Threshold Graphs and Related Topics, Volume 56 (Annals of Discrete Mathematics)

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P and the maximal difference subgraphs of B(P). For this end we need the following results.

Proposition 8.6.2 B(P) satisfies 1. xx' ~ E for all x E X ; 2. N(x) - N(y) if and only if x - y; 3. N(x) D N(y) if and only if x < y. P r o o f . Part 1 is trivial, and so is the "if" part of Part 2. For the "only if" part of Part 2, note that if y ~ x, then y' C N(x) - N(y), and hence N(x) 5r N(y); thus if N(x) - N(y), we must have y < x and similarly x <_ y, i.e., x - y. For the "if" part of Part 3, assume that x < y and z' e N(y). Then z :~ y and h e n c e z :~ x, and so z' e N(x). Thus x < y implies N(y) C_ N(x), and equality does not hold by Part 2. For the "only if" part of Part 3, assume that N(y) C N(x). We cannot have x - y; if we had x :~ y, then we would have x' e N(y) and hence x' e N(x), contradicting Part 1; therefore we must have x < y. 9 8.6.3 Let H - ( X , X ' ; F) be a maximal difference subgraph of the bipartite graph G - (X, X'; E) and let xy' C E - F. Then there exists some z e X such that NH(Z) D NH(X) and z y ' ~ E.

Lemma

P r o o f . Assume that some x C X violates the lemma. Among all such x, choose one with a maximal NH(X). There exists a y' E X ~ such that every z E X satisfying NH(z) D NH(X) must also satisfy zy' E E. Moreover, zy' C F , for otherwise z would also violate the lemma, contrary to the m a x i m a l i t y of NH(X). Since xy' is a nonedge in the maximal difference subgraph H, adding it to H creates an alternating 4-cycle. In other words, there exist z E X and w' C X' with zw' C F and zy', xw ~ ~ F. Since the neighborhoods NH(Z) and NH(X) are comparable and w' C N H ( Z ) - NH(X), we must have Nu(z) D Nu(x). But then by assumption zy' E F , a contradiction. .. 8.6.4 Let P - (X, R) be a strict poset and H - (X, X'; F) a maximal difference subgraph of B(P). For every x, y E X, if NH(x) D NH(y), then xy ~ C F.

Lemma

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P r o o f . Assume that x violates the lemma and is chosen so that NH(X) is maximal. Then there exists a y C X with NH(X) D NH(y) and xy' ~ F, and every z C X satisfying NH(Z) D NH(X) must also satisfy zy' C F. By lemma 8.6.3 xy' ~ E, where B(P) = (X,X'; E). Therefore by definition of B(P) we have y < p x, and hence by Proposition 8.6.2 NB(p)(y) D NB(p)(X). Since NH(X) D NH(y), the set S = {w C X : xw' E F, yw' ~ F} is nonempty, and each w C S satisfies yw' C E because xw' E E and NB(p)(y) D NB(p)(X). If we add all these edges yw', w E S to H, H remains a difference subgraph of B(P) since the neighborhood of y has been increased to equal the neighborhood of x. But this contradicts the maximality of H. 9 L e m m a 8.6.5 Let P = ( X , R ) be a strict poset and let H be a maximal difference subgraph of B(P). Define a strict poset L = (X, T) by

x

y

N.(x)

N.(y).

(8.8)

Then L is a linear extension of P. Consequently x = y

N.(x)=

(8.9)

P r o o f . To show that L is complete, assume that, on the contrary, x IlL Y. Since g is a difference graph, we have Ng(x) = NH(y), and as xx', yy' are nonedges of H, both xy' and yx' must be nonedges of H. By the definition of B ( P ) , one of xy' and yx', say xy', must be an edge of B(P). By Lemma 8.6.4, for each w satisfying NH(W) D Nil(x) = NH(y), wy' is an edge of H. Therefore H remains a difference subgraph of B(P) even after xy' is added to it, contradicting its maximality. This shows that L is a strict linear order. To show that L is an extension of P, assume that x

xy~ C F ,~-~, x < L y. Then H is a maximal difference subgraph of B(P).

(8.10)

2-Threshold Graphs

216

P r o o f . We assert that x
the maximal difference subgraphs H of B(P) and the linear extensions L of P given by Lemmas 8.6.5 and 8.6.6 are inverses of each other. Consequently, if H and L correspond under these mappings, then for every x, y C X, the following conditions are equivalent: 1.

x
Y;

2. xy' is an edge of H;

s. N.(x)

N.(y).

P r o o f . Starting with a maximal difference subgraph H of B(P), define a linear extension L of P as in Lemma 8.6.5, and from L define another maximal difference subgraph H' of B(P) as in Lemrna 8.6.6. For every x, y C X, if xy' is an edge of H', then x
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Recognizing Difference Dimension 2

217

L e m m a 8.6.8 For every strict poset P, d o ( P ) - udimd(B(P)). P r o o f . Let H1,...,Hk be difference subgraphs covering B(P). We may assume that they are maximal, and define the corresponding linear extensions L1,...,Lk of P - ( X , R ) as in Lemma 8.6.5. For every x,y E X, if y f p z, then x9' is an edge of B(P) and hence also an edge of some Hi, and by Proposition 8.6.7 we have x
udimd(B(P) < do(P).

"

Lemma 8.6.8 reduces the recognition of partial order dimension of at most 2 to the recognition of union difference dimension of at most 2. However, ~ve need a reduction in the opposite direction. Before achieving it, we need to define a certain induced subgraph of B(P). Given a strict poset P, we say that an ordered pair (x, y) of incomparable elements is covered by a linear extension L when x
8.6.9 Let P - (X, R) be a strict poset and consider its bipartite

2-Threshold Graphs

218

representation B ( P ) - (X,X'; E). The reduced bipartite r e p r e s e n t a t i o n B ' ( P ) of P is the subgraph of B(P) induced by U U V', where U -- {x C X " (x,y) is an unforced pair for some y C X} V' -- {y' C X " (x,y) is an unforced pair for some x E X}. The following lemma says that, in a sense, B'(P) carries as much information as B(P) does. Lemma

8 . 6 . 1 0 For every strict poser P, -

P r o o f . Since B'(P)is an induced subgraph of B(P) and an induced subgraph of a difference graph is a difference graph, it is clear that udime(B'(P)) <_ udime(B(P)). To show the reverse inequality, let G 1 , . . . , Gk be difference subgraphs covering B'(P). Extend them to maximal difference subgraphs HI,..., Ilk of B(P). Each Hi leads to a linear extension Li of P according to L e m m a 8.6.5. We show that each edge xg' of B(P) is an edge of some Hi, which gives the required inequality udime(B(P)) <_ k. By definition of B(P) we h a v e x < y o r x l l Y . If x < p y, then x
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S t e p O- If G has distinctvertices z and w with N(w) identify them (this does not change udimd(G)).

N(z) (twins),

S t e p 1" Add (xi, xj) to R if N(xi) D N(xj). S t e p 2" Add (yi, yj) to R if N(yi) C N(yj). S t e p 3- Add (yj, x~) to R if z~yj ~ E. S t e p 4: Add (xi, yj) to R if U(yj) N X C_ U(xi) N X, where U(x) denotes the set of all y satisfying (x, y) C R. Figure 8.22 illustrates this construction. Figure 8.22" A bipartite graph G - (X, Y; E) with X - { X l , X 2 , x 3 } , Y - {yl, Y2, y3} and the Hasse diagram of the strict poset P constructed from it by the reduction.

Xl

Yl

x2

Y2

X3

Xl

Yl

y2

x3

Y3

1

X2

G

P

The crucial properties of the construction are given by the following lemma. L e m m a 8.6.11 In the construction above, where G is after Step O,

1. P is a strict poset; 2. G is isomorphic to an induced subgraph of B(P); 3. B'(P) is isomorphic to an induced subgraph of G.

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2-Threshold Graphs

P r o o f . 1" Clearly R is irreflexive, and we have to show that it is transitive. This is done by showing that each step preserves its transitivity. Step 0 leaves R empty. Steps 1 and 2 use containment relations to add pairs to R, and so they maintain transitivity. Assume that, if possible, adding the pair (yj,xi) to R in Step 3 violates its transitivity. At this stage no arc is directed from X to Y, and so there are two types of possible violation" (i) (xi, xk) C R, (yj, xk) ~ R and (ii) (yk, yj) C R, (yk, xi) ~ R. These types are similar, and we deal with (i). Since (x~, xk) has been added to R in Step 1, we have g(x~) D N(xk); since (yj,xk) has not been added to R in Step 3, we have yj C N(zk); since (yj,z~) has been added to R in Step 3, we have yj ~ N(x~). But these conditions are contradictory. Assume that, if possible, adding the pair (xi, yj) to R in Step 4 violates its transitivity. Then there exists a w such that (i) (w, x~) is in R but (w, yj) is not, or (ii) (yj, w ) i s in R but (xi, w)is not. We deal with (i) first. Since U(yj) n X C_ U(xi) N X and xi e (U(w) - U(xi)) n X, we have w r yj. If w C X, then U(w) n X D U(x~) n X because R was transitive after Step 1, and hence (w, yj) should be added to R in Step 4, contrary to assumption. If w E Y, then U(w)NX D U(x~)NX D__U(yj)NX (the first inclusion holds since R was transitive after Step 3 and it is strict since xi E ( U ( w ) - U(xi))n X). Therefore, by Step 3, for every x E X, xyj ~ E implies xw ~ E, but not conversely; in other words, N(w) C N(yj). But then (w, yj) was added to R in Step 2, contrary to assumption. Now assume that the violation is as in (ii). We must have w -r xi, for otherwise xi e U(yi) n X C_ U(xi) N X, which is impossible. I f w C X, t h e n w E U(yj) N X C_ U(x~)NX, and so (xi, w) was added to R in Step 1, contrary to assumption. If w E Y, we have U(w) C U(yj) by the transitivity of R after Step 3, hence U(w)O X C_ U(yj) N X C_ U(xi)N X, and hence (xi, w) should be added to R in Step 4, contrary to assumption. 2" G is a bipartite graph with color classes X, Y and B(P) is a bipartite graph with color classes X U Y, X~U Y~. For every x C X and y E Y, we have by Step 3 and the definition of B(P) that xy C E if and only if y :~p x if and only if xy ~is an edge of B(P). Therefore G is isomorphic to the subgraph of B(P) induced by X U Y'. 3: by Definition 8.6.9 B'(P) the subgraph of B(P) induced by U U V', where U

-

{w " ( w , z ) i s an unforced pair in P for some z},

V' -

{ z " ( w , z ) is an unforced pair in P for some w}.

We now show that every unforced pair (w, z) in P belongs to X x Y. Conse-

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Recognizing Difference Dimension 2

221

quently U C_ X, V' C_ Y', and since G is isomorphic to the subgraph of B ( P ) induced by X U Y', it follows that B'(P) is isomorphic to a subgraph of G, as required. The unforced pair (w, z) cannot belong to X x X, for otherwise, by w II z, N(w) and N(z) are incomparable sets (they are unequal because of Step 0). In particular there exists a y C N(w) - N(z). By Step 3 we have (y, z) C R and (y, w) ~ R. By transitivity and w II z we have (w, y) ~ R. Therefore (w, y) is a pair of incomparable elements, and it forces (w, z), a contradiction. Similarly (w, z) cannot belong to Y x Y. The unforced pair (w, z) cannot belong to Y x X, for otherwise, by Step 4, there exists an x C (U(w) - U(z))N X. Then (x, z) ~ R by transitivity and w II z, and so (x, z)is a pair of incomparable elements, and it forces (w, z), a contradiction. .. T h e o r e m 8.6.12 Let G be a bipartite graph and P the strict poset constructed in Steps 0-4 above. Then udimd( G) - do(P). Proof. The proof comes from the following equalities and inequalities d o ( P ) - udimd(B'(P)) < udimd(G) < udimd(B(P)) - do(P),

where the first equality follows from Lemmas 8.6.8 and 8.6.10, the next inequality from Part 3 of Lemma 8.6.11, the next inequality from Part 2 of Lemma 8.6.11, and the last equality from Lemma 8.6.8. 9 In recognizing threshold dimension 2 in the next section, we need variations of the recognition problems for poset dimension 2 and union difference dimension 2. These variations are called "restricted" problems. D e f i n i t i o n 8.6.13

1. Let P be a strict poset and R1,R2 disjoint sets of incomparable pairs of elements of P. The r e s t r i c t e d poset d i m e n sion 2 p r o b l e m asks whether there exist linear extensions L1, L2 containing R1, R2, respectively, whose intersection is P.

2. Let G be a bipartite graph and El, E2 disjoint sets of edges of G. The r e s t r i c t e d u n i o n d i f f e r e n c e d i m e n s i o n 2 p r o b l e m asks whether there exist difference subgraphs H1, H2 containing El, E2, respectively, whose union is G.

T h e o r e m 8.6.14 There exists a polynomial-time reduction of the restricted poset dimension 2 problem to the poset dimension 2 problem.

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2-Threshold Graphs

P r o o f . We regard P as a set of ordered pairs and begin by finding pairs that must be put into L1 or L2. First of all, L1 must contain the transitive closure of P U R1 and L2 must contain the transitive closure of P U R2. Hence we put Pi = P U Ri for i = 1, 2 and replace each Pi by its transitive closure. If any Pi contains a directed cycle, we stop with a No answer, and otherwise the current Pi are strict posets. Next, if a pair (u, v) belongs to P~ but not to P2, then it is incomparable in P, and thus the opposite pair (v, u) should belong to L2. We then add (v,u) to P~ and replace P2 by its transitive closure. Again, if the new P2 contains a directed cycle, we stop, and otherwise it is a strict poset. We do similar steps of adding pairs to P1, and repeat these steps as long as possible. We then obtain two strict posets P1 and P2 having the same set R* of incomparable pairs, unless we stopped previously with a No answer. We know that if L1 and L2 exist, they should contain P1 and P2, respectively. Therefore we should also have P1 N P2 = P, for otherwise there cannot be extensions L1 and L2 with L1 N L2 = P and we stop with a No answer. Now we solve the poset dimension 2 problem for P1. Assume first that the answer is Yes and we obtain a partition of R* into R~ and R~ such that P1 U R~ and P1 U R~ are strict linear orders. Note that R~ and R~ have opposite pairs, i.e., (u, v) C R~ if and only if (v, u) C R~. We assert that P2 U R~ is also a strict linear order, and so L~ - P~ U R~ and L2 - P2 U R~ are strict linear extensions of P whose intersection is P, as required. If the assertion is false, then P2 U R~ contains a directed cycle. Let C be a shortest such cycle. C cannot have length 2, since P2 is acyclic, R~ has no cycles of length 2, and its elements are incomparable pairs for P2. Assume that C has length 3 and it consists of the pairs (a, b), (b, c), (c, a). At most one of these three pairs can be in P2, since two consecutive pairs in P2 imply a shorter cycle by transitivity. Since R~ is contained in the strict linear order P1 U R~, at least one of the three pairs must be in P2. Therefore one of the three pairs, say (a, b), is in P2 and the other two pairs are in R~. Since (b, c) and (c, a) are in R~, the opposite pairs (c, b) and (a, c) are in R~. The pair (a, b) cannot belong to P1, otherwise C would be a cycle in the strict linear order P1 U R~. Hence the opposite pair (b, a) is either in P1 or is incomparable for P1, namely is in R*. But the latter possibility contradicts (a, b) C P2, since R* is the set of incomparable pairs for P2 as well. Therefore (b,a) C 1'1. Thus the opposite cycle is contained in the strict linear order P1 U R~, a contradiction. Assume now that C has length k > 3 and let its pairs be (a~, a2), (a2, a 3 ) , . . . , (ak, al). If (al, a3) C R~, then we have the shorter cycle

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Recognizing Difference D i m e n s i o n 2

223

(al, a3), (a3, a 4 ) , . . . , (ak, al). If (al, a3) e R~, then (a3, al) e / ~ and we have the shorter cycle (al, a2), (a2, aa), a3, al). Otherwise al and a3 are comparable for P2, and (al, a3) or (a3, aa) is in P2, giving rise again to one of the above shorter cycles. This contradiction proves the assertion. Finally, assume that the poset dimension 2 problem for P1 has a No answer. In that case there cannot be strict linear orders whose intersection is P. For if there were, they would have to be of the form P1 U R~ and P2 U R~, where (R~, R~) is some partition of R* into sets of opposite pairs. But from our assertion in the previous paragraph, P1 U R~ is also a strict linear order, and so P1 U R~ and P1 U R~ are strict linear orders whose intersection is P1.

T h e o r e m 8.6.15 ([Ma94]) There is a polynomial-time reduction from the restricted union difference dimension 2 problem on a bipartite graph without twins to the restricted poset dimension 2 problem. Proof. Let the input of the restricted difference dimension 2 problem be the bipartite graph G = (X,Y; E) and disjoint sets of edges E1 and E2. Construct the strict poset P by Steps 0 through 4. Since G has no twin vertices, Step 0 does not apply and hence does not eliminate any edges of E1 or E2. Set R1 = {(x, y): xy' C El}, R2 = {(x, y): xy' C E2} and solve the restricted poset dimension 2 problem for P, R1 and R2. We now show that the answer of the restricted union difference dimension 2 problem is Yes if and only if the answer of the restricted poset dimension 2 problem is Yes. Assume that P is the intersection of strict linear orders L1 and L2 containing R1 and R2, respectively. Let M1 and M2 be the corresponding maximal difference subgraphs of B(P), and let H~ be the subgraph of M~ induced by X 12Y'. By Lemma 8.6.11 G is isomorphic to the subgraph of B(P) induced by X U Y' and we identify these two graphs. By this identification, the Hi are maximal difference subgraphs of G. Since Li contains Ri, we have that Hi contains Ei by Proposition 8.6.2. We show that HI and//2 cover G as follows. Let xy be any edge of G. By Step 3 (y, x) q~ P, and since LI N L2 = P, we must have x
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Ri by Proposition 8.6.2. To show that L1 N L2 = P, it is enough to show that if w lip z, then w
8.7

Intersection Threshold Dimension

2

In this section, based on Ma [Ma93], we present the first polynomial-time algorithm for recognizing threshold dimension 2. The problem is to recognize whether a given graph is the union of two threshold subgraphs, or equivalently whether its complement is the intersection of two threshold graphs on the same vertices. Threshold graphs are (special) interval graphs, and therefore an intersection of two threshold graphs is a boxicity-2 graph, namely the intersection graph of rectangles whose sides are parallel to the coordinate axes in N 2. This is so because two rectangles intersect if and only if their projections on each of the two axes intersect. Threshold graphs can be characterized as those interval graphs that have an interval model in which every interval has its left endpoint at the origin or is a point (an interval of length 0) and all the points are distinct from each other and from the endpoints of the intervals. We call such an interval model a threshold interval model. A graph with a threshold interval model is clearly a threshold graph, since the vertices represented by the intervals form a clique K, the vertices represented by the points form a stable set I, and the neighborhoods of the vertices of I are nested according to the order of the points on the line. Thus the graph satisfies Condition 3 of Theorem 1.2.4. Conversely, every threshold graph has a threshold interval model, constructed as follows. Represent the vertices v C K by intervals i(v) whose left endpoint is the origin and such that for all u, v E K we have N[u] C N[v] if and only if i(u) C i(v); represent each vertex v C I by a point slightly to the left of the right endpoint of the largest interval i(u) such that u is adjacent to v (or by a point to the right of all the intervals if v is isolated). Clearly this can

8.7

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Intersection Threshold Dimension 2

be done so that all the points are distinct. Figure 8.23 illustrates a threshold interval model for a threshold graph. Figure 8.23: A threshold graph and its threshold interval model. The oval around a, b, c, d indicates that these vertices form a clique.

e

d C

I b

f g

a

g

Consider now the intersection G = T1 fl T2 of two threshold graphs T1 and T2 on the same vertices. We can construct a threshold interval model for T1 and T2 along the positive x and y axes, respectively. Thus each vertex of G is represented by an interval or a point along the x axis and by an interval or a point along the y axis. In the Cartesian product of the two threshold interval models, the vertex is represented by the Cartesian product of its two representations. We call this a threshold rectangle model (TRM) for G. More specifically, a T R M consists of the following kinds of geometrical objects in the first quadrant: boxes

Cartesian products of an interval along one axis and an interval along the other axis; they are rectangles with sides parallel to the axes and one corner at the origin;

s e g m e n t s Cartesian products of an interval and a point; they are line segments having an endpoint on one of the axes and parallel to the other axis; p o i n t s Cartesian products of two points; they are single points.

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Graphs

We denote boxes, segments with an endpoint on the x and y axes and points by the letters b*, x*, y*, p*, and their corresponding vertices in the intersection graph by the letters b,x,y,p, respectively. The sets of these geometrical objects and also of the corresponding vertices are denoted by B, 5'~, Su, P, respectively. A TRM is specified by the following key-points: for every box b* its corner c(b*) opposite the origin; for every segment s* its base b(s*) (the endpoint on the axis) and its tip t(s*) (the other endpoint); and for every point the point itself. Figure 8.24 illustrates a TRM and its intersection graph. Figure 8.24: A threshold rectangle model and its intersection graph. The boxes are drawn with dashed lines. The oval around bl, b2, b3 indicates that these vertices form a clique.

Pl

b~ I I

y~

I I I I

b~ Yl

I I

b; . . . .

I I I I I x* 1

I I I

I

Xl

y2

x2

Y3

I I I I

I

I

.

y4

~g2

We note that the (vertices corresponding to the) boxes of a T R M form a clique, since all of them contain the origin. Also the points intersect only boxes, since the points of a threshold interval model are distinct. For the same reason no two segment have the same base, and so the segments form a bipartite graph. No segment is on the same line as a side of a box, since the points of a threshold interval model are not endpoints of intervals. The question we now have to answer is this: given a graph, does it have a TRM? To facilitate this, we observe that every TRM can be transformed into a normalized threshold rectangle model (NTRM) without changing its

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Intersection Threshold Dimension 2

227

intersection graph by the following normalization steps (each step is applied as long as possible before the next step; the five steps are then repeated until no longer posssible; the process terminates because segments can only become points or boxes and points can only become boxes). These steps are also discussed in Sterbini [Ste94]. 1. Every segment that intersects only boxes becomes a point near its base. Every remaining segment therefore intersects other segments. 2. If a point intersects all the boxes, it becomes a small box near the origin. This can be done only once, since the new box will not intersect any remaining point (or segment). 3. Every point is moved near the corner of the intersection of all the boxes that contain it (without leaving these boxes). .

If there are vertical segments that intersect all the boxes, the leftmost of them becomes a narrow box with the same height as the segment. (There can be no obstruction to this step, since an obstruction would have to be a horizontal segment that does not intersect the vertical segment but is lower than its tip; such a horizontal segment has already become a point in Step 1 and moved out of the way in Step 3; it could not have become a box in Step 2, because such a box would not meet the vertical segment, so Step 4 would not apply to the latter.) After this operation no vertical segment meets the new box. An analogous operation is done on the lowest horizontal segment that intersects all the boxes.

5. If a box b~ intersects all the segments and points that intersect a box b~ and yet b~ does not include b~, the corner of by is moved horizontally or vertically until it lies just inside b* (by does not lose any point contained in it, even if the point was moved in Step 3). Figure 8.25 illustrates the NTRM resulting from the TRM of Figure 8.24. We can reformulate our recognition problem as follows: given a graph G, does G have a NTRM? Two consequences of the normalization are that the boxes form a maximal clique (Steps 2 and 4) and that every segment intersects other segments (Step 1). Hence if it is known which vertices of G are to be represented by boxes, the points can be identified as the vertices that become isolated after

228

2-Threshold Graphs

Figure 8.25: The normalized threshold rectangle model obtained from the threshold rectangle model of Figure 8.24. (a) After the first round of Steps 1-5, segment y~ becomes a point near its base and then moved near the corner of the intersection of b~ and b~, point p~ moves near the corner of b~, segments y~ and y~ become narrow boxes, and box b~ moves inside box b~. (b) After the second round, segment y~ becomes a point near its base and then moved near the corner of box x~.

y~

y~ x 1

--lXl

--'1 I

y;

I

b~

I

_A

m

_e--.

,

b2

1

I I I

I

_

--1" I

I

I I

I

--'I

m

b;

_.]

-

pT i I

-I-t

-1

II

I

I I

b;

I

--t

I -

I

__

m

I A I I I I I I -- -- - - I - t

I

p~ ---1 I

II

I I

X2

(a)

(b)

the boxes are removed, and the segments are the rest of the vertices, which should induce a bipartite graph. Proposition 8.7.1 below shows t h a t a graph on n vertices having a N T R M has at. most n 2 m a x i m a l cliques. Since it is straightforward to generate the m a x i m a l cliques one by one in polynomial time per clique, we can try each m a x i m a l clique in t u r n as a candidate for B, identify the corresponding P and S, and see if it is possible to build a corresponding N T R M . If the answer is negative (for example, because S does not induce a bipartite subgraph of G), we consider the next m a x i m a l clique of G. If G has more than n 2 m a x i m a l cliques, the answer is a u t o m a t i c a l l y negative. P r o p o s i t i o n 8.7.1 Let G be a graph on n vertices having a N T R M . G has at m o s t n 2 m a x i m a l cliques.

Then

P r o o f . It is easy to see that each set of geometrical objects (boxes, segments, points) forming a clique has a c o m m o n point. Hence we can obtain all max-

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Intersection Threshold Dimension 2

229

imal cliques by considering, for each point in the plane, the set of objects containing that point. Clearly it is sufficient to consider only the origin, the bases of the segments, the intersections of segments, and the points. Let s~,sv, p denote the number of vertical segments, horizontal segments, and points in the NTRM, respectively. Then s~ + sy + p _< n and the number of maximal cliques is bounded by 1 + s~: + s v + s~sy + p. This is bounded by 1 + n + max s~sv, where the maximum is taken subject to s~ + s v - n. It is easy to see that the bound is not more than n 2 for n > 1. 9 We remark that the estimates in the proposition are crude. Additionally, one can show that if G has a NTRM, it has one in which the boxes form a m a x i m u m clique, that G has at most n 2 / 4 maximum cliques, and that these cliques can be obtained efficiently. See [Ma93] for the details. From now on we assume that the vertices of G are already partitioned for us into boxes B, segments S and points P, with B forming a maximal clique, the vertices that become isolated in G - B forming the set P, and the remaining vertices forming the set S and inducing a bipartite graph. We ask whether we can position these boxes, segments and points to form a NTRM so that their intersections represent the adjacencies of G. We may now assume that no two boxes are twins, namely have the same closed neighborhood, for otherwise we can delete one of them without affecting the answer (if we find a N T R M without this box, we can then add it as an exact or approximate copy of its twin box). Similarly we may assume that no two points are twins, namely have the same open neighborhood. As for the segments, we assume for the moment that the bipartite graph that they induce is connected, and therefore the segments partition uniquely into two sets that we may assign arbitrarily as horizontal and vertical segments. We may then assume that no two horizontal and no two vertical segments are twins, for similar reasons. The search for a NTRM is done in two stages. In the first stage we ignore the tips of the segments and try to place the other key-points (corners of boxes, points, and bases of segments) in the plane so as to reflect the adjacencies of the corresponding vertices, except for adjacencies between segments. In the second stage we arrange the segments, without disturbing their intersections with the boxes, to reflect their adjacencies with each other. Let us define a binary relation ~ on Ii{2 as follows" ( x l , y e ) __ ( x 2 , y 2 ) , when X 1 ~ X 2 and y~ _< y2. We say that ( x i , y l ) d o m i n a t e s (x~,y2), written as (xl,yl) -~ (x2, y2), when (xl, yl) ~___ (x2, Y2)and (xl,Yl) -r (x2,y2). Two different points that do not dominate each other are said to be incomparable. Clearly ~ is a partial order on Ii{2, and it is 2-dimensional, since it is the

230

2-Threshold Graphs

intersection of the linear orders of the separate z and y coordinates. The strict partial order -~ is also 2-dimensional. We are interested in finding domination relations between some of the key-points in every NTRM of the given graph G, namely between the corners b of the boxes, the bases s of the segments and the points p (since we are now in the first stage, we ignore the tips of the segments). It turns out that to a large extent these can be deduced directly from the neighborhoods of G. Every NTRM of G must satisfy the following, where we use the letters b, s, p to denote both the key-points of the geometrical objects and the corresponding vertices of G" 9 For a box corner b, a base s and a point p we have s-.< b < > s c:_ N(b)

(8.11)

p--< b ~

peN(b).

(8.12)

N[bl] C N[b2]

(8.13)

9 For box corners bl a n d b2 we have b1 ~ b2 ~

(the ~ the ~

part follows from b~ C b~ and from the absence of box twins; part follows from Step 5 of the normalization of the TRM).

9 For a point p and a box b we have b-.~ p ~

Yb' E N(p)

(8.14)

b-~ b'

follows from Step 3 of the normalization of the TRM). Hence (the < by (8.13) (8.15) b -~ p r Vb' e N(p) N[b] C N[b']. 9 For points pl and p2 we have

Pl ~ P2 ~

(8.16)

N(pa) D N(p2)

(the ~ follows from the absence of point twins; the < Step 3 of the normalization of the TRM).

follows from

9 For a point p and a segment s we have s -.~ p ~

(the ~

N(p) C_ N ( s ) N B

follows from Step 3 of the normalization of the TRM).

(8.17)

8.7

Intersection Threshold Dimension 2

231

9 For segment bases 81,82 E ~x or 81,82 E ~y we have the implication 81 "~ 82 ,r

N ( S l ) N B D N(s2) N B.

(8.18)

Let Kb, Kp, Kx, Ky be the sets of box corners, points and bases of the vertical and horizontal segments, respectively. Let R be the binary relation on K = Kb O I(p O K~ O Ky, R such that (a, b) C R if and only if a -~ b according to (8.11)-(8.18). The relation R can be computed from the neighborhoods of G. If G and the partition of its vertices into B, P, S~, Sy have a NTRM, then P = (K, R) is a strict poset of dimension 2, which we know how to check in polynomial time. If it turns out that P has a realizer, we can use the positions of the elements of K in the two linear extensions of P as coordinates of the corresponding key-points in R 2. However, we want to ensure that the bases of the vertical segments are lower than any other key-point and that the bases of the horizontal segments are to the left of any other key-point. Therefore we solve the restricted poset dimension 2 problem given by P and the following two sets of incomparable pairs of P:

{(a,b) : a C Kx, b ~ Kx,(a,b) ~ R} R2 = {(a,b) : a C Ky,b ~ Ky,(a,b) ~ R}

1~ 1

:

(i.e., we ask whether R is the intersection of two linear extensions containing RU R1 and Rt.J R2, respectively; this can be determined in polynomial time by Theorem 8.6.14). If the answer is negative, the current B is discarded. If the answer is positive, the two linear extensions are used to place the box corners, the points and the segment bases in the plane. Since the bases of the vertical segments are lower than any other key-point, we can lower them down further to the x axis without changing their intersections with the boxes or any domination relations, and similarly for the bases of the horizontal segments. We then obtain a NTRM for G except for the intersections between segments, since we have not yet placed the segment tips. Now we enter the second stage and try to arrange the segments bases and tips without changing the intersections between segments and boxes, so as to reflect the intersections between segments. Let us ignore the intersections between segments and boxes for the moment. We say that a bipartite graph has a segment intersection model when it is the intersection graph of vertical and horizontal segments in the first quadrant whose bases are on the coordinate axes. The question is then whether the bipartite subgraph of G induced by the segment vertices has a segment intersection model.

