I-colouring possible for graph G. This number was originally expressed by G . Birkhoff [1912] with determinants. An excellent treatment of chromatic polynomials, due to Read [1968], serves as the basis for this presentation. G 7

U

-

b

-

C

Fig. 15.11

EXAMPLE 1. Colour the graph G in Fig. 15.11 with ,? colours; vertex b can be coloured first with any of the 1 colours, then vertex a or c can be

353

CHROMATlC NUMBER

coloured with any of the remaining (1. - 1) colours. Hence, P(G;A) =

A(A -

1)2.

EXAMPLE 2. Let G be the n-clique K,. Vertex x1 can be coloured first with any one of the I colours. Then vertex x2 can be coloured with any one of the remaining A - 1 colours. Then vertex xg can be coloured with any one of the remaining 1 - 2 colours, etc. Hence,

P(G ; A) = A(A

- 1) (A - 2) ... ( A - n + 1) .

This polynomial is often denoted by [A],,

where [A],, = 0 for

A < n.

Property 1. Let a and b be two non-adjacent certices in graph G. Let c be the graph obtained from G by joining a and b by an edge, and let G be the graph obtainedfrom G by contracting a and b into a single uertex. Then,

A)

P(G ;A) = P(&;

+ P(G; A) .

Recall the principle of contraction-connection mentioned in Section 1 : Clearly, P(G ;A) equals the number of A-colourings in G for which a and b have different colours plus the number of A-colourings of G for which a and b have the same colour. The formula follows. Q.E.D.

If the connection and contraction operations performed on graph G terminate with complete graphs HI, H,, H 3 . ..., then

P(C ;A) =

1P ( H , ;A) . i

For example, the contractions and the connections on the graph G in Fig. 15.2 terminate with five distinct cliques: one 5-cIique, three 4-cliques and one 3-clique. Hence

P(G ; A) =

[A], + 3 [ 4 , + v i 3 .

Corollary. If G is a graph of order n, the chromatic polynomial P(G ;A) is a polynomial of degree n in A. The coejirient of An is 1, and the coeficient of Ao is 0.

+

Clearly, P(G ;A) is the sum of terms of the form [Alp, where p assumes the value n exactly once. Q.E.D. Property 2.

If graph G has p connected components H I , H 2 , ..., H p , then P(G ; A) = P(H1 ;A) P(H2 ;A) ... P(Hp ;A).

Clearly, each connected component can be coloured separately, and the Q.E.D. formula follows.

354

GRAPHS

Property 3. I f graph G contains an articulation set A that is a k-clique uith qpieces H 1 , H , , ..., H, relative to A , then

P(Hl ; A)

P(G ; A) =

... P(H, ; A).

Colour A in one of the [AIk possible ways. Then there are P(Hi ;2) differen, PI, ways to colour piece H i , and the formula follows. Q.E.D. Theorem 10. The coeficients of P(G ;A ) are alternately non-negatice and non-positice integers.

This result is a direct consequence of the Mobius function theory developed by G. C. Rota [1964]. We shall give here an inductive proof on the order n of G. If n = 1 , 2, the theorem is obvious. If the theorem is valid for all graphs of order < n, we shall show that it is also valid for a graph G of order n. Clearly, this is trivial if G has only one edge. If the theorem is true for all graphs of order n with less than m edges, we shall show that it is also true for a graph G of order n with m edges. If a and b are two adjacent vertices in c", let G = G" - [a, b ] , and denote by the graph obtained from G by contracting { a, b } into a single vertex. From Property 1 , P ( C ; A) = P(G ; A) - P(G ; A) Since the theorem is assumed to be valid for all graphs of order n with less than m edges, there exist integers a, 2 0 such that P(G ;A) = A" - a ,

A"-'

+ a2 A"-'

- .*.

Furthermore, since the theorem is assumed to be valid for all graphs of order < n, there exist integers 6, 2 0 such that -

P(G ; A) = An-' - b,

A""-' + b2 A"-3 -

Hence,

P(E ;A) =

- (al

+ 1) 2 n - l + (a2 + b,)

-

s.1

Thus, the coefficients are alternately non-negative and non-positive integers. Q.E.D. Corollary. If G is a graph of order n with m edges, then the coefficient of An-1 in the polynomial P(G; A) equals - m.

If m

=

1, then P(G; A)

=

An-l(A - l), and the result follows. If m > 1,

355

CHROMATIC NUMBER

it suffices to observe in the proof of theorem 10 that the addition of an edge increases the absolute value of the coefficient of In-’ by one unit. Q.E.D. Theorem 11. A graph G of order n is a tree i f , and only i f ,

P(G ; 1) = #?(A - 1>.-1

.

1 . First we shall show that if G is a tree of order n, then the formula is valid. Clearly, this is true for n = 1 and n = 2. If the formula is valid for all trees of order < n, we shall show that it is also valid for a tree G of order n. Since G has a pendant edge (Theorem 2, Ch. 3), Property 3 can be invoked with the pendant edge as HI and with the remaining tree of order n - 1 as H 2 . Thus,

P(G ;A) =

A().

-

1) #?(I- I)’’-’

= A(A

- I)”-’ .

2. Let G be a graph of order n with P ( G ; A) = ,I( -# 1)n-l. ? Graph G is connected: Otherwise, it has p 2 2 components H I , H , , ..., H , and no constant terms appear in the polynomials P ( H , ; A) by the corollary to Property 1 ; by Property 2, P(G; 2) equals their product and, therefore, could not have a non-zero coefficient for A. Since ;5e coefficient of 2 - l is - (n - l), graph G has n - 1 edges (by the corollary to Theorem 10). Hence, G is a tree. Q.E.D. The roots of chromatic polynomials have been notably studied by Berman and Tutte [1969] and by Tutte [1970].

5. Vertex colourings of planar graphs (abstract)

Recall from Example 2, Section 1, that colouring the countries on a map corresponds to colouring the vertices of a planar graph. The four colour conjecture, i.e., a planar graph is 4-colourable, has been the subject of many works starting with an incorrect proof by Kempe in 1875. We shall not study here the recent developments of this extensive topic; however, let us mention the following results: ( 1 ) A planar graph that does not contain four cycles of length 3 is 3-colourable (Grunbaum [1963]). This result is in some sense the best possible because there exist graphs (for example, K4)with chromatic number 4 that contain exactly four cycles of length 3. This is an extension of the Grotzsch theorem: A planar graph without cycles of length 3 is 3-colourable (Grotzsch [1958]).

356

GRAPHS

( 2 ) The maximum wlire of the chromatic number of graphs that can be embedded in a orielitable surface of genus g 2 1 equals

W F -I This result, known as the Heawood conjecture, has been the object of much mathematical research; a proof was recently discovered by Ringel and Youngs [1969]. For the case of a plane or of a sphere ( g = 0), this formula cannot be used to prove the four colour conjecture. However,

Theorem 12. Every planar graph is 5-colourable. We shall assume that this is true for all planar graphs of order < n, and we shall show that it is also true for a planar graph G of order n. From Corollary 2 to Theorem (2, Ch. 2), there exists a vertex xo with degree < 5. Colour the subgraph Gx-(xo)with the five colours a l , a,, a,, a4, as. The colour for vertex x,, can be chosen without difficulty unless dG(xo)= 5 and the 5 vertices y l , y a , y,, y , , y 5 adjacent to xo (in clockwise order) all have different colours. Suppose that y, has colour xi for i = 1,2, ..., 5. In the bicolour graph G(al, a,), the connected component that contains y1 also contains y , . (Otherwise, colours a1 and a3 could be interchanged in this component, and xo could be coloured a1 .)

Fig. 15.12

Similarly, the component G(al, a4) that contains y, also contains y,. See Fig. 15.12. Hence, vertices y , and y s cannot be connected by a chain in the bicolour

351

CHROMATIC NUMBER

graph G(a,, a s ) ; if the colours a, and a5 are interchanged in the component of this graph that contains y,, then x,, can be coloured a,. Therefore, G is 5-colourable. Q.E.D. Corollary. If G is a planar graph of order n, then

Let (Sl, Sz,S,,S,,S5)be a 5-colouring of G. Then, 5

n = i= 1

I & I < 5a(G),

and the corollary follows. Q.E.D. Theorem 13. I f G is a simple planar graph whose faces are triangles, and ifall the degrees are multiple of 2 (respecticely, multiple of 3), then y(G) < 4.

This result is a reformulation of Corollary 1 (respectively, Corollary 2) of Theorem (8, Ch. 12). There is an enormous literature on the four colour problem. The reader is referred to Ringel [I9591 and to Ore [1967].

EXERCISES 1. Use the Brooks theorem to show that a regular graph G with

Y(@

+Y

C =n

+ 1,

is either (1) n isolated vertices, or (2) an n-clique, or (3) an elementary cycle of length 5 . 2. A graph G with n vertices is said to be of type T3(n,p, q), whefe n 2 pq, if there exist two partitions (Cl, C 2 , ..., C,)and ( D l , Dz, D,) of the vertlces such that (1) max I D1 I = q, (2) max I Cl I = P , (3) I C1n D, 1 G I for all C j , (4) two vertices belonging to the same Cl are adjacent, ( 5 ) two vertices belonging to the same D, are non-adjacent. Show that if G is of type T&, p , q), then G is of type T3(n,p , q ) and that

...,

Y(@=p

=.

Y m = 4 .

358

GRAPHS

Show that for any two integers p and q such that pq n, p t q S n S 1 ,

there exists a graph G of type T&z, p, q) with y ( G ) = p and y ( C ) = 4 . (Finck 119661)

3. Show that in a Ccolourable simple graph G, the edges can be coloured red and blue such that each triangle contains two blue edges and one red edge. 4. Show that in a simple planar graph the edges can be coloured red and blue'such that each triangle contains two blue edges and one red edge. (SchPuble [1968J) 5. Consider an infinite graph whose vertices are the pairs ( p , q) of integers with p < q, and with an edge linking ( p , q ) and (r, s ) if q = r or if p = s. Show that this graph contains no triangles and that its chromatic number is infinite. (Erdos, Hajnal [1960]) 6. Consider an infinite family of sets ( A l / i E I ) where Al C { I , 2, ...), 1 A, I < co. Form a graph G whose vertices are the sets A, and with an edge joining every pair of vertices

whose corresponding sets meet. Show that for each i, the complementary graph of the subgraph of C generated by r&)has chromatic number < I A, 1. Show that an infinite graph G = (I, r ) represents a family of finite subsets of { 1, 2, 3, ... } if, and only if, for each i E I,

Y(G-dfd <

(Kreweras [1946])

'xf

7. Consider the graph of the queen's moves in a chess game: the vertices correspond to the squares of an n x n chessboard, and two vertices are joined by an edge if a queen placed on the square corresponding to the first vertex controls the square corresponding to the second vertex. Call this graph G?. Show that

r(G9 = 4 , r(G'i?) = 5 , r ( G f )= 5 . r(G3 = n r(G2) = n or n n < y(Gz) < n + 3

+1

if n is not divisible by 2 or by 3, if n 1 is not divisible by 2 or by 3, in all other cases.

+

It has been conjectured that y(G2) = n or n

+ 1 for n > 3. (M. R. Iyer, V. V. Menon [1966])

8. If Gz is the graph of the king's moves on an n x n chessboard where n 2 2, show that y(Ci) = 4. 9. If G,R is the graph of the rook's moves on an n x n chessboard, show that

r(Gt)

=n

,

10. If Cf: is the graph of the knight's moves 'on an n x n chessboard where n > 2, show that y ( C z ) = 2.

11. If Cg is the graph of the bishop's moves on an n x n chessboard, show that y(Gz) = n

if n is odd.

12. Show that if G is an elementary cycle with n vertices, then

P(G ; A) = (A- 1)" f (- 1)''

(A- 1) .

359

CHROMATIC NUMBER

13. Show that if G is connected, then the absolute value of the coefficient of A' in

(Read [1968]) 14. Show that the smallest integer r such that A' has a non-zero coefficient in P(G;A) equals the number of connected components in G. 15. Use Theorem 4 to prove the following result: If p and q are two integers, and if G is al-graph without loops and with r ( C ) > pq, and iffis a real valued function defined on the vertex set X,then there exists a path [ao,a l , ..., a,] with

f ( a 3 d f ( a , ) < .*. or there exists a path [bo, bl, ..., b,] with

c f(a,)

m3 > f ( b A > - - * > f(b4) (Chvhtal, Komlbs [1971])

+

Show that if ( x l , x 2 , ..., x ~ ~ is+a ~sequence ) of p q 1 distinct integers, then this sequence contains an increasing sequence with p members or this sequence contains a (Erdos, Szekeres [1935]) decreasing sequence with q members. 16. Show that the simple graph G with y ( G ) = k and with the maximum number of edges is unique. Show that this graph is the complementary graph of graph GnVk defined (Tomescu) in Theorem 5, Chapter 13.

CHAPTER 16

Perfect Graphs

1. Perfect graphs

For a simple graph G = ( X , E), let a(G) denote the stability number, O(G) denote the minimum number of cliques that partition X, y(G) denote the chromatic number, w(G) denote the maximum cardinality of a clique. Clearly, a(G) < 6(G) since a stable set S can have at most one vertex in each clique of the partition. Similarly, o(G) 6 y(G). Graph G is defined to be a-perfect if

(A c X) .

a(G,) = O(G,) Graph G is defined to be y-perfect if y(GJ = a ( G A )

,

( A c X).

EXAMPLE 1. If G is a bipartite graph, then we know that a(G) = O(c) * from Corollary 2 to the Konig theorem, Chapter 7. Thus, a bipartite graph is a-perfect. A bipartite graph G is also y-perfect because if it has an edge, then Y ( G ) = 2 = NG) (since from Theorem 4, Chapter 7, G contains no triangles), and if G has no edges, then y(G) = 1 = o(G).

EXAMPLE 2. If G consists of an odd cycle of length 2 k + 1 > 3 without chords, then G is not a-perfect because a(G) = k and 6(G) = k + 1. (A minimum partition of G consists of k 2-cliques and one 1-clique,) Moreover, G is not y-perfect because y(G) = 3 and o(G) = 2. The concepts of a-perfect and y-perfect graphs were introduced (Berge [1961], 19621) to conjecture that a graph is a-perfect if, and only f, it is y-perJect . 3 6@

361

PERFECT GRAPHS

An analytic formulation of this conjecture is due to Fulkerson [1971]. In 1970, Olaru and Sachs presented an interesting approach and finally, Lovhsz proved the conjecture. (A proof is given in Chapter 20.) An easy result is the following:

Theorem 1. A simple graph G is a-perfect if, and only if, its complementary graph G is y-perfect. Clearly, @ ( G A ) = W ( ~ A9 )

Thus, a(GA) = e(GA)is equivalent to o(GA)=

y(GA). Q.E.D.

Corollary. If either graph G or its complementary graph contain an odd elementary cycle of length > 3 without chords, then G is neither a-perfect nor y-perfect. Let A be the vertex set of such a cycle of G. Then, from Example 2, a(GA) # d(GA), o ( G J # y(GA). Thus, G is neither a-perfect nor y-perfect. If the complementary graph G contains such a cycle, then it is neither aperfect nor y-perfect, and, from Theorem 1, G is neither a-perfect nor yperfect. Q.E.D. This result, and the study of various classes of perfect graphs (Berge [1963], [1967], [ 19691) suggest the following conjecture: The strong perfect graph conjecture. For a graph G, the following conditions are equit>alent: (1) G is a-perfect, (2) G is y-perfect, (3) G contains no set A such that G A or GAis an odd elementary cycle of length > 3 without chords,

We have shown above that (1) 3 (3), (2) rj (3). If (3) 3 (I), then (3) => (2). If (3) 3 (2), then (3) 3 (1). This conjecture is still unproved. Theorem 2. If G is a connected graph with an articulation set A that is a clique, and if each piece relative to A is a y-perfect graph, then G is y-perfect. Let G be a graph that satisfies the hypothesis of the theorem. It suffices to show that

4c;) = r(c).

362

GRAPHS

If w(G) = k, then there exists a k-clique in at least one piece G' relative to A, and y(G') = w(G') = k. Each other piece G" relative to A satisfies y(G") = w(G')

< k.

Thus, G is k-colourable, and

< k.

k = w(G) < y(G)

Hence, w(G) = y(G). Q.E.D. Theorem 3. If G is a connected graph with an articulation set A that is a clique, and if each piece relutive to A is an a-perfect graph, then G is a-perject.

Let G be a graph that satisfies the hypothesis of the theorem. It suffices to show that a(G) = f3(G).

Let C1, C,, and let

..., C, denote the connected components of subgraph G X - A ,

Two cases must be considered:

CASE1.

UYElAt # A. Then, there is a vertex a e A with W C * " { . ) ) = @%,>+ 1 (i = ..., PI. '

1 9 2 9

Therefore, a maximum stable set S, in Gcf,(,) satisfies

( Set So =

Up,l

I Si I = 4 G C J { a

} c Si c

+1

9

ci u { a } .

S, is stable in G,and P

'I

So I =

c

i= 1

Wc,)

+1

*

Moreover, since a partition V of G into cliques can be formed with A and with the B(G,,) cliques of a minimum partition of C,, for i = 1, 2, . . . , p , we have

363

PERFECT GRAPHS

Hence, a(G) CASE 2.

=

B(G).

Upzl A, = A . Then, for all i, u Al)

=

a(GCl)

.

Otherwise, there would exist a stable set S, c A , u C, with

I

Si

I

=

~ ( G c ,+ ) 13

and there would exist a vertex a E A , that is non-adjacent to some maximum stable set of Ci, which contradicts the definition of A t . Hence,

Hence, a(G)

=

B(G).

Q.E.D. The next three sections treat particular classes of a-perfect and y-petfect graphs. 2. Comparability graphs

A simple graph G = ( X , E ) is called a comparability graph if it is possible to direct its edges so that the resulting 1-graph (A', U ) satisfies: (x,y ) € (x,Y ) E

u, (.v, z ) € u u

=>

(x, z ) E u cV,x)4U

(transitivity) (ant i-symmetry).

Clearly, a bipartite graph is a comparability graph. Furthermore, each subgraph of a comparability graph is a comparability graph.

A

The graph in Fig. 16.1 is not a comparability graph because edges ab, bc, cd, be, ce can be directed appropriately, but then it is impossible to direct .f edge eJcorrectly.

a

/

Fig. 16.1

364

GRAPHS

Theorem 4. Every comparability graph is a-perfect.

Consider a I-graph G = ( X , U ) whose arcs represent an order relation. From Corollary 2, to Theorem (14, Ch. 13), a(G) equals the smallest number of elementary paths that partition the vertex set. Because of transitivity, each path generates a clique of G , and, by the corollary to Theorem (4, Ch. 10) each clique is the vertex set of some elementary path of G. Hence, m(G) = O(G). Q.E.D.

Theorem 5. Each cotriparability graph is y-pe$ect. It suffices to show that if G = ( X , U ) is the I-graph of an order relation, then y(G) = w(G). Let t ( x ) denote the length of the longest path from x plus one. Since G has no circuits, t ( x ) < co. If max t ( x ) = k, then there exists a k-clique. There I)-cliques (because this clique would contain a path passing exist no (k through all its vertices, and the longest path contains only k vertices). Thus

+

o(G) = k

.

Consider k colours denoted by 1,2, ..., k , and colour each vertex x with colour t ( x ) . Two adjacent vertices cannot have the same colour, because if there is an arc directed from x toy, then t ( x ) > t(y). Thus Y(G)d k

.

Since y ( C ) b o(G) = k, we have

y(G) = k = o ( G ) . Q.E.D.

Theorem 6. Let G = ( X , U ) be a transitive I-graph, and let E denote the set of all puirs of a4acent certices. Then, the simple graph ( X , E ) is a comparability graph. We shall remove from G all the loops and one arc in each multiple edge, so that the resulting I-graph (A', V ) satisfies the transitivity property. Let xl, x 2 , ..., x, be the vertices of G , and let.x, = x j if either i = j or f j (Xi,X j ) E ( x , , Xi) E u . From the transitivity of G , the relation = is an equivalence. Define set V as follows : (1) If x, and x, are in the same equivalence class, and if i > j , let ( x i ,x j ) E V, ( 2 ) If xi and x, are in different equivalence classes, and if ( x i ,x,) E U, let 7

(XI,X I ) E

v.

7

3 65

PERFECT GRAPHS

Hence, we have:

*

(x, z ) E V

=>

( y , X) $ V

(transitivity) (anti-symmetry). Q.E.D. Theorem 7 (Ghouila-Houri [1962]). A relation > is said to be a semi-order relation if nie have

(x,y ) E V , ( y , z ) E V (X,Y)E v

a > b, b a>b

>c

* *

a > c or c not b > a

>a

(semi-transitivity) (anti-symmetry)

A simple graph can be directed so that its arcs represent a semi-order relation

5f, and only if, it is a coniparability graph. It suffices to show that i f G = ( X , U ) is a I-graph whose arcs represent a semi-order relation, then there exists an orientation of the edges of G that represent an order relation. Assume that the theorem is true for all graphs of order < n ; we shall show that it is valid for a 1-graph G = ( X , U ) of order n whose arcs represent a semi-order. First, note that if three distinct vertices a, b, and c satisfy

( a , b) E U , ( b , c>E u, ( c , a>E u , then each other vertex is adjacent with either 0 or 2, or 3 of these vertices. Suppose that G is not transitive; then there exist three vertices x l , x2, x3 satisfying (XI

2

x*)E

u,

( x 2 , x3) E

u,

(x3, X I ) E

u

*

Two cases must be considered:

CASE1. Each itertex x # xl, x 2 ,x3 adjacent with one of these three vertices is adjacent with all three.

Remove vertices x2 and xg and direct the edges of the resulting subgraph so that it is transitive. (This is possible from the induction hypothesis.)

A

Fig. 16.2

366

GRAPHS

Then, direct each edge of the form [x,x,] or [x,x3],where x # xl, x 2 , x3, in the same direction as [x, x l ] ,and direct transitively the triangle formed by x,, x,, x3. Thus, the obtained graph is transitive. CASE2. There exists at least one i?ertexa # x, , x,, xBthat is adjacent with exactly two of these vertices, say, x2 and x3.

Let A be the set of vertices y that satisfy C ~ , X Z ) E Uand

(x,,y)~U.

Thus, x, E A. Furthermore, since a is not joined to x l , then a E A. We shall show that i f x 4 A, then either x is adjacent with aI1 the pertices of A, or x is not adjacent with set A.

Fig. 16.3

Let b E A , and suppose that x is joined to a but not to b. By considering the triangle formed by a, x 2 ,x 3 , we see that x is joined to at least two of these three vertices and thus to x2 or x s . By considering the triangle formed by b, x, , x 3 , we see that x is necessarily joined to x, and x3. Since x is not adjacent with b, it follows that

( x , x2) E U

and

( x g ,x)

E

U.

This contradicts x $ A . Now, direct transitively the subgraph generated by A and the subgraph obtained by removing from G the vertices of A other than x1 (this is possible from the induction hypothesis), then direct each' edge of the form [x,a], with x $ A and a E A , in the same direction as [x,x,]. The resulting graph is transitive. Q.E.D. A conjecture due to A. Hoffman can now be proved:

Theorem 8 (Gilmore, Hoffman [1964];Ghouila-Houri, [1962]). A necessary and suficient conditionfor a simple graph G = (X,E ) to be a comparability

367

PERFECT GRAPHS

graph, is that for each pseudo-cycle [a,, a,, ..., a,, , ,, a,] of odd length, there exists an edge of the form [a,, a,,,] (where the addition is modulo 29 1). A pseudo-cycle is a sequence of vertices starting and ending with the same vertex such that any two consecutive vertices are adjacent. For example, graph G in Fig. 16.1 contains a pseudo-cycle

+

[a, b , c , d, c , e , f , e, b , a1

of length 9, but no edges of the form indicated. Thus, this graph is not a comparability graph. 1. Necessity. If the graph ( X , U ) of an order relation contains an odd pseudo-cycle [a,, a 2 , ..., a,, + a,] without any edge of the form [acJa, + J , then when this pseudo-cycle is traversed, arcs are alternately encountered along and against their direction. But, this is impossible if the pseudo-cycle is odd.

2. Sufficiency. Consider a simple graph G the simple graph H = ( Y , F ) defined by (l)

(

-

Y b,Y’lEF Y E

c>

=

( X , E ) , and associate with G

y = (a, b), a , b E X , (a, ~ ] E E { Y J ’ } ={(a,b),(b,c)}, [a,cI#E.

It is easy to show that the hypothesis implies that H contains no odd cycles, and therefore, by Theorem (4, Ch. 7), H is bipartite. Thus, the vertices of H can be partitioned into two classes Yo and Y - Y o ,with

Y,Y‘E y o Y , Y ’ E Y - yo

=. =>

b,Y’l#f;, Iv,Y‘l#F.

Direct each edge [a, b] of G from a to b (and write a > b) if (a, b) E Yo: From (l), note that in graph H the vertices y = (a, 6) and y’ = (b, a) are adjacent (because [a, a] $ E, i.e. G has no loops); thus, these vertices belong to two distinct classes, and each edge of G has received exactly one direction. Furthermore, we have =>

b>c

[

(a, b) E yo

or ( b , c> E yo

]

*

=>

[(a, b ) , ( b ,C

[a,c]~E

-

) ] U

=>

[

c>a.

368

Therefore, the relation comparability graph.

GRAPHS

> is a semi-order relation, and by Theorem 7, G is a

Q.E.D. Other characterizations of comparability graphs have been discovered by Gallai [1967].

3. Triangulated graphs A graph G is defined to be a triangulated graph if each cycle of length > 3 possesses a chord (i.e. an edge joining two non-consecutive vertices of the cycle). The concept of triangulated graphs is due to Hajnal and Surhyi [1958]. A subgraph of a triangulated graph is also a triangulated graph because, otherwise, it would have a cycle of length > 3 without chords, and G would also have a cycle of length > 3 without chords

1. A tree is a triangulated graph. EXAMPLE

EXAMPLE 2. A cactus with only cycles of length 3 is triangulated because it contains no cycles of length > 3.

3. Graph L(G) is defined to be the graph in which each vertex EXAMPLE Zi represents an edge ei of G and with two vertices joined together if, and only if, they represent adjacent edges in G. If G is a cactus with only cycles of length 3, then the graph L(G) is triangulated: Otherwise, L(G) contains a cycle [Fl,F,, ..., g k , F1] without chords and with k > 3, and this cycle corresponds in G to a cycle ( e l , e , , e 3 , ..., ek, e l ) of length > 3, which contradicts the definition of G . The structure of triangulated graphs can be clarified by the following theorem, due essentially to Hajnal and Sursinyi [1958].

Theorem 9. If G is a triangulated graph, then each minimal articulation set of G is a clique.

Let A be a minimal articulation set of G. Removing A creates several connected components C, C', C", .... Each vertex a E A is joined to each of these components. (Otherwise, A - { a } would be an articulation set of G, which contradicts the minimality of A.) Let a , and a, be two vertices in A . There exists a chain p = [a,, cl, c, ..., c,, a,], where c1,

cz,

...)cp E c .

369

PERFECT GRAPHS

Assume that p is a chain of this type with minimum length. There also exists a chain p’ = [az,c;, c;, ..., cb, all, where c;,c;, ..., C ; E C

Assume that /I’ is a chain of this type with minimum length. The cycle p p’ has no chords of the following types:

+

- [a,, cil

( i # 1)

- [ci, cj]

(i # j )

- Ca,,cil

(i #

- Cci,

P)

- [a,, cjl

( j # 1)

- [ci,

(i # j )

- [a,,

,

because C and C‘ are two distinct connected components of Gx-a

c~I c;]

1

because p would not be the shortest chain

41 (i #

because p’ would not be the shortest chain.

4)

+

Since the graph is triangulated, cycle p p’, which has a length of at least 4, possesses a chord. This chord must necessarily be [a,, az]. Thus, any two vertices of A are adjacent, and A is a clique.

Q.E.D. Corollary 1 (Berge [1960]). A triangulated graph is y-perfect. Clearly, this is true for graphs with 1, 2, or 3 vertices. If this is true for all graphs with < n vertices, we shall show that it is true for a graph G with n vertices. Suppose that G is neither disconnected nor a clique (because then the proof would follow immediately). From Theorem 9, G contains an articulation set A that is a clique. Each piece relative to A is y-perfect by the induction hypothesis. Thus, from Theorem 2, G is y-perfect. Q.E.D.

Corollary 2 (Hajnal, Surinyi [1958]). A triangulated graph is a-perfPct. The proof is identical to the proof for Corollary 1, with Theorem 3 replacing Theorem 2. 4. i-Triangulated graphs A graph G is defined to be i-triangulated if each odd cycle p of length > 3 has a set of chords E(p) that form with p a planar graph whose unbounded

310

GRAPHS

face is the exterior of p, and whose bounded faces are all triangles. This concept is due to Gallai [1962]. The following theorem gives several characterizations of i-triangulated graphs.

Theorem 10. For a graph G,the following properties are equivalent: (1) G is i-triangulated, (2) For each elementary cycle p of odd length in G, and for each edge e of p, there exists a vertex of p that forms a triangle with edge e, (3) Each elementary cycle of G of odd length k has k - 3 chords that do not cross one another, (4) Each elementary cycle of G of odd length greater than 3 has at least two chords that do not cross one another, (5) If two distinct vertices x and y on an elementary cycle p of odd length are non-adjacent,, then p contains two other vertices z and i (distinct from x and y ) that are adjacent and such that the four uertices x, z, y , t are encountered on 11 in this order, ( 6 ) For each elementary cycle p = [x,,

x1, x21

.-.,X2qr X o l

of odd length and for each index i such that [ x i - 1 , x i + 1 ]$ E, there exists an index j # i - 1, i, i 1, such that [xi, x,] E E.

+

The proofs for (1) immediate.

(6)

(1)

3

-

=>

(2), (2)

=$-

(4), ( 3 )

=>

(4), (1)

3

(5), (5) + (6) are

(4) Let p = [x,, xl, x l , ..., xlp, x,] be a cycle in G of odd length > 3. If there is an edge [xi,x,] E E w i t h j # i - 1, i + 1, i - 2, i + 2, then one of two cycles determined by this edge and by p is of odd length > 3 and has a chord; since this chord does not cross chord [x,, x,], condition (4) follows. On the other hand, if each chord is of the form [x,, X , + ~ ]condition , (4) also follows. (3) Let p be a cycle of odd length n. Let m be the number of chords of E(p). Let f denote the number of triangles in this triangulation of p. From Euler’s formula, we have n - (n m) ( f 1) = 2. On the other hand, we have 3 f + n = 2(m n) . Hence, m = n -3.

+ + + +

(4)

=>

(1) Let p be a cycle of odd length, and let E,, denote a maximum set of chords that do not cross one another. The edges of p and of E,,

371

PERFECT GRAPHS

form a planar graph. Each bounded face of this planar graph is either a triangle or an even face. Since p is odd, at least one of the faces is a triangle. If not all of the bounded faces are triangles, then an even face can be found adjacent to a triangle. By removing the edge that separates them, an odd face having at least five sides is obtained. Since this face has at least two chords that do not cross one another, this contradicts that E,, is maximum. Q.E.D. Gallai [1962] proved : Eiiery i-triangulated graph is a-perfect. The original proof was very involved, but later, L. Surhyi [1968] presented a simpler proof. In fact, Gallai proved a stronger result: I f G is a connected i-trianguluted graph, at least one of the foliowing three conditions is satisfied:

( I ) G contains an articulation set that is a clique, (11) There exists a set A c X such that G A is bipartite and Gx-,, is com-

plete, and such that

aEA, b E X - A

3

[a,b]EE,

(111) The relation ‘‘x = y or [x, y ] c$ E” is an equit3alencerelation.

Other classes of perfect graphs have been investigated, and the following results have been proved : (1) I f each cycle of odd length > 5 has two chords that cross one another, then the graph is a-perfect (Olaru; in Sachs [1970]). (2) If each odd cycle contains an edge such that each maximal clique that contains this edge contains three certices of the cycle, then the graph is a-perfect (Berge, Las Vergnas [1970]). The proof for this result in a different form is presented in Chapter 20. 5. Interval graphs

Consider a family &’ = ( A , , A 2 , .. ., A,) of intervals on a line. The representative graph of &‘ is defined to be a simple graph G in which each vertex a, corresponds to an interval A , , and with two vertices joined together if, and only if, the two corresponding intervals intersect. Such a graph is also called an intercalgraph. G. Haj6s 1957 and N. Wiener were the first to study interval graphs, and so far two topological characterizations have been found; the first is due to Lekkerkerker and Boland [1962], and the other is due to Gilmore and Hoffman [1964]. For bipartite interval graphs, see also Kotzig [1963].

372

GRAPHS

EXAMPLE 1. Each student visits the university library once a day, and at the end of the day he submits the list of the names of the students met in the library while he was there. The problem is to find the order in which the students entered the library. Construct a graph G in which each vertex represents a student, with two vertices joined together if, and only if, the corresponding students were present in the library at the same time. This graph is an interval graph, because it represents the intervals of time during which the students were present in the library. The theorems of this section will give all the possible solutions. EXAMPLE 2. In genetics, tests can be performed to determine if two chromosomes overlap one another, and the problem is to prove or disprove that a set of chromosomes are linked together in linear order. Construct the graph G whose edges are the pairs of overlapping chromosomes; if this graph is not an interval graph, it follows that the chromosomes cannot be linked in linear order. EXAMPLE 3. Consider the following problem that occurs in psychology: Given a finite number of points x, , x2, , ., x, on a line and an infinite family s2 of intervals, two points x, and x j are said to be indistinguishable if there is an interval in family Q that contains both of them. The indistinguishable pairs determine a graph, and we may ask for the characteristic properties of these graphs. In fact, such a graph represents a family of intervals I,, Z,, ..., Z, , where interval ZIcorresponds to point xland is defined by

.