232

2-Threshold Graphs

P r o p o s i t i o n 8.7.2 A bipartite graph has a segment intersection model if and only if it is the intersection of two difference graphs with the same color classes. Proof. " O n l y if"" Consider a segment intersection model and its bipartite intersection graph Q. Consider the grid defined by the lines containing the segments. The horizontal and vertical lines of the grid correspond to the rows and columns of the adjacency matrix A of Q, and A has a 1 in a given position if and only if the corresponding horizontal and vertical segments intersect. Thus A is the intersection (component-wise product) of two adjacency matrices A1 and A2:A1 has l's in the positions corresponding to the grid points on the horizontal segments, and similarly for A2 and the vertical segments. Therefore Q is the intersection of the the bipartite graphs/-/1 and /-/2 whose adjacency matrices are A1 and A2, respectively. Now //1 is a difference graph, since every two rows of A1 are comparable by inclusion according to the lengths of the corresponding segments. Similarly//2 is a difference graph, since every two columns of A2 are comparable. Figure 8.26 illustrates this argument. "If": We assume that the bipartite graph Q is the intersection of difference graphs //1 and //2 having the same color classes. Let the color classes be X : { X l , . . . , x k} and Y : {yl,. 9 9 , Yl}, where N , l (xl) D " ' " D N , 1 (Xk)

(8.19)

N , 2 ( Y l ) _~ " ' " __~ NH~(y,).

(8.20)

Consider the adjacency matrices A of Q, A1 of HI and A2 of//2 as having columns corresponding to Xl,..., xk from left to right and rows corresponding to y l , . . . , y t from bottom to top. By (8.19) and (8.20), the l's in every row of A1 are consecutive from the first column on, and the l's in every column of A2 are consecutive from the first row on. We construct horizontal segments corresponding to the l's in the rows of A1 and vertical segments corresponding to the l's in the columns of A2. The i-th vertical segment intersects the j-th horizontal segment if and only if AI(j, i) = A2(j, i) = 1, which is equivalent to A ( j , i ) = 1 since A is the component-wise product of A1 and A2. In other words, the i-th vertical segment intersects the j-th horizontal segment if and only if xi and yj are adjacent in Q. Therefore we have constructed a segment intersection model for Q. ,, Proposition 8.7.2 reduces the recognition of bipartite graphs having a segment intersection model to the recognition of intersection difference di-

8.7

233

Intersection Threshold Dimension 2

Figure 8.26: A segment intersection model and its intersection graph Q. The adjacency matrix A of Q is the intersection of adjacency matrices A1 and A2 of difference graphs.

Y4

Yl

Ys

Y2

Y3

y

Y2

Y4

Yl X1

Xl

X2

X3

X2

X3

y4

0

Y4

1

1

0

y4

0

1

0

ys

0

Y3

1

0

0

y3

1

1

0

y2

1

Y2

1

1

1

y2

1

1

1

yl

0

yl

1

1

0

yl

1

1

1

Xl

X2

X3

Xl

X2

X3

Xl

X2

A

X3

A1

A2

2-Threshold Graphs

234

mension 2 (or equivalently to the recognition of union difference dimension 2 of the bipartite complement). In turn, this problem is reduced by Theorem 8.6.12 to recognizing poset dimension 2. However, we must also take into account the domination relations between segment bases imposed in the first stage by the intersections of segments with boxes. We partition the sets I(= and h'y of segment bases on the x and y axes into equivalence classes called windows. Two elements of I(~ or two elements of I(y are in the same window when they intersect the same boxes. After the first stage we are free to exchange the positions of segment bases only if they are in the same window; if 81,82 E I(= are in different windows, then the first stage has imposed a definite order relation between them, say sl -4 s: (because N(Sl)M B D N(s2) M B). As the proof of Proposition 8.7.2 shows, the segments are induced by the ordering of the roots, whereas the roots on each axis are ordered by the neighborhood containment relations in one of the two difference graphs. We can therefore formulate the problem of finding a segment intersection model for the bipartite subgraph Q = (S=, Sy; E) of G induced by the segment vertices as follows. The color classses S= and Sy are partitioned into windows X0,... ,Xp and Y0,..., Yq, respectively (corresponding to the partitions of the segment bases into windows). We seek difference graphs HI a n d / / 2 with color classes S=, 5'y such that H1 ('1/-/2 = Q and such that

Va C Xi, b E Xi+l,

NH, (a) D___NH, (b)

Va C Y/, b C Y/+I, NH2(a)D___NH2(b). Equivalently we can consider the bipartite complement Q - (S=, Sy; E) of Q and seek difference graphs H1 and H2 with color classes S=, Sy such that H1 U/-/2 - Q and such that

Va e X,, b e X,+,,

Va C

b C Y,+,,

NT, ' (a) C_ NT, ' (l,)

(a) C_

(t,).

Construct a new bipartite graph Q' - (S= u U, S v u V; E U Eu U Ev), where U = {ua,...,uq} and V = {v~,...,vp} are sets of new vertices and

Eu = { u ~ y : y C Y j , i < _ j } Ev

=

{vix:xCXj,i<_j}.

8.7

Intersection Threshold Dimension 2

235

N

It is easy to see that H1 a n d / / 2 exist if and only if Q~ can be covered by two difference subgraphs, one containing Ev and the other containing Eu. This is precisely an instance of the restricted union difference dimension 2 problem. Moreover, Q' does not have twin vertices, since twin vertices of Q' would have to belong to the same window (due to Eu and Ev), and they would then have the same intersections not only with the segments but also with the boxes. In other words, twins of Q~ would be twin segment vertices of G, which do not exist according to our assumption (they were eliminated before the first stage). Therefore by Theorem 8.6.15, we can reduce our instance of the restricted union difference dimension 2 problem in polynomial time into an instance of the poset dimension 2 problem. This completes the description of the second stage of the algorithm. We have one more assumption to be taken care of. We have assumed that the bipartite subgraph Q of G induced by the set S of segment vertices is connected, so that S partitions uniquely into two color classes that we assigned arbitrarily as S~ and Sy. If Q is a bipartite graph with k connected components, we can run the second stage separately on each connected component, but we have the choice of switching color classes (i.e., horizontal and vertical segments) in each component independently. This leads to 2 k possibilities, too many to exhaust in polynomial time. Let C 1 , . . . , Ck be the connected components of Q. The vertices of each Ci partition uniquely into two color classes that we denote as Ci~ and Cib. We may assign the segments corresponding to Cia to S~ (i.e., as vertical segments) and those of Cib to Sy (horizontal segments), or vice versa. Each segment of Ci~ intersects some segments of Cib but no segments from other connected components, and similarly for each segment of Cib. Therefore, no m a t t e r whether the segments of Ci~ are vertical or horizontal, the segments of each Ci occupy an area of IR2 that looks like a hook, and these hooks are disjoint. Thus for every two hooks, one of them is closer to the origin than the other one, as in Figure 8.27. We can tell from the graph G which hooks are closer to the origin as follows. The intersections of B with the segments of Ci~ are linearly ordered by set inclusion. If u* and v* are the (innermost and outermost) segments of Ci~ with the m a x i m u m and minimum intersections with B, respectively, we define max(Cia) and min(Cia) to be NB(u) and NB(V), respectively. In the same way we define max(Cib) and min(Cib). For components Ci and Cj, we write Ci
2-Threshold G r a p h s

236

Figure 8.27: Three hooks generated by the connected components of the segments.

".'.'.'.'.'.'.-.'.'.-.'.'.-.'.-.-.'.'.'.-.-.'.'.'.'.-.'.

i iiiiiiiiiiiiiiiiiiiiiiiii~ 9 .'.'.-...-.-...-.-

....-...........

i:i:i:i:i:i:i:i:i:i:i:i:i:i:i:i~ 9 .'

".'.'.'.'.'.

i iiiii~

closer to the origin than Cj. If neither C~ ~_ Cj nor Cj <1 C~ holds, there is no N T R M for the present choice of boxes. If both Ci <1 Cj and Cj <1 Ci hold, we have min(C/~) - max(Ci~) - min(Cj,) - max(Cj~), min(Cis) - max(Cib) -- min(Cjb) -- max(Cjb ) or

min(C,o) - m~x(C,o) - m~n(Cj~) - m~x(Cj~), min(Cib)

--

max(Cib)

--

min(Cja

) -

max(Cja

)

and we may place either Ci or Cj closer to the origin. Thus we may assume that the Ci have been reindexed so that Ca _<1 ... _<1Ck, as in Figure 8.27. To decide which of Ci~ and Cib will be the horizontal segments, we process the components from the origin outward. When we process Ci, we try making Ci~ the horizontal segments and check if this yields a consistent N T R M with the boxes, the points and the previous segment components. If it does, we accept this assignment and continue with Ci+l without checking the other

8.7

Intersection Threshold Dimension 2

237

possibility that Cib are the horizontal segments. Only if it does not do we check the other possibility, and if it fails too, we conclude that B does not yield a NTRM. Thus we check at most 2k possibilities instead of 2 k. The following lemma and proposition justify this greedy procedure. L e m m a 8.7.3 Assume that both placements of Cia as horizontal segments and of Ci= as vertical segments yield a N T R M with B, P and C I , . . . , Ci-1.

Then there exists a partition of B into subsets B1 and B2 such that 1. no box of B1 intersects any segment of Ci; 2. each box of B2 intersects each segment of Ci-1; 3. c(b~) -~ c(b~) for each bl C BI, b2 C B2, i.e., each corner of a box of BI dominates each corner of a box of B2. P r o o f . [Mar94a] Consider a N T R M of B, P and C1,..., Ci such that flipping Cia and Cib also yields a N T R M for these objects. The lines through the vertical and horizontal segments of Ci-1 with the largest z and y coordinates and those of Ci with the smallest z and y coordinates delimit nine zones, as illustrated in Figure 8.28. No box corner can lie in zone E, because such box intersects some horizontal segments of Ci, and by our assumption it should also intersect them after the flip of Cia and Cib, but this cannot be achieved, since the vertical segments of Ci-1 stand in the way. Similarly no box corner can lie in zone I. If zones B and D also contain no box corners, then we put the boxes with corners in zone A into B1 and the boxes with corners in zones C, F, G, H into B2 and we are done. Otherwise we may assume without loss of generality that zone D contains some box corners. Let b0 = (z0, y0) be a box corner in the union C U D with x0 as large as possible. If each box corner is comparable with b0, we put the boxes whose corners dominate b0 into B1 and the boxes whose corners are dominated by b0 into B2. Conditions 1 and 3 of the lemma are then satisfied, and so is Condition 2 (trivially if b0 lies in zone C, and by its definition if it lies in zone D). Hence we may assume that some box corners are incomparable with b0. Let bl = (xl,yl) be a box corner incomparable with b0 with yl as large as possible. Note that each box corner in zone F is comparable with b0, otherwise the flip cannot be done, as b0 stands in the way. This fact and the fact that zone E has no box corners imply that bl lies in zone A, B, C or

238

2-Threshold Graphs

Figure 8.28" Nine zones used in the proof of Lemma 8.7.3.

E

F

G

I:i:i:i:i:i:i:i:i:i:i:i:i:i:i:il!i:i:i:i:i:i:i:i:i:i:ili~il

B .

.

.

.

.

.

H

D

I

OiL!i!:l

I::::::::: .

C

.

.

.

.

A

.

.

.

.

.

.

D. Each box corner in zone H is comparable with bl; this is obvious if bl is in zone A or D, and follows from the impossibility of the flip because bl stands in the way if it is in zone B or C. Since b0 dominates every box corner in zone F , bl dominates every box corner in zone H, no box corner lies in zone E or I, and by the definitions of b0 and bl, it follows that z = (x0, y~) lies in zone C or D and is comparable with every box corner. Now we place each box whose corner dominates z into B1 and each box whose corner is d o m i n a t e d by z into B2 (if z itself is a box corner, place its box arbitrarily into B1 or B=). As before, B1 and B2 satisfy the required conditions. ,, P r o p o s i t i o n 8 . 7 . 4 The greedy procedure of processing C1, C2,... in order, and assigning Ci~ as the horizontal segments if this is consistent with the already made assignments for C1, . . . , Ci-1 and as vertical segments otherwise will find a N T R M for B, P, S if one exists. P r o o f . For convenience, we may assume that in every N T R M considered, the o u t e r m o s t horizontal and vertical segments of each Ci have the same

8.7

Intersection Threshold Dimension 2

239

distance from the origin (this can be achieved by continuous transformations of the axes, which do not affect our algorithms, because the latter only specify ordinal domination relations between key-points). Assume that a NTRM M exists and let N be the model being constructed by the greedy procedure. Let Ci be the first connected component for which the horizontal and vertical segments are assigned in opposite ways in M and N. Then Ci satisfies the assumption of Lemma 8.7.3, and consequently its conclusion holds for it. Change M as follows: reflect about the line x = y all the segments of Ci, Ci+l,..., all the boxes of B2, and all the points contained in these boxes or in their reflected images. By the properties of B1 and B2 and our assumption at the beginning of this proof, the new M has the same intersection relations of geometrical objects as the old one, i.e., M remains a NTRM. But now M agrees with N on the assignment of horizontal and vertical segments for C1,..., Ci. By induction on i, it follows that N is a NTRM. ,,

Chapter 9 The Dilworth Number 9.1

Introduction

We begin by introducing some definitions from [FH78a]. Recall that for a graph G - (V, E), the vicinal preorder on G is a binary relation ~ on V defined by x~y if and only if N(x) U { x } _ D N ( y ) . We say that x dominates y if x ~ y. It is easily verified that ~ is reflexive and transitive and hence is a preorder. If x ~ y but y ~ x, then we denote this by x ~- y (not to be confused with our notation for strict majorization). I f x ~ y and y ~ x, then we denote this by x ~ y. We say that x,y are comparable if x ~ y or y ~ x, incomparable otherwise. Thus two vertices are incomparable if and only if they are opposite vertices in some alternating 4-cycle. A chain is a set of mutually comparable vertices, or equivalently a sequence of vertices X l , . . . , xk such that xi ~ xj whenever 1 < i < j < k. The Dilworth number of a graph G - (V, E), denoted by D(G), is the least number of chains into which V can be partitioned. By the well-known Dilworth Theorem [Dilh0], the Dilworth number equals the size of a largest set of mutually incomparable vertices. Such a set is called an antichain. Observe that x ~ y in G if and only if y ~ x in the complement G, and hence D(G) - D(G). Clearly the graphs of Dilworth number 1 are precisely the threshold graphs. The graphs of Dilworth number 2 have been characterized in [BHd85b]. In [HM89], the graphs of Dilworth number 3 are shown to form an important subclass of perfect graphs. Payan [Pay83] notes that the graphs of Dilworth number at most 4 are perfect graphs, 241

242

The Dilworth Number

but graphs of Dilworth number 5 need not be perfect. The graphs with the property that in every induced subgraph the Dilworth number is strictly smaller than the number of vertices were studied in [MT85]. These graphs are called domination graphs. The Dilworth number of a graph can be computed in O(n3)-time [Mah84]. We report these and other related results in this chapter. T h e o r e m 9.1.1 ( [ M a h 8 4 ] ) puted in O(n3)-time.

The Dilworth number of a graph can be com-

P r o o f . Let G = (V,E) be a graph with V = { x ~ , . . . , x ~ } . transitive digraph G' on V whose arc set is given by

Construct a

E(G') = {(xixj) : (xi ~- xj) or (xi ~ xj and i < j)}. Clearly each clique of G' forms a chain of G and conversely. Thus the Dilworth number of G is the minimum size of a clique-partition of G'. for transitively orientable graphs, such a partition can be obtained in O(n2) time [Go180]. The construction of G' itself can be done in O(n3)-time since for each pair x, y of vertices one needs to scan through the rest of the vertices to determine if x, y are comparable in the vicinal preorder. 9 The above result is useful, as we show below that the Dilworth number gives an upper bound for some important parameters such as the diameter, domination number and threshold dimension, each of which is NP-hard to compute for an arbitrary graph. T h e o r e m 9.1.2 ( [ F H 7 8 a ] )

The diameter of a graph G is at most D ( G ) + I .

P r o o f . If the diameter of G is k, then G contains an induced path on k + 1 vertices. Clearly no two intermediate vertices of such a path are comparable, so the Dilworth number is at least k - 1. 9 P r o b l e m : Characterize the graphs for which the diameter is exactly one more than the Dilworth number. It is easy to see that a tree has such a property if and only if it is a caterpillar, i.e., a tree with a path P such that every vertex not on P is adjacent to a vertex on P. D e f i n i t i o n 9.1.3 A d o m i n a t i n g set (not to be confused with the domination defined with respect to the vicinal preorder) is a subset S of vertices such that every vertex not in S is adjacent to some vertex in S. The D o m i n a t i o n n u m b e r of G, denoted by 7(G), is the minimum size of a dominating set of G.

9.1

Introduction

243

T h e o r e m 9.1.4 ( [ F H 7 8 a ] ) If G is a graph with no isolated vertices, then

a) <_ D( a) . P r o o f . Let S be a minimum dominating set such that the subgraph of G induced by S has the largest possible number of edges. We assert that S is an antichain. Indeed, assume that for two distinct vertices x and y of S we have x ~ y. If x were adjacent to a vertex of S, the set S - {x} would be dominating, contradicting the minimality of S. Therefore, x is not adjacent to any vertex of S. But since G has no isolated vertices, x is adjacent to a vertex z C V(G) - S. Since x ~ y, y is also adjacent to z, and (S - {x})U {z} is a minimum dominating set inducing a larger number of edges than S, a contradiction. Thus 7 ( G ) = IXl _< D(G) since S is an antichain. ,, P r o b l e m : Characterize the graphs G for which 7(G) = D(G). In [FH78a] it was shown that for split graphs, the threshold dimension is bounded by the Dilworth number. We show below that this inequality remains valid for all C4-free graphs. T h e o r e m 9.1.5 For every C4-free graph G, t(G) <_ D(G). P r o o f . Let V ( G ) be partitioned into d = D(G) chains X x , . . . ,Xd. For each i, let Yi be a maximal clique in the threshold graph Ti induced by Xi. Then for every x E Xi - Yi and y C Yi, we have y )-- x in Ti and hence also in G. Let Gi be the spanning subgraph of G consisting of all edges incident to the vertices in Yi. We show below that G i , . . . , Gd are threshold graphs covering G, proving the result. If Gi is not threshold for some i, then it contains an alternating 4-cycle with edges ab, cd and nonedges bc, da. Since every edge of Gi has an end in Y~, and since Y~ forms a clique, we must have {a, c} C_ Y~ or {b, d} C Y~. But neither a, c nor b, d forms a pair of comparable vertices, contradicting the fact that Xi is a chain. Hence Gi must be a threshold graph for each i. If G has an edge xy that is not covered by any Gi, then there exist i -~ j such that x ~_ Xi - Yi and y E X j - Yj, for otherwise x, y ~_ Xi - Yi for some i, contradicting the stability of Xi - Yi. By the maximality of the cliques Y~ and Yj, there exist an x' C ~ not adjacent to x and a y' E Yj not adjacent to y. We have x' 5r y' since i 5r j. But then x' ~- x and y' ~- y, and hence x, y, x t, y' induce a C4 in G, a contradiction. Therefore every edge of G is covered by some Gi. 9

244

The Dilworth Number

Figure 9.1" A graph G with 3 - t(G) > D(G) - 2. A

A v

v

A v

R e m a r k . The inequality in Theorem 9.1.5 is not true in general. For example the graph G in Figure 9.1 has t(G) > D(G). P r o b l e m " Characterize the graphs G for which t ( G ) - D(G). Recall from Theorem 2.1.6 that a graph is a cointerval graph if and only if it is both a comparability graph and 2K2-free. Also recall that the boxicity of a graph G is the least number of interval graphs whose edge-intersection is G. In other words, it is the least number of cointerval subgraphs whose edgeunion is G. The boxicity of a graph G is denoted by b(G), as in Section 6.3. T h e o r e m 9.1.6 For every graph G, b(G) < D(G). m

P r o o f . It is enough to prove that G is the union of d - D(G) cointerval graphs. Let V(G) be partitioned into d chains X 1 , . . . , Xd. For each i, let Gi be the spanning subgraph of G consisting of all edges incident to the vertices in Xi. We show below that G 1 , . . . , Gd are 2K2-free comparability graphs, and hence are cointerval graphs by Theorem 2.1.6. Thus G is the union of d cointerval graphs, as required. Assume that for some i, Gi contains a 2K2 with edges ab, cd. Since every edge of Gi has an end in Xi, we may assume without loss of generality that a, c C Xi. But a and c are incomparable in the vicinal preorder, contradicting that Xi is a chain. Thus Gi is 2K2-free. To see that each Gi is a comparability graph, let Xl ~ "" ~ xk be all the vertices of Xi. Orient the edges of Gi so that xi is a source vertex in the subgraph G i - {Xl,... ,Xi-1}. In particular, no arc of Gi has its tail outside Xi. Now if (a, b), (b, c) are arcs of the orientation, then a, b C Xi and hence a ~ b. Therefore ac is an edge of G~, and it must be oriented as (a, c) whether or not c C Xi, proving that the orientation is transitive. .. P r o b l e m " Characterize the graphs G for which b(a) - D ( G ) .

9.2

245

Graphs of Dilworth Number 2

9.2

G r a p h s of D i l w o r t h N u m b e r 2

Benzaken, Hammer and de Werra [BHd85b] characterized the graphs of Dilworth number at most 2 by forbidden subgraphs and proved them to be equivalent to the threshold signed graphs defined below. The split graphs of Dilworth number 2 were characterized in [FH77a, BHd85a]. We present these results in this section.

9.2.1

Threshold Signed Graphs

In this subsection we define the threshold signed graphs and prove them to be equivalent to the graphs of Dilworth number at most 2. Definition 9.2.1 A graph G - (V, E) is called a t h r e s h o l d signed graph, or simply a TS graph, if there is a mapping w 9 V ~ R and two positive real numbers S, T such that 1. Iw(x)l < min(S,T) for all x E V; 2. for x 5r Y, xy E E if and only if

I~(x) + ~(~)1 > s

o~

I w ( x ) - w(y)l > r.

Let G, w, S, T be as in Definition 9.2.1. Clearly, if w(x) - O, then x is an isolated vertex. Thus if xy E E ( G ) , then w ( x ) w ( y ) 5r O. The following lemma shows that the "or" in Condition 2 of Definition 9.2.1 is exclusive. L e m m a 9 . 2 . 2 Let G, w, S, T be as in Definition 9.2.1. If xy E E(G), then exactly one of the following conditions holds:

Iw(x) +w(y)l Iw(x) - w(y)]

>_ S, >_ T.

(9.1) (9.2)

P r o o f . Since xy is an edge in G, at least one of Conditions (9.1) and (9.2) is satisfied. If w ( x ) w ( y ) > 0, then

]w(x) - w(y)] < max(iw(x)], Iw(y)l) < rain(S, T),

T h e D i l w o r t h Number

246

and hence Condition (9.2) is not satisfied. If w(x)w(y) < 0, then

Iw(x) + w(y)] < max(Iw(x)l , [w(y)l ) < min(S, T), and hence Condition (9.1) is not satisfied. We call an edge xy an S-edge if Iw(x) + w(y)l >_ S and a T-edge if ] w ( x ) - w(y)l >_ T. Put X = {x E V : w(x) >_ 0} and Y = {x C V : w(x) < 0}. From the proof of Lemma 9.2.2, an edge ab is an S-edge if and only if a, b are in the same set X or Y and a T-edge if and only if a, b are in different sets X, Y. As we have observed in Theorem 2.4.1, the difference graphs are precisely those TS graphs that have no S-edges. L e m m a 9.2.3 The TS graphs have Dilworth number at most 2. P r o o f . Let ( X , Y ) be a partition of V as defined above. We assert that each of X and Y is a chain in the vicinal preorder. Indeed, let x, y C X with w(x) >_ w(y). Then" ( i ) i f yz is an S-edge, then w(x) + w(z) >_ w(y) + w(z) >_ S and hence xz is also an S-edge; and (ii) if yz is a T-edge, then w(x) - w(z) >_ w(y) - w(z) >_ T and hence xz is also a T-edge. Thus x ~ y, proving that X is a chain. Similarly Y is a chain, and the graph has Dilworth number at most 2. " The main result of this subsection is the converse of Lemma 9.2.3. L e m m a 9.2.4 ( [ B H d 8 5 b ] )

The graphs of Dilworth number at most 2 are

TS graphs. P r o o f . Let G - (V, E) be a graph of Dilworth number at most 2 with its vertex set partitioned into two sets X - { X l , . . . , x m } and Y - { y l , . . . , y n } such that Xl ~- " " ;'- xm and y, ~ "'" ~- Y~. We shall obtain distinct positive integral weights ai - w(xi) for i - 1 , . . . , m, distinct positive integral weights bj - - w ( y j ) for j - 1 , . . . ,n and positive integral thresholds S , T such that G is a TS graph with w, S, T. Let GI be the graph induced by X. Clearly G1 is a threshold graph. By Condition 4 of Theorem 1.2.4, there exists a permutation a of { 1 , . . . , m} such that each xo(k) is either an isolated vertex or a dominating vertex in the subgraph induced by x~(1),... ,x~(k). We construct S, a~(a),..., a~(,~) as follows. Initially set S "- 2 and a~(1) "- 1. At step k, we have constructed S and aa(1),..., aa(k-1). If xo(k) is isolated, set

S "- S + 2,

aa(i)

"-- a,7(i)

+ 1 for i < k,

a~(k) "-- 1,

9.2

Graphs of Dilworth Number 2

247

and if x,,(k) is dominating, set S " - S + 2,

ao(i) " - aa(i) + 1 for i < k,

a~(k) "- S + 1.

After m steps we have integers S > a l > ' ' ' > am > 0 such that x i x j is an edge of G1 if and only if ai q- aj > S. We may assume that S - al >_ 2 (otherwise multiply all these integers by 2), hence we can insert an integer T such that S > T > a l > ' " > am > O. We now continue the construction to obtain the bj's. Let G2 be the graph induced by Y. G2 also is a threshold graph, hence there exists a permutation r of { 1 , . . . , n} such that each yr(j) is either an isolated vertex or a dominating vertex in the subgraph induced by y~(~),..., yr(j). The initial step of this part of the construction is as follows. If y~(1) is not adjacent to any xi, set S " - S + 2,

T " - T + 2,

ai " - ai q- 1 for all i,

br(1) "- 1,

and if yr(1) is adjacent to X l , . . . ,Xj and non-adjacent to x j + l , . . . ,x,., then do not change S, T and the ai, and put br T - aj. We continue by constructing br b,(~). Note that we may assume for each j that v ( 1 ) , . . . , r ( j ) are consecutive indices among 1 , . . . , n. Therefore at the general step of this part of the construction, we have constructed thresholds S, T and weights a l , . . . , a m , bk+l,...,bs-1, and must now construct either bk, where the next vertex yk to be introduced is adjacent to each of y k + l , . . . , y~-l, or b~, where the next vertex y~ to be introduced is adjacent to none of them. C a s e 1: yk is introduced. It is adjacent to X l , . . . , x i and non-adjacent to Z i + l , . . . , Zm for some i, 0 <_ i <_ m. Let us begin with the generic case 0 < i < m. The weight bk should satisfy the following inequalities: (9.3)

bk
>

bk+l

(9.4)

bk + b~_l

>_ S

(9.5)

bk + ai

>_ T

(9.6)

<

(9.7)

bk + ai+l

T.

Inequalities (9.3)-(9.7) are solvable for a real bk if and only if

S-

bk+l

<

T - ai+l

(9.8)

bs-1

<

T - ai+l

(9.9)

<

T-ai+l.

T-ai

(9.10)

248

The Dilworth Number

Inequalities (9.8) and (9.10) are already satisfied by the previous construction. Therefore a suitable real bk exists if and only if S-T

<

bs-l-ai+l,

(9.11)

in which case a real value for bk is given by bk

max (bk+l + ~, S -

--

1

bs-1, T -

ai).

If this value for bk is not integral, multiply all integers constructed so far by 2. Suppose that (9.11) is not satisfied. Then we change the weights and thresholds as follows" .

-

4bj + 2

j-k+l,...,s-1

a} "-- 4aj + 2

j-1,...,i

a~ "-- 4aj + 1

j-i+l,...,m

S' "- 4S + 1 T"-4T+4. It is straightforward to verify, using an integrality argument, that these new weights and thresholds are suitable for all the vertices introduced so far. With them, it is possible to solve for bk if and only if S'-T'

!

!

< bs_ 1 - a i + l ,

which is equivalent to

S-

T < bs-1 - a i + l -4- 1.

(9.12)

Inequality (9.12) is a relaxation of (9.11). If it is still not satisfied, we repeat the above steps of changing the weights and thresholds enough times, and eventually will be able to solve for bk. In the extreme case that i - 0, the argument is the same, except that Inequalities (9.6) and (9.10) are omitted. Similarly in the extreme case i - m, the argument is the same if we take am+l - 0. C a s e 2" y~ is introduced. It is adjacent to X l , . . . , x i and non-adjacent to X i + l , . . . , Xm for some i, 0 <_ i _< m.

9.2

249

G r a p h s of D i l w o r t h N u m b e r 2

This case is very similar to Case 1 above. We begin again with the generic case 0 < i < m. We want bk to satisfy the following inequalities: b~ >

0

(9.13)

bs

<

bs-1

(9.14)

b~ + bk+l

<

S

(9.15)

b~+ai

>

T

(9.16)

bs + ai+l

<

T.

(9.17)

Inequalities (9.13)-(9.17) are solvable for a real b~ if and only if (9.18)

T-ai

<

bs-1

T - ai

<

S-

T - ai

<

T - ai+a.

bk+l

(9.19) (9.20)

Again, inequalities (9.18) and (9.20) are already satisfied by the previous construction. Therefore a suitable real b~ exists if and only if S-T

<

bk+l-ai,

(9.21)

in which case an integral solution is given by bs - T - ai.

If inequality (9.21) is not satisfied, we change the weights and thresholds as follows" b~ -

4bj + 2

j-k+l,...,s-1

a~ " - 4aj + 3

j-1,...,i

ajI " - - 4 a j + 2

j-i+l,...,m

S"-4S+4 T'.-4T+

1.

Once again, these new weights and thresholds are suitable for the vertices introduced so far. With them, the condition for determining a real b~ is !

!