I , = [xt,

+~~]~U{W,/O~ER,O~~X~}.

If xi < x j , point x, and point x, are indistinguishable if, and only if, xI E I t , which is equivalent to It n Zj # 0. Theorem 11. Every interval graph G is triangulated. Suppose that there is a cycle [al, a,, ..., a,, a,] without chords. Let At be the interval corresponding to vertex al, since interval A k does not overlap with interval A k - 2 , the initial endpoints of the A* constitute a monotone sequence; and therefore, A , cannot overlap with A , , which contradicts that [a,, a,] is an edge of G. Q.E.D.

Remark. The converse is not true: Graph G in Fig. 16.4 is triangulated, but we shall show that G is not an interval graph. Clearly, the intervals A , , A , , A , are pairwise disjoint and may be placed in this order on the line. But, then, interval B, that intersects intervals A , and

PERFECT GRAPHS

373

"2

Fig. 16.4

A3 must also intersect interval A2, which contradicts that vertices a2 and b3 are non-adjacent.

Corollary. Every interval graph G is a-perfect and y-perfect. Since G is triangulated, G is a-perfect (Corollary 2 to Theorem 9), and yperfect (Corollary .1 to Theorem 9). Q.E.D.

APPLICATION 1 (Gallai).

If&'

is afinitefamily of intervals on a line, and if

k is the maximum possible number ofpairwise disjoint intervals in -c9, then there exist k points on the line such that each interval contains at least one of these points. Let graph G represent these intervals. Each clique in G corresponds to a family of intervals having one point in common, by the Helly theorem.") The result follows. APPLICATION 2. If& is a family of intervals on a line, and k is the maxintum number of interiials that together have a non-empty intersection, then the intervals can be coloured with k colours such that no two intervals with the same colour intersect. The first application can be related to Example 1. The minimum number of photographs of the library that are needed so that each student is photographed at least once equals the maximum number of *students who were pairwise not present together in the library.

We shall now study characterizations of interval graphs. Lemma 1. If G is an intercal graph, then its complementary graph comparability graph.

is a

(l) Helly's theorem for intervals can be stated as follows: If a family of intervals does not contain two disjoint intervals, then all the intervals have a common point. A simple proof is given in Chapter 17, Section 3, Example 1.

3 74

GRAPHS

If graph G represents a family of intervals d,two vertices x and y are ' . linked together in G if, and only if, they represent disjoint intervals of a Direct edge [ x , y ] from x to y if interval y is to the right of interval x on the line. This produces a I-graph ( X , V ) such that:

* *

(X,Y)EU

( x ,Y ) E u, (v,4 E Thus,

u

(Y,X)#U. (X,

4E u.

C is a comparability graph. Q.E.D.

The following lemmas treat a simple graph G with the following properties:

(1) Euery cycle of length 4 has a chord, ( 2 ) the elementary graph C is a comparability graph. Let G = (A', E ) be a graph with these properties. Let V denote the family of the maximal cliques of G. Let C = ( X , U) denote the complementary graph of G, assuming that the edges of G are directed transitively.

c

Lemma 2. Let C1,C2E %'; there exists an arc in that joins together set C1 and set C,. Furthermore, all the circs in G that join Cland C, hare the same direct ion. If C, and C2are two distinct maximal cliques of G , then G contains two non-adjacent vertices a E C, and c E C2.(Otherwise, C1U C2 would be a clique.) For example, let (a, c ) E U. Let bd be another edge of C with b E C1, d E C2 and a # 6. If c = d, then edge bd has the same direction as edge ac (from C , to C,) because, otherwise, G would not be a comparability graph. If c # d, then either ad or bc is an edge of (since, otherwise, the cycle [a, d, c, 6, a] of G would contain no chords). Without loss of generality, let this edge be ad. Then, (a, d ) E U. Hence,

(6, d)E u . Thus, in both cases, edge bd is directed from C1to C2,as is edge ac. Q.E.D.

c,

c2

@::+-;:@ h

- - - -- - - - - - - Fig. 16.5

-- --- Arc EdgeofofGG

3 75

PERFECT GRAPHS

Lemma 3. Let H be a I-graph whose vertices represent the cliques of $? and with an arc from C, to C, if there exist two certices a E C, and c E C, in G with (a, c ) E U. Then, H is a complete, transitive I-graph.

By Lemma 2, H i s complete and anti-symmetric. It remains to show that if (C,, C,) is an arc of H and (C,, C,) is an arc of H , then (C,, C), is an arc of H. We shall suppose that (C,, C,) is not an arc of H (and we shall show that this leads to a contradiction). Then, (C,, C,) is an arc of H .

Edge of G

Arc of

G

Fig. 16.6

For example, suppose that: a,bEC1, c , d € C 2 , e,fEC3 (b, c) E

u,

(d,4 E

u, (f,4 E u .

I n this'case, ad is an edge of because, otherwise, the cliques { a, d } and C, would contradict Lemma 2. Thus, from Lemma 2, (a, d ) E U. Since U is a transitive relation, (a, e) E U . But, then, (A a) E U and (a, e) E U implies that (f,e) E U. This contradicts that both vertices e and f belong to C,. Q.E.D. Theorem 12. (Gilmore, Hoffman [1964]). A simple graph G is an interval graph if, and only if, the following two conditions hold:

(1) every cycle of length 4 has a chord, (2) the complementary graph C is a comparability graph. Necessity. Condition (1) is necessary because a graph that represents a family of intervals is triangulated (Theorem 11). Condition (2) is necessary from Lemma 1. Suficiency. Let G = ( X , E ) be a simple graph that satisfies conditions (1) and (2). Let k? = { C,, C2,..., C, }be the family of the maximal cliques of G. As in Lemma 3, form the I-graph H . Graph H is complete, anti-symmetric

376

GRAPHS

and transitive. From Theorem (5, Ch. lo), H contains a unique hamiltonian path p. Suppose that the cliques of%' are indexed so that p = [ C , , C,, . .., C,]. Consequently, (Ci, C,) is an arc of H if, and only if, i < j . Note that G represents the sets

z,

=

{ i/

ciE w, ci 3 x )

9

because two vertices x and y are joined together in G if, and only if, I,nI,#

@.

To show that I, is an interval, it suffices to show that

::q

==.

XEC,.

p
If x 4 C, there exists a vertex y 16.7).

E

C, such that xy is an edge of

(see Fig.

----- Edge of

(JCr

\

Fig. 16.7

Clearly, (x, y ) E U (since r > p ) and (y, x) E U (since r < 4); this gives the required contradiction. Q.E.D. 6. Cartesian product and Cartesian sum of simple graphs

In Chapter 14, three operations were defined for 1-graphs: normal product, Cartesian product, and Cartesian sum. If G = ( X , E ) and H = ( Y , F ) are two simple graphs, the Cartesian sum of graphs G and H is defined to be a simple graph G H whose vertex set is X x Y, and with two vertices xy and x'y' joined together if, and only if,

+

either x

= x', [y, y ' ] E

F, or [x, x'] E E, y

=

y' .

311

PERSECT GRAPHS

Let the Kronecker delta be defined on graph G as follows: &(x, x’) = 1 if x # x‘, and &(x, x’) = 0, otherwise. The number of edges that join vertices xy and x’y’ in graph G + H can be written as ~zG+H(xY,

x ’ v ’ ) = &Ax, x‘) m d y , Y’)

+ b ( y , Y’) mc(x, x‘) .

The Cartesian product of graphs G and H is defined to be a simple graph G x H whose vertex set is X x Y, and with two vertices xy and xfi’ joined together if, and only if [x, x‘] E E and [y,y’] E F. The number of edges joining these two vertices in graph G x H can be written as

m, x H ( V 7 x’ r’>= 4 x 9 x’) ’%f(YY

r’).

The nornialprodtrct (or simply, product) of graphs G and H is defined to be a simple graph G.H whose vertex set is X x Y, and with two vertices xy and x’y’ joined together if and only if, either

x = x’,

[ y , y’] E F

3

or [X,X‘]EE,

or [ x ,X ’ I E E Y

y=y’7

[v,4”l E F

*

The number of edges joining these two vertices can be written as nzG.H(X,

y> = = mG(X, x’) mHb, y’)

+ dG(x, x‘) nrH(Y, Y’) + mG(x, x’) 611(y,y’)

*

Note that these operations are commutative. If the definitions are extended to more than two graphs, it can be shown that the operations are associative and distributive (C. Picard [1970]). These definitions were first introduced to study the chromatic number and the stability number. The relationships between these operations and the main fundamental numbers are described in this section.

Proposition 1. Let G and H be two graphs. Then

o(G

+ H ) = max { o(G),o ( H ) ) .

Let C1, Cz, ..., C, be the maximal cliques of G, and let D1,Dz,..., D,be the maximal cliques of H ; then the maximal cliques of G H are the sets C, x { y , } and the sets { xi } x D,. Hence,

+

o(G + H) = max{ I { xi 1 x D j I , I C, x { y j } I }

= max { 4 G ) , OW)}.

Q.E.D.

3 78

GRAPHS

Proposition 2 (Vizing [ 19631, Aberth [ 19641). Let G and H be two graphs. Then Y(G

+ H> = m a { Y ( G ) 7 Y ( H ) ) .

1. Let r = max { j ( G ) , y ( H ) }; colour G and H with r colours 0, 1, 2, r Let g(x) = k if vertex x is coloured k. For each xy E X x Y, let

- 1.

+

Consequently, g(xy) defines a colouring of G H because if xy and x‘y’ are adjacent and have the same colour, then either x = x’ and vertices y and y‘ are adjacent in H, or y = y’ and vertices x and x‘ are adjacent in G. In the first case,

A4 + g

o

= A4 + gD’>

(mod. r )

and g ( y ) = g(y’), which is impossible because [ y ,y’] E F. A similar result follows for the second case. Hence, G H is r-colourable.

+

+

+

2. Graph G H cannot be (r - I)-colourable, because, graph G H contains a subgraph isomorphic to G and a subgraph isomorphic to H . Hence, y(G H ) = r. Q.E.D.

+

G + H

Y

Y-

P-

H

aI

I

a

h

Fig. 16.8

Proposition 3. Let G and H be two graphs. Then a(G

+ H ) 2 a(G) a(H) .

I

c

d

X

PERFECT GRAPHS

319

If S is a maximum stable set of G, and if T is a maximum stable set of H, then the set S x T is a maximum stable set of G x H. Hence, a(G

+ H> 2

IS x T ! = a(G)a(H) Q.E.D.

In Fig. 16.8, the vertices of a maximum stable set are circled, and it is easily seen that

a(G

+ H ) = 4 > a(G)a(H) = 3 .

Proposition 4. Ifgraphs G and H respectively have orders n(G)and n(H),then a(G

+ H ) < min { a(@ n ( H ) , a(H) n(G)} . +

Let So be a maximum stable set of G H. Its intersection with set X x { y , } cannot have more than a(G) vertices; hence, 1 So I ,< Y I a(G). Similarly, I So I < I X 1 a(H). The formula follows. Q.E.D. Proposition 5. Ifgraphs G and H respectively haw order n(G) and n(H), then

B(G

+ H) < min { n(G) O(H), n ( H ) ' X G ) } .

Let $9 = (C,,C,,..., C,) be a minimum partition of G. The sets Cix { y, are cliques of G H and cover X x Y. Thus,

+

O(G + H )


x

I Yl

= n(H)B(G).

Q.E.D.

Remark. Vizing [I9651 proved a similar inequality for the dominance numbers, i.e. : B*(G + H ) < min { P*(G) n ( H ) , B * W ) n(G)) , He also conjectured:

Conjecture.

p*(G

+ H ) 2 B*(G)P*(H).

Theorem 13. If only q colours are available, then the maximum number of vertices in G that can possibly be coloured with these q colours so that no two adjacent oertices have the same colour is equal 10 cc(G + K,) . Let K, = ( Y , F) be a q-clique and let G = ( X , E ) be a simple graph. We shall show that a stable set of G + K , determines a set A c Xof vertices in G

380

GRAPHS

that can be coloured with q colours. (This result is illustrated in Fig. 16.8 for q = 3.) Since the sets { xi } x Y are all cliques in graph G K4, a stable set So of G K,, has at most one vertex in each of them. Put

+

+

, A = { x / x € X ; S , n ( ( x }x Y ) # @ j . For x

E A,

put g ( x ) = j if, and only if,

Function g(x) is a q-colouring of G, because two adjacent vertices x and x' in GA cannot have the same colour (since So is stable). Conversely, if g(x) is a q-colouring of a subgraph GA,then the set of vertices xygtx,with x E A is a stable set of G Kq. Thus, there is a one-to-one cor-

+

respondence between the stable sets of G of G

+ K4 and the partial q-colouring Q.E.D.

Corollary. A graph G of order n is q-colourable if, and only id ci(G The proof follows immediately.

+ Ka) =

11.

For example, graph G in Fig. 16.8 is 3-colourable, since

+

LX(G K3) = 4 = n(G) .

We shall now present a similar result for the Cartesian product G x H :

Proposition 6. Let G and H be two graphs. Then y(G x H )

< min { y(G), y ( H )1 .

Let q = min { y(G), y ( H ) } , and suppose that y(G) = q. Let g(x) be a q-colouring of graph G. Put d x y ) = g(x)

(XY E

x x r>

f

Thus, g(xy) is a q-colouring of G x H because if xy and x'y' are two adjacent vertices, then g(xy) # g(x'y') (since x and x' are adjacent in G). Thus, y(G x H ) < q. Q.E.D.

.

The following similar results are available for the normal product G H.

Proposition 7. Let G and H be two graphs. Then y(G.

1-

B max { Y(G) Y(H) 9

381

PERFECT GRAPHS

Let q = max { y(G), y ( H ) }, q = y(G); then q colours are necessary to colour G. H since G. H contains a subgraph isomorphic to G. Q.E.D. Proposition 8. Let G and H be two graphs. Then W(C.

H j = O(G) W(H)

.

Let C be a maximum clique of G, and let D be a maximum clique of H . The set C x D is a clique of G .H because if x , x’ E C and y , y’ E D, and if xx‘ # yy’, one of the three following cases occurs: x = x‘ and [y, y ’ ] E F, and then, xy and x‘y’ are neighbours in G . H , or = y’ and [x, x‘] E E, and then, xy and x’y’ are neighbours in G. H , or [x, x’] E E and [ y ,y’] E F, and then, xy and x’y’ are neighbours in G . H .

y

Hence,

o(G.H)

>ICx

DI

=

w(G) o ( H ) .

.

Conversely, if Co is a maximum clique of G H, let the projection of Co on X be C, and let the projection of Co on Y be D. Since C and D are cliques in G and H , the set C x D is a clique in G . H. Hence, C, = C x D (since Co is a maximal clique). Thus, w(G.H) = I Co I = I C I x

I D I < o ( G )o ( H ) .

The required equality follows.

Q.E.D. Proposition 9. Let G and H be two graphs. Then cr(G.H) 2 cc(G) a ( H ) . If S and Tare maximum stable sets respectively in G and H , then

I SI

= .(G),

I TI

= U(H).

The Cartesian product S x T is a stable set in G . H , and, consequently, we have o((G.H) 2 I S x TI = I SI x I TI = a(G)cc(H). An upper bound for a(G .H ) will be given in Chapter 19, 92.

Q.E.D.

Proposition 10. O(G.H) < O(G) O(H) .

Let (Cl, C2,..., C,) be a minimum partition into cliques of graph G. Let (Dl, D2,..., DJ be a minimum partition into cliques of graph H

382

GRAPHS

Ingraph G.H, set C, x D,is a clique, for i = 1,2, ..., p , and j = 1,2,..., q. The cliques C,x D,partition graph G.H. Therefore, O(G.H)

< p q = O(G) 8 ( H ) . Q.E.D.

APPLICATION (Shannon

[1956]). Consider a transmitter that can emit five

signals, a, b, c, d, e, and a receiver that can interpret each of these signals in two different ways. Signal a can be interpreted as either p or q, signal b can be interpreted as either q or r, etc., as shown in Fig. 16.9. What is the maximum number of signals that can be used for a code so that there is no possible confusion on reception? This problem reduces to finding a maximum stable set S of a graph G shown in Fig. 16.10, in which two vertices are adjacent if and only if they represent two signals that the receiver can confuse. For example, we can take S = ( a , c } , since graph G in Fig. 16.10 has stability number a(G) = 2.

S

u(G)=2

e Fig. 16.9

Fig, 16.10

C.G=G2

b

~

t

b

c

a(G2)=5

Fig. 16.11

d

e

383

PERFECT GRAPHS

Instead of single letter words, we could use a code of two letter words, provided that no two letter words of this code can lead to confusion on reception. Thus, the letters a and c which cannot be confused can form the code: aa, ac, ca, cc which has a vocabulary of (a(G))2 = 4 words. But an even richer code is: aa, bc, ce, db, ed. It is easily seen that no two of these words can be confused by the receiver. This gives a vocabulary of 5 words. Note that the words xy and x'y' can be confused if, and only if, these two words are adjacent vertices in the normal product G G = GZ.A code consisting of 2-letter words has a maximum vocabulary of a(G2) words. More generally, the maximum possible vocabulary for a code of k letter words is the stability number of the product.

.

Gk = G.G.G ... G . k

With this as motivation, the capacity of graph G (or, "zero-error capacity") is defined to be the number c(G) = sup

73).

The capacity of the graph G in Fig. 16.10 is known to be between 2 and 3; however, its exact capacity remains unknown. Furthermore, Ljubich [19641 has shown that IC4a(G)"tends to c(G)when k tends to infinity.") An important problem is the characterization of the graphs whose code is improved when longer words are used. We shall prove the following result, due to Shannon : I f a graph G satisfies a(G) = B(G), then the code cannot be improved by using longer words. From Propositions 9 and 10, we have

.

Fekete's theorem states that if a sequence of numbers al, as,. . is sub-additive (ie., urn+,,"u,+un), then

4n n

+

inf n

an

-.

"

If we let a, = -log a(G"), then a sub-additive sequence is formed because a(Gmtn) 3 a(Gm)a(Gn), log a(Gm+n)2 log a ( P ) + log a(Gn).

Thus,

and

384

GRAPHS

Hence,

Thus, a(G)

=

B(G) implies that c(C) =

and the code cannot be improved..

Q.E.D. In particular, when each signal is determined by its modulation frequency, two signals can be confused if, and only if, the corresponding frequency

intervals of the two signals intersect (“linear noise”). In this case, graph G represents a family of intervals, and therefore, by the corollary to Theorem 11, u(G) = B(G). Hence, ifnoise is linear, then a code cannot be improved by using longer words.

EXERCISES 1. A graph G is defined to be vertex-critical a-imperfect if it is not a-perfect, but if the removal of any vertex makes the resulting graph a-perfect. Show that such a graph G satisfies the following properties: ( I ) for each vertex, there exists a maximum stable set that does not contain this vertex, (2) for each vertex, there exists a minimum partition of G into cliques such that one of the cliques contains only this vertex. E. Olaru has shown that each vertex x belongs t o a cycle of length 3 5 such that all the chords of the cycle are incident to x. (Sachs [ 19701) 2. Graph G is defined t o be edge-critical a-imperfect if G is not a-perfect but if the graph G e is a-perfect for each edge e of G.

-

Show that such a graph is a-critical (see Chapter 13). Show that a connected graph G is edge-critical a-imperfect, if, and only if, C is an odd cycle of length 2 5 without chords. Hint: Use Corollary 3 to Theorem (6, Ch. 13).

c,

3. Let C, be a cycle of length 7, and let G = be its complementary graph. Show that (1) G has no odd cycles of length > 3 without chords, (2) G is vertex-critical a-imperfect. 4. Consider a simple graph G = (X,E) with the following property: if x l r x z , x 3 , x4 are four distinct vertices with [ X I , xz] E E, [ x z , x 3 ] E E, [x3, xq] E €, then [xlr x3] E E, or [ x z , x4]E E. Show that graph G contains a vertex a that is adjacent to every other vertex of the graph. Show (by induction on n) that the complementary graph of G is a comparability graph. (Wolk [1965]) 5. Show that a simple graph G is the transitive closure of an arborescence if, and only if, it satisfies the two following properties:

385

PERFECT GRAPHS

(1) G is connected, (2) Ifxl, xa, x3, x4are four distinct vertices with [xl, xp]E €, [xl, x31 E E, ( ~ 3xr] , E €, then [xl, X J E € or [xa,x4JE E. Hint: The sufficiency can be proved by induction on n, and by using Exercise 4.

(Wolk [1965]) 6. Show that if q = w(G), then NG

+ 4)

= n(C).

7. If p ( G ) denotes the number of connected components of graph G, show that p(G

+ HI

= p(G)dH).

8. Show that if G and H are two simple graphs with hamiltonian cycles, then graph

+

H has a hainiltonian cycle. Show that if both G and H have eulerian cycles, then graph G cycle. G

+ H has an eulerian (Aberth [1964])

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PART TWO

Hypergraphs

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CHAPTER 17

Hypergraphs and Their Duals

1. Hypergraphs Let X = { xl, x,, ..., x,} be a finite set, and let 8 = ( E L /i E Z ) be afamily of subsets of X.The family € is said to be a hypergraph on X if (1)

Ei

#

la

(iEI)

UEi=X ief

The couple H = (X,S)is called a hypergraph.. I X I = n is called the order of this hypergraph. The elements xl, x,, ..., x, are called the vertices and the sets E l , E,, ..., Em are called the edges. A hypergraph is shown in Fig. 17.1. An edge Ei with I El I > 2 is drawn as a curve encircling all the vertices of El. An edge ELwith I Ei I = 2, is drawn as a curve connecting its two vertices. An edge Ei with I Ei I 3 1, is drawn as a loop as in a graph. If the edges Ei are all distinct, the hypergraph is said to be simple, and 8 is a set of non-empty subsets of X.If I Ei I = 2 for all i, and if the hypergraph H is simple, then H is a simple graph without isolated vertices. The main purpose of the theory of hypergraphs is to generalize results from graph theory. In a hypergraph, two vertices are said to be aa’jacent if there is an edge El that contains both of these vertices. Two edges are said to be adjacent if their intersection is not empty. The incidence matrix of hypergraph H = ( X , 8)is a matrix ((a:)) with m rows that represent the edges of H and n columns that represent the vertices of H , such that

4

[

= 1 =0

if if

xjeEI xi$&.

Each (0, 1)-matrix is the incidence matrix of a hypergraph if no row or column contains only zeros. To each hypergraph H = (X;E l , E,, ..., Em)there corresponds a hyper389

390

HYPERGRAPHS

graph H* = ( E ; Xl, X 2 , ..., X,) whose vertices are points e l , e 2 , ..., em (that respectively represent E l , E,, ..., Em)and whose edges are sets XI, X,, ..., X,, (that respectively represent x l , xz, ...,x,), where, for allj,

X,

= {ei/i<

m, E,3x,}.

Thus, X , # and U, X j = E, and H* is a hypergraph. Hypergraph H * is called the dual hypergraph of H. The incidence matrix of hypergraph H * is the transpose of the incidence matrix ((a{))of hypergraph H, and consequently, (H*)* = H . If two vertices x j and X , in H a r e adjacent, then the corresponding edges X , and X , in H * are adjacent. If two edges Ei and Et in H a r e adjacent, then the corresponding vertices e, and e, in H* are adjacent. In hypergraph H, the rank r ( S ) of a non-empty set S c X is defined to be the positive integer

r(S) = max I S n El I . I

The number r ( X ) is called the rank of hypergraph H. If E1 = r ( X ) for each i, then H i s called a uniform hyp’ergraph of rank r(X). Thus, each simple graph without isolated vertices is a uniform hypergraph of rank 2. The partial hypergraph of H = ( X , 8 )generated by a family F = 8 is defined to be the hypergraph ( X g , F ) ,where X F = UEiegEi. The subhypergraph of H = ( X , 8 ) generated by the set A c X is defined to be the hypergraph H A = ( A , cYJ, where &A

= { E i n A / E l E 8 ; E i n A#

0).

Proposition 1. If H = ( X , F ) is a hypergraph with rank function r(S), the subhypergraph H A = ( A , gA)generated by a set A c X is a hypergraph with rank function rA(S),where P ” ( S ) = r(S)

(S c A).

The proof is immediate. Given an integer k > 0, the k-section of hypergraph H = ( X , 8)is defined to be the couple H(,) = ( X , b,,,) formed by X and the set &(k)

=

{F/F

c

x ; 1 G I F I < k ; F c Ei

for some E , E 8

}.

Proposition 2. If H is a hypergraph with rank function r(S), the k-section H(,) of H is a simple hypergraph with rank function r(,)(S) where r(k)(S) = min

{ k , r ( ~1.)

HYPERGRAPHS AND THEIR DUALS

391

The proof is immediate. Thus, the 2-section H,,, of a hypergraph H i s a graph with the same vertices as H and with a loop attached to each vertex. Let denote this 2-section without its loops. Clearly, ( H ) z is a simple graph. 2. Cycles in a hypergraph

In a hypergraph H = ( X , S),a chain of length q is defined to be a sequence (x,, E l , x z , Ez,..., E,, x q + d such that (1) xl, x2, ..., x, are all distinct vertices of H,

(2) E l , Ez,..., E, are all distinct edges of H, (3) X k , X k + l E Ek for k = 1, 2, ..., 4. If q > 1 and x,, , = xl, then this chain is called a cycle oflength q. If H is a graph, these definitions correspond to those for an elementary chain and an elementary cycle (except that a loop is not considered as a cycle in the theory of hypergraphs).

Fig. 17.1. Connected hypergraph (A’:E l , E,, E 3 , E,, Es, Ee) of order 8 with a E, I - 1) = 2 1 2 2 1 -k 0 = 8. unique cycle and with

c(I

+ + + +

If there is a chain in the hypergraph that starts at vertex a and terminates at vertex b. then we shall write a = b. Proposition 3. The relation a = b is an equicalence relation, whose classes are called the “connected components” of the hypergraph. If Clis a connected component that intersects edge E, then C, contains E. It suffices to show that this relation is an equivalence (1) a F a, because there exists a chain of length 1 beginning and terminating at vertex a, (2) a = b implies b = a,

392

HYPERGRAPHS

(3) a = b, b = c implies a = c, because if there exists a chain of length p from a to b and a chain of length q from b to c, then the edges of these chains form a chain of length < p + q from a to c. The theorem follows. Q.E.D.

Proposition 4. If H is a hypergraph with n certices, ~ ? edges i and p connected components, it contains no cycles i f , and only $ m

Consider the bipartite graph G ( H )whose vertices represent the vertices and edges of H , where the vertex representing x, is joined to the vertex representing Ei if, and only if, x, E Ei. Graph G ( H ) has p connected components, m + n vertices, and zr=lI Ei I edges. Hypergraph H contains no cycles if, and only if, G ( H ) is a forest, i.e. if m

or m

Q.E.D.

Proposition 5. If H = (X,8)is a hypergraph with G: = (E, / i E I), then H has no cycles $ and only if,

1 . If H contains a cycle ( a l , E l , a , , E,, { 1, 2, ..., q } , we have

..., E4, al), then, by letting Q

=

and the inequality of the proposition cannot hold. 2. If H = (A’,8)contains no cycles, then a partial family (El / i €1)also contains no cycles. If this family forms a hypergraph with p connected components, then by Proposition 4,

Q.E.D.

393

HYPERGRAPHS A N D THEIR DUAL3

Proposition 6 . A hypergraph ( X , 8) with p connected components and n wrtices has a unique cycle if, and only if,

c (IEJ-l)=n-p+l. m

i= 1

The bipartite graph G ( H ) defined in Proposition 4 has p connected components, nr + n vertices, and IfI Ei1 edges. It has exactly one cycle if, and only if, its cyclomatic number is ni

Q.E.D. Theorem 1 (LovAsz [1965]). r f a hypergraph H = (X,8)of order n with p connected components contains no cycles of length 2 3, and no loops, and if

I Ein E j 1

<2

for E, # E,,

then ni

C (IEiI -

2 ) < n - ~ *

i= 1

1. If n = 2, then H has exactly one connected component, and, therefore, the condition holds. Suppose that the theorem is true for all hypergraphs with less than n vertices, n > 2. Assume that it is not true for a hypergraph H with n vertices, and we shall show that this leads to a contradiction.

2. Hypergraph H is connected because, otherwise,

for each connected component C,, and by summing over k, the inequality is obtained.

3. We shall show that H contains no vertex whose removal disconnects H (“articulation vertex”). If such a vertex xo existed, and if its removal created two connected components C, and C 2 ,then by letting D1= C, u { xo }, we would have:

C

(I Ei I

- 2)

< I D,I - 1

3

EicDi

Since H has no loops, there is no edge contained in both D , and

394

HYPERGRAPHS

D2.Furthermore, each edge intersecting D , (resp. D,)is entirely contained in D, (resp. DJ.Thus,

x,

4. We shall show that no vertex x1 belongs to onlyone edge. If x1 E El and 4 Ei for i # 1, and if I El I 2 3, then the hypergraph

-

(El { x1 1, E2Y ... E m ) contains no cycles of length 2 3 ; then, by the induction hypothesis, Y

m

2 (IEJ

- 2) - 1 < (n

- 1) - 1 ,

i= 1

which implies that the inequality of the theorem holds. If I EiI = 2, the same result is obtained by considering the hypergraph (E,, E3, ..., Em). 5. We shall show that I Ein E, 1 = 0 or 2 for Ei# E,. Suppose that El n E2 = { xo}. Since I Ei I 2 2, consider a vertex a E El with a # x o , and a vertex b E E2 with b # xo. Since xo is the only vertex of E, n E,, it follows that a # b. Since xo is not an articulation vertex from Part 3, there exists a chain (a, E ; ,

4, E ; , ..., x;,

E;, b)

connecting a and b that does not use vertex xo. If there existed two indices i a n d j such that E,' = El and E; cycle (XO,E!,Xf+l,EI+l,

...,E[i,X

=

E2, then the

0)

+

would be of length 2 3 (because i 1 # j ) . If there existed an index i such that E; = El and no index j such that E,' = E,, then the cycle

(~g,Ef,x;+l,E;+ - .~. ,,E i ,b,Ez,Xo) would be of length 2 3. If there existed no index i such that E( = El or E; = E,, then the cycle (aYE;,x;Y *-*YE;,byJ%,Xo,E1,a) would be of length 2 3.

395

HYPERGRAPHS A N D THEIR DUALS

In all cases, there is a cycle of length 2 3 in H , which contradicts the hypothesis. 6. Consider the family 9 = { F / F = E n E’, E, E’ E 8,E n E‘ # 0 } where each set contains exactly two vertices. We shall show that 9is a partition of X . Clearly, U F E F= ~ X because, from Part 4, each vertex x E Xis covered by at least two edges Ei and E, E 8,and E, n E, E 9. Furthermore, let F and F‘ be two different sets of 9with

F = El n E z ,

F’

=

E; nE ; .

If there exists a vertex xo E F n F’, we can write: F={xo,xl},

F ’ = { x ~ , x ; } , xi # x ; *

If El = E ; , hypergraph H contains a cycle (xo, E,, xl, E l , x i , EL, xo) of length 3, which contradicts the hypothesis. If El # EI ,E4 ,and if E2 # E; , E & let E l n E; =

{.yo,

a, }

E , n EL

=

{ x o , a, } .

Clearly, a, # a, because, otherwise, El n E2 = Ei n E i . Therefore, we may assume that, say, a, # x i . Thus, there is a cycle (xo, E l , al, E ; , xi, €;, xo) of length 3, .which contradicts the hypothesis.

7. Construct a bipadte graph G, whose vertices represent the m edges of 8 in one class and the sets of 9 in the other class, and where the vertex representing Ei E d and the vertex representing F? E F are joined if, and only if, E, n F, # 0. Note that, in this case, F, = Ei because, otherwise, F, = E, n E; with E,, E; # E,, and therefore Ei n Ej and F, could not be two different classes of 9. Thus, the number of edges in graph G is

m+n From Part 6, the number of vertices in G is n(G) = 7 . 8. Note that graph G contains no elementary cycle [F,, El,F,, E,, ..., E k , F,], because then k > 4, and if we select one x, in each Fi,the cycle ( x , , E l , x,, E,, ..., xl) would be of length 2 3, which contradicts the hypothesis. From Proposition 4, graph G satisfies

E,

m(G) < n(G) - 1 .

396

HYPERGRAPHS

Hence.

and

I

Q.E.D.

In particular, Theorem 1 shows that if H is a uniform hypergraph of rank 3 without cycles of length 2 3, then

m
3. Conformal hypergraphs

Consider a simple hypergraph H = (X,6 ) where 8 = { E l , E,, ..., E m } . Consider the 2-section ( H ) z ,which is a simple graph. Each edge of ( H ) 2is contained in at least one of the cliques E l , E2, ..., Em.Hypergraph H is said to be conformal if the set of all the maximal cliques of graph ( H ) , is equal to the set G of maximal edges of H . If H i s a conformal hypergraph, each subhypergraph of H is also conformal. Lemma. Hypergraph H is conformal if, and only ift each clique of graph ( H ) z is contained in an edge of H. If H is conformal, it is evident that each clique of ( H ) z is contained in a maximal clique of ( H ) z and thus in an edge of H . To prove the converse, let Vmsxbe the family of maximal cliques of graph ( H ) z , and let gmaX be the family of maximal edges of H . If C E Vmaxr there exists an edge E E L,,, and a clique C' E % with

CcEcC'. Hence, C = E. Thus, Conversely, if E E B,,,,

em,,c t?,,,. there exists a clique C E,,V,

and an E'

E

8 with

E c C c E'. Hence, E = C , and emax c .,& ,, formal hypergraph.

Therefore, d ,,,

=

em,,, and H

is a conQ.E.D.

Theorem 2 (Gilmore [1960]). A necessary and su$cienr

condition f o r a

391

HYPERGRAPHS AND THEIR DUALS

hypergraph H = ( X , 8)to be couformal is that f o r any t h e e edges E l , E2,E3 of H , there exists an edge of H that contains the set

(El n E2) u (€2 n E3) U (E3 n

.

Necessity. If H is conformal, each vertex of El n E2 is adjacent to each vertex of E, n E3 and to each vertex of E3 n E l . Thus, set

(El n E2) u (E2 n E d

U ( ~ 5 3n El)

is a clique of graph ( H ) 2 .From the lemma, this clique is contained in an edge of H . Suflciency. We shall show that each clique C of ( H ) 2 is contained in an edge of H . If I C I < 2, this follows from the definition of the 2-section. Suppose that this is true for each clique C’ with I C’I < k ; we shall show that it is also true for a clique C with I C I = k 2 3. Let x,, x , , x 3 E C, and let C, = C - { X , }. By the induction hypothesis, there exists an edge E, 3 C,. Thus, there exists an edge Eo of H such that

C = (C, n C2)u ( C , n C3) u (C2 n C,) c c (El n E2) u ( E , n E 3 ) u ( E 2 n E3) c E o . Q. E.D. A hypergraph G = (El 1 i E I ) is said to satisfy the HeIly property if J c I and Ein E, # 0for all i,.j E J implies that

nEjZ0.

jt-J

The following is the dual result to Theorem 2. Theorem 3. A hypergraph H is conformal if, and only if, its dual H * satisfies the Helly property. 1. Suppose that H * satisfies the Helly property. We shall show that a clique { xl, x 2 , ...,xk ] of graph (H ) , is contained in an edge of H . Any two vertices x, and x t in this clique are both contained in some edge E, of H . Consider the edges X,= { e, / j Q nz, Ej E X , 1 of the dual hypergraph H * . Clearly, enE X,n X,. The sets X , , A’,, ..., X , are pairwise intersecting, and therefore, by the Helly property, k

n X i # @ . i= 1

398

HYPERGRAPHS

If e, is a vertex in this intersection, then

E,

3 { x ~ , x ~ , . . * ~ X ~ } ,

and E, is the required edge. 2. If H is a conformal hypergraph, consider a family (XI, X , , ..., Xk) of edges in H* which are pairwise intersecting. This family corresponds to a clique { xl, x 2 , ..., xk } in ( H ) , , and there exists an edge E, of H that contains this clique. Hence, k

e,E

n Xi.

i= 1

Thus, H* satisfies the Helly property. Q.E.D.

Corollary. A family (Xl, X,,..., X,)ofsubsetsofaset E = { e l , e,, ..,, e m ) satisjies the Helly property i f , and only is, for each triple (e, , e,, ek), the family of the X , that contain at least two of these three vertices has a non-empty intersection (ifthis family, is not empty). Cdnsider the hypergraph H = ( E ; XI, X,,..., A',). From Theorem 3, the Helly' property is satisfied if, and only if, the dual hypergraph H* = (X;E l , E 2 , ..., Em)is conformed, therefore, from the Gilmore theorem, if for each triple ( e f ,e,, e k ) there exists an index s such that

E,

=3

(Ein Ej) u (Ej n Ek)u (Ekn Ei).

Let P be the set of indices p such that X , contains at least two of the three vertices e,, e,, ek. Thus, the above condition is equivalent to x,EE,

(PEP)

for some index s, which is equivalent to

n x,#ra.

Pep

Q.E.D. EXAMPLE 1. Let E = { e,, e,, ..., em1 be a finite set of points on a line. We shall show that a family (XI, X,,..., X,)of intervals of points satisfies the Helly property. Consider three points e l , e,, e 3 , with el < e2 < e 3 . If interval X, contains

HYPERGRAPHS A N D THEIR DUALS

two of these three points (write this as p fore,

E

399

P),then it contains e2, and, there-

nx P z a

PEP

Thus, from the above Corollary, the family satisfies the Helly property. EXAMPLE 2. Consider the hypergraph (X; El,E,, ..., Em)that is the dual of a hypergraph ( E ; XI, X,,..., X,)defined as in Example 1. We shall show that this dual hypergraph also satisfies the Helly property. Consider three vertices X1,XZ,Xj E X .

Let P denote the set of indices p such that E, contains two of these three points. We shall show that

n E,M.

POP

We may assume that none of the intervals Xl, X,,X 3 contains another, because if X 3 2 X I , then each e, that belongs to two of the three intervals belongs to X 3 , and x3e

n E,.

PEP

Among XI, X,,X 3 , Idt X I be the interval whose initial endpoint is the leftmost, and let X, be the interval whose terminal endpoint is rightmost. Consequently, X, 2 XI n X 3 , and each e, that belongs to two of the intervals also belong to X,.Hence, x2€

n E,.

PEP

Thus, the E, satisfy the Helly property.

EXAMPLE 3. Consider a tree G = (X,U), and let (A, / i E I> denote the family of subsets A , of X such that the subgraph G,, is also a tree. We shall show that this family satisfies the Helly property. Consider three vertices a, b, c of G. Let P be the set of indices p such that A , contains two of these three vertices. If the three vertices belong to the same elementary chain of G, then one of the vertices, say 6, is between the other two in the chain. Thus, b E npPPqp. If the three vertices do not belong to the same elementary chain, then there exists a vertex z of G that separates any two of them. Thus,

400

HYPERGRAPHS

n

# @. Thus, from the Corollary to Theorem 3, the Ai In both cases, satisfy the Helly property.

4. Representative graph of a hypergraph Let H = ( E ;Xl ,X2, ..., X,) be a hypergraph with n edges. The representative graph of H is defined to be the simple graph L ( H ) of order n whose vertices xl, x2, ..., x, respectively represent the edges XI, X,, ..., X , of H and with vertices xi and x j joined by an edge if, and only if, XI n X , # 0. For each simple graph G, there exists a hypergraph H such that G = L(H). More precisely, we have the following: Proposition 1. Let G be a simple graph with vertex set X , and let ( E l , E2,..., Em)be a family of subsets of X with the following properties: (1) each ELis a clique of graph G, (2) each vertex and each edge of G is covered by at least one E l . Then, G is the representative graph of the dual H of the hypergraph

H* = ( X ; E l , E z ,

..., Em).

Conversely, if G is the representative graph of a hypergraph

xi,xz,.-*,xJ, = (X.; E l , E2, ..., Em)satisfiesproperties H=(E;

then the dual H*

(1) and (2).

G = L ( H ) means that vertices xi and x, are adjacent if, and only if, there exists in H a vertex e, E Xt n X,. This is also equivalent to saying that there exists an edge Ek in H * that contains both xt and x,, or to saying that G

= (H*)Z.

Clearly, this can be expressed as H* satisfies (1) and (2).

Q.E.D.

For example, consider the graph G shown in Fig. 17.2. Graph G is the representative graph for each of the following hypergraphs: (1) the multigraph H , (the Et are the maximal cliques of G), (2) the hypergraph H2 (the Et are the edges of G), (3) the simple bipartite graph H 3 , etc. Let Q(G) denote the minimum order of the hypergraphs H for which G = L(H). For graph G in Fig. 17.2, we have R(G) = 2 because G = L(Hl). The following proposition shows that the determination of Q(G) reduces to the determination of a chromatic number.

401

HYPERGRAPHS AND THEIR DUALS

G

d

e

Fig. 17.2

B

E

H , = (A. B. C, D. E )

Order

=

6

Fig. 17.3

Proposition 2. Let G be a simple graph with vertices xl, x2,...,x,,,none of which is isolated. Let G be a graph, each of whose vertices represents an edge of G, with two certices corresponding to edges [a,b] and [ x , y ] of G joined together if, and only i f , { a, 6, x,y } is not a clique in G. The minimum order Q(G) of the hypergraphs for which G is a representative graph equals the chromatic number

?)(GI.

402

HYPERGRAPHS

The graph G corresponding to graph G in Fig. 17.2 is shown in Fig. 17.4. bc

Cd

de Fig. 17.4

G

ce

1. We shall show that each q-colouring s2,..., s,) of G with q = y(C) colours yields a hypergraph ( E ; X , , X2,..., X,) of order q for which

(sl,

graph G is the representative graph. The set Sl of vertices of C with the i-th colour is a stable set. If [a, b] is an edge of G belonging to Si,then vertex a is either identical or joined to each of the endpoints of an edge of S,. Therefore, the endpoints of the edges in generate a clique Eiof G. The hypergraph ( X ; E l , E,, ..., E,) has the property that each edge and each vertex of G are covered by at least one of the E l . From Proposition 1, G is the representative graph of the dual hypergraph ( E ; X I , X2,..., X,) which has order I E I = q. Hence,

s,

Q(G)

G Y (GI

*

2. We shall show that a hypergraph H = ( E ; X,, X,, ..., X,,) of order q = Q(G) for which G = L ( H ) can determine q-colouring ( S , , S2, ..., 3,) of the vertices of G. Let A k denote the set of vertices of H that belong to exactly k of the sets XI. Then, = I E I=

I ~1 I + I AZ I + I ~3 I + ... .

} where e E To each To each vertex e E Al , associate the I-clique { and X,(e),associate the vertex e E A,, that belongs to exactly two sets x,(,) 1 of G. To each vertex e E A3 that belongs to exactly three 2-clique { xlCe), X,(e),X k ( e )associate , the 3-clique { of the sets x , ( ~ X) ,k ( e ) } of graph G, etc. Since H i s of minimum order, we have A' = 0. In this way, a family ( E l , E,,..., E,) of q-cliques in G is defined. Clearly, each edge [ x , , x,] of G is covered by at least one of these cliques (because Xin X , contains a vertex of E ) . Let 3' denote the set of edges of G contained in clique E l . Let S2 denote the set of edges of G covered by clique E2 that

403

HYPERGRAPHS AND THEIR DUALS

s3

are not covered by El. Let denote the set of edges of G covered by clique E3 that are not covered by either El or E,, etc. The family ($, sa,..., S,) clearly is a q-colouring of the vertices of G. Hence, y(c)< Q(G), and SO Y(G)= Q(G). Q.E.D.

Proposition 3. Let G be a simple graph without isolated vertices and without triangles and with m edges. The minimum order of the hypergraphs H for which G = L ( H ) is R(G) = m. Graph defined as in Proposition 2 is an m-clique. Therefore, the required minimum order is R(G) = m. Q.E.D. Proposition 4 (Erdos, Goodman, P6sa [1966]). If G is a simple graph of order n without isolated vertices, then

For each n, this bound is the best possible. 1. From Theorem (5, Ch. l l ) , the edges and vertices of G can be covered by a family of 2-cliques and 3-cliques (El, E,, ..., &) where k < n2 From

[ - I.

Proposition 1, G is the representative graph of the dual of hypergraph ( X ; El, E,, ..., E J . Since this dual has order k , the minimum required order is

< k < [ %] . 2

R(G)

2. We shall show that for each n, there exists a graph G of order n such that

If n = 2 k is even, let G be the complete-bipartite graph Kk,k. Since graph G contains neither isolated vertices nor triangles, Proposition 3 can be applied to G. Thus,

R(K,,,)

=

k2

n2

=4 =

[ -3n42

If n = 2 k + 1 is odd, then let G be the complete-bipartite graph Kk,k+l. From Proposition 3, 2 1 1 - 1 n+l n2 I Q ( K k , k + i ) = k ( k + 1) = 4 2 .2 - 4 Q.E.D.

-[?I.

404

HYPERGRAPHS

The following proposition characterizes the representative graphs of the maximal cliques of a graph. Proposition 5 (Roberts, Spencer [1969]). A simplegraph G is ihe representative graph of ihe maximal cliques of some graph i f , and only i f , ihere exists in G a family of cliques (Ei / i E I ) such that (1) each edge of G is covered by an Ef, (2) (Ei i E I ) satisfies the Helly property.

Clearly, we may assume that G contains no isolated vertices. 1. Let G be a representative graph of the maximal cliques Xl , X , , ..., A',, of some graph. These maximal cliques determine a conformal hypergraph H = ( E ; X I , X,, ..., X J , where n

€ = (J

xi.

i=1

Let E l , E,, ... denote the edges of the dual hypergraph H * . By Proposition 1, these are cliques of G and satisfy condition (1); since H is conformal, they also satisfy condition (2), by Theorem 3. 2. Conversely, consider a family of cliques (Ei / i E I ) in G that satisfy conditions (1) and (2). Cliques E,' = { xi } for i = 1, 2, ..., n can be added to this family without violating conditions (1) and (2). Clearly, H" = ( X ; E l , E,, ..., E;, E;, ..., EJ is a hypergraph. Let H = ( E ;Xi,

9

XJ

be its dual. Condition (1) implies that G is a representative graph of H by Proposition 1. Condition (2) implies that H is a conformal hypergraph by Theorem 3. Moreover, edge Xi contains vertex e; because xi E Ei = { x, }, and this is the only edge of H that contains vertex e ; . Hence, the edges of H are all maximal edges. Thus, H is the family of maximal cliques of graph (H),. Q.E.D. To show that graph G in Fig. 17.2is the representative graph of the maximal family of cliques of a graph (namely, H3), it sufficesto consider El = { a, 6, c } and E, = { c, d, e } as the family of cliques satisfying conditions (1) and (2). The following result which is a new generalization of a result of Krauz, characterizes the representative graph of the k-cliques of a graph.

Proposition 6. A simple graph is the representative graph of the k-cliques of

HYPERGRAPHS AND THEIR DUALS

405

Some graph i f , and only if, there exists a family (El / i E I ) of cliques in G such that (1) each edge of G is covered by an E,. (2) each vertex of G belongs to exactly k of the E,, ( 3 ) each partial family ( E j j E J ) formed froni k of the cliques E, that are pairwise intersecting, has an intersection of cardinality one.

1. Let G be the representative graph of the family V k = (XI, X , , ..., X,,) of k-cliques of a graph (E, r);assume that each vertex or edge that does not belong to a k-clique has been removed from (E, I-). Family%, determines a uniform hypergraph H = ( E ; XI, X,, ..., Xk) of rank k, and

G

= (H*),

Let E l , E,, ... denote the edges of the dual hypergraph H*. Since G = (H*),, these edges are cliques of G and satisfy condition (1). They also satisfy condition (2) because an edge of Hcontains k vertices, and thus a vertex of H * belongs to k edges. Condition (3) is also satisfied because if k vertices e, of H are pairwise adjacent, then there exists a unique edge of H that contains all k vertices (since H is conformal).

2. Conversely, let (Et / i E I) be a family of cliques of G that satisfy conditions (l), (2) and (3), and let H * = ( X ; E l , E,, ..,, En) be the hypergraph whose edges are these sets. By condition ( l ) , G = ( H * ) 2 . By condition (2), each edge of the dual hypergraph (H*)* = H contains exactly k vertices. Finally, the k-cliques of graph ( H ) , are the edges of H, and by condition (3), graph ( H ) , contains no other k-cliques. Thus, G is the representative graph of the k-cliques of graph (H)Z. Q.E.D. Corollary (Krauz [1943]). A simple graph G is the representative graph of some simple graph H $ and only if, G contains a family of cliques (E, / i E I ) such that (1') each edge of G belongs to exactly one El, (2') each vertex of G belongs to exactly two E1. Let k = 2 in Proposition 6 ; then conditions (1) and (3) imply condition (1'), and vice versa, and condition (2) becomes condition (2').

Q.E.D. Note that this corollary gives only a global characterization of an L(H). We shall now present a local characterization that is easier to verify; first some new definitions are required.

406

HYPERGRAPHS

G1

f

A

a. x

Y

Fig. 17.5

A triangle of a graph G = (A’, E ) is defined to be normal if each vertex of the graph is adjacent to an even number of vertices of the triangle. Otherwise, the triangle is defined to be specia!. Note that a normal triangle is a maximal clique of G.

Lemma. If G possesses two special triangles abc and abd with [c, d ] 6 E, then G contains as a subgraph at least one of the graphs shown in Fig. 17.5. We shall consider separately the following cases. CASE1. There exists a vertex of G that is adjacent to an odd number of vertices of triangle abc and to an odd number of certices of triangle abd. Let x be this vertex. Two subcases must be considered : CASE1’. Vertex x is adjacent to exactly one vertex of each triangle. I f x is adjacent to a, then subgraph GI is obtained with vertices x, c, d, a. I f x is adjacent to c and to d, then G2 is obtained. CASE1”. Vertex x is adjacent to all the vertices of one of the two triangles. Then, x must be adjacent to a, b, c, d, and G3is obtained. CASE2. Each certex of G is adjacent to an even nuniber of vertices in either one of the triangles abc or abd. Let x be a vertex of G adjacent to an odd number of vertices of abc; let y be a vertex adjacent to an odd number of vertices of abd. Thus, x # y.

HYPERGRAPHS AND THEIR DUALS

401

First note that x I S not adjacent to both c and d (because, otherwise, x would be adjacent to a, b, c, d and then the conditions for Case 2 would not be satisfied). If x is not adjacent to c, then x is adjacent to exactly one of the vertices a or b, and consequently, to d. Thus, three subcases must be considered. I. x is adjacent to a, b, c but not to d, 11. x is adjacent to c, but not to a, 6, d, 111. x is adjacent to a, d, but not to b, c. By considering the conditions on x and y nd whether or not x and y are adjacent, one of the nine graphs of Fig. 17.5 can be obtained. The details are left to the reader. Q.E.D. Theorem 4 (Beineke [1968]). For a simple graph G, the following conditions are equicalent: (1) G is the representative graph of a simple graph, (2) G contains none of the subgraphs shown in Fig. 17.5, (3) thefamily 5f of all the maximal cliques of G that are not normal triangles

corers each certex at most twice, and corers each edge at most once, and G contains ti0 subgraph of the type G, , (4) there exists a family B of cliques such that each rertex belongs to exactly two cliques of B and each edge belongs to exactly one clique of PP. (1) * (2) From the Krauz theorem, there exists a partition of the edges of G into cliques such that each vertex of G is covered by at most two cliques. If G contains a subgraph of the form G,,then this partition induces in G, a partition of the edges of G1with the same property. It is easily verified that none of the graphs G, of Fig. 17.5 admits such a partition. (2)

(3)

-

(3) If two maximal cliques C and D which are not normal triangles have a common edge [a, b],then there exist two vertices c E C and d E D that are not adjacent. If C # { a, b, c }, then the triangle abc is special. I f C = { a , b, c } , then the triangle abc is also special. From the lemma, the special triangles abc and abd with [c, d ] E, yield that there exists a subgraph of G of one of the types shown in in Fig. 17.5. This contradicts condition (2). (4) Let % be the family of all the maximal cliques that are not normal

triangles. We shall successively add new cliques to this family to form family 9.

408

HYF'ERGRAPHS

Let [a, b] be an edge that is not covered by a clique of V. This edge necessarily belongs to a normal triangle abc. No clique of %' covers only one vertex of this triangle (because, otherwise, the triangle would be special). Similarly, no clique of V covers the three vertices of this triangle. Two cases must be considered: CASE1. No vertex of triangle abc is covered by a clique of %'.There are two subcases to consider. CASE1'. Triangle abc intersects with no other normal triangle. Form 9 by adding the cliques { a, b, c }, { a }, { b } and { c } (or, by adding the cliques { a, b 1, { a, c 1, { b, c 3). CASE1'. Triangle abc has a common edge with another normal triangle (and with only one, because GIis forbidden). Let abd be this new triangle. Noother normal triangle meets abed (because then it would intersect each triangle at two vertices). Triangle abd does not meet any clique of V (because then a or b would be covered by a clique of V). Form B by adding the cliques { a , b, c}, { a, b, a>,{ c 1, { d ) .

Case 2"

Case 1" Fig, 17.6

CASE2. A vertex of triangle abc is covered by a clique C E W. In this case, exactly two of the vertices are covered by clique C (because, otherwise, the triangle abc would not be normal). Without loss of generality, let b and c be these two vertices. CASE2'. No other clique of V meets triangle abc. Form 9 by adding the cliques { a, b } and { a, c }. CASE2". One other clique C' of %' meets triangle abc. In this case, C' meets the triangle at two vertices (otherwise, abc would not be normal), and these two vertices must be u and c from the definition of edge [a, b]. No other

409

HYPERGRAPHS AND THEIR.DUALS

clique ofV meets triangle abc. Thus, form B by adding the clique { a, b }. See Fig. 17.6.

(4) * (1) by the corollary to Proposition 6. Q.E.D. 1

H

568

78

Fig. 17.7

Remark. Note that Theorem 4 presents an efficient method to determine if a graph G is the representative graph of some simple graph. An example is given in Fig. 17.7. The graph H represented by G can be constructed from the family 5f of the maximal cliques of G that are not normal triangles (and which are circled by a continuous line in Fig. 17.7). In graph G, the only normal triangles are 234,456, and 678, which are not to be included in family V. Family B can be formed from 9? by adding the 2-clique { 7 , 8 } and the I-clique { 1 }. Hence, the required graph H , shown in Fig. 17.7, is obtained.

Proposition 7 . A simple graph G is the representatiue graph of some multigraph without triangles and without loops if, and only i f , each vertex of G belongs to at most two maximal cliques. 1. Suppose that G = L ( H ) is the representative graph of multigraph H. A vertex x of G represents an edge [c, d ] of H , and T,(x) U { x } represents the set of edges of H incident to c or to d. Let C, denote the set of vertices corresponding to the edges incident to c, and let 0,denote the set of vertices corresponding to the edges incident to d. Then T,(x) u { x }

=

c, u D, ,

and C, and D, are two cliques of G. Since H contains no triangles, the sets C, - D, and D, - C, are not joined by an edge.

410

HYPERGRAPHS

Let C be a maximal clique in G that contains x . Then, C c C, u 0,.If y E C and y E C, - D,, then each vertex y’ E C that is adjacent to y belongs to C, ;hence, C c C,, and it follows from the maximality of C that C = C,. Therefore, each maximal clique that contains x equals either C, or D,, and consequently, there are at most two maximal cliques that contain x . 2. Conversely, let X’ denote the set of vertices in G that belong to exactly one maximal clique, and let X”denote the set of vertices in G that belong to exactly two maximal cliques. Assume that X = X’ u X”. Consider the hypergraph H * = ( X ; E l , E,, ..., { x i }, { xk }, ...) whose edges are the maximal cliques of G, and the cliques { xi 1 for all xi E X’. Consider the hypergraph H = ( E ; X,,X,,..., X,) dual of H*. Clearly, graph G is the representative graph of H . Hypergraph H i s a multigraph without loops, because each vertex x, of H* belongs to exactly two distinct edges. Suppose that H contains a triangle el e2 e,; let x , y , z be the vertices of H * that represent respectively edges [ e l , e,], [e,, e,], [e,, e l ] . Then,

xEE1 n E 2 ; x#E3

Y E En~E3 ; Y #El

Z E En~El ; z#E2

-

Let C be a maximal clique of G that contains triangle xyz. Then,

c z E,,E,,E3 Consequently, vertex x belongs to three distinct maximal cliques E l , E2, C, which is a contradiction. Q.E.D.

Proposition 8. A simple graph G is the representatice graph of some bipartite multigraph if, and only if, the two following conditions hold: (1) each tlertex of G belorigs to at most two maximal cliques, ( 2 ) each odd elementary cycle of G contains two sides of a triangle. 1 . If G = L ( H ) is the representative graph of a bipartite multigraph H, then condition (1) follows from Proposition 7. Condition (2) also holds, because if [xl, x2, ..., x , , x,] is an elementary odd cycle of G that does not contain two sides of a triangle, then p > 3, and the corresponding edges X , , X,, ..., X,, X I of multigraph H also constitute an odd elementary cycle. This contradicts that H is bipartite.

2. Conversely, consider a graph G that satisfies conditions (1) and (2). From condition (l), there exists a multigraph H = ( E ; Xl, X2,..., X,,) without triangles such that G = L ( H ) . It remains to show that multigraph H contains no odd elementary cycle [ e , , e 2 , ..., e,, e l ] with p > 3. Suppose that H contains such a cycle. Let Xi be an edge of multigraph H

41 1

HYPERGRAPHS AND THEIR DUALS

that joins e, and e,,, in the cycle. The sequence [x,, x l , x,, ..., x,] is an elementary odd cycle of G. From (2), there exists a triangle formed from three consecutive vertices of this cycle, say x l , x,, x3. In multigraph H, edges X , , X,, X3 are pairwise adjacent and, since H contains no triangles, they have a common endpoint. But this contradicts that these three edges lie on an elementary cycle. Q.E.D. This theorem shows why graph G shown in Fig. 17.2 is the representative graph of a bipartite multigraph H 3 .

Remark. Consider a graph G1 that is a triangle, and a graph G, that is three edges with a common endpoint. Clearly, L(Gl) and L(G,) are both isomorphic to a triangle. H. Whitney [1932] has shown: Zf two connected simple graphs have isomorphic representative graphs, then they are isomorphic, except for G, and G,. This result has been extended to hypergraphs by Berge and Rado [1972]. Two hypergraphs H = ( X ; E l , E2,..., Em) and H ’ = ( Y ; F,, F,, ..., F,) are said to be isomorphic if they have the same number m of edges, and if there exists a bijection cp: X - t Y and a permutation 71 on M = { 1,2, ..., m } such that (i = 1, 2, ..., m) . cp(El) = F,,,, H and H’ are said to be strongly isomorphic if there exists a bijection cp: X - t Y such that

(i = 1 , 2, ...,m). cp(E,) = Fl In this case, we write H r H’. Clearly, two simple graphs (El / i E M ) and (6 / i E M ) have the same representative graph if, and only if, (E,, E,) (El, F,) for all i,j . The main results for hypergraphs are the following:

I. For m > 1, there exist two uniJorm hypergraphs K,(M) = (E, / i E M ) and L,(M) = (F, / i E M ) which are not strongly isomorphic and which satisfy

( E , / i E M - { k } ) g ( F , / i E M - {k})

EM)

The vertices of K,(M) arepoints x,, for all J c M with I J I = m modulo 2, and its edges are the sets El = { x, / i E J }for all i < m. The vertices of L,(M) are points y , for all J c M with I J I = m + 1 modulo 2, and its edges are the sets 6 = {y,/iEJ}. 11. Let p be an integer, p > 1, p < m ; let H = ( X ; El,E2, ..., Em) and H‘ = ( Y ;F1, Fa, ..., F,) be two hypergraphs such that f o r each set J c M with cardinalityp, thepartial hypergraphs (Ei/ i E J ) and (F, / i E J ) arestrongly

412

HYPERGRAPHS

isomorphic. Then, Hand H' are strongly isomorphic, except if there exists three sets A c X , B c Y, Z c A4 with I Z I = p and ( A n Ei / i €11)

( B n Fi / i E Z) E' Lp(Z),

Kp(Z),

or vice versa.

For the proofs, see Berge, Rado [1972].

EXERCISES 1. A hypergraph (X,8)is said to be hereditary if A€&,

B c A , B # O

3

BE&'.

Let 2 be the set of all subsets of 8 of the kind { E l l ,E l z ,..., El, } with Ell c Eta c C €in.

Show that the derived hypergraph (8, X ) is conformal. (In algebraic topology, a hereditary hypergraph is an abstract simplicia1 complex, and the derived hypergraph is its first barycentric subdivision.) 2. Let el,e2, ..., em be points on a line, and let XI, X z , ..., X,,, be intervals such that no interval is contained in another. Show that the dual hypergraph H* of H = (XI, Xz,..., X,,) is also a family of intervals. Also, show how to place the points x l , x z , ..., x,, and the intervals E l , E z , ..., Em on a line such that el E X j if, and only if, x j E E,. '

3. Let G be a graph such that the maximum cardinality w(c) of the cliques of G is < 3. Show that G is the representative graph of the maximal cliques of some graph if, and only if, the maximal cliques of G satisfy the Helly property. Show that G is the representative graph of the maximal cliques of some graph if, and only if, G contains no partial subgraph with six vertices a, b, c, d, e, f a n d with edges (Roberts, Spencer [1969]) ab, ac, bc, bd, be, ce, cf, de, ef. 4. Calculate the minimum cardinality d(n) of a set A such that each graph of order n

is the representative graph of n distinct subsets of A . Show by induction on n that

d(2) = 2 ,

d(3) = 3 , d(n) =

[TI

if

n 2 4. (Erdos, Goodman, P6sa [1966])

5. Consider a family of non-empty sets of the form

( x / x = ( x * , x 2 ,...,x * ) E W ; a * <

XI

Q

b * , a 2 d x2 d bz, ..., at G xk C bk ).

Show that this family satisfies the Helly property.

413

HYPERGRAPHS A N D THEIR DUALS

6. Show that if G is the representative graph of

(1) G has

6)

K,,, then

vertices,

-

(2) G is regular of degree 2(n 2). (3) any two non-adjacent vertices of C have exactly 4 common neighbours, (4) any two adjacent vertices of G have exactly n - 2 common neighbours. A. Hoffman [1960] has shown that this necessary condition is also sufficient except for n = 8. For n = 8, there exist three graphs different from L(&) that satisfy these conditions.

7. Show that if G is the representative graph of K,,,.,, then (1) G has tnn vertices, (2) G is regular of degree rn + n - 2, (3) any two non-adjacent vertices have exactly 2 common neighbours, (4) there are n

(3

pairs of adjacent vertices in G that have exactly

neighbours, and there are m

n - 2 common neighbours.

Q

tn

- 2 common

pairs of adjacent vertices in G that have exactly

Moon [1963] and Hoffman [1964] have shown that this necessary condition is also sufficient except for m = n = 4. FOPthis case, there does exist a graph different from L(K,,,) discovered by Shrikhande [1959] that satisfies conditions (l), (2), (3) and (4). 8. Show that a connected graph G is the representative graph of a tree if, and only if, each block of G is a clique and if no three blocks have a common vertex.

(Ramachandra Rao [1969]) 9. Show that a graph G is the representative graph of the blocks of some graph if, and only if, each block of G is a clique. (Harary [1969])

CHAPTER 18

Transversals

1. Matchings and c-matchings For a hypergraph H = (A', a), a family E, c d is defined to be a matching if the edges of 6,are pairwise disjoint. To simplify notation, we shall assume that H is a simple hypergraph, and d and 8, are sets. Let 9 = E - 6,. An alternating sequence relative to matching 8,is defined to be a sequence CT = ( F l , E l , Fa, E,, F3, . ..) of distinct edges with the following properties: (1) Fl is a member of F, (2) El is a member of 8, - { E l , E,, ..., El-l } and Ein

UFj#Da,

j
(3) F , + , is a member of 9 - { F,, F,,

..., 4 } and

Sequence CJ is said to be odd (respectively, even) if its length is odd (respectively, even). Sequence CT is said to be maximal if no more edges satisfying properties (2) or (3) can be added to it. The following result characterizes maximum matchings. Theorem 1. Zn a hypergraph H, a matching 8, is maximum if, and only if, there exists no odd maximal alternating sequence relative to 8,. This follows from Theorem (2, Ch. 13) because each maximum stable set of the representative graph L ( H ) of hypergraph H corresponds to a maximum matching of H. Q.E.D. Note that if hypergraph H i s a graph, an odd maximal alternating sequence defines an alternating chain connecting two distinct unsaturated vertices, and Theorem 1 reduces to Theorem (1, Ch. 7). 414

415

TRANSVERSALS

Consider a simple hypergraph H

=

( X , d), and let V c 8.Let

'Xx= { E / E e U , E 3 x ) .

To each vertex x, associate an integer c(x) such that Cb)

G 18, I

*

A family % c d is defined to be a c-matching of hypergraph H if for each vertex x,

I g x I G 4x1 If c(x) = 1 for every x, then a c-matching is a matching. Every subfamily of a c-matching is a c-matching. We shall present a characterization of maximum c-matchings, that extends Theorem 1. An alternating sequence relative to the c-matching V is defined to be a sequence a = (F, , E l , F,, E2,F3, ...) of distinct edges that alternate between 9 = E - % and V and have the following properties: (1) F, is a member of 9 = 8'- V. (2) Given the odd sequence oi = (F, , El,F2,. .., F,), let V i be the family obtained from % by removing edges El,E2,...,Ei - and by adding edges F,, F,, ...,F,, i.e., %1

= (U

- V) .

- q)u (CTi

Then El must belong to V - atand contain a vertex x with I V i I > c(xj. (3) Given the even sequence CTI = (F,, E l , F,, ..., F,, Ei), let 9; be the family 9* = (9- a;) U (ol - F).Edge F i + , must belong to 9 - a; and contain a vertex x such that I 9 :I > c(x), and (F,, Fa, ..., F,,,) must be a c-matching. An alternating sequence o is said to be maximal if no more edges can be added to it. In fact,a is maximal if (and only if) one of the followingconditions holds : (i) % = B (I fails). (ii) CT = a, is odd, and%,is a c-matching. Clearly, if V* is not a c-matching, then there exists a vertex x with

I Ui I > c(x)

9

and since (F,,Fa, ...,6)is a c-matching, we have

I (W -

CTJx

I = lV;j

- 1 (a*-

J > c(x) - c(x)

= 0.

Therefore, there necessarily exists an edge El E 9? - CT, that contains vertex x and which can extend the sequence, by (2).