S' - T ' > bk+ 1 - a i

,,' ;, 4(S - T) -4- 3 > 4(bk+l - ai) - 1 < ;- S - T + 1

>bk+l-a~.

(9.22)

250

The Dilworth Number

Inequality (9.22) is a relaxation of (9.21), and by repeating these steps as necessary, we eventually find a suitable value for b~. In the extreme case i = m, the argument is the same with am+~ = 0. In the extreme case i = 0, the argument is even easier, since inequality (9.16) drops and there is no lower bound for b~ except for the strict lower bound 0. Therefore we change the existing weights and threshold as follows: aj bj b~ S T

"""-

aj + 1 bj -]- 1 1 S+2 T+2.

j- 1,...,m j-k+l,...,s-1

This completes the construction in all cases. 9 The next Theorem follows directly from Lemma 9.2.3 and Lemma 9.2.4. T h e o r e m 9.2.5 A graph is a TS graph if and only if its Dilworth number is at most 2. C o r o l l a r y 9.2.6 Every TS graph is a comparability graph. P r o o f . Let G - (V, E) be a graph of Dilworth number at most 2 with its vertex set partitioned into two sets X - { x ~ , . . . , x m } and Y - { y ~ , . . . , y ~ } such that x l ~- " . ~ xm and yl ~ "'" ~- y~. Orient each edge ab as (a, b) if a - xi, b - yj for s o m e i , j , o r a - x i , b - xj w i t h i < j , o r a yi, b - yj with i < j. The resulting orientation is transitive. 9 9.2.2

Forbidden

Subgraphs

By the Dilworth Theorem, a graph has Dilworth number at most 2 if and only if it does not contain an antichain of size 3. Thus by obtaining all the minimal graphs with an antichain of size 3, we get a characterization of TS graphs by forbidden subgraphs as stated below. A proof of the next theorem can be found in [BHd85b]. T h e o r e m 9.2.7 A graph G is a TS graph if and only if it does not contain any of the 21 graphs listed in Figure 9.2 as induced subgraphs. Using Theorem 9.2.7, one can characterize the split graphs of Dilworth number 2 as follows.

9.2

251

G r a p h s of D i l w o r t h N u m b e r 2

Figure 9.2: The forbidden subgraphs for TS graphs; G2i+l - G2i, i - 1,..., 10.

_

i I

G1

G2

G4

G6

I I I

X>

alo

G8

a12

. k/Z/_ v

v

w

Q14

v

w

G18

G16

w

v

w

w

G2o

v

252

The Dilworth Number

T h e o r e m 9.2.8 ([FH77a, B H d 8 5 b ] ) For each graph G, the following ditions

con-

are equivalent:

1. G and G are interval graphs; 2. G is a split TS graph; 3. G contains as induced subgraphs neither 2K2, C4, C5, nor Gs, Gg, G14, G15 of Figure 9.2. Proof. 1) =~ 3)" As we have mentioned after Definition 1.4.1, G is an interval graph if and only if G is triangulated and G is a comparability graph. Since G and G are interval graphs, G contains neither C4, C5 (being a triangulated graph) nor Gs, G14 (being a comparability graph) nor their complements 2K2, Gg,

G15. 3) =~ 2)" Since G does not contain 2K2, C4, C5, it is a split graph by Theorem 5.2.1. For the same reason, the only induced subgraphs of Figure 9.2 that G might possibly contain are Gs, Gg, G14, G15. But since by assumption G does not contain them, it is a TS graph by Theorem 9.2.7. 2) =~ 1)" Since G is a split TS graph, it is both triangulated and a comparability graph (by Corollary 9.2.6). But G is also split, and it is a TS graph by Theorem 9.2.5. So G is also triangulated and a comparability graph. Thus G and G are triangulated and cocomparability graphs, and therefore both are interval graphs by the above-mentioned characterization of interval graphs.

9.3 T h e D i l w o r t h N u m b e r and P e r f e c t G r a p h s In this section we examine the relationship between the perfectness of a graph and its Dilworth number. A graph is called minimally imperfect if it is not perfect, but every proper induced subgraph is perfect. L e m m a 9.3.1 If G is a minimally imperfect graph, then

D(G) -IV(G)I. Proof. The proof amounts to showing that V(G) is an antichain. Assume that, if possible, V(G) is not an antichain, and let x ~ Y- If x and y are

9.3 T h e D i l w o r t h N u m b e r and Perfect Graphs

253

not adjacent, then color the perfect subgraph G - {x} with w(G) colors, and then assign to x the color of y, proving that x(G) - c0(G), a contradiction to minimal imperfectness. If x and y are adjacent, then they are not adjacent in G, which is also minimally imperfect [Lov72], and for which y ~ x, so the previous argument applies. Therefore V(G) is an antichain, as required. 9 C o r o l l a r y 9.3.2 ([Pay83]) If D(G) <_4, then G is perfect. P r o o f . If not, G contains an induced minimally imperfect graph H. Now D(H) <_ D(G) _< 4. Hence by Lemma 9.3.1 Iv(a)L <_ 4, which is a contradiction since the smallest imperfect graph is Cs. The pentagon C5 shows that if D(G) is 5 or more, G need not be perfect. Recall that a graph is called perfectly orderable if its edges have an acyclic orientation such that in every P4, at least one of the two end-edges is oriented towards the endpoint of the P4. Such an orientation is called a perfect orientation. An important subclass of perfectly orderable graphs is the class of brittle graphs. These are the graphs with the property that in every induced subgraph there exists a vertex that is not an endpoint of any P4, or not a midpoint of any P4. A perfect orientation for such graphs is easily obtained as follows: if x is not an endpoint (resp. a midpoint) of any P4, orient all the incident edges away from x (resp. into x), then delete x and repeat. A simplicial vertex in a graph is a vertex whose neighborhood forms a clique. Such a vertex clearly cannot be a midpoint of any P4. The following theorem therefore shows that the graphs with Dilworth number at most 3 are brittle. T h e o r e m 9.3.3 ( [ H M 8 9 ] ) If G is a graph with D(G) < 3, then G or its

complement G contains a simplicial vertex. P r o o f . Since D(G) _< 3, the vertex set of G can be partitioned into three chains X1, X2, X3 (possibly empty). Linearly order the elements of Xi by
1. ai to be the largest vertex in Xi, 2. ci to be the smallest vertex in Xi, and 3. bi to be the smallest vertex in Xi with the property that all vertices (of Xi) larger than bi form a clique.

254

The Dilworth Number

The vertices ai, bi, ci need not be distinct. Note that within each Xi, no vertex smaller than bi is adjacent to bi and the vertices smaller than bi form a stable set. Let {i, j, k} - {0, 1, 2}. If c~ is not adjacent to any vertex outside X~, then ci is simplicial in G. Thus, we may assume that ci is adjacent to some vertex in Xj, and hence aj is adjacent to every vertex in Xi. If either Xk is empty, or ck is also adjacent to some vertex in Xj, then clearly aj is simplicial in G. From the above remarks, we may assume now that each ct has a neighbor in Xt+l and no neighbor in Xt+2 (subscripts are taken modulo 3). If some ct is not joined to any vertex smaller than or equal to bt+l in Xt+l, then clearly ct is simplicial in G. Hence we may assume that each ct is adjacent to some vertex smaller than or equal to bt+l in Xt+l. Now, we assert that each at is simplicial in G. To justify the assertion, note that ct+2 is adjacent to some vertex in Xt; therefore at is adjacent to ct+2 and hence to all vertices of Xt+2. It follows that the non-neighbors of at are of two kinds" those in Xt are smaller than or equal to bt and form a stable set, and those in Xt+ 1 a r e smaller than bt+l and also form a stable set. Moreover, there are no edges joining vertices of different kinds. Thus at is simplicial in G as asserted. 9 P r o b l e m . Characterize the graphs G such that for every induced subgraph H of G, H or H has a simplicial vertex. The graphs of Dilworth number 1 are precisely the threshold graphs, the graphs of Dilworth number at most 2 are completely characterized in Section 9.2, the graphs of Dilworth number at most 3 are brittle graphs as shown above, and the graphs of Dilworth number at most 4 are perfect graphs. We define below another interesting class of perfect graphs, using the concept of the Dilworth number. D e f i n i t i o n 9.3.4 A graph G is called a d o m i n a t i o n g r a p h if every induced subgraph H of G with more than one vertex satisfies D ( H ) < IV(H)I, or equivalently, if every induced subgraph with more than one vertex has a pair of comparable vertices. It is easy to see that D(Ck) - k for k _> 5. Therefore the domination graphs contain neither cycles of size 5 or more nor their complements as induced subgraphs. The graphs that contain neither cycles of size 5 or more nor their complements as induced subgraphs are called weakly triangulated,

9.3 The Dilworth N u m b e r and Perfect Graphs

255

and Hayward [Hay85] has shown that the weakly triangulated graphs are perfect graphs. Thus the domination graphs are perfect graphs. Hayward also obtained the smallest known weakly triangulated graph that is not a domination graph; it has 24 vertices. We c o n j e c t u r e that the graphs that are both weakly triangulated and perfectly orderable are domination graphs. A result supporting this conjecture is proved below. We first need the following concepts and lemma. A bipartite graph with no induced cycles of size 6 or more is called a chordal bipartite graph. An edge xy in a bipartite graph is called a bisimplicial edge if every neighbor of x is adjacent to every neighbor of y. The following lemma is due to Golumbic and Goss (see Theorem 12.10 of [Go180]). Lemma

9.3.5

Every chordal bipartite graph with at least one edge has a

bisimplicial edge. T h e o r e m 9.3.6 ( [ M T 8 5 ] ) The weakly triangulated comparability graphs are domination graphs. Let G be a comparability graph with no induced cycles of size 5 or more. Since every induced subgraph of G has the same properties, it is enough to show that G itself contains a pair of vertices that are comparable in the vicinal preorder. Assign a transitive orientation to the edges of G. A vertex x of G is called a source if there is no vertex 9 adjacent to x with the edge x9 oriented from 9 to x. Let S be the set of all sources of G and let T = V(G) - S. Associate with G a bipartite graph H on V(G) with color classes S and T having an edge between a vertex x C S and a vertex y C T if and only if G has an edge oriented from x to y. It is easy to verify that H is a chordal bipartite graph since G is weakly triangulated. If H has no edges, then neither does G, and the result follows. Otherwise by Lemma 9.3.5, H contains a bisimplicial edge xy with x C S and Y C T. We now distinguish two cases as follows: C a s e 1: y has a neighbor x' 5r x in H. Then since x9 is a bisimplicial edge, every neighbor of x in H is adjacent to x' in G. Therefore x' dominates x in G. C a s e 2: y has no neighbors other than x in H. We assert that in this case x dominates y in G. Indeed, let z be any neighbor of y in G other than x. If the edge yz is oriented from y to z, then by transitivity G has an edge oriented from x to z. If y z is oriented from z to y, then G has a directed path from some source to z and then on to y along the edge yz. Since we are in Proof.

256

The Dilworth N u m b e r

Case 2 and by transitivity, that source must be x, and again by transitivity G has an edge oriented from x to z. Thus in either case there is a comparable pair of vertices in G. ..

Chapter 10 Box-Threshold Graphs 10.1

Introduction

We recall from Condition 5 of Theorem 1.2.4 that the threshold graphs are characterized by the absence of incomparable vertices under the vicinal preorder. The box-threshold graphs form a larger class of graphs in which incomparable vertices are allowed under a certain condition. We denote the fact that vertices x and y are incomparable under the vicinal preorder by x I1 Y, and the fact that vertex x is larger than vertex y under the vicinal preorder by x ~- y. D e f i n i t i o n 10.1.1 A graph G is called a b o x - t h r e s h o l d ( B T ) graph when every two vertices x, y of G satisfy x

Ily~degx=degy,

(10.1)

or equivalently degx>degy~x~-y.

(10.2)

The boxes of a graphs are defined as the blocks of its degree partition, i.e., the equivalence classes of vertices under equality of degrees. Thus Definition 10.1 requires that incomparable vertices belong to the same box. The analogy between BT and threshold graphs will become clear below: it turns out that if every box of a BT graph is shrunk into a single vertex, a threshold graph results. Figure 10.1 exhibits the adjacency matrix of a BT graph, with rows and columns sorted by vertex-degrees. Note that the only incomparable pairs are the two vertices of degree 4 and the two vertices of degree 3. 257

258

Box-Threshold Graphs

Figure 10.1" The adjacency matrix of a BT graph. 116544331 1 1

1

1

1

1

1

1

1

1

1

1

1

1 1 1

Condition 10.2 was introduced by Rawlinson and Entringer IRE79], who applied it to a problem of a communications network. They also found the minimum number of edges of a BT graph with given lower bounds for its degrees. The topic was developed further by Peled and Simeone [PS84], on whose work this section is based. Further work on the subject was done by R.I. Tyshkevich and A.A. Chernyak [TC85a], who used the decomposition method to enumerate the BT graphs, as reported in Chapter 17. In Section 10.2 we present elementary properties of BT graphs. Section 10.3 introduces a model of a symmetric transportation network with priority constraints to characterize the degree sequences of BT graphs. Section 10.4 introduces the concept of the frame of a graph to capture the box interconnection structure, and exhibits the close analogy between the frames of BT graphs and threshold graphs.

10.2

Elementary Properties

It follows immediately from (10.1) that every regular graph is BT, and from 10.2 that adding or deleting an isolated vertex to a BT graph results in a new BT graph. Rawlinson and Entringer [RE79] actually considered only connected BT graphs, but the following proposition shows that every nonconnected BT graph results from adding isolated vertices to some connected BT graph or to some regular graph. A

10.2.1 Let G be a B T graph and let G be obtained from G by deleting its isolated vertices. Then G is connected or regular.

Proposition

10.2

Elementary Properties

259 A

Proof. Since all connectedcomponents of G have edges, every two vertices in different components of G are incomparable. Since G is BT, it follows that every two vertices in different components of G have equal degrees. Therefore G is connected or regular. 9 In Section 11.3 we define matrogenic graphs and characterize them by the absence of the configuration 9C(A, B, C, D, E) consisting of vertices A, B, C, D and E, edges A B , A C and ED, and nonedges E B , E C and AD. The following result is then immediate.

Proposition 10.2.2 The class of B T graphs properly contains that of matrogenic graphs. Proof. By Definition 10.1.1, if G is not BT, then it contains incomparable vertices A and E with deg(A) > deg(E). By the incomparability, there exist vertices B and D such that A B and E D are edges and AD and E B are nonedges. Since deg(A) > deg(E), there e x i s t s a vertex C such t h a t AC is an edge and E C is a nonedge. Thus G contains the configuration .T'(A,B, C , D , E ) and is not matrogenic. An example of a non-matrogenic BT graph is illustrated in Figure 10.2. 9 Figure 10.2: A non-matrogenic BT graph.

I I I I The following two results for BT graphs generalize similar results for threshold graphs.

Proposition 10.2.3 The complement of a B T graph is BT. Proof. Complementation does not alter vertex incomparabilities or equality of degrees. 9

Theorem 10.2.4 B T is a forcible property of a degree sequence. In other words, if G and H are graphs with the same degree sequence and G is BT, then H is also BT. We may thus unambiguously define a degree sequence to be B T if (all) its realizations are B T graphs.

Box-Threshold Graphs

260

P r o o f . Since there is a degree-preserving bijection between the vertices of G and H, we may assume that G and H have the same vertices and that each vertex has equal degrees in G and H. Then by Corollary 3.1.11, H can be obtained from G by a sequence of transfers, where a transfer involves dropping the two edges xu and zy and adding the two nonedges x z and yu of an alternating 4-cycle. We may therefore assume that H is obtained from G by a single transfer as above. Moreover, by the symmetry of the four vertices of the alternating 4-cycle, it is enough to show that if x II P in H, where p is a fifth vertex, then deg(x) = deg(p). Since x II p in H, x and p are opposite corners of an alternating 4-cycle A in H. If neither x u nor x z is a side of A, then A is also an alternating 4-cycle in G and the conclusion follows since G is BT. If both o f x u and x z are sides of A, then x II Y II p i n G, and hence deg(x) = deg(y) = deg(p). If xu is a side of A and x z is not, then the configuration in G and H is as illustrated in Figure 10.3, where the vertices of A are x, u, p, q. Figure 10.3" Configurations in the proof of Theorem 10.2.4. q

x

z

q

x

z

T

9

9

9

i I I

I I I

I I I

I I I

I ,k v

v

p

u G

y

I

I

i

9

~k

.L

v

v

v

p

u

y

H

If pz is an edge, then x I] P in G; otherwise x II Y It P in G. Therefore the conclusion follows in both cases as before. Finally, if xz is a side of A and x u is not, we repeat the argument with pu replacing pz. 9 The next theorem gives an easy test of the BT property. To state it, we use the following definitions. D e f i n i t i o n 10.2.5 A bipartite graph with bipartition ( X , Y ) is said to be c o l o r - r e g u l a r ( C R ) if all the vertices of X have the same degree and all the vertices of Y have the same degree. Two disjoint sets X and Y of vertices in a graph G are said to be c o m p l e t e l y c o n n e c t e d , C R c o n n e c t e d , or

10.2

Elementary Properties

261

c o m p l e t e l y d i s c o n n e c t e d when the edges joining them in G form a com-

plete, a CR, or an edgeless bipartite graph, respectively. A similar definition applies in the case of a connection of X to itself, where the subgraph induced by X should be complete, regular, or edgeless, respectively. Note that complete connection and complete disconnection are special cases of CR connection. D e f i n i t i o n 10.2.6 Assume that a graph G has the degree sequence

d? i.e., exactly m i vertices have degree di, with d l > of degree di constitute the i - t h b o x Bi of G.

"'" > d~. The rni vertices

is B T if and only if for each subscript i there is some k(i) such that Bi is completely connected to B I , . . . ,Bk(i)_l, Ct~ connected to Bk(i), and completely disconnected from

Theorem

10.2.7 A graph with boxes B I , . . . , B ~

Bk(0+l,. 9 9 B~. P r o o f . " O n l y if"" Recall that N(x) denotes the set of neighbors of a vertex x. For every vertex x C B~, let k(z) be the largest k such that N ( z ) N Bk r ~. Then by the BT property (10.2), {x} is completely connected to B 1 , . . . , Bk(x)-I and completely disconnected from Bk(x)+l,..., B~. Moreover, if y is any other vertex of B~, then since deg(z) = deg(y), we must have k(x) = k(y) and IN(x) N Bk(~)l = IN(y)N Bk(y)l. This means, first, that k is a function of i alone and not of the particular x in Bi, and may be denoted by k(i); and, second, that for j = k(i) and trivially for each j 5r k(i) all the vertices x E Bi have the same IN(x)N Bj]. In particular, all the vertices x C Bk(~) have the same IN(z)N Bil. Therefore Bi and Bk(~) are CR connected. " I f " : Let x II Y for some x C B~, y E Bj. We have to show that i = j. First of all, k(i) = k(j), for if, say, k(i) > k(j), then N(y) C_ N(z), contradicting non-comparability. Let us denote by p the common value of k(i) and k(j). The only way that x and y can be non-comparable is for each of them to have a neighbor in Bp that is not a neighbor of the other. Therefore Bi and Bj are neither completely connected to Bp nor completely disconnected from Bp. This implies that k(p) is equal to both i and j. " Note that a BT graph is a threshold graph if and only if every two boxes are either completely connected or completely disconnected. This means that

Box-Threshold Graphs

262

the adjacency matrix is equal to the corrected Ferrers diagram of the degree sequence. Deleting a vertex of degree 1 from the graph of Figure 10.2 results in a non-BT graph. This shows that BT is not hereditary, and hence cannot be characterized by forbidden configurations. However, the next theorem shows that the BT property is preserved when an entire box is deleted from the graph. The results of Section 10.4 can be interpreted as a characterization of the BT property in terms of forbidden configurations of boxes. T h e o r e m 10.2.8 If G is BT and H is obtained from G by deleting an entire

box, then H is also BT. P r o o f . let the vertices of the deleted box have degree d in G. Assume that, if possible, H is not BT. Then it has vertices x, y, z such that degu(x ) > degH(y), yet z is adjacent to y but not to x. Since G has the same configuration and is BT, dega(x ) _< dega(y). Hence d e g a ( y ) - degH(y) > dega(x ) - d e g u ( x ) , which means that in G, Y has more neighbors of degree d than x has. By Theorem 10.2.7, y is adjacent in G and H at least to all vertices u such that dega(u) > d, whereas x is adjacent in H at most to these vertices. Hence degu(y ) > degH(x), which is a contradiction. ..

10.3

A Transportation M o d e l

Theorem 10.2.4 raises the question of characterizing BT sequences without constructing an arbitrary realization and testing it for the BT property. Here we give such a characterization, which does not even assume that the sequence is realizable. A transportation network consists of suppliers A1,...,A~ with supplies al,. 9 a~ of some commodity, consumers B 1 , . . . , B~ with demands b l , . . . , b~, and capacities c~j of the route from A~ to Bj. A flow is a matrix (f~j) such that 0 <_ f~j <_ c~j. We say that the flow f satisfies the supplies (resp. demands) when it satisfies the equations Ej f i j -- ai (resp. Ei f i j -- a j ) . We define a special flow r called the supplier-priority flow, by the following recursion:

jl ~)ij --

min { ai

-- E

} r

cij

9

(10.3)

k=l

This has the interpretation that each supplier A i sends as much as possible to the first consumer B1, limited only by its own supply ai and the capacity eil,

10.3

A Transportation Model

263

regardless of the demand bl; then to B2, limited only by the residual supply and the capacity ci2; and so on. Clearly, under the feasibility conditions ~-'~j cij > ai, ~ satisfies the supplies. Similarly, we define the consumerpriority flow ~ by the recursion

r

- min {bj - ~ ~zkj, cij} ,

(10.4)

k=l

which satisfies the demands under the feasibility conditions ~ i cij > bj. The network is said to be symmetric when r = s, ai = bi for all i, and cij = cji for all i, j. T h e o r e m 10.3.1 In a symmetric transportation network satisfying the feasibility conditions, the following conditions are equivalent:

1. r satisfies the d e m a n d s ; 2. r

= Cji for all i, j;

3. r 1 6 2 P r o o f . 1) =~ 2): Assume that r < Cji for some i,j. Since Cji <_ cji = cij, we have r cij, and hence by (10.3) elk = 0 for all k > j. Similarly, since 0 < Cji, we have Cki = cki = cik for all k < j. Therefore

bi

=ai

=

k

k<j

k<j

k<j

and so r does not satisfy the demands. 2) =~ 3): We show by induction on i that r r

=

min{bj, Clj }

--

k

~)ij. For i = 1 we have

=

min{aj, Cjl } = Cjl = r

If ~kj = Ckj holds for all k < i, then

min{

min{a k=l

=

=

k=l

3) =~ 1): This is obvious, since r satisfies the demands. W i t h every sequence d l 1.-- d~mr, dl > ... > dr, of non-negative integers we associate a symmetric transportation network given by

ai = bi = dimi,

cij = m i ( m j - ~ij),

264

Box-Threshold Graphs

where 1, i f i - j O, if not.

5ij -

T h e o r e m 10.3.2 A sequence d~ ~ . . . d ~ ~, dl :> "" :> dr, of non-negative integers is B T if and only if dl < ~ i mi and the associated symmetric transportation networks is such that the conditions of Theorem 10.3.1 hold and the eli are even integers. Before proving the theorem, let us remark that the "if" part does not assume that the sequence is graphical. Another remark is that the condition dl < ~-~imi is equivalent to the feasibility condition ~ j c~j = m ~ ( ~ j m j - 1 ) _> midi = ai, and thus guarantees the equivalence of the three conditions of Theorem 10.3.1. An example of a BT degree sequence is (23)2(20)4(13)3(11)56446. The vector (ai) is (46, 80, 39, 55, 24, 24) and the matrices (cij) and (r are shown in Figure 10.4 Figure 10.4: Capacity and flow matrices for the sequence (23)2(20)4(13)3(11)~6446.

(Cij) --

2 8 6 10 8 12

8 12 12 20 16 24

6 12 6 15 12 18

10 20 15 20 20 30

8 16 12 20 12 24

12 24 18 30 24 30

2 8 6 10 8 12

(r

8 6 10 8 12 12 12 20 16 12 12 6 15 0 0 20 15 0 0 0 16 0 0 0 0 12 0 0 0 0

P r o o f . To prove the necessity, assume that the BT graph G has degree sequence d~ 1 . . . d~ ~. Clearly dl < ~-,i mi, because G has ~-~i mi vertices in total. To prove the other conditions, let xij be the number of edges connecting Bi and Bj for i r j, and let Xii be twice the number of edges within Bi. Thus xij = xji and Xii is even. Moreover, ai = midi is the sum of the degrees of the vertices of Bi, and cij = r n i ( m j - (~ij) is the upper bound for xij reflecting the absence of multiple edges and loops. Hence by Theorem 10.2.7, j-1 X ij --

min

ai-

E

} X ik, Cij

-- (~ij.

k=l

To prove the sufficiency we need the following two facts.

10.4

Frames of BT Graphs

265

F a c t 1 If f is a common multiple of the positive integers p and q and f <_

pq, then there exists a biregular graph with degree sequence ( f /p)V(f /q)q. P r o o f . To construct such a graph, take p vertices x 0 , . . . , Xp_l and q vertices yo,...,yq-1, put r - f / p , and join x0 with y 0 , . . . , y ~ - l , Xl with y~,...,y2~_l, and so on in a cyclic fashion, where the indices are taken modulo q. This gives to each xi the degree r for a total of f edges. Since f is divisible by q, it also gives to each yj the same degree f / q . This proves Fact 1. F a c t 2 If f is an even integer divisible by the positive integer p and f <_ p ( p - 1), then there exists a d-regular graph on p vertices, where d - f /p. P r o o f . Note that d _< p - 1 by assumption. We distinguish the cases p even and p odd. For p even, the complete bipartite graph I(p/2,p/2 is a union of p/2 edgedisjoint perfect matchings (if the vertices are x0,. 9 xp/2-1 and y0,. 9 Y p / 2 - 1 , then the i-th matching joins xy with yj+i with indices modulo p/2). If d _< p/2, take the union of any d matchings, and if d > p/2, take the complement (with respect to Kp) of the union of any p - 1 - d matchings. For p odd, d must be even as f is even. Take p vertices x 0 , . . . , xp-1 and join xi with X i • Xi+d/2. This proves Fact 2. To complete the proof of Theorem 10.3.2, we show that the conditions are sufficient by constructing a BT graph with the given degree sequence. Put mi vertices in cell Ci for each i. Since ai and cij are multiples of mi, so is r For every i r j, since r - Cji, this number is a common multiple of mi and rnj, and furthermore r <_ cij - mimj. Therefore by Fact 1, we can construct a biregular connection between Ci and Cj such that each vertex of Ci has r neighbors in Cj and each vertex of Cj has r neighbors in Ci. Similarly for each i, r iS an even integer divisible by rag, and r <_ ci~ - m i ( m ~ - 1). Therefore by Fact 2, we can construct a (r regular graph on Ci. In the resulting graph G, all the m i vertices of Ci have the same degree, and the sum of these m i degrees is ~ j r - a i - m i d i because r satisfies the supplies. Thus the boxes of G are the Ci and its degree sequence is d~n l . - . d m~. Finally, G is BT by Theorem 10.2.7. 9

10.4

F r a m e s of B T G r a p h s

In order to represent the box-interconnection structure of a graph we use the concept of a frame. The frame of a graph G can be thought of as a graph obtained by shrinking each box of G into a single vertex and coloring

266

Box-Threshold Graphs

the resulting edges to indicate whether or not the boxes were completely connected. Definition 10.4.1 Let G be a graph whose boxes are BI , . . . , Br. The f r a m e of G is a graph F on { B I , . . . , B~} in which Bi and Bj are joined by a black edge, a red edge, or no edge if they are completely connected, partially but not completely connected, or completely disconnected in G, respectively. This includes the case that i - j , and so F has black and red loops. In the case that IBil - 1, there is an ambiguity whether Bi should have a black loop or no loop, since Bi is both completely connected to itself and completely disconnected from itself in G. We resolve this ambiguity by giving to Bi a black loop in F when Bi is connected in G to some box Bk with i > k, i.e., when some vertex of Bi has a neighbor with smaller degree in G. The set of all black (red) edges and loops is denoted by B (R). We denote the box containing a vertex x by 2. In this section we characterize when G is BT in terms of its frame F, and also characterize when a given F is the frame of some BT graph G. For this we use the following definition. Definition 10.4.2 A m a t c h i n g w i t h loops is an undirected graph with loops, no two of whose edges are incident to the same vertex, i.e., its adjacency matrix does not have two 1 's in the same row or column. A t h r e s h o l d g r a p h w i t h loops is an undirected graph with loops such that the rows (and columns) of its adjacency matrix are totally ordered by set-inclusion. An equivalent definition is that it is the underlying graph with loops of a symmetric Ferrets digraph (Section 2.1). If the loops are removed from a threshold graph with loops, it becomes a threshold graph by Theorem 2.1.3, and every threshold graph can be obtained in this way by Theorem 2.1.~. T h e o r e m 10.4.3 Let F be the frame of G. Then G is B T if and only if 1. F is a threshold graph with loops; 2. the black edges of F form a threshold graph with loops; 3. the red edges of F form a matching with loops.

P r o o f . " O n l y if"" The adjacency matrix of F is obtained from the flow matrix r upon replacing r by

10.4

F r a m e s of B T G r a p h s

9 a b l a c k 1 if r

-

cij

> 0 or if

267

i -

j,

eli

-

cii -

0

and ~bik > 0 for some

k>i; 9 a red 1 if 0 < r

< Cij;

9 a 0 in all other cases. Since (r is symmetric by Theorem 10.3.2, and by the definition (10.3) of r F satisfies Conditions 1 through 3. "If"" We assume that deg a x > deg a y and show that z ~ y in G. Some third vertex z is adjacent to z but not to y. We assert that for each vertex w of G, i f t ~ C B, then wz C B. Indeed, 22 C B U R and 2~ ~ B. If zx C B, then t~2 C B by Condition 2 as asserted. If 22 C R, then 29 ~ R by Condition 3, i.e., 2~ is not an edge of F, hence xw C B U R by Condition 1. But t~ 7~ 2 because t ~ is an edge of F and 2~ is not, hence xw ~ R by Condition 3, and therefore 2t~ C B as asserted. Returning to the proof that x ~- y, we assume that, if possible, w r x, y, wy is an edge of G and wx is not. Then t~2 ~ B, so t ~ ~ R by the assertion, hence wy C R. Therefore t~2 ~ R by Condition 3, i.e., t~2 is not an edge of F. Since zx is an edge of F, we have 2 :/- t~, and since wy C R, we have zy ~ R by Condition 3. Hence zy C B by Condition 1, and this contradicts the fact that z is not adjacent to y in G. " T h e o r e m 10.4.4 Let F be a graph with loops whose edges are colored black or red. Then F with its colors is the frame of some B T graph if and only if 1. F is a threshold graph with loops; 2. the black edges of F form a threshold graph with loops; 3. the red edges of F form a matching with loops; ~. every two vertices of F differ in their number of incident black edges or in their number of incident red edges. We say that the vertices differ in their black degree or their red degree, where a loop contributes 1 to the degree. Moreover, in that case no three vertices of F can have the same black degree.