416

HYPERGRAPHS

(iii) 0 = 0; is odd, and family 9‘ is a c-matching. (iiii) 0 = 0; is odd, and family 9‘ is not a c-matching, but for each set F,,, E 9 - D that contains a vertex x with I 9; I > c(x) (which necessarily exists), family (Fl , F2, ..., F, + is not a c-matching. We shall now show that a c-matching V is maximum if, and only if, there exists no odd maximal alternating sequence.

Lemma. Let d be a hypergraph, and let a and%? be two disjoint c-matchings; let = % - 39, = 99 - V, d = 2 ug . Then, 3 a n d e are two c-match- ings of hypergraph 8, and each alternating sequence D

=

(Bi,C1, B2, C2, .*.>

of 8 relative to c-matching 4 is an alternating sequence of d relative to the c-matching %. Besides, i f 0 is an odd maximal alternating sequence in 8,then D is an odd maximal alternating sequence in 8. The proof is easy and is left to the reader.

Theorem 2 (Ray-Chaudhuri [1960]). For a hypergraph H = (X,b), a c-matching V is maximum if, and onIy if, there exists no odd maximal alternating sequence relative to the c-matching %. 1. Suppose that there exists an odd maximal alternating sequence 6.Then % is not a maximum matching, because the family

v‘=

(%-0)U(D-W),

is also a c-matching, and contains one more edge than%. (Otherwise, sequence is not maximal, from (2).)

D

2. Let % be a c-matching for which there exists no odd maximal alternating sequence, and suppose that V is not maximum. We shall show that this leads to a contradiction. Consider the maximum c-matching 9? that minimizes the “distance” d(V,W ) = I V

- W I + IW - %I .

Then, d(V, B) > 0 (because otherwise, % = 9?, and c-matching V is maximum). Let @ = % - 9? and 3 = 39 - V, and consider the hypergraph

417

TRANSVERSALS

If we let d(x) = c(x) - I (8n 9% 1, then 8 and 3 are two d-matchings of d with G? n = @ and I Z?J I > I 9 1. Moreover, there exists a B1 E (because I 3 I > I @ I 2 u). Form from B, a maximal alternating sequence a = (B, , C, ,B2,C,, ...) in hypergraph 8 relative to the d-matching g. Sequence u is even (otherwise, from the lemma, u would be an odd maximal alternating sequence in hypergraph H relative to V?). Thus, sequence u is of the form:

,

There exists a set B,, E 3 - (u n 3,because I 3 I > I 8 I. Furthermore, each such set B,,.l forms a d-matching ( B , , B2,..., B,, B,,,) of 8.Since sequence u is even and maximal, and since condition (iiii) for the maximality of an alternating sequence does not apply, it follows that condition (iii) holds. Thus,

-

af’= (3 - a) u (a - Z)

is a d-matchingof 8.Let

93’ = 9’u(9n %‘). Then,

I a; I < 4x1 + I (an

w x

I = 4x1 -

Therefore, 8‘is a c-matching of 8,and since I 9’ 1 = I 9 I, family 8‘ is a maximum c-matching. But then, d(%‘, 9’)< d(V, @, which contradicts the definition of a. Q.E.D. The deficiency of a c-matching V is defined to be the integer

If c(x) E 1, then S(%) equals the number of unsaturated vertices. If hypergraph H is uniform of rank h, then a maximum matching V is also a matching of minimum deficiency. The following theorem shows how the determination of a c-matching with zero deficiency can reduce to the determination of a matching with zero deficiency.

418

HYPERGRAPHS

H

Fig. 18.1

Theorem 3. Construct a hypergraph R corresponding to hypergraph H = ( X , 8)as follows: For each x E X , consider two disjoint sets

and let

Let the elements of

B be the vertices of E, and let each set of the form E,

=

{L J ~ / x E E ~ }

and each set of the form { a!, b:} be an edge of E (see Fig. 18.1). Then, to each matching g of R that saturates all of the b:, there corresponds in H a c-matching %, and vice versa Furthermore, to each matching g of B with zero deficiency, there corresponds in H a c-matching % with zero deficiency, and vice versa. 1. I f @ is a matching in R that saturates every b:, then Thus, V is a c-matching in H .

419

TRANSVERSALS

Conversely, if V is a c-matching in H, then V induces in that can easily be completed so that it saturates every b:.

a matching

2. Let A: denote the set of vertices in A, that are saturated by matching

0.Then, - 6(W)

=

2 (-

xex

C(X)

+I

wx

I + 18, I) -

c

xex

18,

I

Therefore, V is a c-matching with zero deficiency if, and only if, $7 is a matching of H with zero deficiency. Q.E.D. Given a non-negative integer d ( x ) for each x to be a d-coaering if

IF x I

4x)

E

A', a family 9 c 6 is said

( X E X ) .

If d ( x ) = 1 for each vertex x , then a d-covering is a covering in the usual sense. The following theorem shows that a minimum d-covering problem reduces to a maximum c-matching problem.

Theorem 4 (Ray-Chaudhuri [1960]). Consider a hypergraph 6 with vertex numbers c(x) and d ( x ) 2 0 such that

44 + 4 x 1

= I Q,

I

( X E X ) .

A family $9 of edges is a maximum c-matching i f , and only i f , the family 9 = 6 - $9 is a minimum d-covering.

< c(x) is equivalent to I FxI = I 8, I - I w, I 2 I 8, I -

Note that I V?, I

= 4x1

*

Thus, Y: is a c-matching if, and only if, 9 is a d-covering. Let V, be a maximum c-matching and let 9, be a minimum d-covering; then

Since V,, is maximum, the c-matching d - 9, is also maximum. Thus, the above inequality becomes an equality, and I 6 - V, I = FoI. Thus, d - V, is a minimum d-covering. Q.E.D.

I

420

HYPERGRAPHS

2. Transversal number

A transversal of a hypergraph H = ( X ;El, E,, set T c X such that

T n Ei# /zr

( i = 1, 2,

..., Em)is defined to be a

...,m ) .

The transversal number is defined to be the minimum number of vertices in a transversal; the transversal number of hypergraph H is denoted by t(H) = min] TI.

Many combinatorial problems involve the calculation of the transversal number. EXAMPLE 1. Determination of a maximum stable set in a graph. Let G = (A', E ) be a simple graph. From Chapter 13, we know that a set S c X is stable if, and only if, each edge of G has at least one of its endpoints in X - S, i.e., if, and only if, T = X - S is a transversal of graph G. Thus, the determination of a maximum stable set reduces to the determination of a minimum transversal, EXAMPLE 2. Determination of the minimum absorbant set in a 1-graph G = ( X , r).Consider a hypergraph H on X whose edges are

Ex = { X } u r ( x ) . Each absorbant set of G is a transversal of H, and vice versa. Hence the determination of a minimum absorbant set reduces to the determination of a minimum transversal. Dual to the minimum transversal problem is the minimum covering problem. For a hypergraph H = ( X , 8 ) a covering is a family F of edges whose union is X . The cardinality of a minimum covering of H is denoted by p(H). Since each transversal of H corresponds to a covering of the dual H * , and vice versa, it follows that p ( H ) = z(H*). EXAMPLE. Quine's problem (switching theory).

Consider the boolean variables q,z2,..., z, (that take values 0 and I), and a function rp(zl, z2,..., z,) defined on [0, I]", that takes only the values 0 and 1. A boolean sum,boolean product and boolean complement are defined as follows:

1

9 Fig. 19.4. Good 3-colouring of Kle

421

TRANSVERSALS

=;

if z=z’=O otherwise,

=O = 1

if if

z + z’( -

z=1 z=O.

Clearly, cp can be written as a polynomial in zl, z2,..., z,, F2, ..., 2,. Quine’s problem is to express cp as the sum of the smallest possible number of monomials. To show how this problem reduces to the minimum covering problem, consider a hypergraph H whose vertices are the 2m vertices of an m-dimensional hypercube and whose edges are the k-faces of this hypercube for k = 0, 1, 2, ..., m. For example, consider a function cp of 4 variables whose non-zero values are given in the “truth table” below.

0 1 1 1

1 0 1 1

1 1 0 0 1 1 1 1 1 0 1 1 0 1 0 1 1 1 0 0 0 0 0 0 0 0 0 1 0 0 1 1 _ _ ----- - -1 1 1 1 1 1 1 1 1 1 Let A be the 10 vertices of the hypercube that correspond to the non-zero values of function cp (circled in Fig. 18.2). The vertices .of HA can be covered 01 I 1

001

/

1111

101

’

b 0

II Fig. 18.2

’1 101

422

HYPERGRAPHS

by exactly 5 edges of H A (which,are hatched in Fig. 18.2). Edge { 0101,0100) corresponds to the monomial Z1z& that takes the value 1 only at the vertices covered by this edge. Similarly, the edges { 0111 }, { 1010, 1110, 1100, 1000 }, { 1000, 1100, 1101, 1001 }, { 1001, 1011 } correspond respectively to the monomials 2 1 ~ 2 . ~ 3 ~z,Z,, 4, zl&, ~ 1 2 2 ~ Hence, 4 . cp(z1,

z2,

z3,

z4)

= z1 z2 z3 z4

+ z,

z2.3

+ z1 22 z4 +

z1 z 4

+ z, . z1

Before introducing a method to find the minimum transversals of a hypergraph H or, more generally, of a family of sets, some additional definitions are needed. Let &’ = (Ai / i E I) and &7 = (B, / j E J ) be two sets of subsets of X = { x 1 , x 2 ,. . . , x,I; let

d v 93 = ( A , u B, / ( i , j )E Z x J ) d A 93 = ( A f n B j / ( i , j ) E Z x J ) . Let 2 ( t h e sieve of d ) denote the set of all the transversals of a?. Let C1d denote the set of all sets that contain at least one At E d. Let Min d denote the set of all minimal sets of d . Finally, let Tr G? denote the set of all the minimal sets of sieve 2:It follows that Tr d = M i n ( J ) ,

8’=C I T r d . Proposition 1. - w e hare

..

2’=CIi’= (CId).

Proposition 2. - We have C1 d = C1 Min d , Proposition 3. - w e haue (d G a>= 2 n 2 .

Proposition 4. - We have Min (C1 d n C1 a) = Min (d v

a).

Clearly, CEMin(CldnC193)

=-

==

CIA C3B

CDAUB

forsome A E for some B e 93

=.

Furthermore, D = A’uB’ E C J d n C l B .

~

C x D ~ M i n ( dv

a).

423

TRANSVERSALS

Thus, D contains a C' E Min (Cl d n C l q . Since C minimal, it follows that

3

D

3

C' and C is

C = C ' = D E M i n ( d v W). Hence, Min (C1 d n C1 a)c Min (d v g). Since the inclusion is satisfied in the opposite direction, the equality follows. Q.E.D. Proposition 5. We have: Tr (d u

a)= Min (Tr JXZ v

Tr a ) .

By successively applying Propositions 3, 1, 2, 4, we have: T r ( d u W ) = Min(;n%)= = Min (Cl

Min(Cl2nClG)

Tr d n C1 Tr W)

= Min (Tr d v Tr &?)

.

Q.E.D.

Algorithm to construct Tr d . First, determine M i n d = { A l , A 2 , ...,A k ) . Next, successively determine the following families : d91={A1)

-b

d,= d,u { A z ]

+

Tr{A, 1 = ({.>/..A,) T r d , = M i n ( T r d , v Tr{ A , } ) Tr d 3= Min (Tr d, v Tr { A , }), etc.

+ d,= d,u { A 3 } Proposition 5 shows how Tr df+ is obtained from Tr d i .If Min &' has k members, then the algorithm constructs Tr d = Tr dkin k steps. Maghout [1959] has expressed this algorithm in terms of Boolean operations. This algorithm calculates all minimal transversals and, therefore, all minimum transversal sets. If the size of the problem is too large, or if only one minimum transversal set is required, or if only a transversal set of cardinality less than some fixed number is required, then variations can be employed. The reader is referred to Roy [1970] and Lawler [I9661 for the details. Consider a hypergraph H with a transversal number .r(H). If v(H) denotes the maximum cardinality of a matching in H , then some simple relations exist between v(H) and r ( H ) . First, note that if L ( H ) is the representative graph of H , then v(H) = m - r ( L ( H ) ) .

424

HYPERGRAPHS

Thus, the maximum matching problem reduces to a minimum transversal problem. Theorem 5. Let H be a hypergraph; then V(H)

< T(H)

+

Clearly, if bois a matching of H and if T is a transversal of H, then

IT1 B

1 ~ 0 1 .

Hence, min I 7'I 2 max I 8oI . T

60

Q.E.D.

meorem 6. Let H be a hypergraph with rank r ( X ) = h, then T ( H ) < hv(H). Let d = (Et / i E I ) ; let b, = (Et / i E J ) c b be a maximum matching. Then, I J I = v(H), and UtElEt is a transversal. Therefore,

Q.E.D.

A hypergraph H

=

(X; El, E,, ..., Em)is defined to be r-critiial if

z(H - Ei) < T ( H ) (i = 1, 2, ..., m). Let f ( h , k ) denote the maximum possible number of edges in a z-critical hypergraph of rank h with z(H) = k + 1. If H is a graph, then H is 2critical with r ( H ) = k + 1 if, and only if, H is a-critical with a(@ = n k - 1, (see Chapter 13), since the complement of a maximum stable set is a minimum transversal, and vice versa. Theorem (10, Ch. 13), can be ex- * pressed as f(2,k)=

2+k

(7) *

Erdos and Gallai [1961] conjectured that for each h,f(h, k) = The following theorem presents a proof for this conjecture. Theorem 7 (Jaeger, Payan [1971]). I;fH = (X,8)i s a .r-criticuf hypergraph I , then the number of edges in H is smaller with rank h and with z ( H ) = k

+

than or equal to

(h

L

425

TRANSVERSALS

.Equality holds for a hypergraph ( X , 8)with 1 X

1

=

h

+ k and d = Ph(X).

1. We may assume that H is uniform of rank h ; otherwise, by adding h - I Ei I auxiliary vertices to each edge Ei with I E, I < h, we obtain such a hypergraph.

2. Let H = ( X ; E l , E,, ..., Em). By hypothesis, H - Ei contains a transversal of cardinality k . Let Fi be such a transversal. Since ? ( H ) = k + 1, set F, cannot be a transversal of H, but F, n E, # fz! for a l l j # i ; it follows that F , n E i = @. Let 2 be the set of ordered pairs ( A , B ) of subsets of X such that IAI=h,

IBJ=k,

A n B = @ .

Consider the graph G whose vertex set is Z and with vertices ( A , B ) and (A’, B’) joined together if A n B’ = fz! or A’ n B = @. There exists a subset Y of X with cardinality h + k (because I XI 2 I El U Fl I = h + k), and for such a set Y, let S y = { ( A , B ) / ( A , B ) E Z ,A U B =

Y}.

3. We shall show that Sy is a stable set in G. Consider two distinct vertices z = ( A , B) and z’ = (A’, B’) in Sy; we have AVB A nB

= A ’ U B’ = = 2’n B’ =

Y,

0.

Consequently, A n B’ # @ and A‘ n B # @, which implies that z and

z‘ are non-adjacent in G. Thus, Syis a stable set in G. 4. We shall show that Sy is a maximum stable set in G. Let p be a complete ordering of X,and let C(p) denote the set of vertices ( A , B) E Z such that u p b for all u s A and all b E B . C(p) is a clique in G, because if ( A , B) and (A’, B’) are two distinct vertices of C(p), and if A n B‘ # 0, a vertex c in A n B’ satisfies a ‘ p c for each a’ E A’ and c p b for each b E B ; thus A’ n B = 0,and vertices ( A , B ) and (A’, B’) are adjacent in G. It is easy to see that for each Y and each p,

C(P)n SY # 0 Moreover, for all z E Z , the number of cliques C(p) that contain z has a constant value d 2 1. Therefore, the total number t of distinct cliques C(p) satisfies dlS,I = t .

426

HYPERGRAPHS

Since d I S I 6 t for each stable set S, it follows that Sy is a maximum stable set. 5. The vertices ( E l ,Fl), (E2, F2),..., (Em, F,) constitute a stable set in G, and consequently, a(G) 2 m. Since N(G)= I S,

I=

(

h

Q.E.D. EXERCISES 1. Let H = (X,8)be a uniform hypergraph of rank h. Let 8, be a maximum matching in H. Let A(&,) denote the set of unsaturated vertices in matching 8,.For S c X,let H(s, denote the family of the E, E 8 such that El n S = 0 , and let p h ( H )denote the number of connected components of hypergraph H that have cardinality 1 modulo h. Show that

(N. Sauer [1969] has given an additional condition that implies equality.) 2. Let X be a set of points on a line. Let E l , E a , ..., Em be subsets of X such that each El is the union o f t disjoint intervals. If the intersection of any t 1 of the E, is nonempty, show that there exists a transversal of the El with cardinality t.

+

3. A. GyArfhs and J. Lehel have made the following conjecture: "Let G be a tree, and let A1, A z , ..., A, be sets of the vertices of G such that for each i, the subgraph GAl is a forest with t trees. If the intersection of any t 1 of the A, is non-empty, then there exists a transversal of the A, with cardinality t." Show that this conjecture is true for t < 3.

+

4. Let G be a tree, and let A l , A z , ..., A, be sets of vertices of G such that for each i, G.c,

is a bi-tree (a forest with two trees). Show that if the intersection of each pair of A, is nonempty, then there exists a transversal of the A, with cardinality 3. (GyBrfAs, Lehel [1970]) 5. Consider a set X (possibly infinite) and two families d and I of non-empty subsets of X. If d and 9 are hereditary families (i.e. d .= CI d and I = C1 I), show that the following properties hold :

427

TRANSVERSALS

6. If d is a hereditary family that is closed with respect to intersection (i.e. A, B E d implies A n B E d), and if d does not contain the empty set, then family d is called a filter. For example, the family = C1 I xu1 is a filter, and the family of neighbourhoods

of x o in a topological space is a filter. Show that if d is a filter, then d c d. 7. If a? is a filter, then each family I with C1 I = d is called a basis of filter d.Let d < 93, if for each B E I,there exists an A E d with A c B. Show that if I is the basis of a filter, and if d is a family such that d < I and I < .rB, then a? is also a basis of a filter. 8. Using Exercise 7, show that if 9 is the basis of a filter, then the following conditions are equivalent: (1) for each set A c X , either I < { A } or D < { X - A }, (2) I < $and < I, (3) I'< 93 implies I < 9'. (A basis of filter I with this property is called an ultra-filter basis.)

9. If X i s the edge set of the complete bipartite graph K,.,, and if B is the family of sets of edges of KP,*that form a graph K,,,, then show that H = (X,8)is a hypergraph

with

(L. W. Beineke [19701) 10. If X is the set of edges of the complete graph K,, ,and if 8 is the set of cycles of K,,, then show that H = (X, 6') is a hypergraph with

(Chartrand, Geller, Hedetniemi [1970]; Guy [1967]) 11. Repeat Exercise 10 with the complete bipartite graph K p . qin place of K,,; show that

V(H)=

[;I

"(H)=

[n] [:]

[i]

if

+

[

pq is even,

if

4

pq is odd.

(Chartrand, Geller, Hedetniemi [1970])

12. Show that if H i s a hypergraph formed from the maximal cliques of a simple graph

G, then T ( H )equals the dominance number /3*(G).

CHAPTER 19

Chromatic Number of a Hypergraph

1. Stability number and chromatic number of a hypergraph Given a hypergraph H = (X,b), a set S c X is defined to be stable if it contains no edge Eiwith I EiI > 1. The stability number of H is defined to be the maximum cardinality of a stable set of H. The chromatic number x ( H ) is defined to be the smallest number of colours needed to colour the vertices of H such that no edge Ei of H with 1 EiI > 1 has all its vertices with the same colour. This concept was introduced by Erdos and Hajnal [1966]. A q-colouring is defined to be a partition of X into q stable sets

s1, s,

*-*,

s q

9

each corresponding to a colour. A hypergraph for which there exists a qcolouring is said to be q-colourable. ci

Fig. 19.1 Projective plane with 7 points

EXAMPLE. The projective plane with 7 vertices forms a uniform hypergraph H of rank 3 with 7 edges aef, adg, abc, fdb, fgc, ced, beg. See Fig. 19.1. Set { a, c, e, g } is a maximum stable set. Its chromatic number is x ( H ) = 3, and a 3-colouring of H is shown in Fig. 19.1. 428

429

CHROMATIC NUMBER OF A HYPERGRAPH

Proposition 1 (Erdos, Hajnal, [1966]). I f H is a uniform hypergraph of rank 3 with n vertices that satisfy

IEinEII < 1

(i#j),

then each maximal stable set S satisfies

Let S be a maximal stable set with s vertices. Since S is maximal, each X - S belongs to an edge Ex such that Ex - { x } c S. From the hypothesis, xE

IE,nE,.I G 1

( x , y ~ X S - ;x f y ) .

Thus, the sets Ex n Shave cardinality 2 and are pairwise different; hence

It is easily seen that this1 implies s2

[JG].

Q.E.D. Proposition 2. Let H be a hypergraph of order n with chromatic nitmber x ( H ) and with stability number ,B(H); then X(H)B(H) 2 n 9 x(H) + B ( H ) G + 1* Consider a q-colouring (Sl, S,, ..., S,) of H with q Then,

c I Si I G 4BW)

=

x(H) colours.

4

n =

i= 1

= X(H)B(H) Y

and the first inequality follows. Let S be a maximum stable set of H. Colour the vertices of S with a first colour, and colour each of the remaining vertices with n - B(H) additional colours. Hence x ( H ) G [. -P(H)] and the second inequality follows.

+1 Q.E.D.

Given a hypergrap,h H = (X, &), the degree of a vertex x is defined to be the maximum number of edges different from { x } that form a partial family (E,/ j E J ) with EinE, = { x }

( i , j ~ J i;# j ) .

430

HYPERGRAPHS

Let the degree of x be denoted by d,(x). Clearly, d,(x) = 0 if, and only if, the only edge that contains x is { x } . The following theorem shows that the relationship between the chromatic number of a hypergraph and its degrees is analogous to the relationship between the chromatic number of a graph and its degrees:

Theorem 1 (Tornescu [ 19683). Let (S,, S,, .. ., S,) be q-colouring of H , and let dk = max d,(x) . X€sk

Then, x(H)

< max min { k, dk + 1 } . kbq

1. We shall show that there exists an r-colouring (Si,Sh, ..., S:) of H with I

min ( k,r )

( S; is a maximal stable set in X -

U Sir . i < k

Clearly, if S, is not a maximal stable set in X , then vertices can be added to S, until a maximal stable set S, is obtained. If S, - S; is not a maximal stable set in X - S ; , then vertices can be added to it until a maximal stable set 5'; is formed, etc. This gives the required colouring of H. 2. Let x $ UlklS!, and l e t j < k ; then from the maximality of S; there exists an edge E; with

Since x E E;, for all j

< k , the family ( E ; , E i , ...,Ell.) satisfies E;nE;={x}

(i#j;i,j
From the definition of dH(x), this implies that

>k.

~H(x)

-

3. Let i(x)-denote the index of the set Si that contains x . From part 2, i(x) 2

By letting k

=

i(x)

k

+1

- 1, it follows that i(x) < d&)

+1

dH(x)> k . ( x E X) .

431

CHROMATIC NUMBER OF A HYPERGRAPH

< k (from part l), and i(a) < max i ( x ) < max (d&) + 1)

Let a E Sk,then i(a)

d, + 1 ,

I

XSsk

XSsk

Hence, x(H)

< max i(x) < max min { k, dk + 1 } , k XPX

Q.E.D. Corollary 1. I f H is a hypergraph with x(H) = q + 1, and fi the hypergraph .‘to has chromatic number x(Ho) = q, then

Ho obtained by removing vertex d&O) 2 4.

S2, ..., S,) be a q-colouring of Ho = Hx-( x,, ). If dH(xo)< q, then Let (S1, by considering the (q I)-colouring ( S , , S,,..., S,, { xo I), it follows that

+

x ( H ) < max min { k , dk k < q + l

which contradicts x ( H ) = q

+1}

Q q,

+ 1. Q.E.D.

Corollary 2. Ifg is a positive integer such that

I { x / x E X , dH(x) 2 4 ) I < 4 , then

x(H)

< 4.

Index the vertices such that

d&i) 2 ddxz) 2 d&)

2

**.

Z d,(xn).

Then, for k > q,

+ 1) < q .

min{k,dH(xk)

Clearly, this inequality also holds for k < q. For the n-colouring ({ x1 }, { x, }, ..., { x,, }), Theorem 1 yields

x(H)

< max min { k, df,(xk) + 1 ) Q q . k4n

Q.E.D. Corollary 3. If H is a hypergraph of maximum degree do, then ~

. x ( H ) < do

-I- 1 .

The proof follows from Corollary 2 with q

=

dl

+ 1.

APPLICATION. The following theorem is due to T. S. Motzkin [1968]:

432

HYPERGRAPHS

For a simple graph G with maximum degree h, it is possible to colour the vertices with

[;] + 1 colows such that no cycle has only one colour.

To show this, let H = (X,8) be the hypergraph in which X i s the set of vertices in G that belong to at least one cycle and in which d is the family of the sets of vertices that are contained in the same cycle. h The maximum degree of hypergraph H is do < 2 and the Motzkin

[

1,

theorem follows from Corollary 3. We can also state: The edges of a simple graph G can be coloured with q colours such that each cycle has more than one colour when q = max min

{ &(a), d&)

1.

C4,bI E E

To show this result, consider the hypergraph H whose vertices are the edges of G that belong to at least one cycle. The degree in H of an edge ab of G equals

dH(ab)< min{ dG(a)- 1, dG(b)- 1 ) Q q

- 1,

and the result follows from Corollary 3.

2. Cliques of a hypergraph Let H = (X, 8)be a hypergraph of rank h, and let r < h. A set A c X is defined to be a clique of rank r if either I A I < r, or I A I 2 r and each subset of A with cardinality r is contained in at least one edge of H . Clearly, each subset of a clique of rank r is also a clique of rank r. Let w,(H) denote the maximum cardinality of a clique of rank r in H. Since r 6 h, and since an edge E1 with 1 Ei = h is a clique of rank r, then

I

w,(H)

> I Ei 1 = r ( H ) .

If the vertex set X of hypergraph H i s a clique of rank r, then hypergraph H i s said to be r-complete. If H is a uniform hypergraph of rank h, then a “clique of rank h” is often called a clique, and each set with less than h vertices is called a triaial clique. Remark 1. Sauer [1971] has proved the following generalization of the Turan theorem:

CHROMATIC NUMBER OF A HYPERGRAPH

Consider a set X with I X classes A , with

Let H,,, that

I = n and apartition (A,, A 2 , ..., A,)

433

of X into p

,= (X,8)be a hypergraph in which & is the set of subsets E of X such (El=h (,EnAil 6 1

(idp).

Then, H,,, is the only simple uniform hypergraph with n(H) = n, r(H) = h, o,(H) < p , and with the maximum number of edges. Remark 2. Let m(n, h, r ) denote the smallest possible number of edges in a uniform hypergraph of rank h and order n that is r-complete. If E l , E2, ..., Em are the edges of such a hypergraph, then

ij

l=l

Y,(EO

=

Y,(W

Y

and, hence,

Therefore,

Furthermore, equality holds if, and only if, there exists a uniform hypergraph of rank h and order n such that each subset of X with cardinality r is contained in exactly one edge of H. The hypergraphs with these properties are studied in finite geometries and are called Steiner systems.

Remark 3. Chvdtal [I9711 has shown how the numbers m(n,h, r ) are related to the Turdn numbers. Let n, k , b be integers such that 1 < k < b < n, and let T(n, k,6 ) denote the smallest value of t n such that there exists a uniform hypergraph of rank k and order n with m edges whose stability number is smaller than b. The Turdn theorem (Ch. 13, $2) determines the value of T(n, 2,b). The number T(n, k , b) is called the Turbn number in n, k , 6. If a hypergraph H = (Et / i E Z) has a stability number smaller than b, then each set B with 1 B I = b is non-stable and contains some edge E i . Thus,

434

HYPERGRAPHS

an (n - b)-set X - B can always be covered by X - E, for some i, and hence, the hypergraph ( X - Ei / i E I ) of rank n - k is (n - b)-complete. Therefore, T(n, k , b) = m(n, n - k , n - b) . From Remark 2, it follows that

T(n, k, b) 2

(

b)

(n --b

=

-

n! (n - b) !(b - k ) ! (n - k ) ! b !(n - b) !

n! k!(b-k). q)(b)-l. k !(n - k ) ! b! k k

This inequality, which was also discovered by Katona, Nemetz and Simonovits, yields a bound for the stability number P(H) of a hypergraph H with n vertices and m edges such that minter I Et [ = k. If H is uniform of rank k, then

and, therefore,

If H is not uniform, this inequality is satisfied a fortiori.“)

Theorem 2. Let H be a simple uniform hypergraph of rank h, and let K he a clique with k vertices, k 2 h. Then each vertex of K has degree

and the chromatic number of the subhypergraph generated by K is

,

+

x ( H R ) = do 1. 1. The degree of a vertex x E K is the maximum number of pairwise disjoint subsets of K with h - 1 elements that do not contain x. Thus,

For asummaryof previous results for T(n, k, b), see V. ChvatalI19711. More recently J. Spencer [1972] found an improved lower bound for T(n, k, b) for large values of n. He proved that if n l b k ( k - l ) - l , then n” k - ’ . T(n, k , b) 2

43 5

CHROMATIC NUMBER OF A HYPERGRAPH

2. A minimum q-colouring of Hkis obtained from a partition ( S ,7

s, ..*,S,) 7

such that ISiI = h - 1 IS,l G h - 1

(i= 1,2,...,q-

1).

Therefore,

From that part 1, it follows that

k

* h-1

=do

Furthermore, from Corollary 3 to Theorem 1, x(HJ X(HK) = d,

+1.

< do + 1. Hence,

+ 1. Q.E.D.

Remark I . Theorem 2 shows that the bound given by Corollary 3 to Theorem 1 is the best possible: If H is a uniform hypergraph of rank h and maximum degree do, then x ( H ) < do 1, and equality holds for each H that contains a clique whose vertices have degree do. However, x ( H ) = do 1 does not imply the existence of such a clique, as shown by the hypergraph H in Fig. 19.2 due to L. Lovrisz. The edges of H are abc,, abc,, abc,, c, c2 c,, acb,, acb,, acb, , b, b, b, , bca, , bca, , bca, , a, a, a, .

+

+

Fig. 19.2

436

HYPERGRAPHS

Hypergraph H is uniform of rank 3 and contains only vertices with degree do = 2. The chromatic ndmber of H is 3 = do f 1. However, H contains no cliques of degree do = 2. The following result due to Lov6sz [1965] generalizes Theorem (6, Ch. 15). If H is a connected uniform hypergraph of rank h > 1 with maximum degree do > 1 , and i f I U E EEX I < do(h - 1) 1 f o r each x, then x ( H ) = do 1 holds only in the two following cases: (1) h = do = 2, and H is an odd cycle, or (2) H contains a clique of degree do.

+

+

Remark 2. If H is a uniform hypergraph of rank h with a maximum clique Kof cardinality k,then from Theorem 2, x ( H )is bounded below by a function of h and k. However, x ( H ) is not bounded above by a function of h and k. In fact, Erdos and Hajnal [1966] have shown that for each q > 1, there exists a uniform hypergraph H of rank h with x ( H ) = q that contains no cliques of order > h. More precisely, If h, q, lare integers > 1, there exists a uniform hypergraph H(h, q, 1) of rank h with chromatic number q and without cycles of length < 1. A constructive proof has been given by Lov6sz [1968]. The following is a well known result:

Theorem 3 (Ramsey [1930]). Consider three integers p, q, h with p , q 2 h. Thete exists a$nite integer R,(p, q ) such that each uniform hypergraph of rank h with n vertices, n 2 R,(p, q), contains a clique with p vertices or a stable set with q oertices. The “Rarnsey number of rank h in p and q” is dejned to be the smallest number R,(p, q) with this property. We shall show by induction on h that if I X 1 is sufficiently large, then for any partition of y h ( X ) into two cIasses 6 and S,there exists a set A c X with 1 A I = p , Y k ( A ) c 8,or there exists a set B c X with

I B I = 4 , g,,(B) c 9 .

1 . Clearly, the theorem is true for h = 1 , because if 1 X I = p + q - 1 , and if X is partitioned into two classes, with no p elements of X in the first class, then there exist at least q elements of X in the second class. Therefore, (1)

RdP, 4) = P

+4- 1

*

We shall assume that the theorem is true for ranks less than h, and we shall show that it is true for the rank h > 1. 2. Clearly, the theorem is true for p = h because if I X 1 >, q and if X contains no h-set in 8,then each h-set belongs to 9. Therefore, (2) &(h, 4) = 4 *

437

CHROMATIC NUMBER OF A HYPERGRAPH

By the same argument, the theorem is true for q = h, and (2‘)

h)

Rh@,

p

*

+ q = 2 h, we shall prove it by induction

Since the theorem is true for p o n p q.

+

3. Let p, q be two integers withp

+ q > 2 h ; we shall show that

Rhb, q) Q Rh-l[Rh(P -

(3)

=F

Y

41, R h b , 4 - l)]

+

*

Let Pf =

Rh(P

- 1, 41,

q’ = Rh(P,

- 1)

7

and consider a set X with cardinality n > Rh- l(p‘, 4’) + 1. It suffices to show that for any partition (8,P) of Y,,(X), there exists either a set A with [ A I = p and Ph(A) t 8, or a set B with 1 B I = q and Ph(B) c 9. Let a E X,and let X‘ = X - { a }. For E‘ E g h - I ( x ‘ ) , let

1-’;: Since 1 X ’

I

if if

E’ u { a } E 6“ E‘u ( a } E F .