268

Box-Threshold Graphs

P r o o f . " O n l y if"" Assume that F is the frame of a BT graph G. By Theorem 10.4.3, Conditions 1 through 3 hold. To prove Condition 4, assume that ~ and ~ are distinct vertices of F, say deg a z > deg a y, and that ~ and have the same black degree. By Condition 2, for every z, z z E B if and only if ~2 E B. But some z is adjacent to x and not to y in G, hence zy ~ B and 22 E R. By Condition 3, ~,~ ~ R, i.e., 29 is not an edge of F. We assert that no red edge is incident with 9, which shows that 2 and 9 differ in their red degrees. Indeed, if @~ E R, then w x ~ R by Condition 3 and @2 ~ B by the above, and this contradicts Condition 1, thus proving t h e assertion and Condition 4. Moreover, if 2 and r have the same black degree, then by Conditions 3 and 4, their red degrees differ by exactly 1, and so no three vertices can have the same black degree. "If"" We construct a BT graph G whose frame is F. For each vertex v of F, let G have four vertices v0, vl, v2, v3. For each black edge v w E t3 of F with v 7~ w, join each vi to each wj. For each red edge v w E R of F with v -~ w, join each vi to wi and Wi+l (indices modulo 4). For each black loop vv E B of F, join each vi to each vj, i 7~ j . For each red loop vv E R of F, join each Vi to Vi+I and v i - , . This construction is illustrated in Figure 10.5. Note that each red edge or loop of F contributes 2 to the degrees of the vertices of G corresponding to its ends, a black loop contributes 3, and a black edge that is not a loop contributes 4. It is sufficient to show that F is the frame of G, for this implies that G is BT by Theorem 10.4.3. Clearly for each vertex v of F, all the vi have the same degree in G. Therefore it only remains to show that if v and w are distinct vertices of F, then deg a v0 =fi dega w0. If v and w have the same black degree, then by Condition 2, either both have a black loop or both do not, so the black edges incident to v and w contribute equally to the degrees of v0 and w0. Also by Condition 4, v and w have different red degrees, hence deg a v0 r deg a w0 as required. Assume now that the black degree of v is larger than the black degree of w. Then by Condition 2, for each vertex x of F, z w E B implies x v E B , but not conversely. Therefore the black edges contribute more to deg a v0 than to dega w0. We now prove that for each vertex y of F, y w E R implies yv E R U B , which shows that deg C v0 > dega w0 as required. Assume that, if possible, y w E R and yv ~ R U B. Let x be a vertex of F such that x v E B and x w ~ B. Then x-~ y (sinceyv ~ B ) , hence by Condition 3, x w ~ R, i.e., x w is not an edge of F. This contradicts Condition 1. .. The construction used in the "if" part of the proof of Theorem 10.4.4 can be generalized to yield all the degree sequences of BT graphs with a given frame.

10.4

Frames of BT Graphs

269

Figure 10.5: The construction of a BT graph with a given frame.

vo . . . . ~

wo

V0

W0

Vl

Wl

-

Wl

V2 ~

W2

.,.

W2

V3 -

. W3

?-)3

.,.

vwEB

W3

vwER

V2

"

-

131

V2

"

V3

...

-

VO

Y3

.

vv E B

~

Vl

-

vv E R

VO

Chapter 11 Matroidal and Matrogenic Graphs 11.1

Introduction

We have seen in Theorem 1.2.4 that a graph G - (V, E) is threshold if and only if G does not contain any alternating 4-cycle. Therefore, if we call a set of edges I C_ E "independent" when no two edges of I induce an alternating 4-cycle, then a spanning subgraph (V, I) is a threshold graph if and only if I is independent. Obviously the independent subsets of E form an independence system, i.e., a subset of an independent set must be independent. The circuits of this independence system, namely the minimal dependent sets of edges, are the sets of two edges whose endpoints induce an alternating 4-cycle in G. We call these sets couples. We are interested here in those graphs for which the above independence system is a matroid. As is well-known, one definition of a matroid is an independence system such that if C and C ~ are distinct circuits and e C C N C', then C U C ' - { e } is dependent. In the present situation, where every circuit has cardinality 2, this condition simply reads as follows" if a, b, c are distinct edges and {a, b}, {b, c} are couples, then {a, c} is a couple.

(11.1)

An equivalent condition is the following. Define a reflexive and symmetric binary relation R on E by having aRb if and only if a = b or {a,b} is a couple in G. Then our independence system is a matroid if and only if R is transitive. In that case we say that G is a matroidal graph. Although the 271

272

Matroidal and Matrogenic Graphs

corresponding matroids are uninteresting, the matroidal graphs themselves have a nice structure, which permits us to find their threshold dimension and show that they are perfect graphs. Peled [Pe177] introduced these graphs, characterized them by two forbidden configurations, the chordless pentagon and configuration ~ of Definition 11.2.1, and found their structure. We present his results in the next section. FSldes and Hammer [FH78b] introduced a variation of the above concept. Instead of an independence system on the edge set E, they considered an independence system on the vertex set V whose circuits, i.e., minimal dependent sets, are the sets of four vertices that induce an alternating 4-cycle of G. If this independence system is a matroid, the graph G is said to be matrogenic. The matrogenic graphs are characterized by forbidding configuration 9r alone. Furthermore, a matrogenic graph can have at most one chordless pentagon, and hence the structure of matrogenic graphs is almost the same as that of matroidal graphs, even though the two are defined in an apparently different way. We characterize the matrogenic graphs and present their structure Section 11.3. In Section 11.4 we present the work of Marchioro et al. [MMPS84] that characterizes the degree sequences of matrogenic graphs. These results were also obtained independently by Tyshkevich [Tys84].

11..2

Matroidal Graphs

11.2.1

Forbidden Configurations

We begin our study of the matroidal graphs by characterizing them by forbidden configurations. Definition 11.2.1 A configuration ~ consists of vertices A, B, C, D, E such that A is adjacent to B, C but not to D, whereas E is adjacent to D but not to B, C. As with all configurations, unspecified edges (in this case among B, C, D and between A and E) may or may not exists. See Figure 11.1. We denote the fact that A, B, C, D, E generate a configuration jz in this order by 5(A,B,C,D,E).

T h e o r e m 11.2.2 A graph is matroidal if and only if it contains neither the configuration ~" nor a chordless pentagon.

11.2

M a t r o i d a l Graphs

273

Figure 11.1: The forbidden configuration $'(A, B, C, D, E). A

B

E

P r o o f . The "only if" part is immediate: if .T(A, B, C, D, E), then the edge D E forms a couple with both A B and AC, which do not form a couple with each other, contradicting transitivity of the binary relation R; similarly each edge of a chordless pentagon forms a couple with two adjacent edges. To prove the "if" part, assume that R is not transitive, so that there exist edges a, b, c satisfying aRb, bRc, but not aRc. It follows that a, b, c are distinct edges having a total of five or six endpoints, according as a and c have a common endpoint or not.

Five endpoints. Put a = AB, b = CD, c = B E with AC and B D nonedges (Figure ll.2(a)). If E C is a nonedge, we have ~ ( B , A, E, D, C), so assume that E C is an edge. Since b, c form a couple, B C and D E are nonedges (Figure ll.2(b)). Now if AD is a nonedge, then JZ(B, A , E , C, D), and if AD is an edge, then A, B, E, C, D induce a chordless pentagon if A E is a nonedge and $'(A, B, E, C, D) otherwise. Six endpoints. Put a = AB, b = CD, c = E F with AC, BD, CE, D F nonedges (Figure 11.3). Since a r c does not hold, some endpoint of a is adjacent to some endpoint of c. By symmetry we may assume that B E is an edge, and then we have .T(B, A, E, D, C). 9

C o r o l l a r y 11.2.3 The complement of a matroidal graph is matroidal.

P r o o f . This follows immediately from the fact that 5 and the chordless pentagon are self-complementary. 9

274

Matroidal and Matrogenic Graphs

Figure 11.2" Configurations in the proof of Theorem 11.2.2.

c

A a

AT

b

B

a

D

/

B

---

c

b D

,,~,

E

E

(~)

(b)

Figure 11.3" A configuration in the proof of Theorem 11.2.2.

j

A

C

//

B

/%

/

D

/

/

/

/

/

/

c

11.2.2

F

P4-vertices

In this subsection we describe the s t r u c t u r e of those m a t r o i d M graphs for which every vertex belongs to an induced P4 (a p a t h on 4 vertices). 1 1 . 2 . 4 Let vertices A1, A2, B1, B2 induce a P4 with edges Al131, BIB2, B2A2 (Figure 11.~). We then say that A1,A2, B1,B2 i n d u c e a P4 in t h i s o r d e r and denote this fact by 7 9 4 ( A ~ , A 2 , B~,B2). We call the A~ the e x t e r n a l v e r t i c e s and the Bi the i n t e r n a l v e r t i c e s of this P4, and similarly call the edges AiBi the e x t e r n a l e d g e s and the edge B1B2 the i n t e r n a l e d g e of the P4. Definition

1 1 . 2 . 5 Assume that ~P4(A1,A2, B1,B2) in a matroidal graph G, and let X be a fifth vertex of G. If X is adjacent to A1, then X is adjacent to A2, B1 and B2. If X is adjacent to B1, then X is adjacent to B2.

Lemma

11.2

Matroidal Graphs

275

Figure 11.4: P4(A1, As, B1, B2). B1 "

i

"" B2

\

/ \

/ x

/ /

"

\ \ \

A1

A2

P r o o f . If X is adjacent to A1 but not to A2, then we have the configuration S'(A1, B I , X , B2, A2). If X is adjacent to A1 and A2 but not to B1, then in case X is not adjacent to B2 the five vertices induce a chordless pentagon, and in case it is we have ~ ( X , A2, B2, B1,A1). Finally, if X is adjacent to B1 but not to A1, A2, B2, then we have JZ(B1,A1,X, A2, B2). " 1 1 . 2 . 6 In a matroidal graph, if edges a and b form a couple of the form P4 and edges b and c form a couple, then the latter couple is also of the Lemma

P r o o f . We may obviously assume that a, b and c are distinct edges. Assume that P4(A1,A2, B1, B2) with a = AIBI and b = A2B2. Since c forms a couple with each of a and b, there are a fifth vertex X and a sixth vertex Y with c = X Y . If the couple bc is of the form 2K2, then by L e m m a 11.2.5 X and Y are not adjacent to B1, and hence X Y forms a couple with BIB2, contradicting transitivity. If the couple bc is of the form C4, then by L e m m a 11.2.5 X and Y are both adjacent to A2 and B2, a contradiction. " C o r o l l a r y 1 1 . 2 . 7 Let R' be the binary relation defined on the edges of a

graph G so that e R ' f if and only if e = f or e and f form a couple of the form P4. If G is matroidal, then R' is an equivalence relation. The next l e m m a says that if a vertex in a matroidal graph belongs to some P4's, then it is either external in all of them or internal in all of them, and we may call it an external vertex or an internal vertex unambiguously. The same applies to edges too. Lemma

1 1 . 2 . 8 A vertex or an edge in a matroidal graph cannot be internal

in one P4 and external in another P4.

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The result for edges follows from the one for vertices, which we prove below. Assume that P4(A1,A2, B1,B2), and assume that, if possible, A1 is an internal vertex in another P4. Then A1 must have neighbors in the new P4, which by Lemma 11.2.5 are also adjacent to B1 and A2. It follows that B1 cannot belong to the new P4, so A1, having degree 2 in the new P4, must have two new neighbors X and Y in it. But one of X and Y is external in the new P4, and since it is adjacent to A2, A2 is also adjacent to the internal vertex A1 of the new P4 by Lemma 11.2.5 applied to the new P4. This is a contradiction.

L e m m a 11.2.9 In a matroidal graph, all the internal vertices form a clique and all the external vertices form a stable set. P r o o f . We prove the result for the internal vertices, and the result for the external vertices then follows by Corollary 11.2.3. If BIB2 and B2B3 are two internal edges, then B1 is adjacent to B3 by Lemma 11.2.5. Now assume that B1B2 and B3B4 are two internal edges and show that the B i form a clique. If some of B1, B2 are adjacent to some of B3, B4, the result follows from Lemma 11.2.5. If not, let A1 be an external vertex adjacent to B1. By Lemma 11.2.5 A1 is not adjacent to B3 and B4, and therefore B3B4 forms couple with each of B1A1 and BIB2, contradicting the transitivity of R. 9 In order to describe the structure of the equivalence classes of the binary relation R ~ of Corollary 11.2.7, we introduce the following terminology. Definition 11.2.10 Let A 1 , . . . , A p , B 1 , . . . , B p be 2p distinct vertices for some p > 2. We say that these vertices induce (in this order) a net (resp. a n e t - c o m p l e m e n t ) of o r d e r p and denote this fact by A f ( A ~ , . . . , Ap, B 1 , . . . , Bp) (resp. A T ( A ~ , . . . , Ap, B 1 , . . . , B p ) )

if all the Ai form a stable set, all the Bi form a clique, and each A~ is adjacent to Bi and to no other Bj (resp. each Ai is adjacent to all the Bj except for Bi). Figure 11.5 displays a net and a net complement of order 3. If N ' ( A I , . . . , A;, B I , . . . , Bp) holds in G, then ~?(B1,...,/3p, A1, . . . . Ap) holds in G, and conversely; and if 794(A1, A2,/~1~/~2), then Af(A1, A2,/31,/32) and A?(A1, A2,/~2,/~1 ). L e m m a 11.2.11 In a matroidal graph, two or more edges forming an equivalence class under the binary relation R' of Corollary 11.2.7 induce a net.

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277

Figure 11.5: A net and a net-complement of order 3. A1

A1

B

2

_

A2

A3 (a) a net

A2

B1

A3

(b) a net-complement

P r o o f . Let A 1 B 1 , . . . , A p B ; be the distinct edges of the equivalence class with the Ai external and the Bi internal vertices. Clearly if Ai = Aj, then Bi = Bj and as the p edges are distinct, we have i = j. Similarly Bi = Bj implies i = j. Also by Lemma 11.2.8 A~ r Bj for all i,j. Thus we have 2p distinct vertices. By Lemma 11.2.9 the Ai form a stable set and the Bi form a clique. Finally for i -r j, A~ is not adjacent to Bj since we have 794( Ai, Aj, Bi , Bj ). .. Although the equivalence classes of R' are edge-disjoint, they may share vertices. The next lemma is crucial in determining the situation in these cases. L e m m a 11.2.12 In a matroidal graph G, let E and E ~ be distinct equivalence classes under the binary relation R' of Corollary 11.2.7, each having two or more edges. Assume that some vertex is an endpoint of an edge of E and of an edge of E'. Then each of E and E' has exactly two edges and these four edges induce in G a net-complement of order 3. P r o o f . Let us first consider the case that the common vertex is internal. Let 7)4(A1,A2, B1,B2) where AIB1 and A2B2 belong to E, and let the edge XB1 belong to E'. Then X must be external, since it follows from Lemma 11.2.5 that internal edges do not belong to couples, and so X -r B2 by Lemma 11.2.8. Also X r A1 since E -r E', and X :/- A2 since A2 is not adjacent to B1. Therefore X is a fifth vertex, which is adjacent to B2 by Lemma 11.2.5, but not to A1 or A2 by Lemma 11.2.9. By assumption ~P4(X, Y, B~, Z) for some

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vertices Y, Z. By Lemma 11.2.8 Z -~ X, A1, A2 and by definition of a couple Z -~ B1, B2. Thus Z is a sixth vertex, which is adjacent to B1 and B2 by Lemma 11.2.9 but not to X by definition of a P4 (see Figure 11.6). Figure 11.6" A configuration in the proof of Lemma 11.2.12.

al

B1

B2

A2

A

A /

/ \

/ /

u

We assert that Y = A2. Clearly Y r Z, B1, B2 by Lemma 11.2.8; Y -~ A1 since A1 is adjacent to B1 and Y is not; and Y -J= X. Therefore if Y -r A2, then Y is a seventh vertex. In the latter case, since YZ forms a couple with XB1, it cannot form a couple with XB2 by transitivity, and therefore Y must be adjacent to B2. But then Y is adjacent to B1 by Lemma 11.2.5, contrary to the assumption that YZ forms a couple with XB1. This proves the assertion Y = A2, from which it follows that Z is adjacent to A2 and thus also to A1 by Lemma 11.2.5. We have therefore shown that ~(A1,A2, X, B2,B1,Z). Moreover, since A2 was an arbitrary external vertex other than A1 belonging to an edge of E, it follows from the assertion that E has only two edges A1B1 and A2B2. By symmetry E' also has only two edges XB1 and A2Z, and the lemma is proved in this case. We now examine the other possibility, that the edges of E and E' share an external vertex but not an internal one. We show that this is impossible. Let 7)4(A1, A2, B1, B2) where A1B1 and A2B2 belong to E, and let the edge XA1 belong to E'. Then X must be internal, and therefore by assumption X -r A2, B1,B2. By Lemma 11.2.5 X is adjacent to A2, B1,B2. By assumption E' contains another edge that forms a P4 with XA1. This edge cannot contain B1 or B2 by assumption and it cannot be of the form A2Y, since Lemma 11.2.5 would imply that Y is adjacent to A1, contrary to the definition of P4. Thus this edge is of the form YZ, where Y is an internal sixth

11.2

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279

vertex and Z is an external seventh vertex (see Figure 11.7). We now apply Lemma 11.2.5 several times to the two P4's. Since B1 is adjacent to A1, it is adjacent to Z. Since Z is adjacent to BI, it is adjacent to B2. Since B2 is adjacent to Z, it is adjacent to A1. But this contradicts the definition of P4.

Figure 11.7: A configuration in the proof of Lemma 11.2.12. z Y x v

A1

B1

w

B2

v

A2

In order to generalize Lemma 11.2.12 to more than two equivalence classes, let us consider a new abstract graph G whose vertices are the equivalence classes of two or more edges under R'. By Lemma 11.2.11, the edges of G that belong to a single vertex of G induce a net in G. Two vertices of G shall be adjacent when the corresponding two nets of G have vertices in common. L e m m a 11.2.13 For a matroidal graph G, let X l , . . . , 2 q be a sequence of

vertices of G such that X~ is adjacent to some of X 1 , . . . , X~-I for each i 2 , . . . , q, and q >_ 2. Then each )[i contains exactly two edges of G and these 2q edges induce a net-complement in G. P r o o f . The proof is by induction on q. The basis q - 2 is the content of Lemma 11.2.12. Assume that q > 2 and that the lemma holds for q - 1 . Then each of J ( 1 , . . . , Xi-1 contains exactly two edges of G and these 2q - 2 edges induce in G a net-complement N ' ( A ~ , . . . , Ap, B 1 , . . . , Bp). By assumption Xq is adjacent to 2 i for some i - 1 , . . . , q - 1. Hence by Lemma 11.2.12 J~q contains exactly two edges of G, which induce in G a net N(Q, T, R, S), and Xi contains exactly two edges of G, which induce in G a net N(A~, A~, B~, B~). It further follows from Lemma 11.2.12 that these two nets share exactly one internal vertex and exactly one external vertex, and their six distinct vertices induce in G a net-complement of order 3. Hence by renaming Q, T, R, S if ,v

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necessary we may assume that Q = A~ and S = B~, as illustrated in Figure 11.8. Figure 11.8: A net-complement in the proof of Lemma 11.2.13.

Q=Ar

R

w

As

8

v

S = Br

v

T

Clearly T is distinct from B I , . . . , Bp and R is distinct from A I , . . . , Ap. Assume that T is one of A I , . . . ,Ap, say T = At. Then R -r Bs because B~ is adjacent to T and R is not. If R r Bt, then the fact that R is adjacent to A~ but not to At contradicts Lemma 11.2.5 as applied to 7)4(As, At, Bt, B~). Therefore R - Bt and the vertices of G in ) ( 1 , . . . , 32q are just A I , . . . , Ap, B 1 , . . . , Bp, which induce a net-complement of order p in G, as was to be proved. Now assume that T is distinct from A 1 , . . . , Ap. Then R is distinct from A I , . . . , Ap, because by Lemma 11.2.5 R = Bt would imply that R is adjacent to T. Then for each t -r s we can apply Lemma 11.2.5 to P4(At, A~, B~, Bt) to conclude that T is adjacent to Bt but not to At and R is adjacent to both At and Bt. Thus we have a net-complement of order p + 1, ~?'(A1,..., Ap, T, B 1 , . . . , Bp, R), as was to be proved, tt C o r o l l a r y 11.2.14 For a matroidal graph G, the vertices belonging to a

single connected component of G induce in G a net or a net-complement. These nets and net-complements are vertex-disjoint. P r o o f . If a connected component C of G contains exactly one vertex of G, then the corresponding vertices of G induce in G a net by Lemma 11.2.11. If C contains two or more vertices of G, then these vertices can be arranged in a sequence satisfying the conditions of Lemma 11.2.13, and so the corresponding vertices of G induce a net-complement. Assume that C1 and C2

11.2

M a t r o i d a l Graphs

281

are connected components of G such that the corresponding nets or netcomplements N1 and N2 of G share a vertex X. Then X belongs to some P4 contained in N1, hence X is an endpoint of an edge of G belonging to a vertex X1 of G belonging to C1. Similarly X is an endpoint of an edge of G belonging to a vertex X2 of G belonging to C2. By definition of G, X1 is adjacent to 22, hence C1 - C2, hence N1 - N2. 9 Thus we have shown that in a matroidal graph G, the vertices that belong to P4's partition into disjoint sets, which we now call cells, such that each cell induces in G a net or a net-complement, and each P4 in G is induced by the vertices of a single cell. We number the cells as 1 , . . . , k and denote by E i (resp. P) the set of external (resp. internal) vertices of the i-th cell. We say that cell i dominates cell j, where i =/= j, when each vertex of E i is adjacent to each vertex of I j, and no vertex of E j is adjacent to any vertex of I i. L e m m a 11.2.15 The domination relation between the cells of a matroidal

graph is a linear order. P r o o f . By definition the domination relation is irreflexive and antisymmetric. We show that it is transitive and total. Totality. Consider any two cells, say cell 1 and cell 2, and let AiBi be any external edge of cell i, with Ai external and Bi internal (i = 1, 2). We assert that either A1 is adjacent to B2, or else A2 is adjacent to B1, but not both. If neither holds, then by Lemma 11.2.9 we have JP4(A1, A2, B1, B2), contrary to the fact that the vertices of each P4 are contained in a single cell. If both hold, then by Lemma 11.2.5 each vertex of E 1 is adjacent to each vertex of 12 and each vertex of E 2 is adjacent to each vertex of I 1. Let "]')4(Ai, A~, Bi, B~) in c e l l / - 1,2. Then we have 7'4(A1,A2, B;,B~), again contrary to the fact that the vertices of each P4 are contained in a single ceil. This proves the assertion, say A1 is adjacent to B2 and A2 is not adjacent to B1. Then it follows from Lemma 11.2.5 that cell 1 dominates cell 2. Transitivity. In view of the totality, it is enough to show that it is impossible for cell 1 to dominate cell 2, cell 2 to dominate cell 3, and cell 3 to dominate cell 1. Assume that, if possible, this is so, and let A1 E E 1, B2 C 12 , Aa E E 3, and B3 E 13, with A3 adjacent to B3. By Lemma 11.2.9 B2 is adjacent to B3 and A1 is not adjacent to A3. By our assumption we have 7?4(A1, A3, B2, B3), contrary to the fact that the vertices of each P4 are contained in a single cell. 9

282

Matroidal and Matrogenic Graphs

Lemma 11.2.15 allows us to renumber the cells so that cell i dominates cell j if and only if i < j. This determines completely the structure of the subgraph induced by the edges of couples of the form P4 in a matroidal graph. We summarize the results of this subsection by the following theorem. T h e o r e m 11.2.16 In a matroidal graph G, the vertices belonging to couples of the form P4 partition into disjoint sets called cells. Each cell induces in G a net or a net-complement. All the internal vertices form a clique and all the external vertices form a stable set in G. The cells can be indexed so that the external vertices of cell i are adjacent to the internal vertices of cell j if and only if i < j. We call a matroidal graph all of whose vertices belong to couples of the form P4 a cell graph.

11.2.3

2I(2-vertices and C4-vertices

In this subsection we complete the description of those matroidal graphs for which every vertex belongs to some edge in some couple. In particular, we show that every vertex of such a graph belongs either only to couples inducing a P4, or only to couples inducing a 2K2, or only to couples inducing a 6'4.. Moreover, a matroidal graph cannot contain couples of the form 2K2 and 6'4 at the same time. As usual, by a perfect matching inK2 we mean a graph with 2m vertices, all of degree 1. Figure 11.9 displays a perfect matching 3K2 and its complement 3K2. Figure 11.9: A perfect matching and its complement.

3K2

3K2

11.2

M a t r o i d a l Graphs

283

T h e o r e m 11.2.17 Let G be a matroidal graph in which the set M of vertices belonging to couples of the form 2K2 is not empty. Then M induces a perfect matching in G and the vertices of G outside M partition into a clique C whose vertices are adjacent to every vertex of M and a stable set I whose vertices are adjacent to no vertex of M . P r o o f . Let edges A B and C D form a couple of the form 2K2, and let X be a fifth vertex. We observe that X is adjacent to all of A, B, C, D or to none of them. Indeed, if X is adjacent to A, then by the transitivity of R, X A cannot form a couple with CD, and therefore X is adjacent to both C and D (Figure 11.10). Figure 11.10: A configuration in the proof of Theorem 11.2.17. A

C \

/ \

X

/ x

/

\

/

B

D

A repetition of this argument shows that X is also adjacent to B. From this observation it follows that the binary relation "equals to or forms a couple of the form 2K2 with" is transitive, and thus an equivalence relation. By definition each equivalence class induces in G a perfect matching, and from the assumption of the theorem there is at least one equivalence class g with at least two edges A B and CD. Let M' be the set of endpoints of the edges of g. By the above observation the vertices of G outside M ~ partition into a set C of vertices adjacent to every vertex of M ~ and a set I of vertices adjacent to no vertex of M ~. Any two vertices X, Y of I are not adjacent, otherwise the edge X Y would belong to g. Every two vertices X, Y of C are adjacent, otherwise X A would form a couple with both Y C and Y D , contradicting transitivity. Thus I is a stable set and C is a clique. It follows that the vertices of M ~ belong only to couples formed from two edges of g, and that C U I induces a split graph, which has only couples of the form P4, if any. Therefore M ~ = M. 9

Matroidal and Matrogenic Graphs

284

C o r o l l a r y 11.2.18 Under the conditions of Theorem 11.2.17, G has no couples of the form C4, C contains all the internal vertices and I contains all the external vertices of G. P r o o f . We have seen that G has no couples of the form C4. Let B1 be any internal vertex of G. Then we have 794(A1,A2, B1, B2) for some A1,A2, B2, and It follows from Lemma 11.2.5 that the edge B I ~ 2 does not belong to any couple of G. Therefore at least one of B1, B2 must be in C. If B1 is in I and B2 is in C, then A1 cannot be in C since it is non-adjacent to B2, and cannot be in I U M since it is adjacent to B1, a contradiction. Therefore B 1 must be in C and C contains all the internal vertices. It follows from this that A1 and A2 must be in I, so I contains all the external vertices. .. By taking the complement of G we obtain from Theorem 11.2.17 and Corollary 11.2.18 the following analogous results. T h e o r e m 11.2.19 Let G be a matroidal graph in which the set M of vertices belonging to couples of the form C4 is not empty. Then M induces a complement of a perfect matching in G and the vertices of G outside M partition into a clique C whose vertices are adjacent to every vertex of M and a stable set I whose vertices are adjacent to no vertex of M. C o r o l l a r y 11.2.20 Under the conditions of Theorem 11.2.19, G has no couples of the form 2K2, C contains all the internal vertices and I contains all the external vertices of G. The following summarizes the results of this subsection. D e f i n i t i o n 11.2.21 A matroidal graph G all of whose vertices belong to couples is called a s t r i c t m a t r o i d a l g r a p h . T h e o r e m 11.2.22 A graph G is strict matroidal if and only if the vertices of G partition into subsets M and S, M induces a perfect matching or a complement of a perfect matching, S induces a cell graph, and the vertices of M are adjacent to every internal vertex and no external vertex of S. 11.2.4

Threshold

Vertices

We now have to consider those vertices of a graph G that do n o t b e l o n g to any couples. We call these vertices threshold vertices, since they induce

11.2

M a t r o i d a l Graphs

285

a threshold subgraph of G. Recall the definition of the vicinal preorder on the vertices of G. We extend its definition to subsets of vertices so that X~Yifandonlyifx~yforallxCX, yEY. If G is matroidal, then as seen in the previous subsection, its vertices belonging to couples partition into sets M, I 1, E l , . . . , [k E k, where M is the set of vertices that belong to couples of the form 2K2 or C4, I t is the set of internal vertices of cell l, and E l is the set of external vertices of cell l. These vertices induce in G a strict matroidal graph S, relative to which

[k ~ ... ~_ I 1 ~ M ~ E 1 ~ "'" ~- E k.

(11.2)

The threshold vertices of G induce a threshold graph T, and they can be labeled as X 1 , . . . , Xt so that relative to T

X , ~ ... ~ Xt.

(11.3)

Moreover, (11.2) and (11.3) remain true relative to G, because they could only be violated if a vertex of T belonged to a couple of G (there is an exception in the case that Xi ~ Xj relative to T but Xi ~- Xj relative to G; in that case we adopt the convention that i < j). Thus the sets M , I ~ , E 1 , . . . , I k , E k, { X 1 } , . . . , {Xt} are totally ordered by ~ relative to G in a way that is a common refinement of both (11.2) and (11.3). Conversely, assume that S is a strict matroidal graph satisfying (11.2) and T is a threshold graph disjoint from S and satisfying (11.3). If S and T are connected by edges so that relative to the resulting graph G the sets M, I 1, E l , . . . , I k, E k, {X1},..., {Xt} are totally ordered by ~- in a way that is a common refinement of both (11.2) and (11.3), then G is matroidal and its threshold vertices are precisely the vertices of T. Indeed, the Xi are comparable with every vertex of G and hence do not belong to couples of G; the other vertices of G do belong to couples since S is strict; further, R is transitive on S by assumption and hence remains transitive on G. It remains to see how S and T are to be connected by edges so as to satisfy the above necessary and sufficient requirements. By Theorems 11.2.17 and 11.2.19, if a vertex of T is adjacent to some vertices of M, it is adjacent to all vertices of M. By Lemma 11.2.5, if a vertex of T is adjacent to some vertices of E t, it is adjacent to all vertices of E t t_J I t, and if it is adjacent to some vertices of I l, it is adjacent to all vertices of I t. It therefore follows that a vertex of T cannot be equivalent to a vertex of S' under the vicinal preorder. The remaining question is to which sets M, I 1, E l , . . . , I k, F_,k a given vertex of T should be adjacent.