2 Rh-l(p’,q’), there exists (for example) a set A‘

I A’ I = p‘, Since [ A’ I = p‘

=

gh-l(A‘)

t

t

X’with

8’.

R,(p - 1, q), there exists either a set B” c A‘ with

I B” I = 4

ph(p) C F (in which case the above proposition is true), or a set A” c A‘ with 9

I A” I = p -

1,

Ph(A”) c 8 .

In this case, set A” u { a } satisfies

I

IA”u { a } =p,

’ g h ( A ” u { a } ) c 6“

9

and the above proposition is true. Q.E.D.

Corollary. Let p , q

> 2, and let G = (X,E ) be a simple graph of order p + 4 * .( p - 1 *)-

Then there exists in G either a clique of cardinality p or a stable set of cardinality 4.

438

HYPERGRAPHS

From Theorem 3, it suffices to show that

This is true for p = 2, since from (2)

(;)=

4 = R,(2,4)

*

Suppose that the inequality is valid for each R,(p’, q’) withp’ then from (1) and (3), it follows that

u p , 4) < R,(P

- 1,q) + M

4

P - 1+4 p - 2

P ,4

-’)+(P+;-

+ q’


+ q;

- 1) - 1 + 1 - 2 ) = ( P +p -4 ; 2 ) , - 1

Q.E.D. Rz(p,q)

P =2

p = 4

P =3

p=6

p = 5

...

a

p + q = 4

2

1

p + q = 5

3

3

p + q = 6

4

6

1

4

1

~

p + q = 7

5

9

9

5

1

p + q = 8

6

14

18

14

6

(P

+ 4 - 2) P - 1

p = 2

I

p = 3

I

p = 4

p + q = 4 p + q = 5

p + q = 6

p = 6

p = 5

... 1..

I

3 4

I ~

3 6

1 4 -

p+q=7

5

10

lo

p+q=8

6

15

20

5l

l

’

I

6

l5

...

439

CHROMATIC NUMBER OF A HYPERGRAPH

Remark. So far, the numbers R2(p,q ) have been determined only for very small values of p and q. These numbers are almost equal to the binomial numbers

(” P - 1- )’ +

,(see the tables above).

Very little is known about the Ramsey numbers. Sobczyk [1968], Kery and Kalbfleisch have also shown that R2(3,6 ) = 1 8 ; Graver has shown that R2(3, 7) = 23. Moreover, Erdos [1947] has proved that for p 2 3,

Rz(P9P) 2

%@.

Frasnay [1963] has proved that:

R&, p )

< 2ac”-2)+1 .

Giraud [I9681 has proved that for p 2 5,

APPLICATION (Hedrlin [1966]): Let G . H be the normal product of two simple graphs G = ( X , E ) and H = ( Y , F), then a(G. H ) G R,[a(G) 4- 1, a ( H )

+ 11 - 1 .

We have shown (Proposition 9, $6, Ch. 16) that

.

a(G N ) 2 a(G) a ( H )

.

We shall now describe an upper bound for cr(G.H);let

k = R,[a(G)

+ 1, a ( H ) + 13.

.

.

Suppose that a(G H ) 2 k . Then there exists in G H a stable set S c X x Y with I S I = k. Let xy, x‘y‘ E S. If x # x’ and [x, x‘] $ E, join xy and x’y’ by a red arc; otherwise, y # y‘ and [ y ,y ‘ ] $ F, and join xy and x’y’ by a blue arc. Since I S I = k , there exists in the red graph a clique C with [ C I = a(G) I, or there exists in the blue graph a clique D with I D I = a ( H ) 1. In the first case, the set

+

+

x,xy E C } is a stable set of graph G with a(G) + 1 elements, which is a contradiction. In the second case, we obtain the same contradiction with graph H . Hence, a(G . H ) < k. {x /x

E

Q.E.D.

440

HYPERGRAPHS

This result, applied to the Shannon Problem (Ch. 16, §6), shows that the graph G in Fig. 16.10, with stability number a(G) = 2, satisfies: a(G2) < R2(3,3)

- 1=5.

3. Good colourings of the edges of a graph

Consider a simple graph G = (X,E). A good colouring of the edges of G is defined to be a mapping g from E into a given set of colours such that no three edges of a triangle have the same colour. If the set { g(e) / e E E } has cardinality q, then g is called a good q-colouring. Consider a hypergraph whose vertices are the edges of graph G, and whose edges are the triples of edges of G that form a triangle. Let this hypergraph be called the triangle hypergraph of graph G and be denoted by G T. The smallest value q for which the edges of G have a good q-colouring equals the chromatic number x(GT). Let n(q) denote the maximum order of a complete graph that has a good qcolouring. The number n(q) is related to the Ramsey theorem. In fact, the Ramsey numbers R,(p, q ) defined in Section 2 can be generalized to Ramsey numbers Rh(pl,p 2 , ..., p,) of q variables with the following property-:

If I X I 2 R h ( p l , p z ,. . . , p a ) ,for each partition (B1, g2, ..., a,,) of the set Pn(X)of all subsets of X with cardinality h, there exist one class Bi and a set A , c X with

I 4 I = Pi ;

%(A3

= gi.

Thus, it follows that n(q) = &(3,3,

..., 3) - 1 . 4

Proposition 1. Ifn(q) is the maximum order of a clique with a good colouring of its edges, then

Let K be a clique of order n(q), and consider a good q-colouring of its edges. Let a be a vertex of K, and let A , denote the set of vertices linked to a by an edge with the i-th colour. The clique KA,has its edges coloured with a good ( q - 1)-colouring, and therefore

I A , I < n(q - 1).

441

CHROMATIC NUMBER OF A HYPERGRAPH

Hence

Clearly, n(1)

=

2, and we have:

<1 < 2 n(1)

+ 1.

n(q - 1) < (q - 1) n(q - 2) n(q) < q n ( q - 1)

+1 + 1.

n(l) n(2)

.............................

+1

Hence, n(q)

< 1 + q + q(q - 1) + ... + q ! + q ! =

c 4k '!

-L

k=o

Q.E.D.

For q

=

1, 2, 3, the number n(q) is equal to 4

1

For q = 1, we havef(1) = 2, and since K2 has a good 1-colouring, n(1) = 2 . For q = 2, we have f ( 2 ) = 5, and since K5 has a good 2-colouring (shown in Fig. 19.3), n(2) = 5. For q = 3, we havef(3) = 16, and K,, has exactly two non-isomorphic good 3-colourings respectively discovered by Greenwood and Gleason [1955] and by Kalbfleisch and Stanton (19681 (see Fig. 19.4); thus, n(3) = 16. For q = 4, we havef(4) = 65, but so far no good 4-colouring has been found for &5 (see Graham [1968]); only one good 4-colouring has been found for Colour 1 Colour 2 ---

Fig. 19.3. Good 2-colouring of Ka

442

HYPERGRAPHS

Proposition 2. Let q be an integer, and let G be a graph with chromatic number y(G) < n(q); then the edges of G have a good q-colouring.

Consider a colouring g : A'-+ { 1, 2, ..., n ( q ) } of the vertices of G in n(q) colours. Denote the vertices o f K,,,,, by 1, 2, ..., n(q) and consider a good qcolouring of the edges of the clique K,,,,,. If edge [x, y ] of G is coloured with the same colour as edge [g(x),g ( y ) ]in K,,(,), then no triangle of G has all its edges with the same colour. Q.E.D. Proposition 3. Let G be a simple graph, and let a, be a vertex of G. I f H is the graph obtained from G by adding a vertex a, that is joined to each vertex of rc(al), then

X ( m = X(G7

*

1. Let q = x(GT). Consider a good q-colouring of the edges of G, and colour each new edge [a,, x ] with the same colour as edge [al, x ] in G. The resulting colouring is a good q-colouring of H. Hence,

X(HT) G x(G'>

*

2. If q = x(H*), then each good q-colouring of the edges of H determines a good q-colouring of the edges of G. Consequently,

and equality holds.

Q.E.D. Remark. A simple graph G

=

x(GT>= q

(1)

(2)

( X , E ) is defined to be q-minimal if

X((G

+1

- elT>= 4

(eEE).

+

Clearly, each graph G with a good (q 1)-colouring has a partial subgraph that is q-minimal. We shall show that K,,,)., is q-nzinimal. Let n(q) denote the maximum number of vertices of a complete graph with a good q-colouring. Then, the graph G obtained by removing an edge [a,, a,] from K,,,,,., also contains a good q-colouring from Proposition 3. Since K,,,,., has no good q-colouring, it follows that conditions (1) and (2) for q-minimality are satisfied. Other q-minimal graphs have been discovered. Las Vergnas [I9691 has shown that if q 2 2 and n(q) = f ( q ) , then the graphs Kn(q)+ 1

and

Kn(q)+3

- C,

443

CHROMATIC NUMBER OF A HYPERGRAPH

+

(the clique with n(q) 3 vertices less a cycle of length 5) are the only qminimal graphs with less than n(q) + 4 vertices. 4. Generalizations of the chromatic number of a graph

Consider a multigraph G = (X, E ) . Ap-colouring ofdegree t is defined to be a partition ( X I , X,, ..., X,) of X into p classes such that none of the subgraphs Gx, has maximum degree greater than t , i.e., maxdGXl(x)
( i = l , 2 ,.,.,p ) .

Let y,(G) denote the smallest integer p for which G has a p-colouring of degree t. The chromatic number of G is simply y,(H), since the vertices of the same colour determine a subgraph of maximum degree 0. The study of y,(G) reduces to the study of the chromatic number of a hypergraph H = ( X , d)whose vertex set is X and whose edges are the sets A c X such that GA has maximum degree greater than t. A stable set S of hypergraph H = ( X , 4 generates a subgraph Gs of maximum degree G t ; thus

x(H) = YAG). Theorem 4 (Lovisz [1966]). I f G = ( X , E ) is a multigraph without loops and with maximum degree h = max,,, dG(x),and if h , , I t , , .. ., h, are non-negative integers such that h, + h2 + ...+ h, = h - p + 1, then there exists u partition ( X I , X,, ..., X,) of X with maxdGX,(x)
( i = l , 2 ,..., p ) .

We shall prove this result by induction on p. 1. Suppose p = 2. Since h - 1 = h, + h, 2 0, it follows that h 2 1. For h = 1, the theorem is evident. Therefore, we shall assume that h 2 2. Consider a partition (XI, X,) of X that minimizes

We shall show that the maximum degree of G,, is < h,. Let x E X,.The partition ( Yl , Y,), where Yl = Xl - { x ), Y2 = X2 u { x ), satisfies

Since ( ( X I , X,) is minimum, this expression is 2 0. Hence, 2 h z ~ G ( x , X-~2) hl m G ( x , X 2 )6 hz m ~ ( xXI) , m,(x ,X , ) = dG(x) 6 h

+

.

- hl

444

HYPERCRAPHS

It follows that

2(h

- 1) m,(x,

< hz - h, + 2 hhl

X,)

= (h - 1)

+ 2 hl(h - 1).

Since h - 1 > 0, this inequality implies that

A similar argument shows that dGx,(X)

G

h2

(x

x2)

8

Hence, the theorem is true for p = 2. 2. Let p > 2, and suppose that the theorem is true for all p' < p . Let

h; = hl , h; =

h2

+ h3 + *..+ h, + p - 2 .

Thus,

From Part 1, there exists a partition ( Y1,Y,) of X that satisfies

max dGU,(x)< h; . x E Y2

Since

c hi=&-@-

1)+ 1 ,

1 3 2

Yz can be partitioned in ( p

- 1) sets X,, X,, ..., X, with max dGx,(X) G h i , XCXI

by the induction hypothesis. Thus (Yl,X,, X,, ..., X,) is the required partition. Q.E.D.

445

CHROMATIC NUMBER OF A HYPERGRAPH

Corollary. Let G be a multigraph without loops and with maximum degree h;

then there exists a colouring of degree t in

Suppose hi satisfies:

=

[ L] + 1 colours. t + l

t for all i. Then, there exists a p-colouring of degree t if p

p t 2 h - p + l ,

or p(t

+ 1) 2 h + 1 ,

or h f l

P>- t + l ‘ Thus there exists a colouring of degree f i n

[=]h + 1 colours.

I t +

11

*

colours, i.e.

Q.E.D. For t = 0, this corollary reduces to Corollary 2 to Theorem (5, Ch. 15). For t = 1, this result was proved by Gerencstr [1965].

*

*

*

A second generalization of the chromatic number of a graph has been studied by G . Chartrand, D. P. Geller, S. Hedetniemi [1968]. Let f k ) ( G ) denote the smallest number of colours needed to colour the vertices of a simple graph G such that no elementary chain with length k is monocoloured For k = 1, then $l)(G) is the chromatic number y(G). For k = 2, then, from the preceding corollary,

No‘e that f k ) ( G )is also the chromatic number of a uniform hypergraph of rank k 1, whose edges are the sets of vertices that lie on an elementary chain of length k in G. Chartrand, Geller and Hedetniemi [1968] have shown that if G is a simpIe graph, and i f the Iength of the longest elementary chain in G is I, 2 < I < k, then

+

446

HYPERGRAPHS

A third generalization of the chromatic number of a graph can be defined by replacing "chain" with "path" i n the definition of z")(G).

*

*

*

A fourth generalization of the chromatic number of a graph has been studied by S . Hedetniemi [1970]. Let zk(G) denote the smallest number of colours needed to colour the vertices of graph G such that no connected subgraph of order k has only one colour. Thus,

XZW)

= y(G)

9

Clearly, z k ( G )is also the chromatic number of a hypergraph defined in the obvious way.")

*

*

*

Another generalization of chromatic number has been given by H. Sachs and M. Schauble 119671. For a simple graph G = ( X , E ) and a number k 2 2, a coloiiring by k-cliques is defined to be a partition (XI, X,, ..., X,) of X such that no k-clique is contained entirely in one class of the partition (i-e., no k-clique has only one colour). Let zk(G)denote the smallest integerp for which there exists such a partition in p classes. Thus, the chromatic number of G (in the usual sense) is x2(G). Consider the hypergraph H formed from the k-cliques of graph G (which was characterized in Proposition 6, Ch. 17). Clearly,

K W ) = X,(G)

'

Sachs and Schauble have discovered an inductive construction that yields the following result: For each p 2 1 and for each k 2 2, there exists a graph G = G ( p , k) with the following properties:

(I) G contains no (k

+ I)-cliques,

(2) ZAG) = P,

( 3 ) for each colouring by k-cliques that uses p colours, there exist at least p pairwise disjoint ( k - I)-cliques that have only one colour. For k = 2, this construction gives a graph without triangles and with chromatic number p for each integer p (Blanche Descartes [1947]; Zykov [1949]; Mycielski [1955]; Erdos, Rado [1960]). Zn particular, Hedetniemi [1970] has shown that if G is a planar graph, then it is possible to colour its vertices in 4 colours such that none of the 4 subgraphs generated by the vertices of the same colour is connected. Furthermore, if G contains no triangles, then 2 colours are sufficient.

CHROMATIC NUMBER OF A HYPERGRAPH

447

EXERCISES 1. Using the argument for Proposition 1, Section 1, show that if a uniform hypergraph H with rank h 2 3 satisfies

1-4n E j l

< h-2

(i # j ) ,

then the stability number k of hypergraph H satisfies

2. Show that a hypergraph H with chromatic number x ( H ) = q contains a chain of length q - 1. (Tomescu [19691)

3. Consider a hypergraph with n vertices and with edges E, such that (1) I EfI = 3 for all i, (2) for a, b E ,'A a # b, there exists exactly one edge that contains both a and b. Such a hypergraph is called a Steiner system of order n. For example, the projective plane with 7 points (Fig. 19.1) is a Steiner system of order 7. Show that for n 3 7, then x ( H ) 2 3. It is known that a Steiner system exists if, and only if, n I 1 modulo 6, or if n = 3 modulo 6. If n I 3 modulo 6, a Steiner system H of order n with x ( H ) = 3 can be constructed by a well known method due to M. Hall (see, Combinarorial Theory, GinnBlaisdell, Toronto, 1967, Theorem 15.3.2). If n = 1 modulo 6, a Steiner system H with x ( H ) = 3 can be constructed by a method due to A. Rosa [1970]. 4. Given an n x ti chessboard, define the Queen's hypergraph H,O as the hypergraph whose vertices are the squares of the chessboard and whose each edge E, is the set of squares that a queen at square x controls (including square x). Similarly, define the King's hypergraph H Z , Rook's hypergraph H,", and Bishop's hypergraph H!. Show that x(H,f') = ,y(Hf) = x(H,?) = x(H,f) = 2 .

5. In the h-dimensional space Rh, let be a family of R h + l ( p ,q) compact convex sets such that each subfamily of q members, contains h + I members whose intersection is non-empty. Show that there exist p sets of 5 ' whose intersection is not empty. Hint: Use Helly's theorem: If V: is a family of p 2 h 1 convex compact sets in Rh, and if the intersection of any h 1 of them is non-empty, then the intersection of the family W is not empty. (Berge, Espaces Topologiques, Paris [1962]).

+

+

6. In the plane, let S be a finite set of points with 1 S I 3 R 4 ( p , 5), such that no three points are on a line. Show that there exist p points of S that are the vertices of a convex polygon. Hint: Show that (1) if I S I = 5, then there exist 4 points of S that form a convex quadrilateral. (2) if all the quadrilaterals formed from a set o f p points are convex, then thesep points are the vertices of a convex polygon. (Erdos, Szekeres [1935]).

CHAPTER 20

Balanced Hypergraphs and Unimodular Hypergraphs

1. Strong chromatic number

S) be a hypergraph with rank function r(S). A set S is Let H = (X, defined to be strongly stable if r(S) = 1, i.e. if (i = 1 , 2 ,

ISnE, I d 1

..., m).

Note that each strongly stable set is also a stable set. The strong stability number a(H) of hypergraph H is defined to be the maximum number of vertices in a strongly stable set. Clearly,

a(H) = a ( ( W 2 ) . The covering number p ( H ) of H is defined to be the smallest number of edges of H that cover all the vertices of H. Clearly, p ( H ) 2 13((H)2),the smallest number of cliques needed to cover graph (I&. Furthermore, if H is a conformal hypergraph, then p ( H ) = O((H)&. A strong q-colouring of hypergraph H is defined to be a q-colouring of the vertices of H such that no two vertices contained in the same edge have the same colour. Clearly, a strong q-colouring is a partition of X into q strongly stable sets. The strong chromatic number y(H) of H is defined to be the smallest integer q for which there exists a strong q-colouring. Clearly, y(H) 2 r(X). Hypergraph His said to be y-perfect if y(HA) = r(A) for all A = X. Proposition. Euery y-perfect hypergraph is conformal.

Let H = ( X , 8)be a y-perfect hypergraph. To show that H is conformal, it suffices to show that each clique C of graph (H)2 is contained in an edge of H. Since hypergraph H is y-perfect,

Ic I = W

C )

= r(C),

and, consequently, there exists an edge of H that contains C. Q.E.D. 448

BALANCED HYPERGRAPHS A N D UNIMODULAR HYPERGRAPHS

449

Remark. Let V denote the family of maximal cliques of graph G . If graph G can be characterized by a property of family ??, it is often interesting to relate the chromatic number of G to properties of hypergraph H = ( X , 59) since y(G) = y(H). The relationship between the rank of a hypergraph and its strong stability number is described in the following results. Theorem 1. Let H

=

( X , b) be a hypergraph with rank r(A). Then

To demonstrate the first inequality, suppose that S is a maximum strongly stable set. Then, r(S) = 1, and

To demonstrate the second inequality, let k be the smallest number of edges that cover X,and denote these edges by E l , E,, ..., Eli. Thus,

+ + I A n Ek I < kr(A) . Therefore, and

and, finally,

[ -1

A H ) = k 2 max I A I A#@ dA)

*

A cX

Corollary.

If H is a y-perfect hypergraph, then

ACX

Suppose A is a non-empty subset of X . Let q = r(A) = y(HA). If ( A l , A 2 , ..., A4) is a strong q-colouring of H A , then

I A I = 1 A1 1 + ' * *

+ 1 A, 1 < qL%(H)= r(A) cr(H).

450

HYPERGRAPHS

Hence,

and

a ( H ) L max I A I A# 0 r(A) From Theorem 1, it follows that

[--]

*.

a ( H ) = max I A I A#O r(A)

[-3

*.

Q.E.D. Remark. If hypergraph H is a graph, then the inequalities of Theorem 1 are the best possible, since the expression

can attain the value of cr(H) as well as the value of p ( H ) . For example, consider the graph GI with 6 vertices shown in Fig. 20.1.

Fig. 20.1

We have : a(Gl) = 3 = ko(Gl) < p(G,) = 4 .

Let G2be the graph obtained from G, by removing the isolated vertex. For graph G 2 , we have: a(G,) = 2 < ko(G,) = 3 = p(GJ

.

2. Balanced hypergraphs A hypergraph H is said to be balanced if every odd cycle (al, E l , a2, E2, ..., E2p+1, a l ) has an edge Ei that contains at least three vertices a, of the cycle. For example, a family of intervals of points on a line

BALANCED HYPERGRAPHS A N D UNIMODULAR HYPERGRAPHS

45 1

is a balanced hypergraph. Clearly, a multigraph is balanced if, and only if, the multigraph is bipartite.

Proposition 1. Let H be a balanced hypergraph, then ecery partial hypergraph H' is balanced. If H' had an odd cycle with no edge containing three of its vertices, then this sequence would be also an odd cycle of H with no edge containing three of its vertices. This contradicts that H i s balanced. Q.E.D.

Proposition 2. Let H be a balanced hypergraph, then every subhypergraph Hs is balanced.

If Hs had an odd cycle with no edge containing three of its vertices, then this sequence would define for H an odd cycle with no edge containing three of its vertices. Q.E.D. Proposition 3. Let H = (Et I i E I ) be a balanced hypergraph, and let Eo be = (E, u Eo 1 i E I ) is also balanced.

a set. Then the hypergraph H'

Suppose that H' has an odd cycle p with no edge containing three of its vertices. Let this cycle be denoted by p = ( x i , E;, x;, Eh, ..., xk, EL, x i ) .

At least one vertex xi lies in Eo (otherwise, p would define an unbalanced cycle of H ) . Since k 2 3, p has an edge E; w i t h j # i - 1 and j # i, and E; contains x;, x ; + and ~ xi. This contradicts the definition of p. Q.E.D.

Proposition 4. Let H = ( E l , E 2 , . .., Em)be a balanced hypergraph, and let x,, $ U Ei. Then the hypergraph (El u { xo }, E,, ..., Em)is balanced. The proof is immediate.

Proposition 5. Let H = ( X , 6 ) be a balanced hypergraph, and let x1 E X . The hypergraph H' = (XI,a') obtained f r o m H by adding a new vertex xi and by setting X' = x u { x i } if x1$ E, ~'{:zu{x~'} if xlEE, is also balanced. The proof is immediate.

452

HYPERGRAPHS

Proposition 6. Let H Then its dual H* = (e,,

= (x, ,

..., x, ;E l , ..., Em)be a balanced hypergraph.

..., em;XI, ..., X,)

is also a balanced hypergraph.

By definition,

X , = { e i / i < m,E,

3

x,}.

Consider an odd cycle p = ( e l , XI, e 2 , ...,e2p+ xaP+ e l ) in H* ; it corresponds to an odd cycle (xl,E,, x 3 , ..., E z P c l ,X 2 p + l ,E l , xl) in H. Since H i s balanced, some E,contains three of the x,. Therefore some vertex e, of p belongs to three of the X,;or, equivalently, some X, contains three of the e,. Thus, cycle p is balanced. Q.E.D. Proposition I. If H = (Ei / i E I) is a balanced hypergraph, and if H’ = (Ei / i E J ) is a partial hypergraph such that i, j E J implies El n E, # 0,then

nE

,

Z

~

ieJ

(In other words, the edges of a balanced hypergraph satisfv Helly’s property.) The proof is an induction on the number of edges in H‘. The proposition is true for every subfamily with two edges. Assume that it is true for every subfamily with less than p edges; we shall prove it for a subfamily (E, / i E J ) = ( E ; , E6, ..., Ek) wherep B 3. By the induction hypothesis, for every k < p , there exists a vertex a, with

a, E

n E,’ .

t#k

We may assume that the a, are all distinct (otherwise Consider the sequence p = (a,, E;, a3,E ; , a,, E4, 4

n E ; # @). .

If two of the sets €I, EL, El are equal, for instance if €; = €;, then a, belongs to n t G I E ; and , the proof follows. If no two of the sets E; , E;, E; are equal, then p is an odd cycle, and one of these edges, say E; , contains { a , , a2,u3 }. Hence,

n E,.

IEJ

Q.E.D. Propositions 6 and 7 show that a balanced hypergraph is conformal. Theorem 2. A hypergraph H = (X,6 ) is balanced S c X , the subhypergraph Hs is bicolourable.

and only i f , for every

BALANCED HYPERGRAPHS A N D UNIMODULAR HYPERGRAPHS

If every subhypergraph of H cause otherwise, there exists an Ef containing three of the a,, and graph Hs with x(HJ > 2, which

453

is bicolourable, then H is balanced beodd cycle (a,, El, a2, ..., E p , al) with no S = { a,, a2, ..., up } generates a subhyperis a contradiction.

Conversely, let H be a balanced hypergraph that is not bicolourable, with minimum order n = I X I. We shall show that this yields a contradiction.

1. We shall show that each vertex xo belongs to at least two different edges of H with exactly two elements. The subhypergraph Ho generated by X - { xo } is balanced (Proposition 2) and has order n - 1. Therefore it has a bicolouring (Sf,Si).If xo did not belong to an edge with exactly two elements, then (Sp u { xo }, Si) would be a bicolouring of H, which contradicts the definition of H . If xo belongs to only one edge with two elements, and if its other endpoint lies in Sp, then (Sf, SX u { xo )) would be a bicolouring of H. Hence, xo belongs to at least two edges with two elements, say [x0, y ] and [xo, z ] with z # y. 2. Denote by 9 the family of edges with exactly two elements. Consider Since G is balanced (Proposition I), G is a bipartite graph. graph G = ( X , 9). Consider a connected component C of G o . Since G has order 2 3 (from part l), there exists at least one vertex x1 in C which is not an articulation point.

3. Consider the subhypergraph Hi generated by X - { x1 }. It is balanced and has order n - 1. Therefore H I admits a bicolouring (&, SJ, and all the vertices adjacent to x1 in G have the same colour. Let S, be the set of vertices with this colour. Then, (Sl, S, u { x1 }) is a bicolouring of H because each edge of H with two elements is bicoloured (since it is an edge of G), and each edge of H with more than two elements is also bicoloured (because its intersection with X - { x1 } is bicoloured). This contradicts the assumption that H does not have a bicolouring. Q.E.D. Theorem 3. For a balanced hypergraph H = (E, / i E I ) , let k = mi ti,,, I Ei There exist k transversals of H that partition the vertex set X .

I.

Let (Sl, S,, ..., S,) be a partition of X into k classes, and let k(i) be the number of classes which meet edge Ei.If k(i) = k for every i, then X is partitioned into k transversals. If k(i) < k for an index i = io, then k(io) c I E,, 1; hence, there exist indices p and q such that: I S p n E i , I 2 2,

454

HYPERGRAPHS

and IS,nE,,,I = O .

The subhypergraph H' generated by S, u S, is balanced. Thus, by Theorem 2, H' has a bicolouring ( S ; , Si). Let 5'; = S, for j # p , q. The partition (Sl, S;l, ..., SL) determines new coefficients k'(i) with k'(io) = k(i,) 1

+

k'(i) 2 k(i)

( i # io).

By repeating this transformation as many times as needed, we obtain a new partition (Si,S ; , ..., SL) with k'(i) = k for every i. Clearly, this is the required partition of X . Q.E.D.

Theorem 4. A hypergraph H is balanced f, and only i f , y(H') = r(H') for eoery partial subhypergraph H' of H . 1. Let H be a hypergraph, and suppose that the above equality holds for every partial subhypergraph H' of H . If hypergraph H is not balanced, then there exists an unbalanced cycle, say, =

(01,El,

a2, E2,

.-.)a k ,

Ek,

a

The partial subhypergraph H' = (S, 8') defined by S = { a l , a2, ..., ak] and ..., Ek n S ) is a graph consisting of an odd cycle. Hence y(H') = 3, r(H') = 2, which contradicts that y(H') = r(H').

b' = (El n S, Ezn S,

2. Let H be a balanced hypergraph of rank h ; we have y ( H ) 2 h since h different colours are needed to colour a maximum edge; therefore, it suffices to show that there exists a strong h-colouring of H. Consider a hypergraph H' = ( X ' , 6 ' ) obtained from H by adding a set A , with 1 A , I = h - 1 E, I elements for every i and by letting:

X'=

XU

u A' i €1

and €'= ( E , u A i / i E Z ) .

Clearly, H' is a uniform hypergraph of rank h, and by Proposition 4, H' is balanced. By Theorem 3 applied to H', there exists a partition ( T i , T;, ..., 7';) of X' into h transversals. Since I 7';n E, 1 < 1 for every i and everyj, the sets S, = T; n X are the classes of a strong h-colouring of H. Q.E.D.

BALANCED HYPERGRAPHS AND UNIMODULAR HYPERGRAPHS

Corollary. If H is a balanced hypergraph with rank h, and if hl then there exists a partition (A', , X,)of X such that

455

+ h, = h,

r(Xl) = h, r ( X 2 ) = h,. Let (Sl, S,, ..., S,) be a strong h-colouring of H. Let Xl

=

u

s,

l
and x 2

=

u

s,.

h l + 1
Clearly, (X1, X,) is the required partition of X .

Q.E.D. APPLICATION. Let G = ( X , 8 ) be a bipartite multigraph, and consider the dual hypergraph G*. Since G is balanced, hypergraph G* is also balanced (Proposition 6). By applying Theorem 4 to G*, we obtain a well known result: The chromatic index of a bipartite multigraph is equal to its maximum degree (Theorem 2, Ch. 12). Applying Theorem 3 to hypergraph G*, we obtain a result of Gupta [1967]: In a bipartite midtigraph G rrith ininimuni degree k , there exists a partition of its edges into k classes, such that each class covers all the vertices of G. Theorem 5 (Berge, Las Vergnas "19701). Let H be a hypergraph, let v(H) denote the maximum cardinality of a matching and let T ( H )denote the minimum cardinality of a transcersal. Then, H is balanced if, and only if,

v(H')

=

T(H')

f o r ecery partial subhypergraph H' of H. If the above equality holds, it is easily seen that H is balanced (the proof is the same as for Theorem 4). It remains to show that v ( H ) = T ( H )for every balanced hypergraph H. Consider a balanced hypergraph H = (E, / i E Z) with v ( H ) < 7 ( H ) in which I E, I is minimum. Let q = v ( H ) . Thus, z ( H ) > q. We shall show that this leads to a contradiction.

zLE,

1. Clearly, ELQ E, for i # j since Ei c Ej implies that v ( H ) = v ( H - Ej) which contradicts the definition of H.

=

T(H - Ej)

=

T(H),

456

HYPERGRAPHS

2. Also, I Et I 2 2 for every i E I. If El= { x1 }, for instance, then from part 1, x1 $ Ut,l E i . Therefore, ~ ~1)=) T ( H X - < ~+~ )1) = z ( H ) . v(H) = v ( H ~ - { + 3. We shall show that if 8,= (E, / j E J ) is a maximum matching of H , then it partitions the vertex set X of H. Suppose that there exists a vertex a $ U,,,E,, and let E( = EL- { a } . The subhypergraph H‘ = ( E ; / i E I ) generated by X - { a } is balanced (Proposition 2), and from Part 2, it follows that

v(H‘)

=

T(H’) 2 T ( H ) > q

Thus, the matching 86 = (E; / j E J ) is not maximum in H‘, and by Theorem (1, Ch. 18), there exists an odd maximal alternating sequence 0’= (F;, E;, F;, ...,EL, F i + l ) ,with E I , E;, ..., E; E 86 F;, F;, ..., F ; + ~ E &’ 8;. Since the sequence corresponding to CT’ in H is not a maximal alternating 1 with sequence by Theorem (1, Ch. 18), there exist indices kl,k2 < p

+

Fkl

Fkz

=

{ a 1*

Let G be the representative graph of the family of sets ( E l , E 2 , ..., E,, F,, F,, ..., F p + J .Graph G is connected and has an odd cycle. A minimal odd cycle of G defines an odd cycle of H of the form:

(a, 41,xl? Ej,,yl,

Ft2,

x2, E j 2 ,

*..)F k , a > .

No edge of this cycle contains three vertices of the sequence, which contradicts that H is balanced. 4. Let x1 E El, and consider the hypergraph -

R = ({ X I }, EZ, E3, ..., Em)= (El, E2,. .., Em). No cycle of R uses edge El, and therefore R is balanced. From Part 2

v(R) = z ( R ) . Hence v(R)= T(R)2 T ( H ) > 4 .

By the induction hypothesis,

Let zobe a maximum matching of R; then ElE b, because v(R)> q = v(H). is a maximum matching of H , and the union of its edges Thus, do - {El} does not contain x1, which contradicts part 3. Q.E.D.

BALANCED HYPERGRAPHS AND UNIMODULAR HYPERGRAPHS

457

If H is a graph, Theorem 5 yields: In a bipartite graph, the maximunz cardinality of a matching is equal to the minimum cardinality of a transversal (Konig's theorem, Chapter 7).

Corollary 1. then

If

H is a balanced hypergraph with a rank-function r(A), a(H) = p(H)

= A +0

Clearly, a ( H ) = v(H*) and p ( H ) = 7(H*). Since H* is balanced (Proposition 6), Theorem 5 implies that a ( H ) = p(H). From Theorem I, the required equalities follow. Q.E.D.

Corollary 2. Let H be a balanced hypergraph with m edges and with rankfunction r(A), and let k 6 m. A necessary and sufficient condition for rhe existence of a covering with k edges is that (A c X ) . kr(A) - I A I 3 0 This follows from Corollary 1. Theorem 4 and Theorem 5 together yield a new result for graphs: Let G = (X,E ) be a simple graph. Let %? be the family of all its maximal cliques. If A C X and 9 c Gf, let GA.9 denote the graph obtained from G by deleting the vertices of X - A and the edges that do not belong to a clique in 9. Let a(G) denote the stability number of G (maximum cardinality of a stable set), let 8(G) denote the partition number of G (smallest number of cliques which cover X ) , let y(G) denote the chromatic number of G, and let o ( G ) denote the density number of G (maximum number of elements in a clique).

Theorem 6. The three following conditions are equivalent: (1) a(GA,9) = e(G,,

3)

for every A and every 9 ;

(2) y(GA.9) = o(GA.a)for every A and ecery 9 ; (3) every odd cycle in G contains at least one edge with the property that eLtery maximal clique containing this edge contains a third vertex of the cycle. Let H = ( X , 'e) be the hypergraph of the maximal cliques of G. Clearly, (3) is equivalent to (3') H is balanced

and to

(3") the dual H * of H is balanced.

458

HYPERGRAPHS

Also, (1) is equivalent to (1‘) v ( H ’ ) = ~ ( H ’ ) . f oecerypartial r subhypergraph H ‘ of H * ,

Clearly, (2) is equivalent to

(2’) y(H‘)

=

r ( H ‘ ) f o r etlery partial subhypergraph H ‘ of H.

By Theorem 4, (2’) is equivalent to (3’). By Theorem 5, (1’) is equivalent to (3”). Hence (l), (2), and (3) are equivalent. Q.E.D. Lovhsz has studied the conditions under which each partial hypergraph H’ of hypergraph H satisfies v(H’) = r(H‘). The class of hypergraphs with this property is more general than the class of balanced hypergraphs. The chromatic index q ( H ) is defined to be the smallest number of colours needed for a colouring of the edges of H , i.e. q ( H ) = y(H*). Let 8, denote the family of edges that contain vertex x, and let 6,(H) = I & I denote the number of edges that contain x . Let

6 ( H ) = max 6,(H). X € X

Clearly, q ( H ) 2 S(H). A hypergraph H is defined to be normal if q(H‘) = 6(H’) for every partial hypergraph H ’ of H . Each partial hypergraph of a normal hypergraph is normal, but a subhypergraph of a normal hypergraph need not be normal.

Lemma. r f H = ( E l , E2,..., Em)is a normal hypergraph, and i f an edge Eo = El is added to H , then the new hypergraph Ho is also normal. It suffices to show that if H ‘ is a partial hypergraph of H and if HA = H’ Eo, then 6(Hh) = q(H4). We may assume that H ‘ contains E, (otherwise, the equality follows immediately). Let

+

q

=

q(H’) = d ( H ’ ) .

CASE1: Edge El contains a vertex x with 6,(H‘) = q . Clearly,

6(HL) = 6(H’) + 1 = q

+ 1.

Hence, 6(HA) < q(HA) < q ( H ’ ) + 1

=

q

Thus, d(HA) = q(HA) ‘

+ 1 = 6(HA).

BALANCED HYPERGRAPHS A N D UNIMODULAR HYPERGRAPHS

CASE2: S,(H‘)

459

< q - 1for every x E El.

Consider a q-colouring of the edges of H’ and suppose that edge El is coloured red. Each vertex a with d,(H’) = q is incident to a red edge distinct from E l . If 6 denotes the set of red edges distinct from E l , then d(H’ - 8,) = q

- 1.

Since H ’ is normal, this implies that q(H‘ - &J = q - 1, and there exists a (q - I)-colouring of the edges of H ’ - 8,.Since &l u { Eo} is a matching, the edges of this matching can be assigned a q-th colour; this produces a q-colouring of HA. Hence, 6(Hh) < q(H6) < 4 = 6 ( H k ) .

Q.E.D. Theorem 7 (Lovisz [19721). A hypergraph H is normal i f , and only iJ;

Thus, 6(Hh) = q(H&

v ( H ’ ) = r(H’) f o r each partial hypergraph H’ of H . Necessity. To show that each normal hypergraph satisfies this condition, it suffices to show that i f H is a normal hypergraph with v(H) = q, then there exists a transversal T with I T I = q. Since H is normal, its dual H* is y-perfect, and conformal (by Proposition, Section l), and consequently H satisfies the Helly property (Theorem 3, Ch. 17). Thus, v ( H ) = 1 implies that there exists a transversal T with I T I = 1, and the proposition is valid for q = 1. Let q > 1. Assume that the proposition is valid for each normal hypergraph Ill with v ( H l ) < q. We shall show that it is also valid for a normal hypergraph H = ( X , 8)with v ( H ) = q. I f there exists a vertex x E X with v(H - 8,) < q, then there exists a transversal T of H - &,‘ with q - 1 vertices, and T u {XI is a transversal of H with q vertices. Hence, the proposition is valid. Therefore, we may assume that

v(H - 8x)= q (X E X). For each xi E X,consider a maximum matching gi of H - B,, . Let n = I X I. Consider the hypergraph Ho = P1+ g2+ ... + S,,formed with all the edges of the Pi for i = 1,2, ..., n. (Thus, hypergraph Ho contains exactly nq edges.) Since each vertex x, is not covered by Fitit follows that S(Ho) < n .

460

HYPERGRAPHS

Moreover, in a colouring of the nq edges of H , , the same colour occurs at most q times, and consequently, there are at least n different colours. Hence, dH0) 2 n >

wfo)

However, from the lemma, Ho is a normal hypergraph, which implies that q(H,,) = 6(Ho). This contradiction achieves the proof. Suficihncy. Consider a hypergraph H = ( X , 8) such that each partial hypergraph H’ of H satisfies v(H‘) = T(H’). It suffices to show that 6 (H) = q(H)* Consider a hypergraph R whose vertices correspond to the matchings in H, and whose edge E, is the set of all the matchings in H that contain edge E1.A subfamily H ’ = (Ej / j E J ) has a non-empty intersection if, and only if,

El n Elf # 12/ ( j , j ’ €4 (because this implies that T ( H ’ ) = v ( H ’ ) = 1). This is equivalent to

Ej n El, = 0

( j , j’ E J ) .

In other words, h sets of the family (€, / j E J ) have a common vertex in H if, and only if, (E, / j E J ) is a matching in 17. Thus, (1)

q(R)= T ( H ) ,

(2)

6(H) = v(R).

Moreover, the sets of a family (El / j E J ) have a common vertex 6, if, and only if, for each j E J , edge E, belongs to matching go,that is, if, and only if, (El / j E J ) is a matching in H . Thus,

(3)

q(H) =

(4)

6(R)= v ( H ) ,

7(m,

By applying (1) and (4) to a partial hypergraph H ‘ of H and to the corresponding partial hypergraph R‘ of R, it follows that

q(R’)= 6(R’). Thus, R is a normal hypergraph, and from part I , from ( 2 ) and (3), it follows that

v(R) = s(R).Therefore,

6(H) = q ( H ) .

Q.E.D. Fournier and Las Vergnas [i972] have also shown that a normal hypergraph is bicolourable. More precisely, if each odd c.vcIe of H contains three edges with a non-empty intersection, then H is bicolottrable.

BALANCED HYPERGRAPHS A N D UNIMODULAR HYPERGRAPHS

46 1

Theorem 7 has been used by Lovsisz to prove that a graph is a-perfect if, and only if, it is y-perfect. More precisely:

Theorem 8 (LovQsz ([ 19721). Let G = ( X , E ) be a graph such that a(G,) = O(Gs) for each S c X ; then y(G) = w(G). Construct a hypergraph whose vertices are the cliques of G and whose edge X , is the set of all cliques of G that contain vertex xi E X . Clearly, graph G is a representative graph of the edges of H . Thus, (1)

v(H) = 4G)

(2)

q ( H ) = Y(G) *

9

Furthermore, from its definition, hypergraph H satisfies the Helly property. Therefore,

(3)

6(H) = w(G),

(4)

T ( H ) = O(G)

.

Each subgraph G, of G corresponds to a partial hypergraph H’ of H , and this correspondence is bijective. Applying (1) and (4) to G, and to H‘, it follows that v ( H ’ ) = 7 ( H ’ ) . Therefore, from Theorem 8, H is normal. Hence, q ( H ) = 6 ( H ) , and from (2) and (3), we have y(G) = w(G). Q.E.D. We shall now use the above properties of balanced hypergraphs to determine a sufficient condition for a hypergraph to contain k pairwise disjoint transversals. This condition was first shown for k = 2 by Lovasz [1968] and then extended to the general case independently by Lovhsz [1970] and Las Vergnas [1970]. First, a lemma is needed. Lemma. Let k 2 2. Let G = ( X , Y ,

r ) be a

simple bipartite graph such

that

I T(S)I

2 (k

- 1)

1 SI +

1

( S C X, S # @).

If no partial

graph of G (other than G itself) satisfies this condition, then each certex x in X has degree k.

First note that by letting S = { x } , the above condition implies that d,(x) 2 k for every vertex x.

Let d denote the family of all sets A c X such that 1T(A))=(k-l)IA\+l.

462

HYPERGRAPHS

1. We shall show that if A , A’ ES? and if A n A’ #

0, then A n A’ ~

d .

We have :

I T ( A u A’) 1 + I T ( A n A ’ ) I G I T ( A )u r ( A ’ ) 1 + I T ( A ) n r(A’)I = IW ) I + I W ’I ) =(k-l)IA]+l+(k-l)(A’(+l = ( k - I ) ( A U A ’ (+ 1

+(k- 1 ) I A n A ’ I

+ 1.

Moreover, since G satisfies the condition of the lemma, and since A n A’ #

gf,we also have: (T(’4UA’)I 2 ( k - 1 ) I A u A ’ )

+1

[ T ( A n A ’ ) [2 ( k - l ) [ ~ n + ~ 1’. [ Therefore, equality must hold in these two inequalities. Hence, A n A’ ES?.

2. Consider a vertex a in X , and let T(a) = { b l , b2, ..., 6, ). Let G, = ( X , Y, r,)denote the partial graph of G obtained by removing edge [a, b,] There exists a subset At of X such that

I Tf(4) I < (k - 1) 14 I + 1 Hence,

A, 3 a ; Let A

=

At E&’;

6, $ T,(A,).

A l n A 2 n ... n A,. From Part 1 , A

E&.

T(A - { a }) n r(a)= @

From above,

.

Therefore, =

+

+

I T ( A - { a > )I 2 k ( k - 1) I A (k - 1) I A I 1 = I T ( A ) I .

I r(A)I = &(a)

+

- { a >I

Consequently, equality holds in this inequality, and &(a) = k .

Q.E.D. Theorem 9 (Lovsisz [1970], Las Vergnas [1970]). Let k 2 2, and let H (El / i E I ) be a hypergraph such that

I

u

Ei~2(k--l)IJl+l

(Jl=I,Jf0).

1EJ

Then hypergraph H contains k pairwise disjoint traversals.

=

BALANCED HYPERGRAPHS AND UNIMODULAR HYPERGRAPHS

463

Consider a family H' = (El / i E I) such that

E,'

I 1u €J

C

E(

E*'IZ(k-l)lJI+1

(i E Z) (JCZ,J#I),

and that is minimal with respect to the order relation H' < H" defined by: E; c E ; for all i E Z. Consider the bipartite graph C,. = (I, X , r) where r(i)= El. Since this bipartite graph satisfies the conditions of the lemma, I E ; I = k for every i E I. Moreover, for each J c I, J # 0,

From Proposition (5, Ch. 17, $2), this implies that hypergraph H' contains no cycles. Therefore, H' is balanced, and, from Theorem 3, H' contains k painvise disjoint transversals. These sets are also transversals of H. Q.E.D.

3. Unimodular hypergraphs This section treats a class of hypergraphs which, like balanced hypergraphs, generalizes the concept of a bipartite graph. These hypergraphs which are important in integer linear programming give sharper results. First note that the concept of a q-colouring of a graph can be extended to hypergraphs in many different ways. (1) Aq-colouringof a hypergraph H = (Ei/ i E I ) is a partition (Sl,S,, S,) of its vertices into q classes such that iEZ, j 1 * Et Q S , .

...,

The smallest number of classes is the chroriiatic number x(H). See Chapter 19. (2) A strong q-colouring is a partition (Sl, S,, ..., S,) into q classes such that I E , n S , I G 1 ( i ~ z j, d 4). The smallest number of classes is the strong chromatic number y ( H ) . See Section 1. (3) An equitable q-colotiring is a partition (Sl, S,, ..., S,) into q classes such that for each i E Z and for each j , j' d q, - 1 d IE,nS,I

- IElnSi.I G 1 .

The smallest number q 2 2 for which there exists an equitable q-colouring is defined to be the equitable chromatic number k ( H ) of hypergraph H.

464

HYPERGRAPHS

If H is a graph, then these three definitions coincide with the ordinary definition of the chromatic number of a graph. The reader can verify that the hypergraph H defined by the projective plane with 7 points (Fig. 19.1) satisfies

x(H) = 3 ,

y(H) = 7 ,

k(H) = 7 .

The reader can also verify that the hypergraph H with 9 edges in Fig. 20.2 satisfies x(H) = 2 , y(H) = 4 , k(H) = 3 . Proposition 1. For each hypergraph H, X(H) G k ( H )

Y(H)

-

Clearly, each strong q-colouring of H is also an equitable q-colouring of H . Similarly, each equitable q-colouring of H is also a q-colouring of H. The proposition follows. Q.E.D. Proposition 2. If there exists an equitable bicolouring of H , and if GH = ( I , X,r) denotes the bipartite graph of the vertex-edge incidences in H , then for each partial hypergraph H' of H , the graph G,. is either non-eulerian, or eulerian with the number of edges equal to a multiple of 4.

The bipartite graph GH is defined by T(i) = E t . A graph is said to be eulerian if it has only even degrees. If H has an equitable bicolouring, then clearly H' also has an equitable bicolouring. Let (Sl, S,) be such a bicolouring. For k = 1, 2, let Fk denote the set of edges of G H d that are incident to a vertex x E S,. If Gw is eulerian, then each vertex i E I is incident to an equal number of edges of Fl and F,. Thus, I Fl

I = I F, I. Moreover,

I Fl I = 2

dCH' (x).

XESl

is even. Therefore, the number of edges in G H , is I Fl which is a multiple of 4.

1

+ I F2 I = 2 I Fl I,

Q.E.D. If H is a graph, or if H is the dual of a graph, it is easy to show that H contains an equitable bicolouring $ and only %,for each partial hypergraph H' of H, the graph G ., is either non-eulerian or eulerian with the number of edges equal to a niultiple of 4.

BALANCED HYPERGRAPHS AND UNlMODULAR HYPERGRAPHS

465

This result is similar to a theorem of Camion [1965] for totally unimodular matrices. A hypergraph H = ( X , 8)is defined to be unimodular if for each S c X , the subhypergraph Hs admits an equitable bicolouring.

EXAMPLE 1. Clearly, a graph is unimodular if, and only if, the graph is bipartite. Thus, the concept of a unimodular hypergraph is an extension of the concept of a bipartite graph. 2. Consider a set X of points on a line, and a family 8 = EXAMPLE ( E l , E 2 , ..., Em)of non-empty intervals whose union is X . The hypergraph H = ( X , 8)is unimodular because for each S c X an equitable bicolouring of Hsis obtained by alternately colouring red and blue the points of S as they appear in order on the line. Proposition 3. If H is a unimodular hypergraph, each partial sirbhypergraph of H is also imirnodulur:

This follows from the definition of unimodularity. Proposition 4. A unimodular hypergraph is balanced.

Suppose that hypergraph H is unimodular but not balanced. Then H contains an odd cycle ( a l , E l , a2, E,, ..., Ep,a,) of which no edge contains three of the a,. Thus, the subhypergraph H A generated by A = {a,, a,, ..., a,} does not have an equitable bicolouring, which contradicts the unimodularity

of H. Q.E.D. H i

Fig. 20.2

The converse is not true. Consider the hypergraph H in Fig. 20.2. Each odd cycle of H contains three vertices of edge E l . Hence, H is a balanced hypergraph. However, H is not unimodular. Hypergraph H has a unique

466

HYPERGRAPHS

bicolouring of its vertices. (This can be seen by colouring vertex a red, then successively colouring in red or blue vertices b, c, d, e, S, g, h, i, j . ) This bicolouring is not equitable for edge E l . Theorem 10. r f H is a hypergraph without odd cycles, then H is unimodular. Since each subhypergraph of H contains no odd cycles, it suffices to show that a hypergraph H = ( X , 8)without odd cycles has an equitable bicolouring. There exists a function x,(t) such that Ei can be written as { x l ( l ) ,xl(2),

...)xdrt) 1 ,

where

r,

=

I El I .

Let 9, be the set of pairs xi( l)xi(2), xi(3)x,(4), etc. Consider the graph G = (X,a),where G = U 4. We shall show that graph G cannot have an odd cycle. Suppose that G contains odd cycles. Let p = [a,, a,, ..., a,] denote an odd cycle of G with minimum length. Cycle p is elementary; suppose that it contains two disWithout loss of generality, let [a,,a,,,] joint edges in the same class 4. and [a,, a, + ,] be these two edges. Transform 4 by replacing these two edges by edges [as,a,,,] and [a,, a , + , ] . Replace p by the sequence either ( a l , a , , ..., a,, a , , , , ..., a , ) or (a,,,, a , , , , ..., a,, a , , , ) that has odd length. Repeat this process as many times as possible. Upon termination, an odd sequence is obtained that determines an odd cycle in hypergraph H , which is a contradiction. Since graph G contains no odd cycles, there is a bicolouring (S, , S,) of its vertices. Clearly, (S, , S,) is also an equitable bicolouring of H . Q.E.D.

Corollary. A hypergraph H = (Ei I i E I ) contains no odd cycles if, and only if, each hypergraph H' = (El/ i E I ) with E( c Elfor all i E Isatisjies k(H') < 2. The condition is necessary because hypergraph H' contains no odd cycles. The condition is sufficient because an odd cycle of Zf (if it exists) would induce a hypergraph H' that is a graph consisting of an odd cycle. Hence, k ( H ' ) = 3, which is a contradiction. Q.E.D. The following theorem and its corollary are direct extensions to unimodular hypergraphs of a property initially discovered by D. de Werra [I9701 for the dual G* of a bipartite graph: Theorem 11. A unimodular hypergraph H has an equitable q-colouring for each positive inteaer Q 2 2.

BALANCED HYPERGRAPHS A N D UNIMODULAR HYPERGRAPHS

467

For q = 2, the theorem is clearly true. For q > 2, consider a unimodular hypergraph H = (E, / i E I ) , and consider a partition ( S , , S,, ..., S,) of its vertices into q classes. For eachj, k 6 q, let ejk(i) = 1 sj e ( i ) = max

Ei

,.k

I - I SICn Ei 1 el,&)

.

Clearly, e(i) Z 0. If e ( i ) for all i E Z, the partition is an equitable q-colouring of hypergraph H and vice versa. Suppose that there exists an index io with e(io) 2 2, and let r and s be two distinct indices with er,s(io)= e(io). Then, for allj,

I ss n Eio I 6 I S, n Eio I

-

6 I S r n Ei, I Since H is unimodular, the subhypergraph H' of H generated by the set Sru S, has an equitable bicolouring ( S ; , S:). Let S; = S, for j # r, s. The partition (Si,SL, ..., Si) defines as before coefficients e;,(i), and f o r j # r, s, we have : e 2 i 0 ) < e,,(io) - 1 6 e(io) - 1 e:,(io) 6 e,,(i0) - 1 6 e(io) - 1 e:,(io) 6 erj(io) - 1 < e(io) - 1 e;,(io) 6 e,s(io)- 1 ' 6 e(io) - 1 e:,(io) 6 1 6 e(io)- 1 . Since eb,,(io) = ep,q(io)for p , q # r, s, the number of pairs ( j , k ) with e;,(io) 6 e(io) - 1 is greater than the number of pairs ( j , k ) with e,k(io) < e(io) - 1. Moreover, for all i # i o , e;,(i> 6 e(i). By repeating this transformation, a partition with e(i) 6 1 for all i is finally obtained. This partition is an equitable q-colouring of H. Q.E.D. Corollary. I f H = ( E J i E Z ) is a unimodular hypergraph, and if k = mini,, I EiI, then there exists a partition ( T , , T,, ..., T,) of the vertex set X of H into k transversals such that

Note that there always exists a partition of X into k transversals from 'Theorem 3 (since a unimodular hypergraph is balanced). Moreover, from Theorem 1 1 , hypergraph H has an equitable k-colouring ( T I ,Tz,-.*,.Tk), and

468

HYPERGRAPHS

Furthermore, since k = min 1 Ei1, each T,is a transversal of H. Q.E.D.

APPLICATION (de Werra [1970]). If G = (X,E ) is a bipartite multigraph and if q > 1, then the edges of C can be partitioned in q partial graphs GI, Gz,..., G, such that for each j < q and for each x E X ,

It suffices to apply Theorem 11 to G*. Let A = ((a:))be a matrix with m rows and n columns. Matrix A is defined to be totally unimodular if the determinant of each square submatrix of A equals + 1, 0 or - 1. If A is unimodular, then clearly each coefficient aJ of A must be + 1, 0 or - 1, since a: is a minor of order 1. We shall state without proof two fundamental properties of unimodular matrices.

Theorem of Hoffman and Kruskal [1956]. A H n x m matrix A is totally unimodular if, and only if, f o r any integer m-rectors b and b' and for any integer n-vectors a and a', each face of the polytope { X / X E

R";a

< x < a';b C Ax 6 6')

contains an integer point. It follows that each vertex of this polyhedron has integer coordinates.

Theorem of Ghouila-ilouri [1962]. An n x m matrix A = ((a:)) is totally unimodular f, and only f, each set J c { 1,2, ..., n } can be divided into two disjoint sets J1 and Jz such that

1

2 a: - C fEJI

af

Ic

1

( i ~ m ) .

j EJz

A consequence of this theorem is:

Property 1. A hypergraph H is unimodular if, and only if, its incidence matrix ((a;))is totally unimodular. Property 2. If H is a unimodular hypergraph, then its dual H* is also a unimodular hypergraph.

This follows from Property 1 since the incidence matrix of H * is the transpose of the incidence matrix of H .

BALANCED HYPERGRAPHS A N D UNIMODULAR HYPERGRAPHS

469

4. Stochastic functions Let H = (X,6 ) be a hypergraph. A real valued functionf(x) defined on X is said to be a stochastic function associated with H if

0
<1

1 f(x) = 1

(x

EX)

(i = I , 2 , ..., m).

x E Ei

The support of function f is defined to be the set

s, = { X / X € x,f(x)

> 0 1.

Not every hypergraph has a stochastic function associated with it. 'The problem of characterizing the supports S, of H is due to J. Csima [19701.

EXAMPLE 1. A square matrix

i

P:

P:

...

pf

....

Pi ... pi' ... ..................................

i P = ( ( P i ) ) = P:

P1

Pnl

.- .

p;

...

"j P::

is called bi-stochastic if the following conditions hold : pi20

\Ip i

= t

i

/ Ii p ;=. 1

< 1 1 , j < ti) ( j < ti) (i

(i

< n).

Bi-stochastic matrices have been used to generalize convex functions (cf. Berge [1958]). A matrix of order n defines a hypergraph H whose vertices are the cells of the matrix and whose edges are the sets of cells in the same row or in the same column. Note that hypergraph H contains no odd cycles and is therefore unimodular by Theorem 10. Clearly, each stochastic function of H defines a bi-stochastic matrix of order n, and vice versa.

Lemma. If H = ( X , 8)is a hypergraph that has a stochastic function, then the support of this,function contains a strongly stable transrersal of H .

410

HYPERGRAPHS

Consider the incidence matrix of H : Xl

...

...

' 1

a1

...

a:

...

a?

................

xj

x,

...

...

a:

If H possesses a stochastic function f , then the vector p = ( p l , pz, .... p,), where p j = f ( x , ) , satisfies

A p = (1,1, .... 1) ( ( 0 3 0 7 ... , O ) < p < (1,1,

.... 1 ) .

Since matrix A is totally unimodular, the above system of equations has an integer solution p'. The set

T = {xj/p;= l}, which has exactly one vertex in each edge is therefore a transversal and is strongly stable. Q.E.D.

Theorem 12. If H is a imimodulur hypergraph that has a stochastic function, then each stochastic function f ( x ) associated with H can be expressed in the following form: k

f(x) =

1 Pi vi(x)

9

i= 1

k

where q,(x) is a (0, 1)-stochastic .function.

Clearly, the subhypergraph of H generated by the support S, contains a set TI that is strongly stable and is a transversal (from the lemma). Let q D 1 ( xbe) its characteristic function, i.e.,

471

BALANCED HYPERGRAPHS AND UNIMODULAR HYPERGRAPHS

Let p l = m i n ( f ( x ) / x E T l } > 0.

The non-negative functionf(x) - plcpl(x) has more zero values than functionf(x). If this function is not identically zero, then it is proportional to a stochastic function, and therefore its support contains a set T, that is strongly stable and is a transversal. Let p , = min { f(x) - plql(x) / x E T2) > 0. Similarly, define the function cpz(x), and consider the function f(X)

- P1 V l W - P 2 (P2(x) 2 0

9

etc. Finally, consider the function

f ( x ) - P1 91(x)

- P2 q 2 ( x )

- ' * * - Pk ( P k ( X ) =fk(x).

Clearly, k

k

If f k ( x ) is not identically zero, then 1 - z p r > 0, and the function

is stochastic. Therefore, from the lemma, its support contains a set Tk+, that is strongly stable and is a transversal. With the characteristic function (pk+l(x),the procedure can be continued as above, etc. The procedure terminates only when the function f k ( x ) is identically zero, which implies that

Then, functionfis equal to

zf=lpi'pi, where z p i

=

1.

Q.E.D. Corollary. If H is a unitnodular hypergraph, a set S is the support of a stochastic function i f , and only i f , S equals the urriori of strongly stable transrersals of H. 1. Let S be the support of a stochastic function of the form k

f(x) =

C piqi(x), i= 1

0 < pi

< 1,

k

C Pi = 1 i= 1

*

472

HYPERGRAPHS

If we let Ti = { x/cpi(x) = 1 }, then, clearly, S set T, is a strongly stable transversal.

=

Uf=lT , where

each

2. A set of the form S = Uf-l Ti is the support of a stochastic function

where cp,(x) is the characteristic function of set T,.

Q.E.D. APPLICATION 1 (G. Birkhoff, [1946]). Each bi-stochastic matrix P is a barycentre of permutation matrices M i , i.e., k

P =

piMi i= 1

O
c p i = l . i= 1

Consider the hypergraph H associated with matrix P as before (see Example p. 469). The coefficients of P correspond to a stochastic function .f of H. Therefore, Theorem 12 can be applied tof, and the functionscp,(x) correspond to (0, 1) matrices M i , that are permutation matrices. The equality follows. Q.E.D.

APPLICATION 2 (Theorem of Perfect and Mirsky [1968]). Let M be a (0, 1) matrix of order n. There exists a bi-stochastic matrix P that has a zero coeficient wherever M has a zero coeficient, i f , and only if, M is not of the

form

where A is a square matrix of order k, 0 < k < n, and nhere B is a non-null matrix. If matrix kfcan be fit into the above form by a permutation of its rows and columns, then M is not a bi-stochastic matrix ((p:)) because then the sum of the coefficients p;. in A would equal k , and consequently the sum of the coefficients p: in B would equal 0, which contradicts that B is a non-null matrix.

-

BALANCED HYPERGRAPHS AND UNIMODULAR HYPERGRAPHS

473

.i’

i’

Fig. 20.3

Conversely, consider a matrix M = ((rn;)) that cannot be fit into the above form. Let H be the hypergraph whose vertices are the pairs (i, j ) with m: = 1 and whose edges are the sets of cells in the same row of M or in the same column of M . It suffices to show that for each vertex x = (i’, j’) of H , there exists a transversal T, that is strongly stable and that contains x. Clearly, this would imply that X = Uxsx T,, and from Theorem 12, that X is the support of a stochastic function. L e t N = { i / i < n},andforeachiEN-{i‘),let

< i t , j z j ‘ , mi = 1 1 c N - ( j ’ 1 . This defines a bipartite graph ( N - { i’ }, N - { j ’ }, r). r(i> = { j /j

We shall first show that: (1)

IrCI>I2 111

(Ic N - { i ’ ) ) .

Otherwise, there would exist a set I c N - { i‘} with 1 1 I > 1 T(Z) 1, and there would exist a set J of columns with J ~ T ( l ) u { j ’ }and

IJI

=

111.

However, this is impossible since the matrix A obtained from A4 by removing the rows of N - I and the columns of N - J would not be of the form allowed by the theorem. Inequality (1) implies that the bipartite graph ( N - { i‘ 1, N - { j ‘ 1, r )has a perfect matching (Theorem 5, Ch. 7). The set of vertices of hypergraph H that correspond to the edges (i, j ) of this matching form with vertex x = (i’, j‘) a transversal T, that is strongly stable. Q.E.D.

414

HYPERGRAPHS

EXERCISES 1. Consider a hypergraph H = ( X , 8 )where Xis a finite set of integer points in the plane, where the sets of I are of the form

or Show how t o construct an equitable bicolouring for hypergraph H. 2. Show how to construct an equitable bicolouring for the dual of a bipartite multigraph.

3. Consider the hypergraph ( X , 8)where the vertex set Xis the set of vertices of a graph G,and where each edge is either a singleton { x } or a set of vertices on a cycle of G (that belongs to a cycle basis associated with a given forest F of G). Show that H has an equitable bicolouring. 4. Let ( X ; E l , E z , ..., Em)be a hypergraph such that E, n E, # D implies that either E, c E,, or E, c E,.

Let ( Y ; F l , Fz, ..., F,) be a hypergraph with the same property. Show that if X a n d Y are disjoint, then the hypergraph ( X U Y ;El U F l , E2 U F2, ..., E,, U F,) is unimodular. 5. Recall from Theorem (14, Ch. 3) that “if G is a simple connected graph, then for each vertex x o , there exists a spanning tree T and a direction of the edges of G such that: (1) T is an arborescence with root x,,, and

(2) each cycle of the cycle basis associated with T is a circuit.” Let G be a connected graph and let T b e a spanning tree of G. Let E, denote the set of edges on a c y c l ~ belonging e t o the cycle basis associated with T. Show that T can be chosen so that ( E , , E 2 , ...) is a unimodular hypergraph. 6. Let H = (X, 6) be a hypergraph whose vertex set X i s the set of integer points ( x , y ) in the plane with O G x G p , o < y s q

and whose edges are rectangles of the form:

6’) . Show that there exists an equitable bicolouring for hypergraph H. (In general, H is not E(u, a’, 6,6’) = { ( x , y ) / u C x C

U’

;6 C y

Q

unimodular.)

7. Show that a y-perfect hypergraph, uniform of rank h, has an equitable bicolouring. (If the hypergraph is not uniform, then it might not have an equitable bicolouring. See Fig. 20.2.) 8. Let G = ( X , E ) be a simple connected graph, and let d ( x , y ) denote the distance between vertex x and vertex y (see Ch. 4, 94). The chroniatic number yP(G) of distance p is defined to be the smallest number of colours needed to colour the vertices such that no two vertices x and y with 1 < d ( x , y ) G p have the same colour. Show that y P ( G )is the strong chromatic number of some hypergraph. Also show that rp(G) = P 1 if, and only if, one of the following conditions is satisfied:

+

(1)

I XI = P +

1,

BALANCED HYPERGRAPHS A N D UNIMODULAR HYPERGRAPHS

475

(2) G is an elementary chain of length 3 p, (3) G is a n elementary cycle whose length is a multiple of p 1. (F. Kramer, H. Kramer [19691)

+

9. Let H = ( X , b ) be a unimodular hypergraph, and let p , ( j = I , 2, be non-negative integers with b, < c,. F o r K c I, let:

..., n), b , , CI (i E I )

K(x) = { i / i c K ,E 1 3 x } XK,K'- { X / X

E X,

I K'(x) I > I K(.Y) 1 )

.

A. Hoffman has shown that: A necessary and sufficient condition that each vertex x can be assigned a n integer p ( x ) with 0

bi

< p ( x j ) < pi < ~ ( E IQ) ct

I , 2, (iE I ) ,

(j

=

...,n ) ,

is that b ( K ) - c(K') ip ( X , . x e ) for all K , K' c I such that

K n K ' = @ , I

I K ( x ) I - I K'(x) I I < 1

(.u E A').

Show that this condition is necessary even if hypergraph H is not unimodular.

10. Use Theorem 3 to prove the following result: Let X be a set with cardinality kn, and let ( A 1 ,A z , ... , A,) and ( A , , , , A , + z , ..., Azn) be two partitions of X with I A , I = k for all i. Then, there exists a set T such that I T n A , I = 1 for all i. (B. L. van der Waerden, ,,Ein Satz uber Klasseneinteilung von endlichen Mengen," Abh. math. Sem. Univ. Hambirrg, 5 , 1927.)