Matroidal and Matrogenic Graphs

286

We begin to consider this problem by assuming that S is a single cell, i.e.,M-Oandk-1. T h e o r e m 11.2.23 Let S be a cell with external vertices E and internal vertices I. Let T be a threshold graph disjoint from S, whose vertices X 1 , . . . , Xt satisfy (11.3) relative to T. If S and T are joined by edges so that the resulting graph G is matroidal and its threshold vertices are precisely X 1 , . . . , Xt, then the following conditions hold: 1. there is an index r - O , . . . , t such that { X r + l , . . . , X t } is a stable set, or equivalently such that X~+I is not adjacent to X~+2 (provided r + 2 <_ t); 2. there is an index s - 0 , . . . , t satisfying i < s < r and Xs >- X j , where j is the smallest index other than r such that X j and X~ are not adjacent, and i is the largest index such that Xi and X~+I are adjacent (if i or j is not defined, the corresponding condition on s is omitted; Xo :'- XN is considered to be true); 3. each vertex of I is adjacent to X1, . . . , X~, but not to X r + l , . . . , X t / ~. each vertex of E is adjacent to X1, . . . , X~, but not to X s + l , . . . , X t. Conversely, r and s can always be chosen to satisfy 1 and 2, and if S and T are connected according to 3 and ~, then the resulting graph G is matroidal and its threshold vertices are precisely X 1 , . . . , Xt. P r o o f . Assume that G is matroidal and its threshold vertices are X1, . . . , Xt. Then there are indices 0 _< s <_ r _< t such that relative to G we have X1 ~ . . .

~ X s ~- I ~ Xs+ 1 ~ . . .

~'- X r

~- E

~- Xr+ 1 ~ - . . ,

~-- X t .

(11.4) We prove that r and s satisfy 1-4. To prove 3, consider an external vertex A, an internal vertex B adjacent to A, and an internal vertex B' not adjacent to A. By Lemma 11.2.5, since X 1 , . . . , X ~ >- A, X 1 , . . . , X ~ are adjacent to B and hence to every vertex of I. Since A >-- X r + l , . . . , Xt, X r + l , . . . , Xt are not adjacent to B' and hence not to any vertex of I. Condition 4 has a similar proof.Assume that 1 fails and there is an edge connecting two vertices among X ~ + I , . . . , Xt. Then by 3 this edge forms a couple with any internal edge of S, contradicting the assumption that X 1 , . . . , Xt are threshold vertices of G. It remains to prove 2. We assert that I ~ Xj. Indeed, otherwise we would have

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M a t r o i d a l Graphs

287

Xj ;,-- I, and since the vertices of I are adjacent to X~, Xj would be adjacent to X~, contradicting the definition of j. Similarly we see that Xi ~ I. Since Xi ~- I ;,- Xj and s is the largest index such that X~ ~ I, we have i <_ s and X~ ~- Xj relative to T. Conversely, assume that r is chosen to satisfy 1. We assert that s can always be chosen to satisfy 2, in other words Xi ~- Xj and i _< r must hold. To see that X~ ~ Xj, note that if i -r r, then X~ is adjacent to X~ whereas Xj is not, and if i - r, then X~ is adjacent to X~+I, { X i , . . . , X~+I} is a clique, j must be one of r + 2 , . . . , t, and Xi is adjacent to X~+I whereas Xj is not. To see that i _< r, note that by 1 i >_ r + 1 would contradict the definition of i. Thus it is always possible to choose r and s to satisfy 1 and 2, often in different ways. Now assume that this was done and S and T were joined by edges according to 3 and 4. We complete the proof by showing that (11.4) holds relative to G. First we show that if u _< v, then X~ ~ X~ relative to G. Indeed, let B be an internal vertex adjacent to X~. Then by 3 v _< r, hence u _< r and by 3 B is adjacent to X~. A similar use of 4 results in the same conclusion for external vertices. It remains to show that X~ ~- E ~- X~+I and X~ ~- I ~- X~+I. X~ ~ E: Let Z be any vertex adjacent to some vertex of E. Then Z is in I o r T. In the first case Z is adjacent to X~ by 3 and we are done. In the second case we have Z - Xq for some q and q _< s by 4. If j is defined, then by 2 Xq ~ Xj relative to T, hence q < j, and by the definition of j we have that Z coincides with or is adjacent to X~. If j is undefined, then X~ is adjacent to all other vertices in T, so again Z coincides with or is adjacent to XT. E ~- X~+I" Let Z be any vertex adjacent to X~+I. Since s < r by 2, and by 3 and 4, Z must be in T, and so Z - Xq for some q. Then i is defined and satisfies q < i, and by 2 i < s. Therefore q < s and so by 4 Z is adjacent to the vertices of E. Xs ~ I: Let Z be any vertex adjacent to some vertex of I. If Z is in S, then by 3 and 4 and the fact that s < r, Z is adjacent to X~ and we are done. If Z is in T,we have Z - Xq for some q and q < r by 3. If j is defined, we distinguish the cases s 7~ r and s - r. In the first case, since by 2 X~ ~- Xj relative to T and therefore s < j, Xs is adjacent to XT by the definition of j, and therefore X~ coincides with or is adjacent to Xq. In the second case X~ ~- Xj by 2 so that r < j, by the definition of j { X 1 , . . . , X~} is a clique, and in particular since q, s _< r we have that Xq coincides with or is adjacent to X~. Again, if j is undefined, then X~, and hence X~, is adjacent to all

288

Matroidal and Matrogenic Graphs

other vertices in T, so t h a t Xq coincides with or is adjacent to X~. I ~- Xs+l" Let Z be any vertex adjacent to Xs+l. If Z is in S', then by 4 Z is internal, and therefore Z is adjacent to all other vertices of I and we are done. I f Z i s i n T , then Z - X q for s o m e q . Since b y 2 i_< s < s + l , and by the definition of i, X~+I and X~+I are not adjacent. This means t h a t either s + 1 - r + 1, in which case q _< r by 1, or else X~+I and XT+I are distinct non-adjacent vertices, in which case Xq ~- XT+I relative to T, hence again q _< r. Thus in both cases Xq is adjacent to the vertices of I by 3. .. We illustrate T h e o r e m 11.2.23 with the following example. Let T be the threshold graph illustrated in Figure 11.11, relative to which one has X1 ~- X2 "~ X3 ~- X4. Table 11.1 summarizes all the possibilities of m a t r o i d a l graph whose threshold vertices are X1, . . . . X4 and whose other vertices induce a cell. These graphs are illustrated in Figure 11.12 for the case t h a t the cell is a P4. Figure 11.11" A threshold graph.

X3

X1

X2

X4

Table 11.1" A summary of the possibilities for the graph of Figure 11.11. Case

r

i

j

s

(a) (b)

1 2

1 1

4 3

1 1

(c)

3

2

0

(d) (e)

3 4

2 1

1 0

Order of vertices X1 X1 I X1 I

~ ~ ~ ~~-

I I X1 I X1

~ ~ ~ ~~-

E X2 22 X2 X2

}.- 2 2 }.- f_, ,.-, 2 3 ,.~ X 3 ,~ X 3

~ ~ ~.~:,-

X3 X3 ~, E X4

~ ~~~~-

24 24 24 X4 E

Having e x a m i n e d the case where S' is a cell, we now generalize to the case t h a t S is a cell graph. The next theorem describes the situation in this case. Theorem

1 1 . 2 . 2 4 Let S be a cell graph with cells 1 , . . . , k satisfying Ik;, - . . . ~ - I 1 ~- E 1~- . . . ~ - E k

(11.5)

relative to S. Let T be a threshold graph disjoint from S with vertices X 1 , . . . , X t satisfying (11.3) relative to T. If S and T are joined by edges

tl.2

Matroidal Graphs

289

Figure 11.12: Matroidal graphs corresponding to Table 11.1.

X3

Xl

X2

X4

X3

X1

(~)

X3

X1

X2

X4

X2

X4

(b)

X2

X4

X3

X1

(c)

(d)

X3

X1

X2

X4

~v

v

v

v

(e)

Matroidal and Matrogenic Graphs

290

so that the resulting graph G is matroidal and its threshold vertices are precisely X 1 , . . . , Xt, then for each cell l there exist indices rl, st satisfying 1 of Theorem 11.2.23 with respect to I t and E l, and furthermore Sk ~

"'"

< 81

~

rl ~

...

~ rk.

(11.6)

Conversely, r~ and st can always be chosen to satisfy 1 and 2 as well as (11.6), and if the connecting edges satisfy 3 and ~ with respect to I t and E ~, then the resulting graph is matroidal and its threshold vertices are precisely X1, . . . , Xt.

P r o o f . If G is matroidal with threshold vertices X 1 , . . . , X t , then the sets 11, E l , . . . , I k, E k, {X1 } , . . . , {Xt } are totally ordered by ~ relative to G in a way that is a common refinement of both (11.5) and (11.3). For each cell l, I l U E t U { X 1 , . . . , Xt } induces a matroidal graph whose threshold vertices are X 1 , . . . , Xt and whose other vertices induce a cell. By Theorem 11.2.23 there are indices rl and st satisfying 1-4 of that theorem with respect to I t and E I. Using (11.4) applied to rt, st, I t and E t and also (11.5), we obtain (11.6). Conversely, assume that S and T are given as above. Then by Theorem 11.2.23, for each cell 1 it is possible to choose indices rt, st satisfying 1-2 of that theorem. Furthermore, it is also possible to satisfy (11.6), for example by choosing rl . . . . . rk and Sl . . . . = sk. We assert that if the vertices of I t are joined to X 1 , . . . , Xr~ and the vertices of E l are joined to X 1 , . . . , X~ so as to satisfy 3-4 of the theorem, then the resulting graph G is matroidal and its threshold vertices are X 1 , . . . , X t . To show this, it is sufficient to demonstrate that (11.3) and (11.5) continue to hold relative to G and that (11.4) holds relative to G with rl, sl, I l and E t It is immediate to obtain (11.3) from 3 and 4 of Theorem 11.2.23. As for (11.5), 11 ~- E 1 follows from Theorem 11.2.23. To show that 12 ~ 11, assume that Xq is adjacent to the vertices of I ~. Then q _< rl by (3), hence q _< r2 by (11.6), and Xq is adjacent to the vertices of 12 by 3. Similarly all the relations of (11.5) can be proved. It remains to prove the relations Xr z > E t > X~z+1 and Xs~ ~- I l ~ Xs~+1. Let us prove X~.t ~ E l , the other relations being similar. Let Z be any vertex adjacent to the vertices of E z. If Z is in T, then Z coincides with or is adjacent to X~ z by Theorem 11.2.23. If Z is in S, then by definition of a cell graph and by (11.5), Z must be in 11 U ... U I k. By 3 Z is adjacent at least to X 1 , . . . , X m , where m - m i n ( r l , . . . , r k ) . But m - rt by (11.6), and therefore Z is adjacent to X~. 9

Corollary 11.2.25 Let G be a matroidal graph all couples of which are of the f o r m P4. Then G is a split graph.

2

M a t r o i d a l Graphs

291

roof. We use the results and notations of Theorem 11.2.24. Let h be the rgest index such that Sl _< h _< rl and { X ~ , . . . , Xh} is a clique. Then if < rl, Xh is not adjacent to Xh+x. Therefore in all cases { X h + l , . . . , X~ } is a able set. We assert that in fact { X I , . . . , Xh} is a clique and { X h + l , . . . , Xt } a stable set. Indeed, if sl < h, the first statement is obvious; if Sl - h, follows from the facts that X1 >- I x, . . . , X~ >- I ~ and that X~ ~ are adjacent to all the vertices of 11. If h < rl, the second statement obvious; if h - rl, it follows from the fact that X~+I and X~+2 are not tjacent. Since X 1 , . . . , X~ 1 are adjacent to the vertices of 11 U . . . U I k, u U 11 U ... U I k is a clique. Since X~ l + l , . . . , X t are not tjacent to any vertex of E 1 tJ . . . t.J E k, { X h + l , . . . , X t } U E 1 U " " [_J E k a stable set. .. We conclude the description of the matroidal graphs by answering the ~estion which vertices of T are adjacent to M. h e o r e m 11.2.26 Let S be a cell graph and T a threshold graph as in Theo:m 11.2.2~. Let M be a perfect matching inK2 or a complement of a perfect ~atching inK2 disjoint from S and T, with rn > 2. If the edges joining M S and T are such that the resulting graph G is matroidal with threshold ~rtices X1, . 9 Xt , then 1. the vertices of M are adjacent to all the internal vertices and to none of the external vertices of S; 2. if h is the largest index such that Xh is adjacent to the vertices of M (h - 0 if none), then Sl < h < rl, { X l , . . . , X h } U 11 U . . . U I k is a clique and { X h + l , . . . , X t } t2 E 1 tJ . . . U E k is a stable set. ~onversely, it is possible to choose an index h such that S 1 < h < r l , X I , . . . , X h } is a clique and { X h + l , . . . , X t } is a stable set. If the vertices f M are joined to the vertices of { X 1 , . . . , X h } U 11 U . . . U I k and to no ther vertices, then the resulting graph G is matroidal with threshold vertices ~'1,''', Xt.

'roof. Condition 1 has been proved in Subsection 11.2.3, together with the mt that the vertices of S partition into a clique whose vertices are adjacent ) M and a stable set whose vertices are not. If h - 0, then the vertices of { X 1 , . . . , X t } U .~1 I,_J . . . I,.J E k are not djacent to M, hence they form a stable set, Sl - 0, and 2 follows. The

292

Matroidal and Matrogenic Graphs

case h - t is similarly easy, and therefore we now assume that 0 < h < t. Since (11.3) holds relative to G, the vertices of M are adjacent to X 1 , . . . ,Xh. It follows that { X 1 , . . . , X h } t0 I 1 tO ... tO I kis a cliqueand { X h + l , . . . , X t } tO E 1 U ... U E k is a stable set. Since M is comparable with every vertex of S and T relative to G, it satisfies in particular 11 P- M >- E 1 and Xh >- M >Xh+l relative to G. By transitivity of >- we have Xh >- E 1 and 11 >- Xh+l, from which it follows that Sl _< h _< rl and 2 is proved. Conversely, by Corollary 11.2.25 there is an index hi _< h _< rl such that { X 1 , . . . , X h } tO I1 U ... tO I k is a clique and {Xh+~,...,Xt} U E 1 U ..- tO E k is a stable set. Assume that M is joined to the vertices of { X 1 , . . . , X h } U I 1 U " " U I k and to no other vertices of S U T. In order to prove that the resulting graph G is matroidal with threshold vertices X 1 , . . . , Xt, it is sufficient to show that M is comparable with every vertex of S U T relative to G. We show below that each cell l satisfies I 1 >- M >- E t and t h a t X 1 > - M f o r q _ < h a n d M > - X q forq_>h+l. I t >- M: Let Z be any vertex adjacent to some vertex o f M . I f Z i s i n M or in S, then Z is adjacent to all vertices of I t. If Z is in T, say Z - Xq, then q <_ h <_ rl ~ rl and therefore Xq is adjacent to all the vertices of ft. M >- E l" This is proved similarly. Xq >- M for q _< h" Let Z be any vertex adjacent to some vertex of M. If Z is in M, then Z is adjacent to Xq. If Z is in S, then it is internal, say Z C fl, and thus adjacent to Xq since q _< h _< r 1 ~ r I. If Z is in T, say Z - Xp, then p <_ h and Xq is adjacent to Xp because { X 1 , . . . , X h } is clique. M >- Xq for q _> h + 1" This is proved similarly. 9

11.2.5

Properties of Matroidal Graphs

The structure of matroidal graphs has been completely described in the previous three subsections. Each matroidal graph can be decomposed into a cell graph S, induced by the vertices that belong to P4's and described by Theorem 11.2.16, a perfect matching or a complement of a perfect matching M, induced by the vertices that belong to 2K2's or C4's, and a threshold graph T induced by the threshold vertices. The vertices of M are adjacent to all the internal and to none of the external vertices of S'. The edges joining T to S' are described by Theorems 11.2.23 and 11.2.24, and the edges joining T to M are described by Theorem 11.2.26. We can now determine the threshold dimension t(G) of a matroidal graph G, which is the original

11.2

M a t r o i d a l Graphs

293

reason of interest in these graphs. T h e o r e m 11.2.27 Let G be a matroidal graph and let E l , . . . ,Eq be the

equivalence classes of the relation R of page 271. Then

t(G) = max(IE, I,..., levi). P r o o f . Put t = m a x ( l E l l , . . . , IEql). If fewer than t subgraphs of G cover all its edges, then one of these subgraphs H has two edges from the same equivalence class Ei. These two edges form a couple in G and therefore in H too, and so H cannot be a threshold graph. This shows that t(G) >_ t. To prove the reverse inequality, we exhibit t threshold subgraphs of G covering its edges. We construct one of these subgraphs, called H, as follows. Since G is matroidal, it decomposes into S, M and T as above. Consider a cell of S. If it is a net of order p, J~f(AI,...,Ap, BI,...,Bp), then its p external edges AiBi form one equivalence class, and each internal edge BjBk forms an equivalence class by itself. There are p ways to select one edge from each of these equivalence classes, giving p threshold subgraphs of the net covering its edges. We select one of these p threshold subgraphs and put its edges in H. If the cell is a net-complement of order p, A/'(A1,..., dr, B 1 , . . . , Bp)), then each pair AjBk, AkBj, j ~ k, of external edges forms an equivalence class, and each internal edge BjBk forms an equivalence class by itself. The collection of the external edges AjBk with j < k and all the internal edges is a threshold subgraph of the net-complement containing one edge of each of these equivalence classes. This collection and a similar one obtained with j > k cover all the edges of the net-complement. We select one of these two and enter its edges in H. Now consider M, and assume it is a perfect matching inK2. Then each of its m edges forms an equivalence class by itself and is a threshold subgraph of M, and these m subgraphs obviously cover the edges of M. We select one of these edges and put it in H. If M is a complement inK2 of a perfect matching induced by the vertices A 1 , . . . , Am, A 1 , . . . , A" with all edges present except for AiA~, i - 1 , . . . , m, then each edge X Y forms an equivalence class with ZU, where Z is the unique vertex of M not adjacent to X and U is the unique vertex of M not adjacent to Y. The collection of edges AjA'k with j < k and all edges AjAk forms a threshold subgraph of M having one edge of each of these equivalence classes. This collection and its complement, which has a similar form, cover all the edges of M. We select one of these two collections and put its edges in H.

294

Matroidal and Matrogenic Graphs

Each of the remaining edges of G (namely the edges of T and the edges joining S, M, T and the different cells of S) forms an equivalence class by itself. We put all of them in H. The H constructed in this way has exactly one edge of each equivalence class Ei, and it is clear that we can construct t H's of this form that will cover all the edges of G. Moreover, we assert that each H constructed in this way is a threshold subgraph of G, in other words, no two edges a and b of H form a couple in H. Indeed, if both a and b are in the same cell of S, or if both are in M, then the assertion follows by construction. Otherwise a and b do not form a couple in G, because G contains enough edges so that the endpoints of a and b induce a triangle in G, namely the edges of T and the edges joining S, M, T and the different cells of S. But these edges are in H as well, so the endpoints of a and b induce a triangle in H too, and so a and b do not form a couple in H. This proves that t(G) <_ t. 9 As an illustration, consider the matroidal graph displayed in Figure 11.13. Figure 11.13" A matroidal graph.

A

B

E

H

C

D

Here M is induced by A, B, C, D and is a complement of a perfect matching, S is induced by E, F, G, H and is a cell (both a net and a net complement of order 2), and T is empty. Thus the largest equivalence class Ei has two edges. The two threshold subgraphs used in the proof above are shown in Figure 11.14. By computing the Orlin weights (Section 1.3) of these two threshold subgraphs we find a smallest system of linear inequalities separating the incidence vectors of the stable and the non-stable sets of vertices of the graph of Figure 11.13, where in each inequality each coefficient is an integer which is as small as possible: llF+10G+8A+4B+4C+2D+E

<_ 11

llG+10F+SD+4B+4C+2A+H

_< 11.

11.2

M a t r o i d a l Graphs

295

Figure 11.14: Two threshold subgraphs covering the graph of Figure 11.13. A

B

E

A G

H

B H

E

A v

v

C

D

C

v

D

Chvgtal and H a m m e r [CH77] have proved that for each ~ > 0 there exists a positive integer n and a graph G on n vertices satisfying t(G) > (1 - e)n. The situation for matroidal graphs is quite different, as the next theorem shows. 1 1 . 2 . 2 8 If G is a matroidal graph on n vertices, then t(G) <_ n/2, and equality holds if and only if G is a perfect matching, a square or a net.

Theorem

P r o o f . The graph G decomposes into T, S and M as above. Let T have t vertices, let S have k cells of orders p l , . . . , pk, and let M have m vertices. If m = k = 0, then G is a threshold graph, n = t and t(G) is 1 or 0 according as G possesses edges or not. In that case clearly t(G) <_ n/2, equality holding if and only if G is a single edge, which is a perfect matching. Now assume that m + k > 0. Let at be pl or 2 according as cell 1 of S is a net or a net-complement. Also let b be m / 2 , 2, or 0 according as M is a matching, a complement of a matching, or empty. Then n = t + 2(pl + - " + pk) + m and t(G) = max(b, a l , . . . , ak). By definition of b, if m > 0, then b/rn <_ 1/2, equality holding if and only if M is a perfect matching or a square. By definition of at, at/2pt <_ 1/2, equality holding if and only if cell 1 is a net. From this the result follows immediately. ,. Theorem

11.2.29

Each matroidal graph is chordal or co-chordal, and hence

perfect. P r o o f . Let G be a matroidal graph, partitioned into M, S and T as above. If M is empty, then G is split by Corollary 11.2.25 and therefore chordal. If M is a matching, then G is chordal by Theorem 11.2.17. If M is a complement of a matching, then G is co-chordal by the same result. '.

296

11.3

Matroidal and Matrogenic Graphs

Matrogenic Graphs

We begin our study of the matrogenic graphs, based on [FH78b], with the following simple lemma, which uses the notation $-(A, B, C, D, E) of Definition 11.2.1. L e m m a 11.3.1 For a graph G, the following conditions are equivalent.

1. G does not contain the configuration ~; 2. if two alternating ~-cycles of G have exactly three common vertices, then their union induces a chordless pentagon. P r o o f . 2) =~ 1): This is clear, since the vertices of 7 ( A , B, C, D, E) cannot induce a chordless pentagon, yet the configuration contains two alternating 4-cycles with exactly three common vertices, A D E B and ADEC. 1) =~ 2): Assume that G has two alternating 4-cycles on the vertex-sets {A, B, C, D} and {A, B, C, E}. Without loss of generality, we may assume that A B and CD are edges and BC and DA are nonedges. Note that if E A is an edge, then so is EC, otherwise we have ~-(A, B, E, D, C), a contradiction. Similarly, E A is an edge if EC is an edge, and E B is an edge if and only if ED is an edge. Since { A , B , C,E} induces an alternating 4-cycle, the above considerations imply that if EA and EC are edges, then AC is a nonedge, and E B and ED are nonedges. Now if BD is a nonedge, we have ~ ' ( A , B , E , C,D), a contradiction. Therefore BD is an edge, and our five vertices induce a chordless pentagon. If instead EA and EC are nonedges, then AC is an edge, and E B and ED are edges. Now if B D is a nonedge, we have a chordless pentagon, otherwise we have ~ ( D , B, E, A, C). .. We are now ready to characterize the matrogenic graphs. Recall that by definition, a graph is matrogenic if and only if whenever C1 and C2 are distinct sets of four vertices, each inducing an alternating 4-cycle, and sharing a vertex x, then C1 U C2 - {z} contains an alternating 4-cycle. T h e o r e m 11.3.2 A graph is matrogenic if and only if it does not contain the configuration jz. P r o o f l The necessity is clear, since in a configuration JZ(A,B, C , D , E ) ,

C1 = {A, B, E, D} and C2 = {A, C, E, D} induce alternating 4-cycles and share the vertex D, but C1 U C 2 - {D} = {A, B, C, E} does not contain an alternating 4-cycle.

11.3

Matrogenic Graphs

297

To prove the sufficiency, assume that G has no configuration ~- and that C1 and 6'2 are distinct sets of four vertices, each inducing an alternating 4-cycle, and sharing a vertex x. We show below that C1 U C2 induces a matching of three edges, a net of order 3 (Definition 11.2.10), a chordless pentagon, or their complements. Each of these graphs is easily seen to be matrogenic, and therefore C1 U 6'2 - {x} contains an alternating 4-cycle, as required. We distinguish the following cases. C a s e 1: 6'1 and C2 share exactly one vertex. Then the situation is as illustrated in Figure 11.15. Figure 11.15: A configuration in Case 1 of Theorem 11.3.2. E

--- ---I C

i

F

B

Now EC must be an edge, otherwise we have an )r(A, B, E, D, C), contrary to our assumption. Similarly CG must be a nonedge, otherwise we have an 5 ( C , D, G, B, A). But then we have an ~-(E, A, C, F, G), a contradiction. Therefore this case is impossible. C a s e 2 : C 1 and C2 share exactly two vertices, which are adjacent. Let C1 = {A, B, C, D} and C2 = {A, B, E, F} where AB is an edge. We distinguish three subcases depending on how many of the edges CD and E F are present. C a s e 2.1: Both edges CD and E F are present. We may then assume without loss of generality that AD, BC, AF and B E are nonedges. If the edge CF were present, then {A,B, C,F} would induce an alternating 4cycle, but {A, B, C, D, F} cannot induce a chordless pentagon, contrary to Lemma 11.3.1. Therefore CF is a nonedge, and similarly DE is a nonedge. The situation is illustrated in Figure 11.16.

298

Matroidal and Matrogenic Graphs

Figure 11.16: A configuration in Case 2.1 of Theorem 11.3.2.

E

A

c

B

D

i;'" •

F

We assert that {A, C, E} is a stable set or a clique. For otherwise, we may assume without loss of generality that CE is an edge and AC is a nonedge. Then {A, B, C, E} induces an alternating 4-cycle, but {A, B, C, E, F} cannot induce a chordless pentagon, contrary to Lemma 11.3.1. Similarly {B, D, F} is a stable set or a clique. Moreover, {A, C,E} and { B , D , F } cannot both be cliques, because in that case {C, D, E, F} would induce an alternating 4-cycle, but {A, C, D, E, F} cannot induce a chordless pentagon, contrary to Lemma 11.3.1. We conclude that either {A, C,E} and { B , D , F } are both stable, in which case the six vertices induce a matching 3K2, or else one is stable and one is a clique, in which case the six vertices induce a net of order 3. Case 2.2: Both edges CD and EF are absent. We may then assume without loss of generality that AC, BD, AE and BF are edges and AD, BC, AF and BE are nonedges, as in Figure 11.17. It follows that {A, B, C, D} and {A, B, C, F} induce alternating 4-cycles even though {A, B, C, D, F} cannot induce a chordless pentagon, contrary to Lemma 11.3.1. Therefore this subcase is impossible. Case 2.3: Exactly one of the edges CD and EF is present, say CD is present and EF is absent. We may then assume without loss of generality that AE and B F are edges and AF and BE are nonedges. If BD were present, then AD must be absent to avoid the triangle ABD, and this gives an ~ ( B , D, F, E, A), a contradiction. Therefore BD must be a nonedge, and similarly AC, AD and BC are nonedges. Further, if FD were present, we would have an ~ ( D , C, F, B, A), a contradiction. Therefore FD is a nonedge, as illustrated in Figure 11.18.

11.3

Matrogenic Graphs

299

Figure 11.17" A configuration in Case 2.2 of Theorem 11.3.2. E

A

I \ \

/ /

I

\

I

i

I /

X

C \

\

/ /

/

\

\

i

i

\

\

/

X

/

I

\

/

I

I

\

\

I v

F

B

D

Figure 11.18" A configuration in Case 2.3 of Theorem 11.3.2. E

A

A

A

T, I

\

\

I I

I /

/

\ i /

X

/

C \

\

/

/

\

\

i \

/

\

/

X

/

/ \ \

2,,

F \-.

B

--~ D

It follows that {A, B, C, D} and {B, C, D, F} induce alternating 4-cycles even though {A, B, C, D, F} cannot induce a chordless pentagon, contradicting Lemma 11.3.1. Therefore this subcase is also impossible. Case 3 : C 1 and C2 share exactly two vertices, which are not adjacent. Since the alternating 4-cycle, the configuration $', and the chordless pentagon are isomorphic to their complements, it follows from Case 2 that in Case 3, the six vertices induce a complement of a matching 3K2 or a net-complement of order 3. C a s e 4 : C 1 and C2 share exactly three vertices. Then by Lemma 11.3.1 C 1 LJ C 2 induces a chordless pentagon. 9 The following corollaries follow easily from Theorem 11.3.2. C o r o l l a r y 11.3.3 All induced subgraphs of a matrogenic graph are matrogenic.

300

Matroidal and Matrogenic Graphs

Corollary 11.3.4 The complement of a matrogenic graph is matrogenic. In the following we find it useful to refer to a perfect matching with two or more edges or its complement or a chordless pentagon as a core.

Corollary 11.3.5 All cores are matrogenic. We now reduce the problem of finding the structure of matrogenic graphs to the problem of finding the structure of split matrogenic graphs. Specifically, we shall see that the vertices of a matrogenic graph partition into S U K U C, where S U K induces a matrogenic split graph with S a stable set and K a clique, C induces a core, and every vertex of C is adjacent to every vertex of K and to no vertex of S. Conversely every graph with these properties is matrogenic. T h e o r e m 11.3.6 Let G be a graph whose vertex set is partitioned into disjoint sets S U K U C, where S U K induces a matrogenic split graph with S a stable set and K a clique, C induces a core, and every vertex of C is adjacent to every vertex of K and to no vertex of S. Then G is matrogenic. P r o o f . Since the graphs induced by Q = S tJ K and by C are matrogenic, it is sufficient to show that no alternating 4-cycle can meet both Q and C. Assume that, if possible, vertices x, y, z, u induce an alternating 4-cycle and z E C, y E Q. We may further assume without loss of generality that y E K, for if y E S, then one of z and u must be in K, and we can rename it to be y. Then z E S U C, for if z E K, then the alternating 4-cycle contains a triangle xyz, an impossibility. Similarly u E S U C. But now if z, u E S, then {z, z, u} is stable, if z, u E C, then y has three neighbors in the alternating 4-cycle, and if z E S and u E C, then y has three neighbors or z has three non-neighbors in the alternating 4-cycle, depending on whether yz is present or not. In each case we have a contradiction. .. To prove the converse of Theorem 11.3.6, we prove the following two lemmas. L e m m a 11.3.7 Let P be a chordless pentagon in a matrogenic graph G. Then the vertices of G outside P partition into a clique whose vertices are adjacent to every vertex of P and a stable set whose vertices are adjacent to no vertex of P. Moreover, P is uniquely determined by G.

11.3

Matrogenic Graphs

301

P r o o f . Denote the vertices of P by A, B, C, D, E cyclically along the pentagon. Let x be a vertex outside P. We assert that if x is adjacent to A, then it is also adjacent to B. Assume that, if possible, xA is an edge and x B is a nonedge. Then {A, B, C, E} and {A, B, C, x} induce alternating 4cycles, but {A, B, C, E , x } cannot induce a chordless pentagon, since A has three neighbors in this set, contrary to Lemma 11.3.1. It follows that every vertex outside P is adjacent to all or to none of the vertices of P. If two vertices x and y outside P are adjacent to every vertex of P, then they are adjacent to each other, for otherwise G has an 9r(x, A, B, y, D), contrary to Theorem 11.3.2. By working in the complement we then see that if two vertices x and y outside P are non-adjacent to every vertex of P, then they are not adjacent to each other. The uniqueness of P follows from the facts that no chordless pentagon can meet both P and G - P, and that a split graph cannot contain a chordless pentagon (Theorem 5.2.1). 9 L e m m a 11.3.8 Let G be a matrogenic graph and M a maximal induced perfect matching in G having at least four vertices. Then the vertices of G outside M partition into a clique whose vertices are adjacent to every vertex of M and a stable set whose vertices are adjacent to no vertex of M. Moreover, M is uniquely determined by G. The same is true i f M is a maximal induced complement of a perfect matching having at least four vertices. P r o o f . Let A B and CD be distinct edges of M and x a vertex outside M. If x is adjacent to A, then it must also be adjacent to C, for otherwise G has an .7"(A, B , x , D, C). By the same argument x must also be adjacent to B, and therefore to every vertex of M. It follows that every vertex outside M is adjacent to all or to none of the vertices of M. If two vertices x and y outside M are adjacent to every vertex of M, then they are adjacent to each other, for otherwise {A, C, x, y} and {B, C, x, y} induce alternating 4-cycles even though { A , B , C , x , y } cannot induce a chordless pentagon. If two vertices x and y outside M are non-adjacent to every vertex of M, then they are not adjacent to each other, for otherwise we could add them to M, contradicting its maximality. The uniqueness of M follows from the facts that no perfect matching with two or more edges can meet both M and G - M, and that a split graph cannot contain a perfect matching with two or more edges (Theorem 5.2.1). This proves the lemma if M is a perfect matching. If M is a complement of a perfect matching, the result follows by working in the complement of G. 9 We are now ready for the converse of Theorem 11.3.6.

302

Matroidal and Matrogenic Graphs

T h e o r e m 11.3.9 A matrogenic graph is either a split graph or its vertex set partitions uniquely into S O K U C so that Q = S o K induces a split graph, S is a stable set and K is a clique, C induces a core, and every vertex of C is adjacent to every vertex of K and to no vertex of S. P r o o f . If G has an induced core, the result follows from Lemmas 11.3.7 and 11.3.8. If G has no induced core, then it has no induced 2K2, C4 or C5, and therefore it is a split graph by Theorem 5.2.1. [] As we have pointed out, Theorems 11.3.6 and 11.3.9 reduce the problem of finding the structure of matrogenic graphs to the problem of finding the structure of matrogenic split graphs. However, by Theorems 11.3.2, 5.2.1 and 11.2.2, a matrogenic split graph is matroidal, and the structure of a matroidal split graph was given in Section 11.2, namely it is a cell graph with threshold vertices connected to it according to Theorems 11.2.23 and 11.2.24. From the above structure of a matrogenic graph G, we can determine its Dilworth number D(G) (Section 9.1) and threshold dimension t(G) (Section 6.1). T h e o r e m 11.3.10 Let G be a matrogenic graph. Let Q, K, S and C be as in Theorem 11.3.9 if G is not a split graph, and let Q = V(G), C = 0 if G is a split graph. Then

D(G) t(a)

=

max(D(G[C]), D(G[Q])),

=

max(t(G[C]), t(G[Q])).

P r o o f . If G[X] is an induced subgraph of G and A is an antichain of (the vicinal preorder of) G[X], then A remains an antichain of (the vicinal preorder of) G. Therefore D(G) >_ D(G[X]) and in particular

D(G) >_ max(D(G[C]), D(G[Q])). On the other hand, no antichain of G can meet both C and Q, since every vertex of C is comparable to every vertex of Q. Hence

D(G) < max(D(G[C]), D(G[Q])). If T is a spanning threshold subgraph of G and G[X] is an induced subgraph of G, then T[X] is a spanning threshold subgraph of G[X]. Therefore t(G) >_ t(G[X]), in particular t(G) >_ On the other hand, if Tc is a spanning threshold subgraph of G[C] and T o is a spanning

11.4

Matrogenic Sequences

303

threshold subgraph of G[Q], then the spanning subgraph of G consisting of the edges of Tc U TQ together with all the edges of G[K] and all the edges of G joining C to K is a threshold subgraph of G. This can be easily verified by noting that no alternating 4-cycle of this subgraph can meet both C and Q. If we apply this fact to a minimum collection of spanning threshold subgraphs of G[C] covering its edges and similarly for G[Q], we see that

t(G) <_ max(t(G[C]), t(G[Q])).

..

To apply Theorem 11.3.10, we need to know the threshold dimension and the Dilworth number of cores and split matrogenic graphs. It is easy to see that D(Ps) = 5, D(mK2) = D(mK2) = m and t(Ps) = 3. Split matrogenic graphs are matroidal and their threshold dimension is given in Theorem 11.2.27. Their threshold vertices are comparable with every vertex and therefore can be dropped without changing the Dilworth number, and we remain with the problem of the Dilworth number of a cell graph. Since vertices of different cells are comparable, the Dilworth number of a cell graph is the maximum of the Dilworth numbers of its cells. A cell is a net or a net-complement of order p, and it is easy to see that its Dilworth number is p.

11.4

Matrogenic Sequences

In this section we present some results of Marchioro et al. [MMPS84] characterizing the degree sequences of matrogenic graphs, called matrogenic sequences. Since matroidal graphs differ from matrogenic graphs only by the fact that their core cannot be a chordless pentagon, the results characterize degree sequences of matroidal graphs as well. Our task is simplified by the following result. 11.4.1 Every matrogenic sequence is unigraphic, i.e., all its realizations are isomorphic.

Theorem

P r o o f . We assume that G is matrogenic and H has the same degree sequence as G, and show that G and H are isomorphic. We may assume without loss of generality that G and H are graphs on the same vertex set. Then by Corollary 3.1.11, H can be obtained from G by a sequence of transfers. We may therefore assume that H is obtained from G by a single transfer, namely that G has an alternating 4-cycle with edges AB, CD and nonedges AC, BD, and H is obtained from G by dropping edges AB, CD and adding edges AC,

304

Matroidal and Matrogenic Graphs

B D . The required isomorphism from G to H simply exchanges A with D and fixes every other vertex. Showing that it is indeed an isomorphism is equivalent to showing that if X is a fifth vertex, then X A and X D are either both edges or both nonedges of G. This holds, for otherwise G contains the configuration ~ (Definition 11.2.1), contrary to Theorem 11.3.2. 9 As an aid in describing the structure of matrogenic graphs, we now introduce a graph operation of Tyshkevich and Chernyak [TC853], which is used extensively in Chapter 17.

Definition 11.4.2 Let G = G ( K , S ) be a split graph with a clique K and a stable set S, and let H be a graph disjoint from G. The c o m p o s i t i o n G @ H of G and H is the graph obtained by taking G U H and joining every vertex of H to every vertex of K . In Chapter 13, we give a slightly more general definition of graph composition. As we saw in Section 11.3, a matrogenic graph has the form G ( K , S ) Q C, where G is a matrogenic split graph with a clique K and a stable set S, and C is a core, namely a perfect matching or a complement of a perfect matching or a chordless pentagon. Conversely, every such graph is matrogenic. This reduces the problem of studying the structure of matrogenic graphs to the one for split matrogenic graphs. Since a split matrogenic graph is also matroidal, its structure has been determined in Section 11.2, namely it consists of a cell graph and threshold vertices connected to it according to Theorem 11.2.24. That theorem assumed as given the structure of the cell graph and of the threshold graph, and was concerned with the edges connecting them. Here we reformulate this result to describe the overall structure of split matrogenic (equivalently split matroidal) graphs. T h e o r e m 11.4.3 Let G = G ( K , S ) be a split graph with a clique K and a stable set S. Then G is matrogenic if and only if K tO S partitions into disjoint sets C 1 , . . . , Cq, called c h a m b e r s , such that 1. each chamber Ci satisfies one of the following: (a) it induces a net or a net-complement (Definition 11.2.10) with internal vertices Ii = Ci N I( and external vertices Ei = Ci N S, in which case it is called a cell c h a m b e r ; (b) it is contained entirely in K , in which case it is called a clique chamber:

11.4

Matrogenic Sequences

305

(c) it is contained entirely in S, in which case it is called a s t a b l e chamber; 2. for i 7/= j, a vertex of Ci N S is adjacent to a vertex of Cj N K if and only if i < j. When G is matrogenic, the threshold vertices of G are the vertices of the clique and stable chambers, and the vertices of the cell chambers of G induce a cell graph as in the end of Section 11.2. P r o o f . "If"" Assume that G satisfies the conditions in the theorem. We assert that the threshold vertices of G are precisely the vertices of the clique and stable chambers. Clearly, if a chamber induces a net or a net-complement, then its vertices belong to couples and are not threshold vertices. Conversely, consider an alternating 4-cycle of G with vertices x E Ci N S, y E Cj N K, z E Ck n K, u E Cl n S, edges xy, zu and nonedges xz, yu (since G is split, every alternating 4-cycle has this form). Assume that, if possible, Ci c_ S. Then i :fi j, k, hence by Condition 2, k < i < j. But the edge and nonedge at u imply that j _< l _< k, a contradiction. Similarly Ci c_ K leads to a contradiction, proving that Ci is a cell chamber. Dropping the threshold vertices from G cannot affect the matrogenic property, but the resulting graph is a cell graph and is therefore matroidal and matrogenic. Therefore G is matrogenic. " O n l y if"- If chambers Ci and Ci+I a r e both clique chambers or both stable chambers, we may merge them into one chamber without affecting the validity of Conditions 1 and 2. We always assume that this has been done whenever possible. We say that cell chambers C~ and C~ with r < s are consecutive when C ~ + I , . . . , C~-I are not cell chambers, and therefore are alternately clique and stable chambers. This terminology is extended to the case that r - 0 and C~ is the first cell chamber, or s - q + 1 and C~ is the last cell chamber. We prove that Conditions 1 and 2 hold by induction on the number of threshold vertices of G. If G has no threshold vertices, then it is a cell graph and the conditions are clearly satisfied. For the induction step, we assume that G satisfies our conditions and we add to it another threshold vertex x, and show that the resulting graph G ~ still satisfies the conditions. We assume without loss of generality that x is added to S (otherwise work with the complementary graph). Note that if x is adjacent to some vertex of Ii, then it is adjacent to all vertices of Ii, for otherwise x belongs to an

306

Matroidal and Matrogenic Graphs

alternating 4-cycle. For the same reason, x is not equivalent (under the vicinal preorder) to any vertex of Ei. If x is equivalent to some vertex of a stable chamber Ci, then G' satisfies our conditions with the new chambers C I , . . . , C i _ I , Ci u { z } , C i + l , . . . , Cq,

and we are done. Therefore we may assume that x is not equivalent to any vertex of S. Since x is comparable with every vertex of S, there exist consecutive cell chambers C~ and C~ such that

Er~..x~-E ~. It follows t h a t x is adjacent to every vertex of Ci VI K for i >_ s and to no vertex of Ci fl K for i _< r. We now compare x with the stable chambers between Cr and Cs. Since the chambers between Cr and Cs are alternately stable and clique chambers, and since x is not equivalent to any vertex of S, the typical case is that there exists an index j with r < j - 1 < j + 1 < s such t h a t Cj-1 and Cj+I are stable chambers and Cj_ 1 ~ x ~ Cj+l.

(If one or both of Cj-1 and Ci+I is missing, the a r g u m e n t is similar.) It follows t h a t x is adjacent to all the vertices of the clique chambers Ci for i > j + 1 and to no vertices of the clique chambers Ci for i < j - 1. There remains the question of the adjacencies between x and the vertices of Ci. Put - Cj n N ( x ) - Cj n N ( x ) . Both sets C~ and Cy are not empty, for otherwise x would be equivalent to the vertices of Cj-1 or Cj+I. It now follows that G' satisfies Conditions 1 and 2 with the new chambers C1,...,~-I,C],{x],~t,G+I,...,Cq,

as required. 9 We now give an explicit characterization of the degree sequences of BT graphs and matrogenic split graphs. Recall the definitions of the conjugate sequence d*, corrected Ferrers diagram C(d), corrected conjugate sequence d' and corrected Durfee n u m b e r m of a proper sequence d = ( a l l , . . . , dn) from Definitions 3.1.5 and 3.1.6. We use the notation /xk - d~ - dk.

11.4

Matrogenic Sequences

307

Theorem 3.2.2 characterized the threshold sequences d by the property Ak -0 for k - 1 , . . . , n. Recall that the boxes of a graph are the maximal sets of vertices with the same degree. In analogy we say that the boxes of a proper sequence dl >_ ... >_ dn are the maximal sets of indices i over which di is constant. The following theorem generalizes Theorem 3.2.2 in characterizing the BT sequences. Theorem only if

11.4.4 A proper sequence d with boxes B 1 , . . . , B ~

is B T if and

Ai - O for k - 1 , . . . , r. iEBk

P r o o f . Let C - C(d) be the corrected Ferrers diagram of d. The supplierspriority flow for the symmetric transportation network associated with d, as defined in (10.3), is given by Ckl -

and consequently, for k k

E E Cij , iEBk jEBl

1 , . . . ,r,

qSk, -

l=l

E di iEBk

~ l=l

(11.7)

gPlk -- E d:. iEBk

The expression ~i~__1Ckt is equal to the k-th component ak of the supply vector. According to Theorem 10.3.2, d is BT if and only if r satisfies the demands (which are equal to the supplies), that is, if and only if ~t~_-i Ctk - ak for k = 1 , . . . ,r. By (11.7), this is equivalent to ~ i e B k ( d i - d}) = 0 for all k, which is the condition of the theorem. 9 In order to characterize split matrogenic sequences, we need the next two lemmas.

L e m m a 11.4.5 Then

Let G

be a

graph with degree sequence

d -

1. for n > 4, G is a net if and only i f n is even and

A(d)where u - n / 2 ;

( u - - 1, (--1)~'-1, u - 1, (--1)~'-1),

(dl,...,dn).

308

Matroidal and Matrogenic Graphs

2. f o r n > 4, G is a net-complement if and only if n is even and A(d)-

(1 " - '

1-

u, 1 u-1 1 - u)

where u - n/2; 3. for n >_ 2, G is a matching if and only if A ( d ) - (n - 2, 0, ( - 1 ) ~ - 2 ) ;

~. for n > 2, G is the complement of a matching if and only if A ( d ) - (1 n - 2 , 0 , 2 - n);

5. G is a pentagon if and only if

zX(d) - (2, 2, O , - 2 , - 2 ) . P r o o f . To save work, n o t e t h a t C o n d i t i o n 1 is e q u i v a l e n t to C o n d i t i o n 2, a n d C o n d i t i o n 3 is e q u i v a l e n t to C o n d i t i o n 4. Indeed, it is easy to verify t h a t A for t h e c o m p l e m e n t a r y g r a p h G is o b t a i n e d f r o m A for G by n e g a t i n g its c o m p o n e n t s a n d reversing t h e i r order. T h e "only if" p a r t of t h e l e m m a is verified by s t r a i g h t f o r w a r d checking. We prove t h e "if" p a r t s of C o n d i t i o n 1, C o n d i t i o n 3 a n d C o n d i t i o n 5. C o n d i t i o n 1" 1. F r o m t h e f o r m of A, Am is u - 1 or - 1 . In t h e first case we have d m - d'~ - Am - ( m - 1) - (u - 1) _< m - 2, c o n t r a r y to t h e definition of rn. T h e r e f o r e A,~ - - 1 a n d c o n s e q u e n t l y dm - m . 2. If m - n, t h e n A is t h e zero sequence, c o n t r a r y to a s s u m p t i o n . T h e r e fore m < n a n d therefore - 1 - A,~ - d~_ 1 - d n , hence dn _> 1 a n d d~ _> 1. F r o m u - 1 - A1 - d~ - dl - 1 we o b t a i n dl - u a n d t h e r e f o r e :r

~

du+ 1 -

.

.. = dn_ 1 - 0.

3. B y 1 above dm - rn, a n d t h e r e f o r e d~n_ 1 >_ rn. H e n c e by 2, r n - 1 _< u. B u t m c a n n o t be u + 1, since by 1 we have Am - - 1 5r u - 1 - A , + I . T h e r e f o r e m < u.

11.4

Matrogenic Sequences

309

4. F r o m 3 we h a v e - 1 -

A.+I

-

d; - d,+l

-1

-

A,+2

-

d~,+l-du+2

-1

-

A~

-

d;_~-dn.

T h e n f r o m 2 we h a v e du+2 = ".. = dn - 1 a n d 1 _< d , + l - d; - u + 1. F r o m t h e l a t t e r e q u a t i o n we d e d u c e t h a t d; _> u a n d t h e r e f o r e d~ _> u. B u t f r o m 2 we h a v e u - dl >_ d,. T h e r e f o r e d. - u. 5. Since d~ - u, we have m >_ u by definition of m , b u t by 3 we h a v e m _< u. T h e r e f o r e m - u. H e n c e by definition of m d , + l < L,, f r o m w h i c h it follows t h a t d; <_ u. B u t by 4 d; >_ u, a n d so d; - u. T h e first e q u a t i o n of 4 now i m p l i e s t h a t d,+l - 1. 6. T h u s d has r e m 5.4.4, it u v e r t i c e s of a s t a b l e set. u, G is a net

b e e n c o m p l e t e l y d e t e r m i n e d as d - (u ", 1~). B y T h e o follows now t h a t G is a split g r a p h . Since m - u, t h e first G f o r m a m a x i m a l clique, a n d so t h e last u v e r t i c e s f o r m Since t h e last u degrees are 1 a n d t h e first u degrees are of o r d e r u.

Condition a: We h a v e n - 2 - / ~ 1 - d~ - d 1 ~ (?% - 1) - 1. It follows t h a t d~ - n - 1 a n d dl - 1, a n d t h e r e f o r e d - (1 ~) a n d G is a p e r f e c t m a t c h i n g . Condition 5" W e h a v e 2 AI - d ~ - dl. If d 1 - 3 a n d dl - 1, t h e n d - (14, 0), w h i c h gives t h e w r o n g A. T h e only o t h e r p o s s i b i l i t y is d~ - 4, dl - 2. F u r t h e r m o r e , 2 - /k2 d ~ - d2. If d~ - 3 a n d d2 - 1, t h e n we would h a v e t h e a b s u r d d3 > d2. T h e only o t h e r p o s s i b i l i t y is d~ - 4, d2 - 2. T h i s d e t e r m i n e s d as d - (2s), a n d t h e r e f o r e G is a p e n t a g o n . 9 1 1 . 4 . 6 Let G ( K , S ) be a split graph with a clique K , a stable set S, and a proper degree sequence da. Let H be a graph with a proper degree sequence dH. Then the A sequence of G (9 H is given by

Lemma

A - (AK, AH, As), wh e re

A.As

-

k-tKI (/Nk+l(dG),.

. ., /Xk+s(dG)),

-

I 1"

Matroidal and Matrogenic Graphs

310

P r o o f . Put d = dH, e = da, f = da| diagram of e is given by

Since G is split, the corrected Ferrers

-IkQ 0s) where Jk is the all-1 matrix and Ik is the identity matrix of order k, O~ is the all-O matrix of order s, P is k x s, and Q is s x k. On the other hand, the corrected Ferrers diagram of f is given by

C(f) =

(

Jk--Ik

Jkx~

P

)

C(d)

Q

0~•

,

0~

where J, I and 0 are all-l, identity and all-O matrices of the indicated orders, respectively, and n = Iv(g)l. From the structure of C ( f ) , it is apparent that !

!

9 for i - 1 , . . . , k, f[ - fi - ei + n - ei - n - e i - ei; 9 fori-k+l

,..., k+n,

f" . fi . n + d.~ _ k -.n

9 fori-k+n+l,...,k+n+s,f[-fi-ei_

di-k

' k - di-k ; di_

-ei-~.

If d is a proper sequence with boxes B I , . . . , Br and a A sequence A = A(d), we denote by ABk the subsequence of A corresponding to the box Bk. We call a p r i m i t i v e sequence of length p (not necessarily a prime) a sequence of length p of the following types: (p-1,-1,...,-1),

(1,...,1, I-p),

(0,...,0).

We are now ready for the characterization of matrogenic split sequences. T h e o r e m 11.4.7 A proper sequence d is the degree sequence of a m a t r o g e n i c split graph if and only if all its ABk are p r i m i t i v e sequences and 1. the n u m b e r q of non-zero p r i m i t i v e sequences is even; 2. f o r k = 1 , . . . , q/2, the k-th and ( q - k ) - t h are equal.

non-zero p r i m i t i v e sequences

11.4

Matrogenic Sequences

311

P r o o f . " I f " : Let G be a realization of d = (dl,. 9 dn) with boxes B 1 , . 9 9 , B r and A = Aa. The graph G is BT by Theorem 11.4.4, since the sum of the elements of every primitive sequence is 0. We use induction on r, the number of boxes of G. If r = 1, then A = 0 by Condition 1. Hence G is threshold and thus matrogenic split. For the induction step, we distinguish three cases: C a s e 1: G has dominating vertices (dl = n and S = 0. C a s e 2: G has isolated vertices (d~ = 0).

1). In this case, let K = B1

In this case, let K = O and

S=B~. C a s e 3: G has no dominating and no isolated vertices. K = B1 and S = B~.

In this case, let

In Case 3, B~ must be stable, otherwise by the BT property the vertices of B1 would be dominating in G. Similarly, each vertex of B~ has some neighbors in B1, otherwise the vertices of B~ would be isolated in G. Therefore B1 is a clique. Moreover, each vertex of B2 U ... U B~-I is adjacent to each vertex of B1 and to no vertex of B~. In all cases, the subgraph induced by K U S is a split graph G'(K, S), and if H denotes the subgraph induced by v(a)-v(a'), then G = G'(I'(,S)QH. Hence A = (AK, AH,/kS)

by Lemma 11.4.6. In Case 1, AK = 0 since the vertices of K are dominating, and As is empty, hence AH satisfies the conditions of the theorem. Similarly in Case 2, AK is empty and As = 0, and AH satisfies the conditions of the theorem. In Case 3, AK and As are not equal to the zero sequence as B1 and B~ are neither completely connected nor completely disconnected. Hence by Condition 2, Air = As, and again AH satisfies the conditions of the theorem. Since in Case 3 AK and As are equal primitive sequences, G' is a net or a net-complement (or a graph on two vertices) by Lemma 11.4.5. In all cases, H is a split matrogenic graph by the induction hypothesis, and hence G is split matrogenic by Theorem 11.4.3. " O n l y if"" Let G be matrogenic split with boxes B , , . . . , B~. We use induction on r. If r = 1, G must be complete or edgeless, hence A is the zero sequence. For the induction step, we have by Theorem 11.4.3 that G = G' @ H, where H is again a matrogenic split graph and G'(K, S) is a

312

Matroidal and Matrogenic Graphs

split graph which is (a) a complete graph or ( b ) a n edgeless graph or (c) a net or (d) a net-complement. By Lemma 11.4.6, A = (A/,-, AH, As), where in Case (a) As is empty and A/s- = 0, and in Case (b) A/,- is empty and As = 0. By Lemma 11.4.5, both A/s- and As are primitive sequences and if both are non-zero, they are equal. Moreover, in Case (a) in Case (b) B1 U B~, in Case (c)or (d).

B1, v(a')

-

B,,

The result then follows by induction.

Chapter 12 Domishold Graphs 12.1

Introduction

Recall from Definition 9.1.3 that a dominating set of a graph G = (V, E) is a subset S of V such that every vertex not in S has a neighbor in S, and that the domination number 7(G) of G is the size of a smallest dominating set of G. We have seen in that section that the domination number is bounded above by the Dilworth number. Also recall that the threshold graphs are defined as the graphs for which the characteristic vectors of the stable sets are separated from the characteristic vectors of the nonstable sets by a hyperplane. In a similar spirit Benzaken and Hammer [BH78] introduced the domishold graphs as the graphs for which the characteristic vectors of the dominating sets are separated from the characteristic vectors of the nondominating sets by a hyperplane. They characterized the domishold graphs and showed that the threshold graphs form a proper subclass of domishold graphs. They also introduced the concept of domistable graphs. Later, Payan [PayS0] introduced a class of equidominating graphs, and Cherynak and Chernyak [ccg0], introduced a class of graphs called pseudodomishold graphs. We study these results in this chapter. In addition, some enumeration problems and other characterizations of domishold graphs are discussed in Chapter 13.

12.2

Notation and Main Results

We begin with the definitions of domishold graphs and domistable graphs. 313

314

Domishold Graphs

1. A graph G - (V, E) is called a d o m i s h o l d g r a p h if there exist a positive real-valued function w on V and a real 0 such that for all S C V,

D e f i n i t i o n 12.2.1

S is dominating if and only if

>_ o. xES

2. A d o m i s t a b l e g r a p h is a graph all of whose minimal dominating sets

are stable sets. 3. A graph G - (V, E) is called e q u i d o m i n a t i n g if there exist a positive integer-valued function w on V and a positive integer 0 such that for all Q c_ V, Q is a minimal dominating set if

only if

w(x) -- 0 xEQ

Clearly, in the definition of domishold graphs one can assume without loss of generality that w and 0 are positive integers. We refer to w(x) as the weight of x and to 0 as the threshold value. The domishold graphs and the domistable graphs were defined in [BH78] and the equidominating graphs were defined in [PayS0]. The results and concepts in this section are from [BH78]. As usual, N(x) denotes the neighborhood of vertex x and we put M(x) = V - ( N ( x ) U {x}). a stable dominating pair is a set of two nonadjacent vertices each of which is adjacent to every other vertex. Let Sk denote an edgeless graph on k vertices, Kk denote a complete graph on k vertices, and J2k denote the complement of a perfect matching on 2k vertices (kK2). The graph J2k is called a cocktail-party graph. We write Q d o m G to indicates that Q is a dominating set of G. Let us now define a binary relation ~d on the vertex set V of G as follows" D e f i n i t i o n 12.2.2 For x, y E V, x Ld Y if and only if for all Q c V - { x , y}, (Q u {y}) dom G ~ (Q u {x}) dom G. L e m m a 12.2.3 The relation ~d is reflexive and transitive, i.e., it is a preorder.

P r o o f . The reflexivity is obvious. To prove the transitivity, we assume that i, j, k are distinct, i ~e j and j ~d k, and show that i Ld k. Let Q c_ V - { i , k}

12.2

315

Notations and Main Results

be such that ( Q U { k } ) d o m G . I f j ~ Q, t h e n Q U { j } ) d o m G a n d h e n c e (Q u {i}) dom G. If j C Q, put Q' = Q - j. Then Q u {k} = ( ( Q ' u {k}) u {j }) dom G, and hence ((Q'u { k}) U {i}) dom G, and the latter set contains k but not j. So ((Q'U {j}) u {i}) = Q u {i} dom G. Hence i ~d k. 9 In view of Lemma 12.2.3, the relation ~d is called the dominal preorder of G. We now state the main result of this section. T h e o r e m 12.2.4 For every graph G equivalent:

(V, E), the following conditions are

1. G is domishold; 2. the dominal preorder ~d is linear; 3. G can be built from the empty graph by repeated application of the operation G' ~ G", where G" = (G' U S;) | Kq | J2~ and p + q + r > O; in other words, G can be obtained from an empty graph by repeatedly adding an isolated vertex, a dominating vertex, or a stable dominating pair; ~,. V can be partitioned into sets 1/1, 1/2 and 1/3 inducing the graphs SIVll, KIV21, Jig31 respectively, with the following properties: (a) each vertex in 1/2 is adjacent to each vertex in 173; (b) for each i E Vx, g ( i ) N V3 induces a cocktail-party subgraph of G; (c) the neighborhoods of the vertices of V1 are nested; 5. G can be obtained from some threshold graph with split partition (K, S) by substituting cocktail-party graphs for some vertices of K; 6. the graphs of Figure 12.1 are not induced subgraphs of G. Proof. 1) =v 2): It is enough to show that for every two vertices x, y with w(x) > w(y), we have x ~d Y. This is indeed the c~se, since for every

QcV-{x,y}, Q u {y} dom G ==~w(Q) -4- w(y) >_ 0

w(Q) + w(x) > 0 :=~ Q u {x} dom G, where w(Q) - EzeQ w(z). 2) =V 3)" First we show that if x is any maximal vertex in the linear order

Domishold Graphs

316

Figure 12.1" The forbidden induced subgraphs of domishold graphs.

H1

Ha

H2

H4

Hs

~d, Q is any maximal stable set, and x ~ Q, then x is adjacent to every vertex in Q. Since Q is a maximal stable set, Q dom G. Now for every y C Q, Q = ((Q - y ) u {y}) dom G and hence ((Q - y ) u {x}) dom G, since x ~d Y. But the only neighbor of y in ( Q - y ) u {x} is x. Thus x is adjacent to y, as required. We now show that G contains an isolated vertex, a dominating vertex, or a stable dominating pair. The result then follows by induction since removing isolated vertices or a dominating set preserves Condition 2, as is easy to verify. Assume that G contains no isolated vertex or dominating vertex. Let x be a maximal vertex in ~d and y C M(x) (y exists since x is not dominating). We show that X(y) = N(x). Suppose that, if possible, zx C E and zy q~ E. Extend {y, z} to a maximal stable set Q, which will not contain x. Then, as shown above, x is adjacent to every vertex in Q, contradicting xy ~ E and y C Q. This shows that N(x) C_ N(y). It follows that y is also a maximal vertex in Le and hence, by symmetry, N(y) C_ N(x). It follows that every vertex in M(x) has the same set of neighbors as x. Thus if z C N(x) (z exists since x is not isolated), then {z, y} dom G and hence {x, y} dom G as x is a maximal vertex. Thus G has a stable dominating pair {x, y}, as required. 3) =~ 4): Let Go,..., Gt be a sequence of graphs with Go = 0, Gt = G and

G~+I - (G~ u Sp,) | Kq, @ &~,,

i-O,...,t-

i.