CHAPTER 21

Matroids

1. Matroid on a set To study axiomatically the properties of linear independence, Whitney [1935] introduced the concept of a matroid. Let E = { e l , e 2 , ..., em} be a finite set and let 9be a set of subsets of E. Set 9is defined to be a matroid on E if the following conditions hold: (1) { e , } E F ( i = l , 2 , ..., m ) (2) F E Y , F ' # F'c F * FIEF (3) For each S c E, the members of F that are maximal in S have the same cardinality.

a,

is called a matroid (on set E ) . Clearly, a matroid on E The pair M = (E, is a hypergraph.(l) In particular, the rank function r ( S ) of a matroid is defined to be r ( S ) = max I F n S 1 . FE9

Axiom (3) implies that each member of the family 9that is contained in S and maximal in S has cardinality r ( S ) . The elements of E are called the elements of matroid M , and the members of F are called the independent sets of matroid M . The independent sets correspond to the edges of hypergraph (E, 9). The subsets of E that do not belong to 9are called dependent sets. A minimal dependent set is called a circuit. Proposition 1. If M = ( E , F)is a matroid with rank r ( E ) , the maximal independent sets of M form a uniform hypergraph of rank r ( E ) . The proof follows immediately from the definitions. If axiom (1) is eliminated, and if 0 ~ 9 then,F is said t o be a matroid in E (or a pregeomefry in the sense of Crapo, Rota). In this case, the matroid is no longer a hypergraph. Later on, it will be evident that Axiom (1) is the least important. With Axiom (l), .F is somewhat more than pregeometry but is not yet a conzbinatorialgeometry in the sense of Crapo, Rota. A combinatorial geometry would require that each subset of two elements

of E also belong to F. In this chapter, we shall confine ourselves to treating only the combinatorial aspects of matroids considered as hypergraphs. The axiom system used here is due to J. Edrnonds. 476

471

MATROIDS

Proposition 2. If M = (E, 9) is a matroid with rank function r(S), then the subhypergraph SA= { F n A / F E 9, F n A # 0 } of M generated by A c E is a matroid of rank rA(S)= r ( S ) . The proof is immediate. Proposition 3. I f M = (E, 9) is a matroid of rank r(S), then the k-section F ( k ) = { F / l < IF1 < k , F E @ ] is a matroid of rank r(k)(S)= min { k , r ( S ) }. The proof is immediate. EXAMPLE 1. The family B’(E) of all the non-empty subsets of a set E is a matroid of rank r ( S ) = I S 1, and its strong stability number equals 1. 1 is also a The family B<,JE)of subsets of E with cardinality < k and matroid because it is the k-section of the preceding matroid. Its strong stability number equals 1. Each circuit is a subset of E with k + 1 elements. 2. Let E be a finite set of vectors # 0. Let 9be the family of sets EXAMPLE is a matroid and the rank r ( S ) of linearly independent vectors. Then, (E, 9) of a set S = E is the dimension of the linear space generated by S. The strong is the maximum number of vectors of E that are stability number of (E, 9) in the same 1-dimensional subspace.

EXAMPLE 3. Let G = ( X , E ) be a multigraph. Let LFbe the family of sets of edges that do not form a cycle. Then, (E, 9)is a matroid with rank r ( S )equal to the cocyclomatic number of the partial graph generated by S. Each independent set of this matroid is a forest of G. Each circuit of this matroid is an elementary cycle of G. EXAMPLE 4. Let G = ( X , E ) be a multigraph without isthmi. Let 9be the family of sets of edges whose removal from G does not increase the number of is a matroid with rank r(S) equal to the connected components. Then, (E, 9) cyclomatic number of the partial graph generated by S. Each independent set of this matroid is a coforest of G. Each circuit of this matroid is an elementary cocycle of G. EXAMPLE 5 (Edmonds, Fulkerson [1965]). Let G = ( X , U )be a graph without isolated vertices. For each matching V = U,let S ( V ) denote the set of saturated vertices in matching V. We shall show that the sets of vertices of G that are contained in S( V )for some matching Vform a matroid M = ( X , F). Let A = X . Consider two sets Fl and F2 of Fthat are contained in A and are maximal in A . We shall show that I Fl I = I F2 I. Suppose the converse occurs, for example, I F2 I > I Fl I. Then

I F2 - FI I > I FI - F2 I

*

478

HYPERGRAPHS

There exist two matchings Vl and V, such that F, = A n SCV,). F, = A n S(V,), At least one connected component of the partial graph generated by

(V1 - V z ) u ( V z is an open elementary chain the other endpoint b in Fl Therefore,

- V,) = p[a, b] with an endpoint

11

- F,,

a in Fz - Fl and from the lemma to Theorem (1, Ch. 7).

v; = ( V , - P ) u ( V 2 n 14 is a matching that saturates all the vertices in Fl u { a }. This contradicts the maximality of Fl, and completes the proof. EXAMPLE 6. Consider a family ( A , / j E Q ) of subsets of a set E such that: A(Q) = U A j = E . j aQ

A subset T = { t l , t,, ..., tk } of E is called a partial transcersal if there exists an injection j ( i ) from { I , 2, ..., k } into Q = { 1, 2, t . ., q } such that ( i = 1 , 2, ..., k ) . ti € A j ( i , We shall show that the family of partial transversals is a matroid on E with rank r(S) = I Q I min (I A ( J ) n S I - I J I ) .

+

JcQ

This matroid is called the transversal matroid of the family ( A , / j E Q). Consider the bipartite graph (Q, E, r)where r is defined by:

W = Aj

( j e P) *

From Example 5, we know that the sets of vertices that are contained in S( V ) for some matching Vdefine a matroid. The family of partial transversals is a subhypergraph of this matroid. Therefore, the family of partial transversals is also a matroid. From the Konig theorem (Ch. 7, g 3), the rank-function of this matroid is:

r ( S ) = min J C

Q

(I

Q-J

I

+ 1 r(J)n S I) = q + min (I A ( J ) n S I - I J I] . J C Q

EXAMPLE 7. If (C, , C,, ..., C,) is a partition of a set E into p classes, and if cl, c z , ..., c, are integers with 0 6 c1 < I C, I, then the family 9 = { F I F c E , F # @ , IFnC,I
is a matroid on E with rank P

foralli)

479

MATROIDS

Clearly, if a set F E 9is contained in S and is maximal in S, then lFnCiI = min(ci,ISnCi)), and its cardinality is P

IF1

=C

min(ci,lSnCiI}.

i= 1

This completes the proof. The following two propositions will be needed later. Proposition 4. If M = (E, 9) is a matroid, its rank function r ( A ) satisfies the following properties: (1)

40)= 0,

(2) r({ x 1) = (3) A c B r(A) < r ( B ) , (4) r(A) + r ( B ) 2 r ( A u B) + r(A n B ) . 1

9

Properties (I), (2) and (3) are evident. To demonstrate Property (4), let F be an independent set contained in A n B with 1 F I = r(A n B). Let FA be an independent set that contains F and is contained in A , with cardinality 1 FA I = r(A). Let Eo be an independent set that contains FA and is contained in A u B, with cardinality I Eo I = r ( A u B ) . Clearly, Eo n A = FA (since FA is a maximal independent set in A ) . Clearly, Eo n ( A n B ) = F (since F is a maximal independent set in A n B ) . Thus,

I

1

r ( A u B ) = I E, I = (E, n A ) u (E, n O ) = = ~ E o n A ~ + ~ E o n B ~ - ~ E o n A n B ~ ~ ~ ~ A ~ + r ( = r(A) + r(B) - r(A n B ) . Hence, Property (4) follows.

Q.E.D. It can aIso be shown that properties (I), (Z), (3), (4) characterize the rank function and could be used as axioms to define a matroid on E.

Proposition 5. Given a matroid M = (E, F),i f F E .F and if F V { a } 9, then F v { a } contains exactly one circuit. Let F b e a minimum independent set that contradicts the proposition. Then, F u { a } contains two distinct circuits C, and C2, and therefore a E C,, a E C,. From the minimality of C , and C,, there exists an element a, E C1 C2and an element a2 E C2 - C , . 1. We shall show that the set A. = F u { a } - { a , , a, 1 is independent. Otherwise, consider the set F' = F - { a , }, that is independent since it is

480

HYPERGRAPHS

contained in F. The set F’ u { a } contains circuit C, and a minimal dependent set of A o . Therefore, F‘ contains two distinct circuits, and since I F’ I < I F 1, this contradicts the minimality of F. 2. The submatroid generated by F u { a } has rank I F I and contains the independent set A o . Since I A . 1 < I F I, it follows that, for i = 1,2,

A,, u { a , } E 9. This is impossible since C, is a dependent contained in A . u { a, }.

Lemma. I f S is a maximal strongly stable set of matroid M a E E - S. Let

=

Q.E.D. (E, F), then

s E S is adjacent to all

s = { s1, s2,

sp}

*..Y

be a maximal strongly stable set. Consider a vertex a E E - S. Since S u { a } is not strongly stable, there exists a vertex s, E S that is adjacent to a. For k # j , vertex s k is adjacent to { a, s, } because the set A = { a, s j , s k } has rank 2, and an independent set that contains sk is contained in a maximal independent set F such that I F n A I = 2. Therefore, a is adjacent to sk for all k. Q.E.D. Theorem 1. If M = (E, 9) is a matroid with strong stability number a ( M ) 2 3 I E 1, then a ( M ) = p ( M ) , cooering number of M. Consider a maximum strongly stable set

s = { s1, s2, ...

)

s,}

.

Let E - S = { a , , a , ,..., a , } . Thus, q

< p. From the lemma, there exists an edge Fij that contains a, and

s j . Since E can be covered with the p edges Fl, F4.m

F,, ,, ..., F,,,, F,,,,,;

...,

then p(M)


= a(M) *

Since p ( H ) 2 a ( H ) for each hypergraph H , it follows that p ( M ) = a ( M ) . Q.E.D.

Theorem 2. A matroid M = (E, 9) is conformal and only i f , there exists a partition ( S , , S,, ..., S,) of E such that 9is identical to the family of nonempty subsets F of E such that

1F n

SiI

<1

(i = I , 2 , ..., q ) .

48 1

MATROIDS

Let S1= { s1 , s z , ..., s, } be a maximal strongly stable set of a conformal matroid M with rank h. It suffices to show that family F has the form required by the theorem. 1. Let Fl be a maximal independent set that contains vertex sl.Let A=E-S, A , = Fl n A . Thus, I Fl I = h, and I Al I = h - 1. 2. We shall show that A l is a maximal independent set in A . Otherwise, there would exist an a E A with A, u{ a } E B . By the lemma, vertices a and s1 are adjacent and, therefore, are contained in a maximum independent set Fa,sl.Since matroid M is conformal, it follows from Theorem (2, Ch. 17) that there exists an Fo E F such that F,

= [ F n ( A , u { a ])I u [ ( A , LJ { a 1)

rl F O , S , f

u (Fa,s,n F ) =

=

A , u { a ) u { SI}.

+

Hence, I Fo I 2 h 1, which contradicts that matroid M has rank h. 3. Since r ( A ) = h - 1, each maximal independent set F satisfies IFnS,I = 1. Consider maximal strongly stable set S, in the submatroid generated by A (that has rank h - 1). Using a similar argument as above, it follows that I F n S , I = 1. In this way, a partition (Sl, S,, ..., S), of E can be defined. Clearly, each maximal independent set F of M satisfies: ( i = 1 , 2 ,..., h ) . IFnSiI= 1 Hence, (Sl, S, ..., S,) is the required partition. 4. Conversely, each set F that satisfies the above equalities consist of pairwise adjacent vertices. Since matroid M is conformal and has rank k, set F is a maximal independent set. Thus family 9 has the form required by the theorem. Q.E.D.

2. The Rado theorem and related results Let d = ( A l , A , , ..., A,) = ( A , / i E Q ) be a family of subsets of a finite set E. A family of distinct representatives of a? is defined to be a family

482

HYPERGRAPHS

( a ( i ) / i E Q) of elements of E such that

(I) i # j =. a(i) # a ( j ) (i = 1 , 2, ...,4). (2) a ( i ) E A i The element a(i) is called the representative of set A i . Clearly, a set of distinct representatives is a partial transversal of cardinality q, but the converse is not true. Consider the bipartite graph ( Q , E, r)with e E T ( i ) if, and only if, e E Ai. A set of distinct representatives is the image of a matching of Q into E (see Ch. 7, 6 3). For J c Q,let A ( J ) = U A,. Recall from Theorem (5, Ch. 7), that a necessary and sufficient condition for the existence of a family of distinct representatives is that IA(J)I 2 I J I

( J cel.

The following theorems generalize this result.

Theorem 3 (Perfect [1969]). Let M = (E, 9) be a matroid of rank r(E). Let k be an integer < r(E), and let a? = (Al, A 2 , ..., A,) = ( A i / i g Q ) be a .family of q subsets of E. A necessary and suflcient condition f o r M to contain an independent set F = { a ( i ) 1 i E K } , K c Q, I K I = k, such that a(i) E Ai for all i E K, is that r(A(J)) 2 I J

I

+k -q

(J c

Q) .

Necessity. If there exists such an independent set F, then r ( A ( J ) ) > , I F n A ( J ) I >, ) K n J l =

li(l

+ 1.11

-IKnJl 2

+

2 k IJI - 9 . Suficiency. Suppose that the condition of the theorem is satisfied. Consider the families &!I = (Bi / i E Q ) with (1)

B , c Ai

( r(B(J))2 I

J

I+k -4

(i E Q) (J

= Q) .

Let B < &!I' signify that Bi c B; for all i E Q. Clearly, the relation < is an ordering. Consider a family &!I = ( B , , B2,..., B,) that is minimal with respect to this ordering. We shall first show that I Bi I = 1 for all i. Suppose that I B1I > 1. Then, there would exist two distinct vertices 6' and b" in B,. Let B; = B , - { b' )

B; = B ,

- { b" }

B!=B!'=B,

if

j # 1 .

483

MATROIDS

From the minimality of g,there necessarily exist two subsets I, J of Q such that: @’(I)) < I I I + k r(B”(J)) < I J I k - q . Hence, @(I)) r(B”(J)) < I I I I J I 2(k - q) - 2 .

+

+

+

+

Moreover,

B’(I) u W ( J ) = B(I u J ) , B‘(I) n B”(J) =I B(Z n J - { 1 1). Combining these results with Proposition 4 yields r(B’(I)) r(B”(J))2 r(B’(Z) u B”(J)) r(B’(I)n B”(J)) 3 r(B(I u J ) ) r(B(I n J - { 1 })) 2 zu J I ~nJ - { 1 } 1 + 2(k - q) 3 I1I + I J I + 2(k - 4 ) - 1 which is a contradiction. Thus, we have shown that 28 is of the form (( bi } / i E Q). If we let B= {bi/iEQ}, then from (1) it follows that r ( B ) = r(B(Q)) 2 I Q I + k - q = k . Therefore, there exists a subset K of Q with I K I = k and an independent set F = { b, / i E K } contained in B, such that:

+

+I

+

+

9

biEAi

(iEK).

Q.E.D.

The well-known Rado theorem follows immediately from this result: Theorem 4 (Rado [1942]). If A4 = (E, .F) is a matroid, then a f a m i 4 at = ( A l , AP, ..., A,) of subsets of E hare an independent set of distinct representatices if, and only $, ( J c Q) . r(A(J)) 2 I J I The proof follows by letting k = q in Theorem 3.

Q.E.D.

Corollary 1. Two families .d= ( A , , A , , ..., A,) and 93 = ( B l , B,, ..., BJ hace a common set of distinct representatives $ and only if,

IA(J)nB(K)) I J I + I Kl - 4 (J, K c Q ) . Consider the transversal matroid M of family 9 (see Example 6).The rank of this matroid is r ( S ) = q + min (I B ( K ) n S I - I K I). K= Q

484

HYPERGRAPHS

From Theorem 4,there exists a transversal set of .dthat is independent in M if, and only if, for all J c Q,

Q.E.D. Corollary 2. If V = (Cl, C z ,..., C,) is a partition of a set E, and i f c l , c2, ..., c, are integers with 0 < ci < I Ci1 for all i, then a faniily s4 = ( A , , A 2 , . .., A,) has a set T of distinct representatives wfth I T n C, 1 < ci f o r all i, $ and only i f ,

Consider the matroid M formed from the sets F c E with I F n CiI for all i (see example 7), whose rank is

< c,

D

r(S) =

1 min { c i , I S n CiI ] . I= 1

There exists a set of distinct representatives of .d that is independent in M if, and only if, for all J c Q, P

r(A(J)) =

C

min { ci, I A ( J ) n Ci I } 3 I J

I.

i= 1

Q.E.D. The Corollary to Theorem (6, Ch. 7), can be restated in terms of sets of distinct representatives as follows: A necessary and suficient condition .for a set B c E to be contained in some set of distinct representatives of a faniily d = ( A l , A 2 , ..., A,) is that min{IA(J)uBI, q - I B - A ( J ) ( } > I J I ( J cQ ) . We shall extend this result to matroids, but first a lemma is needed. Lemma. Let A4 = (E, 9) be a tnatroid of rank r. Let B E 9, and let q 2 I B I. Then, the family

FGe,q = { F / F c E , F u B E F, I F u B I ,< 4 } is a matroid on E with rank

} - IB - S I. E, and let Sobe a subset of S that belongs to F&.If F satisfies the rGeJS)= min { r(S u B ) , q

Let S c two following conditions:

(1)

F EFB,q

(2)

S o c F c S

485

MATROIDS

then, clearly F satisfies I F I d min { r(S u B) , q ] - I B - S I Thus, it remains to show that equality holds for some F.

Fig. 21.1

Set B u So is independent in M and, consequently, is contained in an independent set F' of B u S with cardinality IF') = r ( S u B ) . Let F" be an independent set such that B u So c F" c F' , I F" I = min { r(S u B ) , q ] . Clearly, set F = F'

A

I FI

S satisfies conditions (1) and (2), and its cardinality is = min{ r ( S u B ) , q } - I B

- SI. Q.E.D.

Theorem 5 (Las Vergnas [1969]). Let M = (E, F)be a matroid with rank r. Let B E F , and let q 2 1 B 1. Then, a family &' = ( A l , A , , ..., A,) ofsubsets of E hace a set T of distinct represenfatiliessuch that T E 9and T 3 B i f , and only if, min{r(A(J)uB),q}-lA(J)-

BI>,IJI

(JcQ).

Consider the family

FB,q = { F / F c E , F u B E 9, I F uB I d 4 1. By the lemma, this family is a matroid on E. If there exists a set Tof distinct representatives of&' with T E .Fand T then I T 1 = q. Therefore, T E FB,q

3

B,

*

Conversely, if there exists a set T of distinct representatives of d without T E FB, , then T u B E F , I T u B l d q , I TI = 4 ,

486

HYPERGRAPHS

and, consequently,

TEB, T 2B. Then, from the Rado theorem, there exists a set T that satisfies the above requirements if, and only if, the rank rB,qof the matroid ( E , SBvQ) satisfies rs,,(A(J))2 I J

I

(J

c

Q)

or

min(r(A(J)uB),q) -

IB-A(J)[

2

IJI. Q.E.D.

3. Image of a matroid Many of the ideas to be presented in this section originated in the work of Edmonds and can be found in Edmonds, [1965], [1970]. Let M = (E, F)be a matroid on E = { e l , e,, ..., en,}, and let cp be a mapping of E onto a set E. Consider the family -

9 = (cp(F) / F E 9). Since cp is a mapping onto E, the pair R = (E, 8 )is clearly a hypergraph, called the image of M . Theorem 6 (Nasi-Williams 119661). Zf R = (E, 9) is the image of a matroid M = (E, 3)by a mappitig cp of E onto E, theti R is a matroid, and its rank is:

I.(E) = min (r(q-' (A))+ I E - A I) . ACE 1. We shall show that

-inax - IF I F € f

Clearly, max

+ I E - A 1).

= min ( r ( c p - ' ( ~ ) ) ACE

I F I is the greatest integer k such that the family

(v-'(G), (P-*(ez), *..>v-'(L)) has a partial set of distinct representatives that is independent in M and has cardinality k. From Theorem 3, this is the greatest integer k such that min (r(q-'(A))+

IE

-

Al)2 k .

ACE

Hence, max I F I = inin ( r ( q - ' ( i ) )+ I E AC

21).

E

2. To complete the proof, it suffices to show that the image of M by cp is a matroid.

481

MATROIDS

Consider a mapping such that

of E = { el, e,, ..., em } onto

'p

E

=

{ Z2,&,

-

..., em1

d e l ) = Z2 'p(ei) =

Zi

for i # 1

,

Mapping cp is said to be an elementary contraction. Since each mapping is a product of elementary contractions, it suffices to show that the image of matroid M by an elementary contraction cp is a matroid. We shall Consider an independent set Fo E .F such that Fo is maximal in 9. show that Fo is a maximum set. From part 1, it suffices to show that there exists a set c E such that

lF0 I

= r(cp-'(A))

+I E -A/.

In order to simplify, let Eo = E - { e l , e2 }. Since Fo is maximal in 9, we may assume that Fo is a maximum set of F. We may also assume that Fo contains both el and e, because, otherwise, Fowould be maximum, since

1 Fo I

= I F,

I = r(E) = r(qD-l(E)) + I E - Zo1 .

Three cases must be considered: CASE1. r(Eo) = r(E). Since

G ' ( E - { el 1) G r(E) 9 it follows that r(E - { el )) = r(E). Since Fo contains both el and e2, then @o)

I F~ - { el 1 1 = r(E) - I < r(E - { el }) . Consequently, there exists a maximum independent set FA that does not contain el and that satisfies: -

Fb

Thus,

-

3

F,

- { e, } = F o .

Fo = F i , and

I Fo I

=

I %I

= I F;

I = r(E) = r(cp-'(E))

+ I E - Eo I .

Hence, the required equality is proved. CASE2. r(Eo) = r ( E ) - 1. Then, each maximum independent set contains either e, or e2. Moreover, I Fo n Eo I = I Fo I 2 = r(E) - 2 < r(E,).

-

Thus, there exists a vertex a E Eo such that (Fon Eo) U { a } is an independent set with cardinality r(Eo) = r ( E ) - 1. Let FA be a maximum independent set that contains this set. Since FA is maximum, it contains el or e2.

488

HYPERGRAPHS

Without loss of generality, we may assume that

>.

Fb = (Fo n E o ) u ( a , e ,

Fo.Therefore, Fo = FA,and 1 F b I = I F, 1 = r(E) = r(cp-'(E)) + I E - E I .

Clearly, FJ2

Hence, the required equality is proved. CASE3. r(Eo) = r ( E ) - 2. Then, each maximum independent set contains both el and e,. Consequently, Fo 2 F, and

I Fo I = I F , 1 - 1 = r(E) - 1 = r(E,) + 1 = r(p-'(&)) + 1 E - E, I Hence, in all cases, Fo is a maximum set of 9. c

Q.E.D. If H 1 = ( X , Sl),H 2 = ( X , S2),..., H P = ( X , 8")are hypergraphs on a set X,their join is defined to be the hypergraph H

=

H 1 v H a v - v H P = (A',

P

V

&'),

1=1

where : P

v

8' = { E'

LJ

E2 v

*--

u E P / E' €8'' ; E 2 € E 2 ; ... : EP E P} .

i= 1

Clearly, H is a hypergraph on .'A

Theorem 7. If (E, Sl), (E, P) ..., ,(E, 9") are matroids whose ranks are respectively rl, r2, ..., rp, then their,join hypergraph is a matroid, and its rank is -

r(E) = min (r'(A) + ... + rp(A)+ I E - A

I).

AcE

Make p identical copies E l , E 2 , ..., E P o f set E. Let ef, denote the element of El that corresponds to ek E E. Consider the mapping cp of Uf=lE' into E that maps e: E E' onto e,. Clearly, M = (U E', V F j ) is a matroid, and its rank is r(U E,) = r l ( E ' ) r2(E2) ... r P ( E P ) .

+

+ +

a,

From Theorem 6, the image of this matroid by cp is a matroid and R is clearly the join Vfml9'Moreover, . the rank of the joint matroid is

F(E)

=

min ( r ( v - ' ( A ) )

+ I E - A I)

ACE

Q.E.D.

489

MATROlDS

Corollary 1 (Edmonds [1968], Nash-Williams [1968]). For a matroid M = ( E , F),the cocering number p(M), i.e. the minimurii number of independent sets that are needed to cotrer E is equal to: p ( ~ =) max

[---I

IAl

*.

r(A)

ACE A # 0

By definition, p ( M ) is the smallest integer k such that the join matroid V M V *.. V M of k identical copies of matroid M has rank I E I. From Theorem 7, p ( M ) equals the smallest value of an integer k such that M

min (kr(A)

+ I E - A I) = I E I ,

ACE

or, equivalently, such that min ( k r ( ~-) I A

I) = 0.

A cE

This is equivalent to: kr(A) = I A

I20

( A c E),

or

k > 'r ( A )

a>.

(A c E , A #

Thus, p ( ~ =) max

[-dI A ) * IAI

ACE A # r'

'

Q.E.D. Corollary 2. For a matroid M = (E, F), the maximum number k , of pairwise disjoint maximal independent sets is

r(A)fr(E)

Clearly, ko equals the greatest integer k such that the joint matroid of k identical copies of matroid M has rank kr(E), where min (kr(A) + 1 E - A

I)

= kr(E)

.

ACE

This is equivalent to rnin (kr(A) - kr(E)

+ I E - A I) = 0,

ACE

or to

-

k(r(E) r ( A ) ) < I E - A I This yields the required equality.

(A c E). Q.E.D.

490

HYPERGRAPHS

Corollary 3. Consider a matroid M of integers szrch that

=

(E, S)and a sequence k l , k,,

..., k,

a

Let k: denore the number of k, that are 2 j . Then, set E can be partitioned into q independent sets Fl , F2,..., F,, with I Ff I = k, for all i, if, and only if,

C

k;<JE-AJ

(ACE).

j>r(A)

Consider the ki-section M ( k l )composed of the family .F(ki)

=

{ F / F E 9, IFI

< ki 1 .

It is a matroid, and this matroid has rank r'(A) = min { r(A), k, }. There exist q sets that satisfy the conditions of the corollary if, and only if, the join matroid M = Vp= M(kl)has rank I E I. This is equivalent to

which is equivalent to 4

1

4

min(r(A),ki}+IE-A121EJ=

C i= 1

i=l

k;

ki=

(ACE)

j>O

or to (Ac E).

Q.E.D. The corollary follows. A!

Corollary 4. Let H M be a hypergraph consisting of the circuits of a rnatroid = (E, 9) of rank r. Then the chromatic number of H M is

A#

0

A set F c E i s independent if, and only if, it contains no circuits. Therefore, a partition ( A l , A 2 , ..., A,) is a colouring of hypergraph H M if, and only if, sets A l , A , , ..., A , are independent. Thus, z ( H M )= p ( M ) , and the corollary follows from Corollary 1. Q.E.D. The above results easily yield several theorems for graphs that were proved earlier by direct but very involved methods.

49 1

MATROIDS

APPLICATION 1 (Theorem of Tutte [1961]). The edge set of a simple connected graph G = (X,E ) contains k pairwise disjoint spanning trees i f , and only i f , for each partition 9 of X,the number mc(P)of edges that join two distinct classes of the partition satisfies

2 k(l 1 - 1 ) . 1. If there exist k spanning trees H I , H , , ..., Hkthat are pairwise edgedisjoint in G , then for each partition B of the vertices, filG(9)

m H , ( 9 )2 191 - 1

( i = 1 , 2 ,...,

k).

Therefore, k

mG(9>

2

mH,(9L) I=1

2 k(l 9 I - 1)

9

which is the required condition. 2. Suppose that the condition of the theorem is satisfied. Consider the where 9is the family of forests F l , F,, ..., F, of graph matroid M = ( E , 9) G. Let r(A) denote the rank of M . If A C E defines a partial graph of G with p connected components that constitute a partition 9'= (XI, X , , ..., X,) of X,then r(E) - r ( A ) = ( n - 1) - (n - p ) = p - 1 = 191 - I , Therefore, from the condition of the theorem, it follows that

1E - A I

> m G ( 9 )> k(l 9 I - 1)

= k(r(E) - r ( A ) ) .

Hence,

Thus, from Corollary 2 to Theorem 7, there exist k disjoint spanning trees in G. Q.E.D. APPLICATION 2 (Theorem of Nash-Williams [1964]). The edges of a siiizple graph G = (X,E ) can be coloured with k colours sirch that no cycle has all its edges with the same colour i f , and only if, for each set A c X , the number mG(A,A ) of edges with both endpoints in A satisfies

mG(A, A ) < k(l A 1 - I ) . In other words, the chromatic number of the hypergraph Gc formed consisting of the cycles of G is

492

HYPERGRAPHS

1. Suppose that the edges are coloured with k colours 1,2, ..., k. Let mi@, A ) denote the number of edges with colour i that have both endpoints in A . These edges form a forest. Therefore, nr,(A, A ) < I A I - 1. Hence,

m,(A, A )

=

rn,(A, A )

+ ... + n1,(A, A ) G k(l A I - 1) ,

and the result follows. 2. Conversely, suppose that the condition of the theorem is satisfied. Conformed by the family of forests of graph G. Let r desider the matroid ( E , 9) note the rank of this matroid. If the partial graph ( X , F) of G generated by F c E has exactly p connected components (XI, Fl),(X2, F2),..., (Xp,Fp) that are not isolated vertices, then kr(Fi) - I Fi I 2 k(l

XiI

- 1) -

rnG(Xi,X i ) 2 0 ,

Hence, P

kr(F) - I F

1

=

1 (kr(Fi) - 1 F iI) 2 0, i= I

and

k2max[!$]*. FCE F+ 0

Thus, from Corollary 4, it is possible to colour the edges of G in k colours such that no cycle has only one colour.

Q.E.D. APPLICATION 3. f l G is a simple graph with maximum degree h, then it is

"ll*+

possible to colour its edges with edges with the same colour.

1 colours such that no cycle has all its

Let G = ( X , E ) be such a graph, and let A = A - { a } . Thus

C

X . If I A

I

> 1. For a E A ,

let

From the Nash-Williams theorem (Application 2), it follows that

493

MATROIDS

G I*+

Therefore, it is possible to colour the edges of G with -

1 colours such

that no cycle of G has all its edges with the same colour.

Q.E.D. 4. Minimum weight basis

A maximal independent set of a matroid M is called a basis. Associate with each element e of matroid M a positive number p(e) called the weight of e. The ueight of a basis F o f M is defined to be P(F) =

c P(4.

esF

We shall consider the problem of determining a minimum weight basis of a matroid.

EXAMPLE 1. Minimum tree in a weighted graph. Let G = ( X , E ) be a connected simple graph. To each edge e associate a “length” p(e) > 0. The minimum tree problem is to find a spanning tree H = ( X , F ) of G whose total length c EP

is as small as possible. This problem due to G. Choquet [1938] and J. B. Kruskal [1956] has numerous applications. For example, n cities in the Benelux countries must be connected by canals, and the total length of the canals must be minimized (and no branching is allowed except at the cities). Since the system of canals must be connected and since its length must be minimum, no cycles are allowed. Thus this problem becomes a minimum tree problem. The minimum tree problem reduces to finding a minimum weight basis for formed by the forests of G. (See Example 3, Q 1.) the matroid (E, 9)

EXAMPLE 2. The following problem studied by Chein [1970] also reduces to a minimum tree problem: Consider a connected simple graph G = ( X , E ) with a positive length y ( e ) associated with each edge e, and a set S c X . Find a tree in G (not necessarily spanning) that contains all the vertices of S and that has a minimum total length. When all lengths are equal to 1, this problem can be solved by treating separately each of the connected components X , , X,, ..., X , of the subgraph Gs, that respectively have spanning trees (XI, F J , ( X 2 ,F J , ..., ( X p ,F,), and by finding a spanning tree F, of minimum weight in the graph obtained from GbecontractingX,, X , , ..., X,.Then,F, u F , u . - u F,istherequiredtree.

494

HYPERGRAPHS

Let F = ( e l , e,, ..., e,) and F’ = ( e i , ek, ..., e:) be two ordered bases of the elements of these r-tuples are indexed such that: matroid M = (E, 9); p(el) 6 p(e2> 6

... < p ( e , ) ,

< p(e;) 6

G p(e:)-.

p(e;)

Let F

< F’ if, and only if, there exists an index k < r such that

i

p(eJ = p(e;)

for

i < k

p ( 4 < pk;) . This order relation < on the set of bases is called a lexicographic ordering: i f the “weight” of a letter is its rank in the alphabet, then the relation F < F‘ between two words F and F’ indicates that word F appears before word F’ in the dictionary.

M

Proposition 1. If F = ( e l , e,, ..., e,) is an ordered basis of a rnatroid = (E, .F),then the following three conditions are equiralent : (1) F is a minimal basis vith respect to the lexicographic ordering, c(, ( 2 ) each ordered basis F‘ = (e;, eh, ..., ek), satisfies p(e,) < p(e,’)for all i, ( 3 ) F is a minimum weight basis.

(1) 3 (2). Otherwise, let F be a minimal basis, and let F’ be a basis that does not satisfy (2). Let k be the first index such that p(eJ > p(ek) A = { e,,e,,

..., e k - * , e ; , e > , ..., e ; } ,

has rank r(A) > k, and consequently there exists an index i set ( e , , e 2 ,..., e , - , , e J ) = F , is an independent set of cardinality k. Moreover, since i

p ( 4 ) < p(ek). contains Fo with I B I = r.

< k such that the

< k, we have:

p(el) 6

Let B be the basis that contradicts the minimality of F.