12.2

Notations

and

Main

317

Results

Putting i

i

i

and using an easy induction, we see that this partition has the desired property. 4) =~ 5)" Identify each vertex of V3 with its unique nonneighbor in V3 to obtain a clique V~. The resulting graph is clearly a threshold graph with split partition (1/2 U V3', 1/1), 5) =~ 6)" The graphs of Figure 12.1 are not threshold graphs, nor do they contain a stable dominating pair. Hence they cannot be obtained from a threshold graph by substituting cocktail-party graphs for some vertices in the clique. The result now follows since if G satisfies Condition 5, so does every induced subgraph of G. 6) => 3)" It is enough to show that G contains an isolated vertex, a dominating vertex, or a stable dominating pair. If G is not connected, then G contains isolated vertex since H1 is forbidden. So assume that G is connected. Then by a well-known result about P4-free graphs (Lemma 14.3.6), G is not connected since H2 is forbidden. If a connected component of G has one vertex, then it is a dominating vertex in G, and if it has two vertices, then they form a stable dominating pair in G. Otherwise, G contains two components, each with at least three vertices. Let xl, x2, x3 be vertices of a component such that x2 is adjacent to the other two. Similarly, let yl, y2, y3 be vertices of another component with y2 adjacent to the other two. Then, depending on the adjacency between xl and x3 and between yl and y3, G contains Ha,/-/4 or Hs as an induced subgraph, a contradiction. 3) =~ 1)" The empty graph is obviously domishold. Assuming that G' = (G u Sp) | Kq @ J2r and that G is domishold, we show that G' is domishold too. Let wl represent the weight of each 1 E V(G), and 0 the threshold for G. Let w* - mint wt/2. We may assume that 2w* < 0, since otherwise G is a clique, and we can change all weights wt for 1 C V(G) as well as 0 into 1 to achieve2w* < 0. Let us also put W - 1 + ~ t e y ( a ) wl and let us define 0 ~ 0 + p W and

i , wi -

W 0 + pW O+pW-w*

v(G),

iESp, i E Kq, i G J2~.

318

Domishold Graphs

The weights w~ and the threshold O' characterize the dominating sets of G'. Indeed, every minimal dominating set D' of G' is of one of the following three forms: 1. D ' = {k}, k e Kq; 2. D' = { x , y } , x C J2r, x r y, y C J2,. U Sp t2 V(G);

3. D' = D tO Sp, where D is a minimal dominating set of G. It is straightforward to verify that these are also precisely the minimal sets Q' of G' satisfying w'(Q') > 0'. tt Since each of Ha,//4, Hs of Figure 12.1 contains an induced 6'4, it follows that all threshold graphs are domishold graphs. The next theorem gives other characterizations of threshold graphs among the domishold graphs. T h e o r e m 12.2.5 For a domishold graph G, the following conditions are equivalent: 1. G is a threshold graph; 2. G is a C4-free graph; 3. G is a split graph; .~. G is an interval graph; 5. G is a domistable graph.

P r o o f . 1) =~ 2,3,4,5): This follows from Theorem 1.2.4 characterizing threshold graphs, Theorem 2.1.6 characterizing interval graphs, and the fact that a minimal dominating set of a threshold graph considered as a split graph G(K, S) is S itself or consists of one vertex from K and all its nonneighbors in S. 3) =~ 2): This follows from Theorem 5.2.1. 4) =~ 2): This follows from Theorem 2.1.6. 5) =~ 2): Since G is a domishold and domistable graph, it must contain a dominating vertex or an isolated vertex (otherwise, as in the proof of 3) =, 4) in Theorem 12.2.4, if x is any maximal vertex with respect to the dominal preorder, y C M ( x ) and z C N(x), then {y, z} is a minimal dominating set that is not stable). Delete any dominating or isolated vertex from G, and the

12.2

Notations and Main Results

319

resulting graph is clearly again a domistable and domishold graph. Hence this process can be repeated until all the vertices are removed. Since no vertex of an induced C4 can become isolated or dominating, it follows that G is C4-free. 2) => 1): This follows from Condition 6 of Theorem 12.2.4 and Condition 2 of Theorem 1.2.4. 9 A linear inequality n

E

WiXi ~__ 0

i=1

with Wl >_ " " >_ Wn > 0 is called domigraphic if W l , . . . , w ~ are weights and 0 is a threshold of a domishold graph. The condition n

y~'~wi >__O i=1

is obviously necessary for the inequality to be domigraphic since V(G) is a dominating set. In the case n - 1, it is sufficient too. For n - 2, this condition and w2 < 0 => wl < 0 are again sufficient. The following theorem gives a recursive characterization of domigraphic sequences for n >_ 3. Theorem

12.2.6 A linear inequality n

E wixi >_ 0 i=1

with W 1 ~ " ' " ~___W n > 0 and n >_ 3 is domigraphic if and only if one of the following conditions holds: 1.

W 1 ~

0 and ~i~2 wixi >_ 0 is domigraphic;

2. w~ < O, E ~ 2 wi < 0 and ~i~2 wixi >_ 0 - Wl i8 domigraphic; 3. Wl < O, W 1 -~- W2 ~__ 0 and ~i~3 wixi >_ 0 is domigraphic. P r o o f . Let the inequality ~-~iL1W i X i ~-- 0 be domigraphic and let G be the associated graph. Let x be a maximal vertex of the dominal preorder. Without loss of generality, we may assume that w(x) - w I (&s in the proof of 1) => 2) of Theorem 12.2.4). If x is dominating, then Condition 1 is satisfied. If x is isolated, then Condition 2 is satisfied. If G has no isolated or dominating vertex, then there exists another maximal vertex y such that {x, y} is a stable

Domishold Graphs

320

dominating pair (as in the proof of 2) ~ 3) of Theorem 12.2.4). Without loss of generality, we may assume that w(y) = w2. Then Condition 3 is satisfied. Conversely, suppose one of the Condition 1 - Condition 3 is satisfied. Let G' be the domishold graph associated with the domigraphic inequality in each case. Then, clearly, the linear inequality of the theorem corresponds to a domishold graph obtained from G' be adding an isolated vertex (Condition 2), or a dominating vertex (Condition 1), or a dominating stable pair (Condition 3). "

12.3

Equidominating Graphs

Payan [PayS0] defined equidominating graphs as follows: D e f i n i t i o n 12.3.1 A graph G = (V, E) is said to be e q u i d o m i n a t i n g when there exist a weight function w: V ~ Z + U {0} and a number t E Z + U {0} such that for every S C_ V, S is a minimal dominating set ~

~

w(x)-

t.

xES

The class of equidominating graphs neither includes nor is included in the class of domishold graphs. As evidence, observe that 2K2 is an equidominating graph but not a domishold graphs. To see that it is equidominating assign a weight of 1 to each vertex of one edge and a weight of 2 to each vertex of the other edge, and let t = 3. From Theorem 12.2.4, 2K2 is forbidden for domishold graphs. On the other hand, K2,z is a domishold graph (Theorem 12.2.4), but not an equidominating 'graph. To see this, let (A, B) be the bipartition of K2,3 into stable sets with ]A I = 2. Then A and any pair {x, y} with x C A and y C B are minimal dominating sets, and hence every vertex should receive the weight t/2. But then B is a minimal dominating set with total weight exceeding t. The following theorem characterizes the equidominating graphs that are also domishold graphs. T h e o r e m 12.3.2 ([Pay80]) The following conditions are equivalent for every graph G: 1. G is both a domishold graph and a equidominating graph;

12.3

Equidominating Graphs

321

2. G has weights and threshold that characterize simultaneously the dominating sets by inequality and the minimal dominating sets by equality; 3. G can be built from the empty graph or from a cocktail-party graph (the complement of a perfect matching) by repeatedly adding an isolated vertex or a dominating vertex; ~. G has no induced graph isomorphic to H i , . . . , H4 of Figure 12.2.

Figure 12.2: The forbidden induced subgraphs for domishold and equidominating graphs.

H1

H2

H3

H4

P r o o f . 2) =~ 1)" This is obvious. 1) =~ 3)" We proceed by induction on IV(G)]. Let G - (V, E) be a domishold graph with weight function w~ and threshold t~, as well as an equidominating graph with weight function w~ and threshold t~. By Theorem 12.2.4, G has an isolated vertex, a dominating vertex, or a stable dominating pair. C a s e 1" G has an isolated vertex x 9 For i - 1, 2, let WG_ ~ x be the restriction of w~ to G - x, and let t ia_~ - t ~ - wb(x). It is easy to check that G - x is domishold and equidominating with these modified weights and thresholds. Hence by induction G - x satisfies Condition 3, and so G does too. C a s e 2: G has a dominating vertex x. The argument is similar to Case 1 with t~_~ - t~. C a s e 3" G does not have an isolated or a dominating vertex, but has a stable dominating pair x, y. Observe that a set S C V(G) is a minimal dominating set of G if and only if S is a minimal dominating set of G - {x,y}, or S - {x,y}, or for some v e V ( G ) - {x,y}, S - {x,v} or 5' - {y,v}. Since we m a y assume that IV(G)I >_ 3, it follows that w~ - t2c/2. Hence S is a minimal dominating set of G if and only if ISI - 2. We assert that G is a cocktail-party graph. Since every two vertices of G form a dominating set, G does not contain any induced K1 tO K2 or an i n d u c e d / 3 . Hence G does not contain any isolated vertices or induced K1,2 or induced K3. Therefore every 9

322

D o m i s h o l d Graphs !

q

connected component of G has exactly two vertices, and hence G forms a perfect matching, and G satisfies Condition 3. 3) =~ 2)" The empty graph trivially satisfies Condition 2 with null w a and ta - O. The cocktail-party graph satisfies Condition 2 with wa - 1 and to - 2. Assume now that x is an isolated or dominating vertex of G such that G - x satisfies Condition 2 with weights wa-~ and threshold t a - ~ . Then we obtain wa and ta as follows: 9 If x is an isolated vertex, then S is a minimal dominating set of G if and only if x 6 S and S - {x} is minimal dominating set of G - x. We then set for all v 6 V -

wa(v)

--

wa_~(v)

wa(x)

=

y~ wa_~(v) + 1 - ta_~ v~V-{~}

ta

=

E w a - x ( v ) + 1. v~V-{~}

{x}

9 If x is a dominating vertex, then S is a minimal dominating set of G if and only if S - {x} or S is a minimal dominating set of G - x. Then we set for a l l v C V - { x }

-

a(x)

-

ta_

ta

--

tG-x.

In each case it can be verified that G is both a domishold graph and a equidominating graph with weights wa and threshold ta, and hence condition 2 is satisfied. 3) =~ 4)" This follows from the fact that the graphs H 1 , . . . , / / 4 neither are cocktail-party graphs nor do they contain isolated or dominating vertices. 4) =~ 3)" Assume that G contains no induced subgraph isomorphic to the graphs of Figure 12.2, and assume further that G contains no isolated or dominating vertices. We complete the proof by showing that G must be a cocktail-party graph. G is connected, for otherwise G contains HI. Hence by a well-known result about P4-free graphs (Lemma 14.3.6), G is not connected since G is H2-free. Each connected component of G has at least two vertices since G has no dominating vertices. If each component of G has exactly two

12.4

P s e u d o d o m i s h o l d Graphs

323

m

vertices, then G is a cocktail-party graph, as required. Otherwise, G has a component with at least two vertices Xl,X2 and another with at least three vertices yl, y2, y3 such that Y2 is adjacent to both yl and y3. But then, depending on whether or not 91 and y3 are adjacent, these five vertices induce Ha or//4 in G, a contradiction. .. The following corollary follows from Condition 3 or 4 of Theorem 12.3.2.

Corollary 12.3.3 The threshold graphs are equidominating (and domishold).

12.4

P s e u d o d o m i s h o l d Graphs

Chernyak and Chernyak [CC90] call a graph G - (V, E) a pseudodomishold graph if there is a function w" V , R + U {0}, w ~ 0, and a nonnegative real t such that for every S c_ V, SdomG

?, ~ w( x ) >_t xES

--(SdomG)

:, ~ w ( x ) <_t. xES

If one of the inequalities above were replaced with a strict inequality, we would have the definition of a domishold graph. Unfortunately, an induced subgraph of a pseudodomishold graph need not be a pseudodomishold graph. This makes the problem of characterizing pseudodomishold graphs a difficult task. Instead, the authors of [CC90] investigated the hereditary pseudodomishold (HP) graphs, namely the pseudodomishold graphs all of whose induced subgraphs are pseudomishold as well. These authors characterized the HP graphs in terms of the composition of the simplest components and by forbidden induced subgraphs. They used this characterization to estimate the number of n-vertex HP graphs, and compare these estimates with estimates of the number of n-vertex domishold graphs. We report their results without proofs in this section. It can be verified from the characterization below that the class of HP graphs strictly includes the class of domishold graphs. We need the following notations. Let M be the class of all graphs H such that either each component of the complement H is a path or a cycle, or H is a Ps. A graph G together with a partition (A,B) of V(G) is denoted by

324

Domishold Graphs

G[A, B]. If G is an edgeless graph on a set V of n vertices, then Q~ denotes G[V,O]. If G is a graph on a set V of a single vertex, then 01 denotes G[O, V]. If G = P4, then we consider P4 as G[A, B], where B consists of the two vertices of degree 1. The composition @ of graphs is defined as follows: Given G[A,B] and H with V(G) N V(H) = 0, G[A, B] Q H - G U H U KA,V(H), where KA,V(H) is the complete bipartite graph with color classes A and V(H). Theorem

12.4.1 For every graph G, the following conditions are equivalent:

1. G is an l i p graph; 2. G contains neither the configurations shown in Part (a) nor the induced subgraphs shown in Part (b) of Figure 12.3; 3. G can be represented in the form G

-

X1 @ ...

@ Xrn

where H E M and Xi C {Q1, Q2, Ol,

@ H with m >_ O,

P4}.

Let tn denote the number of non-isomorphic n-vertex graphs in M without isolated vertices, dominating vertices, or stable dominating pairs (tn -- 0 for n < 4), and put tnX n n>l

For n > 6, it is possible to evaluate tn from the definition of M by partitioning the vertex set of the complementary graph into cycles and paths of length 3 or more. Let Un denote the number of non-isomorphic n-vertex HP graphs, and put u(x) - ~ unx n. n>l

The following result expresses u ( x ) i n terms of t(x). C o r o l l a r y 12.4.2 t ( ~ ) + x - ~3 u(x)

-

1-

2x-

x 2 + x 3-

x 4"

L2.4

325

Pseudodomishold Graphs

Figure 12.3" Forbidden configurations and induced subgraphs for HP graphs.

(a) Forbidden configurations. Dashed lines indicate nonedges and solid lines indicate edges. Other edges may or may not exist.

w

v

"i__ '. i I I I w

w

w

v

w

(b) Forbidden induced subgraphs

w

w

Domishold Graphs

326 An estimate of un is given by the following result.

Corollary 12.4.3

For

n > 6, (2.32) n-1 < un < (2.37) n-1.

We can compare the above estimates with an estimate of the number d~ of non-isomorphic n-vertex domishold graphs. Put

d ( x ) - E dnxn" n>l Then we have the following result.

Corollary 12.4.4 XmX 3 d(x) Further,

dn ~" c r n, w h e r e r ..~

-

1 - 2 x - x 2 + x 3"

2.25

is t h e m a x i m u m

root (in absolute

value)

of the equation x 3 - 2x 2 - x + 1 -

O.

We see that un is larger than dn, consistent with the fact that all domishold graphs are HP graphs, but not conversely.

Chapter 13 The Decomposition Method 13.1

Introduction

In Definition 11.4.2 we saw how to compose a split graph with an arbitrary graph, and in this chapter we investigate the opposite operation of decomposition. This subject was studied by Tyshkevich [Tys84, TysS0], Chernyak and Chernyak [CC90, COS1]and Tyshkevich and Chernyak [TC85b, TC85a]. It turns out that every graph has a unique decomposition into indecomposable components. Tyshkevich and Chernyak [TC85a] also introduced a generalization from the class of split graphs to the larger class of polar graphs that we saw in Section 7.6, and obtained a hierarchy of decompositions and a unique decomposition at each level of the hierarchy. The decomposition method proves to be a unifying approach to the study of many graph problems, since many properties of graphs hold for a graph if and only if they hold for each indecomposable component, perhaps with a small change which is easy to pinpoint. It simplifies the description of the structure of several graph families, leads to their enumeration, and serves as a tool to explore them. Section 13.2 presents the decomposition, the hierarchy, and the unique decomposition result. Section 13.3 applies this theory to threshold and domishold graphs, Section 13.4 applies it to box-threshold graphs and enumerates them, and Section 13.5 does the same for matroidal and matrogenic graphs. 327

The Decomposition Method

328

13.2

The Canonical D e c o m p o s i t i o n

In Section 7.6 we defined the class of polar graphs and its subclasses (c~,/3), and discussed the complexity of their recognition problems. In this section we present a theory of decomposition of polar graphs based on Tyshkevich and Chernyak [TC85b]. To review the definition, a graph G = (V, E) is said to be polar if V can be partitioned into subsets A and B, possibly empty, such that each of the graphs G[A] and G[B] is a union of cliques. This partition is called a polar partition. If in addition every connected component of G[A] (resp. G[B])is (a clique)of size at most c~ (resp./3), we say that G belongs to the class (c~,/3) of polar graphs. The polar partition may not be unique: for example, if G is the path P3 with consecutive vertices a, b, c, then we may take A = {b}, B = {a,c} or A = {a,b}, B = {c}, putting G in (1, 1)in both cases; if G is the cycle C4, we may take A to consist of three vertices, putting G in (2, 1), or we may take A to consist of two adjacent vertices, putting G in (1, 2), or we may take A to consist of two non-adjacent vertices, putting G in (2, 1) again. Other examples are G = C6 with A consisting of three mutually adjacent vertices and G E (1, 3), and G bipartite, with G C ( ~ , 1). The class (1, 1) is precisely the class of split graphs. We indicate a polar graph with its polar partition as G(A, B), and use the (c~,/3) notation for it as well. We say that polar graphs G(A,B) and G'(A', B') are polar isomorphic if there is an isomorphism of G to G' mapping A onto A' (and hence B onto B'). It can be checked that the graph in Figure 13.1 is not polar. Figure 13.1" A non-polar graph.

w

v

w

Clearly if G(A,B) r (o~,/3), then G(B, A) C (~, c~). Another easy result is the following one, establishing a hierarchy among the polar graphs. P r o p o s i t i o n 13.2.1 If a < (~' and/3 < /3', then (a,/3) C_ (a',/3'), and the

inclusion is proper if at least one of the inequalities is strict.

13.2

The Canonical Decomposition

329

P r o o f . The inclusion is trivial. We show that the graph

C = (/3' + I)K~, U (c~' + I )K~,, which clearly belongs to (c~',/3'), cannot belong to (c~',/~'- 1) if /3' > 1. Indeed, if G(A, B) E (c~',/~'- 1), then A must contain at least one vertex of each of the c~' + 1 copies of KZ,. Consequently, G[A] contains a stable set of size c~' + 1, which is not the case by G(A, B) E (c~',/3'- 1). Similarly if c~' > 1, G cannot belong to (c~'- 1, fl). 9 We now define the operation of graph composition, which specializes to the operation of Definition 11.4.2 for the class (1, 1). D e f i n i t i o n 13.2.2 The classes of polar and ordinary graphs are denoted by

7) and G, respectively. We define the c o m p o s i t i o n operator

O:Pxg~g as follows. If G(A,B) C 79 and H C ~ are vertex-disjoint, then G(A, B ) 0 H is obtained from G U H by adding every edge between A and H. The graphs G and H are called the c o m p o n e n t s of the composition. Note that if in the above definition H is polar and H(C, D) E T', then the composition Q = G O H is also polar Q(A U C, B U D). Thus O is a binary operation on the class of polar graphs. In general, composition is associative when all components except possibly the rightmost are polar. Therefore the algebra (T ~, O ) i s a semigroup, and each class (c~,/3) is a subsemigroup. D e f i n i t i o n 13.2.3 A graph G that can (cannot) be written in the form F(A, B ) 0 H is called d e c o m p o s a b l e ( i n d e c o m p o s a b l e ) . If in addition

we require that F(A, B) E (c~,/3), G is said to be decomposable 5ndecomposable) at t h e level (c~,/3). If in addition G is polar, G(C,D) E (c~,/3) and H E (c~,/3), then G(C,D) issaid to be decomposable (indecomposable) in t h e class (c~,/3). It can be verified that Cs is an indecomposable polar graph and the graph of Figure 13.1 is an indecomposable non-polar graph. To state the decomposition result below, we define an equivalence relation on the set of polar graphs as follows: 1. each polar graph G(A,B) is equivalent to itself (but when A and B are non-empty, not to any other polar graph even if they are polar isomorphic);

330

The Decomposition Method

2. every two graphs of the form G(O, B) are equivalent, and every two graphs of the form G(A, rg) are equivalent. It is easy to see that the composition Q is commutative on every class of equivalent graphs. T h e o r e m 13.2.4 1. every graph from the class (a, fl) can be represented as a composition of components (necessarily from (a, fl)) that are indecomposable in the class (a, fl). The decomposition is unique up to permutations of consecutive equivalent components.

2. every graph not from the class (a, fl) and decomposable at the level (a, fl) decomposes uniquely as H(C, D) @ F, where H(C, D) C (a, fl), F ~ (a, fl) and F is indecomposable at the level (a, fl). Before proving the theorem, we use it to make the following definition. D e f i n i t i o n 13.2.5 Let G(A,B) be decomposable in the class (a, fl). Consider its decomposition into indecomposables in this class. If in this composition we replace each maximal set of consecutive equivalent components by their composition, we obtain a representation of G in the form

G(A, B)

-

Xl

(~ . . .

(~ X n ,

X k --

Gk(Ak, Bk) e (a, fl),

(13.1)

where Xk and X k + l a r e not equivalent and each Xk satisfies one of the following three conditions: 1. Ak - O C Bk; 2. Ak r O - Bk; 3. Ak, Bk r 0 and Xk is indecomposable in the class (a, fl). The decomposition (13.1) is referred to as t h e c a n o n i c a l d e c o m p o s i t i o n of G ( A , B ) in t h e class (a, fl). Likewise, if F does not belong to the class (a, fl), its c a n o n i c a l d e c o m p o s i t i o n at t h e level (c~,/3) has the form F

= X 1 (~ . . .

(~ X n

(~

H,

where the Xk are as above, H ~ (a, fl) and H is indecomposable at the level

13.2

The Canonical Decomposition

331

13.2.4. The existence of the decompositions in question follows from the finiteness of the graphs, and what we need to prove is their uniqueness. P A R T 1" Let G(A, B) C (o~,/~). For each a C A, we let d(a) denote IN(a)N BI, and for each X C_ A, d(X) - ~-~aEXd(a). Similarly for each b E B, d(b) denotes IN(b)N AI, and for each Y C_ B, d(Y) - ~be~" d(b). Since the induced subgraph G[B] is a union of cliques of cardinalities at most /3, there exists a unique partition B - B1 tO -.. tO BZ, where G[Bj] ji(j and the nj are non-negative integers. Thus Bj uniquely partitions into cliques of cardinality j" Bj - Bj~ U ... U Byn,. Similarly A - A 1 U . . . U Ac~, where G[Ai] - m i K i and the m i are nonnegative integers, and Ai uniquely partitions into stable sets of cardinality i" Ai -- A l l I..J . . . I..J Aim,. Observe that since the Air are stable sets, for each canonical decomposition (13.1), each All is contained in one of the Bk, and similarly each Bjt is contained in one of the Bk. We represent G(A, B) in the form Proof of Theorem

/..

G(A,B) - X1 6) H(C,D),

Xa - GI(A1,B~),

where X1 is the first term of a canonical decomposition (13.1). We want to show that X1 is uniquely determined by G(A, B), from which it follows that H(C,D) is also determined, and the result follows by induction. We distinguish three exclusive cases. Case 1" A - / ~ and some Bjk satisfies d(Bjk) - O. Denote the union of all such Bjk by N. We must have A1 - 0, for otherwise (0, Bjk) is a further component of X1. Similarly B1 - N, and so G1 - G[N] and X1 is uniquely determined by G(A, B). Case 2" B r e and some Bik satisfies d(Aik) - i . IBI. Denote the union of all such Aik by N. In analogy to Case 1 we must have A1 - N, B 1 - O , G1 - G I N ] a n d X1 is uniquely determined. C a s e 3: The situations under the previous cases do not occur. We may also assume that A, B r O, for otherwise the theorem is trivial, and therefore we have that A1, B1 -r O and X1 is indecomposable. We note that /..

t..

/..

^ ^ d(Bjk) { < j" JAil, if Bjk C Be, j I./~ll, if Bjk ~ B1.

(13.2)

If, for example, the first line of (13.2) is false, then (;g, Bjk) is a right component of X1. The second line follows similarly because in that case Bjk C_ D. Equation (13.2) shows that B1 is uniquely determined by A1.

332

The Decomposition M e t h o d

Assume that G(A, B) can also be represented in the form ~..

G(A, B) - X~ fi) H'(C', D'),

A!

X~ - Gi (A'I, J~l),

where X~ is the first A term ofh another canonical decomposition. We assert that one of the setsA ! B1 and B~ contains the other one. For assume that, if A h possible, x C B a B 1 and y C B ~ B1. Then the assumptions on x imply that h A! ~. x C B~ND' and therefore Aa C_ ANN(x)C_ Aa. Similarly the assumptions on y imply that A1 C_ A 1. Thus A1 A-! A 1. But as noted above, A1 determines B1 h by (13.2), and therefore B1 -- BJ' contrary to assumption. This proves the assertion. Next we assert that .B 1 -- B 1. Assume that, if possible, B 1 C B1. A! A A Then by theA argument ofA ! the t .previous assertion, A 1 C_ A1, Aand since A1 ! determines B1, we ..have A 1 C A1. Moreover, every., vertex of A 1 of adjacent to every vertex of/31 - B[, and every vertex of B~ of non-adjacent to every v e r t e x of A 1 - A 1. This shows that (AI,Ba)is a component of (A1,_B1) , contradictinfi^ the indecomposability of X1. This proves our second assertion that B1 - B~, and therefore D - D'. We also note that

d(Ail)

IDI, if A~, c_ A1, _< i IDI, if Aa ~ m l ,

> i.

(13.3)

for otherwise X1 decomposes. Equation (13.3) shows that D uniquely deterA ~ A! ! mines A1, and thus A1 - A 1. Therefore X1 - X 1. P A R T 2" Assume that G ~ (c~,/3) and G has two decompositions

G - H(C,D) Q F - H'(C',D') (!) F', with H ( C , D ) , H ' ( C ' , D ' ) C (c~,~)and F , F ' ~ (c~,~). It follows that the set R - V ( F ) N V(F') is not empty, for otherwise F ' is an induced subgraph of H and so F ' E (c~,/3), or symmetrically. Assume that , if possible, R' = V ( F ' ) - R 7/=r We represent R' in the form R ' - C1 U D1 with C1 C_ C and D1 C_ D. But then we have the decomposition F ' - F(R')(CI,D1)fi)F(R), which contradicts the indecomposability of F ' at the level (c~,/3). This shows that R ' - e , i.e., V(F') C_ V(F). By symmetry we also have V(F) C_ V(F'), and so V(F') - V(F). Since F is an induced subgraph of G, V(F) uniquely determines F, and also C and D as the sets of vertices of G adjacent to all and to none of the vertices of V(F), respectively. The sets C and D uniquely determine H. 9

13.2

T h e Canonical Decomposition

333

C o r o l l a r y 13.2.6 The canonical decomposition of the graph G(A, B) in the class (a,/~) coincides with its canonical decomposition in every class (c~',/3') containing (a, fl). In other words, a polar graph has a unique canonical decomposition. Proof. This follows from Part 1 of Theorem 13.2.4. C o r o l l a r y 13.2.7 Every non-polar graph G has a unique representation of the form G = H(C,D) 0 F, where H(C,D) is polar and F is an indecomposable non-polar graph. Proof. This follows from Part 2 of Theorem 13.2.4. P r o p o s i t i o n 13.2.8 If (a',fl') D (c~,fl), then there exist graphs indecomposable at the level (a,/~) but decomposable at the level (a', fl'). Proof. We may assume without loss of generality that a' > a. Let F be a non-polar indecomposable graph, such as the graph of Figure 13.1. Define a graph G by its canonical decomposition at the level (a',/Y) as G = S~,(A,O)O F, where S~, is an edgeless graph on a' vertices. We assert that G is indecomposable at the level (a,~). Assume the contrary, and let G1(A1, B1) be the first component of the canonical decomposition of G at the level (c~, fl). Let G2(A2, B2) be the first component in the canonical decomposition of GI(A1,B1)in the class (a', fl'). Then G2(A2, B2)is the first component of the canonical decomposition of G at the level (a',/3'). Since A1 does not contain a stable set of size c~', the same is true of A2. But then G2(A2, B2) cannot be the same as S~,(A, rg), contradicting Theorem 13.2.4. Now we specialize to the class (1, 1), the class of split graphs. Two polar isomorphic split graphs are said to be split isomorphic. We would like to use the canonical decomposition at the level (1, 1) to give a criterion for split graphs to be isomorphic. We need the following lemma, which spells out the freedom in choosing a split partition of a split graph. L e m m a 13.2.9 Let G(A,B) and G'(A',B') be isomorphic split graphs on the same vertices. If they are not split isomorphic, then there exists a vetrex x such that A = A ' U { x } (and therefore B' = B U { x } ) , we have the decompositions

G(A,B) G'(A',B')

= H(A- {x},B)O/(l({X},~), = H ' ( A ' , B ' - {x})O SI( e , {x}),

334

The Decomposition Method

and H ( A - { x } , B ) and H ' ( A ' , B ' - { x } ) are split isomorphic, or the same holds with reversing the role of the primed and unprimed letters. P r o o f . The sets A - A ' = A N B' and A ' - A = A ' N B are both cliques and stable sets, hence their cardinality is at most 1. If they are both empty, then G(A,B) and G'(A',B') are split isomorphic, which is not the case by assumption. If both sets have cardinality 1, say A - A ' = {x} and A ' - A = {y}, then N(x) = N ( y ) = A N A', except that x and y are possibly adjacent; thus x and y are twins and by swapping them we turn (A, B ) i n t o (A', B'), so again G ( A , B ) a n d G'(A',B')are split isomorphic, contrary to assumption. Therefore one of the sets, say A - A', consists of a single vertex x and the other one, say A ' - A, is empty. Then the required properties are seen to hold. ,, We are now ready for characterizing split isomorphism, first proved by Tyshkevich.

Theorem 13.2.10 ([Tys80]) Let G(A,B) G'(A', B')

=

al(A1,B1)

@

-

G'1(A'I, B~) @

am(Am,Bm)

...

@

-..

@ G~n(A'n, B;)

be the decompositions of split graphs G and G' into indecomposable components in the class (1, 1). Then G and G' are isomorphic graphs if and only

if 1. m = n ; 2. Gi(Ai, Bi) and G'i(A~,B~) are split isomorphic for i -

1,...,m-

1;

3. either G,~(Am, Bin) and G~(A~, B~) are split isomorphic, or else one of them is K1 ({x }, ;g) and the other one is S1 (;g, {x }). P r o o f . The sufficiency is clear. For the necessity, we have two cases. If G and G' are split isomorphic, then Part 1 of Theorem 13.2.4 implies that our conditions hold with the first alternative in Condition 3. If G and G' are not split isomorphic, then Lemma 13.2.9 and Theorem 13.2.4 imply that our conditions hold with the second alternative in Condition 3. 9

Theorem 13.2.11 ([Tys80]) Let :

~'

-

~I(A1,B1)@ -

...