Thus, B

< F,

which

(2) 3 (3). This is obvious.

(3) => (1). Let F be a minimal basis with respect to the lexicographic ordering, and let F’ be a minimum weight basis. Since (1) implies (2), basis F satisfies (2), and (i = 1, 2, ..., r ) . p(eJ < p(e9 Since F’ has minimum weight, it follows that p(ei) = p(e{)

(i = 1, 2, ..., r ) .

495

MATROIDS

Thus F' < F ; since F is minimal with respect to the ordering ci, it follows that F' is minimal with respect t o the lexicographic ordering a. Q.E.D. Algorithm 1. To construct a minimum weight basis, first choose an element el of minimum weight, and let Fl = { el }. Then choose any element e2 with e, 4 Fl and Fl U { e2 } E 9, that has minimum weight; let F2 = Fl u { e2 } etc. After r steps, a basis F, is determined. Since F, is minimal with respect to the lexicographic ordering, it is a minimum weight basis by Proposition 1. Dual algorithm. If F i s a basis of matroid A4 = (E,.F) then , its complement E - F is called a cobasis. Clearly, the cobases of M are the bases of some matroid M* called the dual matroid of M . If M is a matroid whose bases are the spanning trees of a connected graph, then the cobases of M are the cotrees of this graph. Instead ofconstructing a minimum weight basis for matroid M with weights p(e), we could construct the maximum weight cobasis or construct a minimum weight basis for the dual matroid M * with weights p*(e) = p o - p(e), where p o is a suficiently large positive number such that p*(e) > 0 (e E E ) . T o construct a maximum weight cobasis for M, first remove from E an element e, of maximum weight. Let E, = E - { e l }. Then, remove an element e2 E Elthat has maximum weight. Set E2 = El - { e, } also has rank r. After m - r steps, a maximum weight cobasis E and a minimum weight basis have been found.

For the special case of constructing a minimum length spanning tree of a graph G, the above algorithms can be improved. First, we need the following proposition : Proposition 2 (Rosenstiehl [1967]). Let G = ( X , U ) be a connected graph with a length p(u) associated uwth each arc u. Suppose that the lengths are all diyerent. Then, the tree of miniminn length is unique. A necessary and suficient condition that arc u belongs to this tree ( X , V ) is that there exists an elementary cocycle o in G such that p(u) = min p(u)

.

UECD

Another necessary and suficient condition that arc u belongs to tree ( X , V ) is that f o r each cycle p,

496

HYPERGRAPHS

1. The minimum tree is unique. Otherwise, if V = ( u , , u,, ..., u,) and W = ( w , , wZ,..., w;) were both minimum trees, then from Proposition 1, (i = 1, 2, ..., r ) p(ui) < p(wJ < d o i ) This contradicts the assumption that the arc weights are all different. 2. If arc u belongs to the minimum tree (X,V ) , then u separates two sets A and X - A , clearly, p(u) = min p ( u ) . u E m(A)

(Otherwise, there would exist an arc u' # u in the minimum tree chosen by the algorithm 1 to link sets A and X - A . ) Therefore, there exists a cocycle w(A) that satisfies (1). 3. If arc L' belongs to a cocycle w that satisfies (l), then u also belongs to an elementary cocycle w(A) that satisfies (l), and u would be the first edge of w(A) to be chosen by Algorithm 1 to construct the minimum spanning tree V. Therefore, t i E V. The proof of condition (2) is left to the reader. Algorithm 2 (Kruskal [1956], Sollin [1962]). A sequence V,, V,, ... of forests can be successively formed in the following way: (I) Let V , = forest ( X , V,) consist of n isolated vertices. (11) If forest ( X , V,) has an isolated vertex x, then join x to its closest neighbour; i.e., if u i C l is the arc of minimum length incident to x, then let

a;

V i + l = V,U { I ' i + l 1 . (111) If forest ( X , V,) has no isolated vertices and is not connected, then

join a connected component to its closest connected component. (IV) If ( X , V,) is connected, then it is a spanning tree and, clearly, this tree has minimum length by Proposition 2. Algorithm 3 (P. Rosenstiehl [1967]). Starting with a cocycle basis { w1,w2, ..., w k } remove from each of these cocycles its minimum weight arc. Then, proceed as in Algorithm 2. Dual algorithm. Starting with a cycle basis { p l , pz, . .., p' } (for example, if G is planar, the contours of the bounded faces), remove successively from each cycle an arc with non-maximum weight. Continue until no cycles remain.

EXERCISES 1. Show that the set 9 of circuits of a matroid satisfy the following: (1) S' c S, S E Y implies S' = S (2) S, S' E 9, S # S, a E S n S' implies that there exists an So E Y with So c S U S' - {a}.

497

MATROIDS

Conversely, show that if a family of sets satisfies (1) and (2), then the family is the set of circuits of some rnatroid. 2. Let Kn denote the complete graph with n vertices, and let K: denote the hypergraph consisting of the cycles of K , . Show that the Nash-Williams theorem implies that

Show how to construct a colouring of the edges of graph K,, in no cycle has all its vertices with the same colour. 3. Let K,, , denote the complete bipartite graph. Show that

4. Let M be a matroid on set E. Show that if 9 is the family of its bases then { E - B / B E } is the set of bases of matroid M . 5. Let G = ( X , P) be a 1-graph, and let Bo be a subset of X with cardinality k. Let d = ( P(x) u { x } / x E X - So}. For a set B C X with cardinality k, show that the following conditions are equivalent : (1) Sets B and Bo can be joined together by k paths that are pairwise vertex disjoint. (2) The set X - B is a set of distinct representatives of family d (see Section 2). 6. Use Exercises 4 and 5 to show the following result: Let G = (X,T)be a 1-graph, and let Bo C X. Then the sets B c Xsuch that I B I = I Bo I and such that there exist 1 Bo I pairwise vertex-disjoint paths joining B and Bocon-

stitute the bases of a matroid on X.

(H. Perfect, Applications of Menger’s Theorem, Math. Anal. Appl., 22, 1968, pp. 96-111.)

7. Let G = (X,T)be a transitive 1-graph, and let S be the family of subsets Fof X such that F 1 ( X - F ) 3 F. Show that P is a matroid on X. 8. Let G be a multigraph. Let 9 be the family of subsets of the edge set that do not contain an even elementary cycle and that d o not contain two edge-disjoint odd elementary (A. Duchamp, Thesis, Caen [1971]) cycles. Show that S is a matroid on E.

9. In the (directed) graph G = (X,U ) , let P be the family of subsets Uo of U = ..., urn} that do not contain an “anti-circuit” (a simple even cycle whose arcs are alternatively directed). Show that 9 is a rnatroid on U.Determine its rank. (A. Duchamp) Hint: Consider the graph H = ( Y , V) with Y = { a,b }, and with m arcs v1, v 2 , ..., urn directed from a to 6. Consider the “product” G 8 H, i.e. the graph on X x Y with an arc from ( x , y) to (x’, y‘) denoted by wk, if (x, x’) = uK and ( y , y’) = Uk. Clearly, each anti-circuit of G is a cycle of G 8 H, and each cycle of G 8 His an anticircuit of G. 10. Show that if (E, F)and (E, 9’) are two rnatroids on E whose ranks are respectively r and r’. then max I S I = [r(A) r’(E - A)] . { ulr ua,

ss9n-F’

y2

+

(This is the “Matroid Intersection Theorem” of J. Edmonds).

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s.

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M. RICHARDSON, Extensions Theorems for Solutions of Irreflexive Relations. Proc. Nut. Acad. of Sciences U. S . A . , 39, 1953, 649-651. B. ROY,Algkbre moderne et thkorie des Craphes. Volume 2, chapter VI, Dunod, Paris, 1970. S. RUDEANU, Notes sur I’existence et I’unicite du noyau d‘un graphe. Revue Franqaise Rech. OpCrat., 33, 1964, 20-26 and 41, 1966, 301-310. C. A. B. SMITH,“Graphs and Composite Games”. A Seminar on Graph Theory (F. Harary-L. Beineke), Holt Rinehart, Winston, New York, 1967,86-111. R. SPRAGUE, Bemerkungen iiber eine spezielle Abelsche Gruppe. Math. Z . , 51, 1947, 82-93.

L. P. VARVAK, Existence of a Kernel in a Product (Russian), Ukr. Mat. Zh., 17, NO. 3, 1965, 112-114. L. P. VARVAK,Games on a Sum of Graphs (Russian). Kiber., 4, No. 1, 1968, 63-66.

V. G. VIZING,A Bound on the External Stability Number of a Graph. Doklady A. N. 164, 1965, 729-731. V. G. VIZING,The Cartesian Product of Graphs. Vych. Sis., 9, 1963, 3 0 4 3 . P. M. WECHSEL, The Kronecker Product of Graphs. Proc. Am. Math. SOC.,13, 1962, 47-52.

C. P. WELTER,The Theory of a Class of Games on a Sequence of Squares, in Terms of the Advancing Operation in a Special Group. Proc. Kon. Nederl. Akad. N. (sCrie A), 57, 1954, 194-198. N. Y. WITHOFF, A Modification of the Game of Nim, Nieuw Arch. voor Wiskunde,

7, 1907, 199-202. CHAPTER 15 : Chromatic Number C. BERGE,Les problkmes de colorations en ThCorie des Graphes. Pub/. Insf. Stat. Univ. Paris, 9, 1960, 123- 160. G. BERMAN, W. T. TUTTE,The Golden Root of a Chromatic Polynomial. J . Combinatorial Theory, 6 , 1969, 301-302. G. D. BIRKHOFF, A Determinant Formula for Coloring a Map. Ann. Math., 14, 1912, 42-46. G. D. BIRKHOFF, D. C. LEWIS,Chromatic Polynomials, Trans. Amer. Math. SOC., 60, 1946, 355451.

R. L. BROOKS, On Colouring the Nodes of a Network. Proc. Cambridge Phil. Soc., 37, 1941, 194-197. V. CHVATAL,A Note on Coefficients of Chromatic Polynomials, J. Combinat. Theory, 9, 1970, 95-96. V. CHVATAL,J. K O M L ~ SSome , Combinatorial Theorems on Monotonicity, Canad. Math. Bull., 14 (2), 1971.

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E. A. NORDHAUS, J. W. GADDUM,On Complementary Graphs. Ann. Marh. Monthly, 63, 1956, 175-177. 0. ORE,The Four Color Problem. Academic Press, New York, 1968. F. P. RAMSEY, On a Problem of Formal Logic. Proc. Lotidon Math. SOC.,30, 1930, 264-286. D. K. RAY-CHAUDHURI and R. WILSON,Solution of Kirkman’s Schoolgirl Problem, Proc. Symposium Pure Math., 19, A.M.S., Providence, 1971, 187-203. R. C. READ,An Introduction to Chromatic Polynomials. J . Combinatorial Theory, 4, 1968, 52-71. G. RINGEL,Farbiingsprobleme aiif Flacheti iind Graphen. Berlin, 1959. G. RINGEL, J. W. T. YOUNGS,Solution of the Heawood Map-coloring Problem, J . Comb. Theory, 7 , 1969, 342-363. G. C . ROTA,On the Foundations of Combinatorial Theory, I, W. V. C., 2, 1964, 340-368. B. ROY,Nombre chromatique et plus longs chemins d’un graphe. Revue AFIRO, I, 1967, 127-132. M. SCHAUBLE, “ Bemerkungen zu einem Kantenfarbungsproblem”. Beitruge ziir Graphentheorie, Leipzig, 1968, 137-142. W. T. TUTTE,On Chromatic Polynomials and the Golden Ration. J. Combinatorial Theory (to appear). K. WAGNER, Bemerkungen zu Hadwiger’s Vermutung. Math. Ann., 141, 1960, 433-45 1. K. WAGNER, Beweis einer Abschwachung der Hadwiger Vermutung. Math. Ann., 153, 1964, 139-141. H. WHITNEY, A Logical Expansion in Mathematics. Bull. Amer. Math. SOC.,38, 1932, 572-579. H. WHITNEY, The Coloring of Graphs. Ann. of Math., 33, 1932, 688-718. C. E. WINN,A Case of Coloration in the Four Color Problem. Am. J . Math., 49, 1937, 515-528. A. A. ZYKOV,On Some Properties of Linear Complexes (Amer. Math. SOC. Transl. No. 79, 1952). Mat. Sb., 66, 1949, 163-188. CHAPTER 16: Perfect Graphs

0. ABERTH,On the Sum of Graphs. Reaue Fr. Rech. OpPrationiielle, 33, 1964, 353-358. M. H. MCANDREW, On the Product of Directed Graphs. Proc. Am. Math. SOC., 14, 1963, 600-606. C . BERGE,Les problemes de coloration en thkorie des graphes. Publ. Inst. Stat. Univ. Paris, 9, 1960, 123-160. C. BERGE,“ Farbung von Graphen, deren samtliche bzw. deren ungerade Kreise starr sind”. Wissenschaftliche Zeitutig, Martin Luther Utiia., Halle Wittenberg, 1961, 114. C . BERGE,Sur line conjectiire relatioe aii problkme des codes optirnaiix, cornm. 13 &meassemblke generale de I’URSI, Tokyo, 1962. C. BERGE,“Perfect graphs”, I. Six Papers on Graph Theory, Indian Statistical Institute, Calcutta, 1963.

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C. E. SHANNON, The Zero-error Capacity of a Noisy Channel. Comp. Information Theory, I R E Trans., 3, 1956, 3-15. L. SURANYI, The Covering of Graphs by Cliques. Studiu Sc. Math. Hmg., 3, 1968, 345-349. V. G. VIZING,The Cartesian Product of Graphs. VyC. Sis., 9, 1963, 3043. P. M . WECHSEL, The Kronecker Product of Graphs. Proc. Am. Math. Soc., 13, 1962, 47-52. E. S. WOLK,A Note on the Comparability Graph of a Tree. Proc. Am. Math. Sac., 16, 1965, 17-20.

CHAPTER 17 : Hypergraphs and their Duals A. ANDREATTA, Sur les graphes finis qui sont des adjoints (Italian). Ist. Lombard0 Ac. Sc. Lett. Rend., A98, 1964, 133-156. L. W. BEINEKE,On Derived Graphs and Digraphs”. Beitruge zur Graphentheorie (H. Sachs, H. J. Voss, H. Walther, ed.), Teubner, 1968, 17-23. C. BERGE,Le graphe adjoinr d’un graphe. Mimeo. Seminaire sur les problbmes combinatoires, I. H. P. 1959. C. BERGE,“The Rank of a Family of Sets and Some Applications to Graph Theory”. Recent progress in Combinatorics (W. Tutte, ed.), Academic Press, New York, 1969, 49-57. C. BERGE,R. RADO,On Isomorphic Hypergraphs, and on Some Extensions of Whitney’s Theorem to Families of Sets (to appear in J . Conib. Theory). J. BOSAK,“The Graphs of Semi-groups’’ Proc. Symposium Smolenice, Prague I‘

1964, 119-125.

R. G. BUSACKER, T. L. SAATY, Finite Graphs and Networks. McGraw-Hill, New York, 1965. G. CHARTRAND, Graphs and Their Associated Line-graphs. Thesis, Michigan State Univ., 1964. K. CULIK,“Applications of Graph Theory to Mathematical Logic and Linguistics ”. Proc. Symp. Graph Theory, Smolenice, 1963, 13-20. P. ERDOS,On Extremal Problems of Graphs and Generalized Graphs. Israel J . Math., 2, 1964, 183-190. P. ERDOS,A. W. GOODMAN, L. P ~ s A The , Representation of a Graph by Set Intersections. Canad, J . Math., 18, 1966, 106-1 12. P. C. GILMORE, A. J. HOFFMAN, A Characterization of Comparability Graphs. Canad. J . Math., 16, 1964, 539-548. B. GRUNBAUM, “Incidence Patterns of Graphs and Complexes”. The Many Facets of Graph Theory (G. Chartrand, S. F. Kapoor, ed.), Springer Verlag, 1969, 115 128. B. GRUNBAUM, Convex Polytopes, Wiley, New York, 1967. A. GYARFAS,J. LEHEL, “ A Helly-type Theorem on Trees”, Combinutorial Theory and its Applications (Erdos, Rknyi, V. T. S ~ Sed.), , North-Holland Publ. Co., Amsterdam-London, 1970. R. K. GUY,E. C. MILNER, Graphs Defined by Coverings of a Set. Acta. Math. Ac. Sc. Hung., 19, 1968, 721-735.

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F. HARARY, Graph Theory. Addison-Wesley, Reading, 1969. P. HELL,J. NESETRIL,Graphs and k-societies. Canad. Math. Bull., 13, 1970, 375381. C. HEUCHENNE, Sur certaines correspondances entre graphes. Bull. SOC.Roy. Sc. Likge, 33, 1964, 743-753. A. J. HOFFMAN, On the Uniqueness of the Triangular Scheme. Ann. Math. Stat., 31, 1960, 492-497. A. J. HOFFMAN, On the Line Graph of the Complete Bipartite Graph. Ann. Math. Stat., 35, 1964, 883-885. J. KRAUSZ,Dtmonstration nouvelle d‘un thCor&mede Whitney sur les rkseaux. Mat. Fiz. Lapok, 50, 1943, 75-85. G. KREWERAS,Counting Problems in Dendroids ”. Combinatorial structures and their Applications, Gordon and Breach, New York, 1970, 223-226. W. KUISHand N. SAUER,“On the Existence of Certain Minimal Regular nsystems with Given Girth”. Proof Techniques in Graph Theory (F. Harary, ed.), Academic Press, 1969, 93-101. L. LovAsz, “Graphs and Set Systems”. Beitrage zur Graphentheorie (H. Sachs, H. J. Voss, H. Walther. ed.), Teubner, 1968, 99-106. V. V. MENON,The Isomorphism between Graphs and their Adjoint Graphs. Canad. Math. Bull., 8, 1965, 7-15. V. V. MENON,On Repeated Interchange Graphs. Am. Math. Monthly, 73, 1966, 986-989. J. MOON,On the Line-graph of the Complete Bigraph. Ann. Math. Stat., 34, 1963, 664-667. A. RAMACHANDRA RAO, Some Extremal Problems and Characterizations in the Theory of Graphs. Thesis, I. S . I., Calcutta, 1969, D. K. RAY-CHAUDHURI, Characterizations of Line-graphs. J. Comb. Theory, 3, 1967, 201-214. F. S . ROBERTS,“Indifference Graphs”. Proof Techniques in Graph Theory, Academic Press, 1969. F. S . ROBERTS,On the Boxity and Cubicity of a Graph. Recent Progress in Combinatorics, (Tutte, ed.) Ac. Press, 1969, 301-3 10. F. S . ROBERTS, J. H. SPENCER, A Characterization of Clique Graphs. J. Comb. Theory, 10, 1971, 102-108. A. C, M. VAN ROOIJ,H. S. WILF,The Interchange Graph of a Finite Graph. Acta Math. Sc. Hiingar., 16, 1965, 262-269. G. SABIDUSSI, Graph Derivatives. Math., 2, 76, 1961, 385-401. S. SESHU,M. REED,Linear Graphs and Electrical Network, Addison Wesley, 1961. S . S. SHRIKHANDE, On a Characterization of the Triangular Association Scheme. Ann. Math. Stat., 30, 1959, 39-47. H. WHITNEY, Congruent Graphs and the Connectivity of Graphs. Amer. J.Math., 54, 1932, 150-168. ‘I

CHAPTER 18 : Transversals L. W. BEINEKE, “ A Survey of Packings and Coverings of Graphs”. The Many Facets of Graph Theory (G. Chartrand, S . F. Kapoor, ed.), Springer Verlag, 1969, 45-53.

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CHAPTER 19 : Chromatic Number of a Hypergraph C. BERGE, Sur certains hypergraphes genkralisant les graphes bipartites, Combinatorial Theory and its applications, Balatonfiired (P. Erdos, A. RCnyi, V. T. Sos ed.), North-Holland Publ. Co., Amsterdam-London, 1970, 119-133. G. CHARTRAND, D. P. GELLER, S. HEDETNIEMI, A Generalization of the Chromatic Number. Proc. Cambridge Phil. S o c . , 64, 1968, 265-271. V. C HV~TAL, Hypergraphs and Ramseyian Theorems, Proc. Am. Math. SOC.,27, 197 1,434-440. V. CHV~TAL, The Smallest Triangle-free 4-chromatic 4-regular graph., J. Combinat. Theory, 9, 1970, 93-94. B. DESCARTES, A Three-colour Problem. Eimka, April 1947 (solution, March 1948). P. ERDOS,Some Remarks on the Theory of Graphs, Birll. A.M.S., 53, 1947,292294.

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Index of Definitions (English, French, German)

The following index gives the page numbers of the terms defined in this text and their French and German equivalents. Synonyms are given in italics. For more specific definitions, the reader is referred to the following excellent glossaries :

J. W. Essem, M. E. Fisher, Some Basic Definitions in Graph Theory, Reviews of Modern Physics, 42, 271-88, April, 1970 (English). R. Pellet, Initiation 6 la ThPorie des Graphes, Entreprise moderne d'Edition, Paris, 1966 (French).

absorbant, externally stable, 303 absorption number, B(G), 303 adjacent t o an arc, 5 to a face, I8 to a set, 4 to a vertex, 4 in hypergraphs, 389 adjoint graph, 38 adjoint matrix, 142 alternating chain, 123 sequence, 275 sequence in hypergraphs,

absorbant, exterieurenient stable nombre d'absorption, de stabilitP externe adjacent

adjazent

adjoint matrice adjointe alternCe

Bogenmittengraph Adjunktmatrix alternierend

anti-noeud antisymCtrique arborescence arc flux ensemble d'articulation point d'articulation

Antiknotenpunkt antisymmetrisch Biischel, Setrbaum Bogen Flus Artikulation, zerfallende Menge Zerfallungsknotenpunkt

matrice associCe nombre associe Bquilibre

Adjazenz Matrix beigeordnete Zahl im Gleichgewicht

Lusserlich stabil lussere Stabilitatszahl

414

anti-node, 30 anti-symmetric, 7 arborescence, rooted tree, 33 arc, directed edge, 3 arc flow, 76 articulation set cut-set, separating set, 8 articulation vertex cut-vertex, separating vertex, 9 associated matrix, 9 associated number, e(x), 61 balanced hypergraph, 450

523

-

524 basis of cocycles, 15 of cycles, 15 of a matroid. 493 bicolouring, 325 bipartite graph, 7 network, 83 block, 175 Boolean complement, 420 Boolean product, 420 Boolean sum, 420 boundary, 18 branch, track, 30 cactus, Husiini Tree, 175 capacity, 76 Cartesian product of I-graphs, 314 of simple graphs, 377 Cartesian sum of 1-graphs, 314 of simple graphs, 376 cell, 285 centre, 61 chain in a graph, 7 in a hypergraph, 391 chord, diagonal, 175 chromatic number of a graph, y(G), 325 of a hypergraph x ( H ) , 428 chromatic index, q(G), 248 chromatic polynomial, 352 circuit, directed circuit, 8

INDEX OF DEFINlTIONS

basis

Basis

bicoloration biparti, bichromatique

Zweifarbung paar, bichromatisch

bloc cornplCmentation booleenne produit booltenne somme booleenne frontibre branche cactus capacitt produit cartCsien

Block Boole-Komplement Boole-Produkt Boole-Summe Rand Ast, Zweig Kaktus Kapazitlt kartesisches Produkt

somme cartesienne

kartesische Summe

cellule centre chaine

Zelle Zentrurn Kantenfolge

corde, diagonale nombre chromatique

Sehne chromatische Zahl

indice chromatique, classe chromafique polynbrne circuit

chromatische Klasse

circuit (in a matroid), 476 stigme clique clique, n-clique of a graph, 7 of a hypergraph, 432 cocircuit, 13 cocircuit cocycle cocycle, coboundary, 13 cocyclomatic number, A(G). I5 nornbre cocyclomatique k-colourable, 325 k-chromatique colouring coloration of edges, 248 g o o d - o f edges, 440 of vertices, 325 for a hypergraph, 428 comparability graph, 363 graphe de comparabilite graphe complCmentaire complementary graph of a I-graph, 189 of a simple graph, 288 complete graph, 7 cornplet complete bipartite graph, 7 biparti-complet component, connected composante component of a graph, 8 of a hypergraph, 391 produit direct composition product, 74 conformal, with a faithful graph conforme representation, 396

chromatisches Polynom geschlossene Bogenfolge, Zirkuit Stigmus Clique Cozirkuit Cozyklus cozyklomatische Zahl k-chromatisch Farbung

Vergleichsgraph komplementarer Graph, Komplement vollstandig vollstlndig-paar Komponente

Kornposi tionsprodukt konform

525

INDEX OF DEFINITIONS

conjugate sequences, 103 connected, 8 h-connected, 164 connectivity, 164 contour, 18 contraction, 31 cotree, 26 covering, cover, 129 covering number, 489 d-covering, d-couer, 41 9 critical, 285, 338, 424 crossing number

conjugut connexe k-connexe connectivit t contour contraction, rktrkcissement coarbre recouvrement nombre de recouvrement d-recouvrement critique nombre de croisement

cubic graph, regular of degree, 3 cut, arc cutset, 82 cycle of a graph, 8 of a hypergraph, 391 cyclomatic number, v(G), 15 dark vertex, 155 deficiency, 41 7 degree, ualency of a graph, 6 of a hypergraph, 429 density, w(G), maximum cardinality of a clique descendant diameter, 6(G),66 6*(G), 73 digital sum, 317 directed distance, 61 dominance number, p*(C), domination number, 305 dual, 390 edge, line, 4 of a hypergraph, 389 edge-connectivity, 182 k-edge-connected, 182

cubique coupe cycle

konjugiert zusammenhiingend k-zusammenhiingend Zusammenhangszahl Kontur Kontraktion Co-Baum uberdeckung, Decke Oberdeckungzahl d-uberdeckung kritisch uberkreuzungszahl, benchneidungszahl kubisch Schnitt Zyklus

nombre cyclomatique tpais dkficience degrt

zyklomatische Zahl dick Fehlmenge Grad. Valenz

densitt

Dichte

descendant diambtre

Nachkomme Durchmesser

somme digitale kart nombre de stabilit6 externe

digitale Summe orientierte Entfernung iiussere Stabilitatszahl

dual arite

dual Kante

connexitk k-arite-connexe

Kantenzusammenhang k-fach kantenzusammenhiingend Weg Elementarzirkuit elementarer Cozyklus elementare Kontraktion Elementarzyklus, Kreis elementare Bahn gleichwertige Fiirbung Eulerscher Fliiche quadratischer Faktor Ferrers-Diagramm

chaine tltmentaire circuit tl6mentaire cocycle tltmentaire contraction tltmentaire cycle Cltmentaire chemin kltmentaire coloration kquitable eultrien face facteur, semi-facteur diagramme de Ferren flot foret conjecture des quatre couleurs fonctionnel, univoque functional, 37 genre genus, g ( C ) girth, length of the h g e s t cycle calibre graph, directed graph, 3 genre elementary chain, 8 elementary circuit, 8 elementary cocycle, 13 elemectary contraction, 350 elementary cycle, 12 elementary path, 8 equitable colouring, 463 eulerian, 228 face, region, 18 factor, 230 Ferrer’s diagram, 103 flow, 76 forest, 24 four-colour conjecture, 280

u

Strom

Wald Vierfarben-Vermutung funktional Art Taillenweite Graph (gerichtet, mit oder ohne mehrfache Bogen, mit oder ohne Schlingen)

526

INDEX OF DEFINITIONS

p-graph, directed graph wiih multiplicity p, 3 Grundy function, 312 hamiltonian, 186 Hamilton-connected, 217 Helly property, 397 hypergraph, set system, 389 inaccessible vertex, 155 incident, 6 incidence matrix, 389 independent set of cycles, cocycles, 15 in matroids, 476 injective, 36 inner demi-degree, denii-degree inward, in-;jaienrc, 6 interval graph, 371 isolated vertex, 4 isomorphic, 41 1 isthmus, 175 kernel, 307 length, 7 light vertex, 155 loop, 3 matching, packing of a graph, 122 of a hypergraph, 414 c-matching, 150 matroid, 476 maximal set, 10 maximum set, 10 minimum set, 10 minimally-connected graph, 30 q-minimal graph, 442 ’ minimum set, 10 mixed vertex, 155 multigraph, 5

p-graphe

p-Graph (gerichtet)

fonction de Grundy hamiltonien Hamilton-connect6 propriCt6 de Helly hypergraphe inaccessible incident matrice d’incidence ensemble indkpendant

Grundy-Funktion Hamiltonscher Hamilton-zusammenhingend Helly Eigenschaft Hypergraph unerreichbar inzident Inzidenzmatrix independente Menge

injectif demi-degrC interieur

injektiv Eingangsgrad, Eingangsvalenz Intervallgraph isoliert isomorph Isthmus. Briicke Kern Lange diinn Schlinge Paarung, Anhiufung

multiple edge, 108 multiplicity, 6 neighbour, 4 network, 77 Nim, 319 node, junction point, 30

ariite multiple multiplicitC voisin r6seau jeu de Nim nceud, carrefour

normal hypergraph, 458 order of a graph, 3 of a hypergraph, 389 outer demi-degree, demi-degree outward, our-valence, 6 partial -graph, 7 -subgraph, 7 - hypergraph, 390 - subhypergraph, 390 path, directed path, 8 pendant vertex, 25 perfect graph, 360 perfect matching, linear factor 122

normal ordre

graphe d’intervalles is016 isomorphe isthme noyau longueur fin boucle couplage c-couplage matrolde maximal maximum minimal connexe-minimal q-minimal minimum mixte multigraphe

c-Paarung Matroid maximal maximal kardinal minimal minimal zusammenhingend q-minimal minimal kardinal gemischt Multigraph, ungerichteter Graph Mehrfachkante Multiplizitat Nachbar Netz, Austauschnetz Nim-Spiel Mehrfachknotenpunkt, Verzweigung normal 0r d n un g

partiel

Ausgangsgrad, Ausgangsvalenz Teil-

chemin

Bogenfolge

sommet pendant parfait couplage parfait

Endknotenpunkt perfekt Linearfaktor

demi-degrC ext6rieur

527

INDEX OF DEFINITIONS

piece, 329 planar, 17 potential, 91 predecessor, 4 product of 1-graphs, 314 of simple graphs, 377 pseudo-cycle, 8 pseudo-symmetric, 239 quadrilateral, elementary cycle of length, 4 quasi-strongly connected graph, 32 radius, 61 Ramsey number, 436 rank, 390 regular, homogeneous, 6 representative graph, line-grdph, 400 root, 32 rosace, 30 saturated, 122 k-section, 390 semi-bipartite, 141 semi-functional, 37 sieve, 422 simple chain, 8 simple circuit, 8 simple graph, ordinary graph,

pike planaire potentiel prtdtcesseur produit normal

Glied, Blatt planar Potential Vorgiinger normales Produkt

pseudo-cycle pseudo-symktrique quadrilaere

Pseudo-Zyklus pseudosymmetrisch Vierkreis

quasi-fortement connexe

quasi-stark zusammenhangend Radius Ramsey-Zahl Rangfunktion regulilr entsprechender Graph

rayon nombre de Ramsey rang rdgulier, h o m o g h e graphe r ep r k n tatif racine rosace saturk k-section semi-biparti semi-fonctionnel, semiunivoque grille chaine simple circuit simple graphe simple

Wurzel Rosette gesilttig!

k-Schnitt semi-paar semi-funktional Gitter Kantenzug einfacher Zirkuit (schlichter) Graph

5

simple hypergraph, 389 simple path, 8 sink, 17 source, 77 spanning tree, 26 stability number, independence number, 272 of a hypergraph, 428 stable, independent, 9 strong stability number, 448 strong q-colouring, 448 strongly connected component strong component, 28 strongly connected, 28 stochastic function, 469 subgraph, section graph, 7 subhypergraph, 390 successor, 3 sum of cycles, 12 of cocycles, 12 support, 469 symmetric graph, 6 tension, potential difference, 91

hypergraphe simple chemin simple sortie entrk arbre maximal nombre de stabilitt, stabilitt interne

(schlichter) Hypergraph Bahn Ausgang, Senke Eingang, Quelle Geriist innere Stabilititszahl

stable, inttrieurement stable nombre de stabilitt forte q-coloration forte composante fortement connexe fortement connexe fonction stochastique sous-graphe sous-hypergraphe successeur somme

innerlich stabil starke Stabilitatszahl starke q-Firbung stark-zusammenhingende Komponente stark zusammenhlngend stochastische Funktion Untergraph Unterhypergraph Nachfolger Summe

support graphe symttrique tension, difftrence de potentiel

Stiitze symmetrischer Graph Spann un g

528

INDEX OF DEFINITIONS

transportation network, 77 rkseau de transport transversal transversal for a graph, 133 for a hypergraph, 420 transversal number, 420 nombre de transversalit6 tree, 24 arbre triangle, cycle of length, 3 triangle triangle hypergraph, 440 hypergraphe des triangles triangulated, 368 triangulC Turin number, 434 nombre de Turin uniform hypergraph, k-gruph, uniforme 390

unimodular hypergraph. 465 matrix, 468 vertex, poinf, 3 thickness topological graph, 18 topological dual, 21 transfer, 152 transitive graph, 310

Transportnetz transversal Transversalitltszahl Baum Dreikreis Dreikreis-Hypergraph trianguliert TurAn-Zahl gleichformig

unimodulaire

unimodular

sommet, p i n 1

Knotenpunkt

6paisseur graphe topologique dual transfert transitif

Dicke topologischer Graph dual Austausch transitiv