QGm(A,~,Bm)Q H

~'1 (A'I, B'I) @ "'" @ Gin (din, B'n) @ H'

13.3

Domishold Graphs and Decomposition

335

be the decompositions of non-split graphs G and G' with Gi(Ai, Bi) and G~(A~,B~) indecomposable in the class (1,1) and H and H' non-split indecomposable at the level (1, 1). Then G and G' are isomorphic graphs if and only if 1. m - n ; 2. G~(A,, B,) and G~(A~, B~) are split isomorphic for i = 1 , . . . , m; 3. H and H' are isomorphic. P r o o f . This is Part 2 of Theorem 13.2.4 specialized to the class (1, 1).

13.3

Domishold Graphs and Decomposition

We apply the theory of canonical decomposition presented in Section 13.2 to enumerate and classify the domishold graphs, which we saw in Chapter 12. First we do the same with the simpler threshold graphs. This section is mainly based on Tyshkevich and Chernyak [TC85b]. We use the notation I<~ = I'(n(A, e ) where A is a clique of size n, and S~ = S n ( e , . ~ ) where B is a stable set of size n. We denote by 7- the class of threshold graphs. From Condition 4 of Theorem 1.2.4, it is easy to deduce the following result. T h e o r e m 13.3.1

1. All threshold graphs belong to the class (1, 1).

2. A graph is threshold if and only if all its indecomposable components in the class (1, 1) have a single vertex. 3. The algebra (7-, Q) is a free semigroup with two generators K1 and

S 1.

According to Part 3 of Theorem 13.3.1, and by considering the canonical decomposition, we see that a threshold graph G on n vertices is determined up to complementation by a composition of n, i.e., an ordered partition Tt = Tt 1 J r ' ' ' + rts of 1"/ into positive integer parts: if G has no isolated vertices, then G = I's @ Sn2 @ I's @ ' ' " (13.4) and if G has isolated vertices, then G = Sn, (D Kn~ (D S,~ Q . . .

(13.5)

336

The

Decomposition

Method

Using in addition Theorem 13.2.10, which gives a criterion for isomorphism of graphs in the class (1, 1) represented as compositions of indecomposable graphs in that class, we obtain the following result. Theorem 13.3.2 1. A graph with no isolated vertices is threshold if and only if it has the form (13.4). 2. A graph with isolated vertices is threshold if and only if it has the form (13.5).

3. For n > 1, graphs of the form (13.4) and (13.5) are not isomorphic. Graphs of the form (13.4) (or (13.5)) corresponding to different compositions n -- rtl - J r " .

Jr- n s

and

n -

ml

-Jr'"-Jr

are isomorphic if and only if s - t + 1, ns - 1, for i - 1 , . . . , t - 1, or vice versa.

mt

1 and ni

n t - - m t --

-

mi

T h e o r e m 13.3.3 Let cm denote the number of non-isomorphic threshold graphs without isolated vertices that have m edges, with Co - 1 by convention. Let D ( m ) denote the set of partitions of m into distinct (positive integer) parts. Then 1.

2.

E cm x m m>O

l-I(l+xJ) 9 j_>l

Proof. The two parts of the theorem are equivalent by the well-known generating function for the number of partitions of integers into distinct parts. We prove Part 1 by exhibiting a suitable bijection [TC85b]. Part 2 has been proved by different techniques by Peled [Pel80], see Chapter 17. The partition m -- ml

- - I - ' ' ' nt- rrtk

with

ml

>

'''

>mk

:>

0

is mapped to the graph K1 @

Sml--m2--1 (~ ''" (~

K1 @

Sink_l--ink--1 (~

K1 @ Sm k C

D(m),

13.3

Domishold Graphs and Decomposition

337

where empty graphs So are understood to be omitted. By Part 3 of Theorem 13.3.2, this mapping is injective. Moreover, for the same reason, every threshold graphs without isolated vertices can be represented in the form K~ Q Sl~ Q . . . Q K~, Q Sl, with positive ni and li, and it is easy to reconstruct the partition that maps to it. Thus the mapping is surjective. .. Turning now to domishold graphs, we denote by D the set of polar graphs in the class (2, 1) that are domishold. We easily deduce the following theorem from Theorem 13.2.4 and Condition 3 of Theorem 12.2.4: Theorem

13.3.4

1. All domishold graphs belong to the class (2, 1).

2. A graph in the class (2, 1) is domishold if and only if all its indecomposable components in this class have one or two vertices. .

o

The algebra (~, 5)) is a free semigroup generated by the polar graphs K l ( { X } , e ) , SI(•, { x } ) a n d S2(;g, {x,y})with x,y non-adjacent. The set of domishold graphs in the class (1, 1) coincides with the set of threshold graphs.

For the classification of domishold graphs, it is convenient to work with their complements, the codomishold graphs, which by Part 3 of Theorem 13.3.4 are the compositions of components of the form KI({X}, e ) , S l ( e , {x}) and M ( e , {x, Y}) with x, y adjacent. Using the notation Mtm - S~ Q M m - S[ Q M TM, we see that the codomishold graphs are the graphs that have one of the following forms of decomposition" M/lrn I @ I(n I @ M/2rn 2 @ tin 2 @ . . . @ MItrnt @ lin t

(13.6)

Mllrnl Q linl Q Ml2rn2 @ l(n2 @ "" Q l(nt_l @ Mltrrtt

(~a.7)

I'(nl @ Ml2m2 @ I'(n2 @ Ml3m3 @"" @ Mltmt @ I~nt

(13.8)

I(nl Q Mt~m~ Q K~2 Q Mt3.~3 Q " Q

K~_I Q Mt,.~,

(~a.9)

with t >_ 0. Here, as above, Kn denotes K~. We call a composition of the form (13.6) or (13.8) with nt >_ 3 or of the form (13.7) or (13.9) with It -t- rat > 2 &

standard decomposition.

338

The Decomposition Method

T h e o r e m 13.3.5 Every codomishold graph with at least three vertices has a unique standard decomposition. P r o o f . Observe that in each standard decomposition, every component with the possible exception of the last two is uniquely determined. For instance, if the graph has isolated vertices or edges, then [1 and ml are the numbers of isolated vertices and edges, respectively, n l is the number of dominating vertices remaining after the previous vertices are deleted, and so on. It follows that in any two standard decompositions, the numbers of components differ by at most 1. The following are the only cases. Case 1" One standard decomposition ends with Mtm q) K~. Then n >_ 3, and n is the number of vertices of degree at least 2 in the graph induced by the vertices of these last two components. Therefore n is uniquely determined, and with it the last two components. C a s e 2" One standard decomposition ends with K~ Q Mzm. Then l + m >_ 2, and n is the number of dominating vertices in the graph induced by the vertices of these last two components. So, again, n and the last two components are uniquely determined. C a s e 3" All standard decompositions have exactly one component, namely Mllml or K ~ . This case is trivial because the graph has at least three vertices. 9 Every threshold graph is a unigraph, i.e., determined up to isomorphism by its degree sequence (in fact, a threshold sequence has a unique labeled realization). In contrast, non-isomorphic domishold graphs can have the same degree sequence. For example, the complete bipartite graph K2,3 and the pentagon with one chord have the same degree sequence (3, 3, 2, 2, 2). The first one is domishold, since its complement has the standard decomposition M2,0 (~ K3; the second is not, since it contains an induced P4, which is forbidden by Theorem 12.2.4. However, a domishold graph without isolated vertices is an edge-unigraph, i.e., determined up to isomorphism by the list of degrees of its edges , where the degree of an edge is defined as the unordered pair of degrees of its endpoints. To show this, we make the following definition. D e f i n i t i o n 13.3.6 Let a graph G have an alternating ~-cycle t with edges ab, cd and nonedges ad, bc, and let deg(a) = deg(c). We then say that G admits a swap t, and the graph t G is obtained from G by swapping the edges and nonedges oft. See Figure 13.2.

13.3

Domishold Graphs and Decomposition

339

Figure 13.2: A swap.

a

d

b

a

c

d I

I

I I Ark

I I

b

d e g ( a ) - deg(c)

c

d e g ( a ) - deg(c)

Chernyak and Chernyak [CC81] have proved that if two graphs without isolated vertices have the same list of degrees of edges, then one can be obtained from the other by a sequence of swaps. Therefore it is enough for us to prove the following result. P r o p o s i t i o n 13.3.7 If G is a domishold graph admitting a swap t, then G

is isomorphic to tG. m

P r o o f . Clearly the complement G also admits t as a swap, and so it is enough to show that G and tG are isomorphic. By Theorem 13.3.5, G has a unique standard decomposition in one of the forms (13.6) - (13.9). Without loss of generality we may assume that the first component of this decomposition contains one of the vertices a, b, c, d of t, for we may delete all preceding components, if any. Considering the polar graph G(A, B), we examine the possible cases of which of the sets A, B contains each vertex of t, taking into account the fact that A is a clique and B induces a matching with isolated vertices. C a s e 1" a, b, c, d C B. Since a and c have the same degree, they must be in the same component, which must also contain their neighbors c and d. But then clearly tG is isomorphic to G. C a s e 2" a,b,c E B, d C A (the case a,c,d C B, b C A is similar). The neighbors a and b are in the same component, and c must be in a later component, since the vertex d of A is a neighbor of it but not of a. Therefore a, b C Miami, hence deg(a) - 1, and it follows that deg(c) - 1. This implies

The Decomposition Method

340

that d E K ~ , nl - 1, and c E Mt2m2. But now swapping a and cgives a graph isomorphic to G and also equal to tG. C a s e 3" b,c,d E B, a E A (the c a s e a , b,d E B, c E A i s similar). Since a E A is adjacent to b E B but not to d E B, d (and with it its neighbor c E B) must be in an earlier component than b. Hence c,d E Mt~n~, and therefore deg(c) - 1, from which deg(a) - 1. Hence b is the only vertex in its component, and this component is the last one. But this contradicts the definition of the standard decomposition, so this case is impossible. C a s e 4: a,b E A, c,d E B (the case c,d E A, a,b E /3 is similar). We must have c, d E Mt~nl, so deg(c) - 1, hence deg(a) - 1. This implies that A - {a, b}. Now if a and b are in the same component, then this component is K ~ and is the last component, contradicting the definition of the standard decomposition. If a and b are in different components, then since deg(a) - 1, the component containing a has no other vertices and is the last component, again contradicting the definition of the standard decomposition. Therefore this case is impossible. C a s e 5" a, c E A, b, d E /3 (the case b, d E A, a, c E B is similar). Since a is a neighbor of b and c is not, the component containing a is earlier than the one containing c. But this contradicts the fact that d is a neighbor of c but not of a, so this case is impossible. 9 We conclude this section by classifying the forcibly domishold degree sequences, i.e., those degree sequences all of whose realizations are domishold. We prove, in fact, that a forcibly domishold sequence must be unigraphic. As usual, we denote a star on n + 1 vertices by KI~. 13.3.8 A forcibly domishold degree sequence is unigraphic, and its realization is of one of the following types:

Theorem

1. a threshold graph; 2. rnK2 U Kin with rn, n > 1; 3. T(A, B ) Q inK2 U Kin with rn, n >_ 1, where T is a threshold graph. Conversely, every graph of the one of the above types is domishold and a unigraph. P r o o f . Let d be a forcibly domishold sequence with a realization G. We have to show that G i s o f t y p e 1 - 3 . I f G E (1,1), then G i s o f t y p e 1 b y Part 1 of Theorem 13.3.4, and we are done. Assume now that G ~ (1, 1).

13.4

Box-Threshold Graphs and Decomposition

341

We show below that if G is indecomposable at the level (1, 1), then it is of type 2, from which it follows that if G is decomposable at the level (1, 1), then it is of type 3. Since all realizations of d are domishold, and since every transfer (cf. Corollary 3.1.11) done on the complete bipartite graph K2,3 yields the nondomishold pentagon with a chord, G has no induced K2,3. Consider the standard decomposition of G. Since G is indecomposable at the level (1, 1), the first component of the decomposition is of the form M0ml, rrtl ~ 1. If there are no other components, then rn~ >_ 2 because G ~ (1, 1), and hence G is of type 2. Assume now that there are other components. Then G is triangle-free, since G is (K2 U K3)-free. Hence in the polar representation G(A,B), IAI _< 2. By the definition of a standard decomposition, the last component must then be of the form Mr,0 with It >_ 1. Further, t - 2 and nl 1 since G is triangle-free. This means that -

-

G

--

M0rrt

I

@ K1 @ M I 2 0

--

rnlK2 U Kll2,

so G is of type 2, as required. To prove the converse, it is enough to notice that if G is of type 1 3, then it is domishold, and every transfer done on it yields an isomorphic graph, hence it is a unigraph by Corollary 3.1.11. 9

13.4

B o x - T h r e s h o l d Graphs and D e c o m p o s i tion

This section is adapted from Tyshkevich and Chernyak [TC85a]. We characterize box-threshold (BT) graphs in terms of their decomposition into indecomposable components at the level (1, 1), and then use these results to enumerate them. All the polar graphs in this section are in the class (1, 1) (split) and the decompositions will always be at the level (1, 1). We use the notation Na(x) for the set of neighbors of a vertex z in a graph G, and ~ a for the vicinal preorder ~ on the graph G. If A and B are sets of vertices of a graph, we use NA(B) to denote A N ( U N(b)), and A ~- B to indicate bEB

that a ~- b for all a C A and b C B, and similarly for A ~ B. L e m m a 13.4.1 A composition is BT if and only if each component is BT.

342

The Decomposition Method

P r o o f . Let F = G(A, B) Q H. We have to show that F is BT if and only if G and H are BT. Note that A ~ V ( H ) ~ B. " O n l y if"" Assume that F is BT. To show that H is BT, consider vertices x,y of H with degH(x ) > degH(y ). Since Na(x) = Na(y), we must have degF(x ) > degp(y). Since F is BT, this implies that x L~ Y, and therefore x ~H Y, so H is BT. To show that G is BT, consider vertices x,y of G with dega(x ) > dega(y ). If x and y are both in A or both in B, then since NH(X) = NH(y), we must have degr(x ) > degp(y). Since F is BT, this implies that x ~ p y, and therefore x ~ a y, as required. If not, then necessarily x C A and y C B, which already implies that x ~ a y. " I f " : Assume that G and H are BT. To show that F is BT, assume that degF(x ) > degF(y ). If y E A, then x E A and dega(x ) > dega(y ). Since G is BT, we have x ~ a y and hence x ~ p y. If y E V(H), then x E V(H) or x C A. In the first case x ~ p y since H is BT, and in the second case x ~ p y since A ~ V(H). If y C B, we similarly reach the same conclusion x LF YTherefore F is BT. .. The next two lemmas identify the indecomposable BT graphs. Observe that if a split graph G(A, B) is biregular, then all the vertices of A have the same degree and all the vertices of B have the same degree. L e m m a 13.4.2 A split graph with more than one vertex is an indecompos-

able B T graph if and only if it is biregular and has no dominating or isolated vertices. P r o o f . " O n l y if": Let G(A,B) be an indecomposable graph in the class (1, 1) with more than one vertex and also a BT graph. By being indecomposable with more than one vertex, G has no dominating or isolated vertex. We show that it is biregular. Let A' be the set of vertices of A with largest degrees, and let B ' = N B ( A - A'). Since G is BT, we have A' ~ A - A' and therefore B' is completely joined to A'. Unless A ~ = A, we then have the decomposition

G(A, B) = G'(A', B - B') Q H ( A - A', B'), contradicting the indecomposability of G. So A' = A and all the vertices of A have the same degree. Similarly all the vertices of B have the same degree. " I f " : Let G(A, B) be a split biregular graph with no isolated or dominating vertices. To show that G is BT, notice that if deg(x) > deg(y), then x C B and y E A, which implies in turn that x ~ y. To show that G is indecomposable, assume the opposite and let G = G'(A', B ~) Q H be a decomposition

13.4

Box-Threshold Graphs and Decomposition

343

of G with G ~ and H non-empty. There must exist a vertex x C A' such that NB,(X) 7/= rg, for otherwise the vertices of B' are isolated in G, and if B' = rg then the vertices of A ~ are dominating in G. Similarly there must exist a vertex y C B' such that N ( x ) is a proper subset of A'. Now pick any vertex h of H, and we must have dega(x) > dega(h ) > dega(y ). But this contradicts the biregularity of G. L e m m a 13.4.3 A non-split graph with more than one vertex is an indecomposable B T graph if and only if it is regular and has no dominating or isolated vertices. P r o o f . " O n l y if"" Let G be an indecomposable non-split graph with more than one vertex and also a BT graph. By being indecomposable with more than one vertex, G has no dominating or isolated vertices. To show that G is regular, let A be the set of vertices with the largest degree in G, let H = (1 N(a), and let B = V(G) - (A U H). For each a E A and x E B U H aG_.A

we have deg(a) > deg(x), and so A ~ BU H. Each b e B has a non-neighbor in A, for otherwise b C H; therefore by A ~- B U H, b has no neighbors in B U H . I f H r 0 , then by A ~- H, A is a clique and G decomposes as a = G ' ( A , B ) O G[HI, where G[H 1 is the subgraph induced by H. This contradicts the hypothesis, and so H = O. If A is a clique, then G is a split graph G(A, B), contrary to hypothesis. Therefore there is a nonedge uv in A. S i n c e u ~- B, v has no neighbors in B. Let b b e a v e r t e x o f B , if any. Since it is not isolated, it has a neighbor z C A. Since A >- b, the closed neighborhood N[z] contains A U {b}. However, N[v] is a proper subset of A, and thus deg(z) > deg(v), contradicting the definition of A. This proves that B = 0 , and so G is induced by A and is therefore regular. " I f " : Let G be a regular graph without dominating or isolated vertices. By being regular, G is BT. To show that G is indecomposable, assume the contrary and let G = G'(A, B ) Q F, where G' and F are non-empty graphs. It follows that A ~- B, and hence deg(a) > deg(b) for all a E A and b E B. Now A r 0 , for otherwise the vertices of B are isolated, and B 7~ 0, for otherwise the vertices of A are dominating. This contradicts the regularity of G. " The next theorem characterizes BT graphs in terms of their indecomposable components at the level (1, 1).

The Decomposition M e t h o d

344

T h e o r e m 13.4.4 Let G = Xx Q ... Q X~ be the decomposition of a graph G into indecomposable components at the level (1, 1). Then G is B T if and

only if 1.

X I

, . . . ,

Xn-1 are split biregular graphs;

2. Xn is a split biregular or non-split regular graph. P r o o f . Note that if Xi has more than one vertex, then it cannot have dominating or isolated vertices by being indecomposable. The result then follows immediately from Lemmas 13.4.1, 13.4.2 and 13.4.3. [] We now enumerate the BT graphs using the above results. Let us denote by ap the number of non-isomorphic indecomposable split graphs on p vertices, by bp the number of non-isomorphic indecomposable non-split graphs on p vertices, and by cp the number of non-isomorphic BT graphs on p vertices. We also use the generating functions of these sequences, namely

a(x) - ~ apx p,

b(x) - ~ bpx p,

p>_l

p>_l

c(x) - ~ cpx p. p>_l

By the uniqueness of the canonical decomposition and Lemma 13.4.1, we have the following relation: 4x) =

1 -a(x)

Below we show how the ap and bp can be computed. C o m p u t i n g ap: Clearly a~ - 1 and for p > 1, by Lemma 13.4.2, ap is the number of non-isomorphic biregular split graphs on p vertices with no dominating or isolated vertices. Therefore ap --

Z Amn, m+n=p m,n>O

where A,~n is the number of non-isomorphic biregular split graphs with no dominating or isolated vertices, having the form G ( A , B ) with IAI = m, It l = S u c h ~ graph G ( A , B ) i s specified by the adjacency matrix M whose rows correspond to the vertices of A and whose columns correspond to the vertices of B. It is an m x n 0/1 matrix with constant row sums r, 0 < r < n, and constant column sums s, 0 < s < m. Let us denote by

13.4

Box-Threshold Graphs and D e c o m p o s i t i o n

345

Ad~n~ the set of all such matrices, and by Sm and S~ the symmetric groups of degrees m and n respectively. We define an action of the direct product Sm x S~ o n . / ~ m n by

(7r,a)M = PTrMP~ -1 ,

~ESm,

aES~,

where P~, Po are the permutation matrices of 7r, a, respectively. Then Amn is the number of orbits of this group action. By Burnside's Lemma, we have 1

A.~ =

~

F(rr, a),

TIt! TL! (Tr,o')e"~mXSn

where F(~r, a ) i s the number of matrices M C JPlmn that are fixed by (re, a), that is (Tr, a ) M = M. To compute F(r~, or), we write rr and cr as products of disjoint cycles 71- - -

where rri is a cycle of length

7f'l

9 9 9 7Ck~

0"

~

O" 1 9 9 9 0 " I ~

mi and aj is a cycle of length nj,

77"t1 :> " ' "

~

?Ytk~

Using the notation m = ( m l , . . . , m k ) a n d n = ( n l , . . . , n t ) , we observe that F(rr, a ) d e p e n d s only on m and n and not on the particular rc and a, and we may denote it by F ( m , n). There are m!/ml...mk permutations rr corresponding to m and similarly for n. Hence rtl >

"'"

~

rtl, m

-- ml

+'..

+ ink,

rt - - rtl + . . .

1

Am,~ - }-~. 77"tl . . .

?Tt k Tt 1 . . .

+ rtl.

r(m, n), TLl

where the summation extends over all the partitions m , n of the integers rn, n into positive parts. To compute F(m, n), consider the action of (rr~, aj) on the submatrix M~j of M consisting of the m i rows cycled by rri and the nj columns cycled by aj. The entries of Mij are partitioned into dij = g c d ( m / , nj) cycles, each one of length rninj/dij (i.e., the least common multiple of m i and nj). Each such cycle meets each row of Mij nj/dij times, and each column of Mij mi/dij times. The matrix M is fixed by (rr, a) if and only if in each submatrix Mij, each cycle is all-0 or all-1. If exactly t cycles of Mij are all-l, then Mij has constant row sums tnj/dij and constant column s u m s tmi/dij. We denote by e~...~k;,~..., , the number of m x n 0/1 matrices M fixed by (rr, a) for which the rows cycled by rri have constant row sums ri and the columns cycled by

The Decomposition Method

346

O'j have constant column sums sj. Then by the above discussion we obtain the generating function

E

"''Yk~kZ~ a ' ' ' z ~ ~ - -

Crl,...,rk;sl,...,sty~l

rl ,...,rk 81 ,...,3l

H

l
tni/di ~ tmi/di3

Yi

zj

,

O
l<_j<_l

and the required F ( m , n ) can be computed from the relation F(m,

n) -

~

e~;~,

O
k

l

where s - m r / n and rk; s t is an abbreviation for r.."-fk'~, r; .~YL~. s 9 s. C o m p u t i n g b v" As before, by Lemma 13.4.3, bp is the number of nonisomorphic regular non-split graphs on p vertices with no dominating or isolated vertices. We may omit the condition "non-split", since it follows from the others. Let Alp be the set of all adjacency matrices of such graphs. We define the action of the symmetric group Sp on M p by 7rM - PTrM P~ -1 ,

7r" C Sp .

Then bp is the number of orbits of this group action. By Burnside's Lemma, we have t,, -

1

where F(Tr) is the number of matrices M E 2t4p that are fixed by 7r, that is 7 r M - M. Write 7r as a product of disjoint cycles 71" - -

71"1 " ' ' T "

k

where 7ri is a cycle of length pi, with pl >_ "'" >_ pk, p - pl nt- ' ' " -t- pk, and use the notation p - ( p l , . . . ,pk). Again, F(Tr) depends only on p, and we denote it by F ( p ) . Thus 1

pl "'" pk where the summation extends over all the partitions p of the integer p.

13.4

347

Box-Threshold Graphs and Decomposition

To compute F(p), consider the action of 7r on the submatrix Mij of M consisting of the pi rows cycled by 7ri and the pj columns cycled by 7cj. The entries of Mij are partitioned into dij = gcd(pi, pj) cycles, each one of length pipj/dij. Each such cycle meets each row of Mij pj/dij times, and each column of Mij pi/dij times. The matrix M is fixed by 7r if and only if in each Mij, each cycle is all-0 or all-1. By the s y m m e t r y of M, the submatrix Mij with i < j determines the submatrix Mji as its image under the symmetry. Thus for i < j, If exactly t cycles of Mij are all-l, then exactly t cycles of Mji are all-1. Therefore Mij has constant row sums tpj/dij and Mji has constant row sums tpi/dij. The number of ways to choose t such cycles is

Let hii,t denote the number of ways to select exactly t out of the dii = pi cycles of Mii to be all-1 so that Mii becomes a symmetric matrix with zero diagonal. When such a choice is made, Mii has constant row sums t. It is not hard to verify that

{0( )

if

pi and t are odd

otherwise

(note in particular that hii,p, - 0). We denote by erl...rk the number of adjacency matrices of order p fixed by 7r for which the rows cycled by ~ri have constant row sums ri. As before, we obtain the generating function E rl ~...~rk

•rl,...,rkXl

rl

rk 9. . X k

X i 1
and the required

Xj

)(

0
t) l
F(p) can be computed from the relation F(p)

--

Crk 9 O
348

13.5

The D e c o m p o s i t i o n M e t h o d

Matroidal and Matrogenic Graphs and Decomposition

In this section, following Tyshkevich [Tys84], we use the decomposition theory to exhibit the structure of the matroidal and the matrogenic graphs and enumerate them. We already know that the threshold graphs are matroidal, the matroidal graphs are matrogenic, and the matrogenic graphs are unigraphs (Theorems 11.2.2, 11.3.2 and 11.4.1). The decomposition theory at the level (1, 1) gives us a convenient way to recognize these graphs. 13.5.1 A graph is threshold, matroidal, matrogenic, or a unigraph if and only if all its indecomposable components at the level (1, 1) are threshold, matroidal, matrogenic, or unigraphs, respectively.

Theorem

P r o o f . To prove the "only if" part of the first three statements, we note that the properties in question (threshold, matroidal, matrogenic) are preserved for induced subgraphs because they are characterized by forbidden configurations. For the "if" part of the first three properties, we note that if a composition F = G(A, B ) Q H contains one of our forbidden configurations (an alternating 4-cycle, the configuration 5 , a chordless pentagon), then this configuration is entirely contained in G or in H. For example, suppose F contains an alternating 4-cycles as in Figure 1.2. If one of its vertices, say a, is in A, then it follows that c C B, d E A, and b E B, so the whole configuration is contained in G. Similarly if a C B, the whole configuration is in G. Otherwise the whole configuration is in H. Since the configuration pc and the chordless pentagon contain several alternating 4-cycles, it follows immediately that if one of their vertices is in G, all are in G. To prove the "if" part of the fourth statement (about unigraphs), we show that the composition of unigraphs is a unigraph. Consider a composition F = G ( A , B ) Q H, where G and H are unigraphs, and let F ' be a graph with the same degree sequence as F. We have to show that F ~_ F ~. By Theorem 3.1.11, F ' can be obtained from F by a sequence of transfers, and we may assume without loss of generality that the sequence consists of a single transfer along an alternating 4-cycle C. By what was shown above, C is entirely contained in G or in H, and in the former case, the transfer preserves the polar partition (A, B). Hence f ' takes the form G'(A, B)(9 H, where G' and G have the same degree sequence, or G(A, B) (9 H', where H' and H have the same degree sequence. Then G _~ G ~ or H _~ H ~ by the assumption, and hence F " F' in either case.

13.5

Matroidal and Matrogenic Graphs and Decomposition

349

To prove the "only if" part of the fourth statement, consider the decomposition of a unigraph G into indecomposable components. We have to show that each component C is a unigraph. Let C ~have the same degree sequence as C. If we replace C by C' in the decomposition, we obtain a graph G' with the same degree sequence as G. By assumption G _~ G ~. Hence by the uniqueness of the decomposition, it follows that C ~_ C ~. .. We. have seen in Theorem 13.3.1 that the only indecomposable components of a threshold graph are the one-vertex graphs. Tyshkevich [Tys84] lists the indecomposable unigraphs, and uses them to find the indecomposable matroidal and matrogenic graphs. Instead, we can rely on the structure of these graphs found in Chapter 11. The result is the following. T h e o r e m 13.5.2 o

.

The indecomposable matroidal graphs are precisely the one-vertex graph, the nets and net-complements (Definition 11.2.10), the perfect matchings of more than one edge, and the complements of these matchings. The indecomposable matrogenic graphs are precisely the graphs listed above and the chordless pentagon.

We can use the results above to enumerate the matroidal and matrogenic graphs as follows. T h e o r e m 13.5.3 Let m~ and l~ denote the number of non-isomorphic ma-

troidal and matrogenic graphs, respectively, and consider the generating functions of the sequences {ran} and {l~}" l(x) -

-

n>l

Z n>l

Then -

2 p ( x ) - 3x + x 4 + x 5

2 p ( x ) - 3x + x 4

1 -p(x)

'

l(x) -

1 -p(x)

wh e re

p(x) - 2x + x 4 +

2x 6 1 B X 2"

350

The Decomposition Method

P r o o f . Let pn denote the number of pairwise non split-isomorphic indecomposable split matroidal graphs. By Theorem 13.5.2 we have Pl -- 2 ( t w o split partitions for the one-vertex graph), p3 - 0, p4 - 1 (the self-complementary net of order 2), and for k _> 3, p2k - 2 (the net and the net-complement of order k) and P 2 k - 1 - - O. Hence the generating function for the p~ is p ( x ) - y ~ p n x n - 2x + x4 + 2 E n>l

x2k-

2x + x 4 +

2x 6 1 -

k>3

x2

Let qn denote the number of pairwise non-isomorphic indecomposable nonsplit matroidal graphs. Then q2k - 2 for k _> 2 (the perfect matching on k edges and its complement) and q~ - 0 for all other n. Therefore the generating function for the qn is q ( x ) -- p ( x ) -- 2X + X 4. Finally, the generating function for the number of pairwise non-isomorphic indecomposable non-split matrogenic graphs is r ( x ) - q ( x ) + x s due to the pentagon. Therefore the generating function for the number of pairwise non-isomorphic indecomposable matroidal graphs (split or not) is p ( x ) + q ( x ) - x - 2p(x) - 3x + x 4 (we subtract x because the two split partitions of the one-vertex graph give isomorphic graphs). By Theorem 13.5.1, this function enumerates the last indecomposable component in the canonical decomposition of a matroidal graph, whereas p ( x ) enumerates each of the previous components. Hence m ( x ) - (2p(x) - 3x + x 4) ~ ( p ( x ) ) y>_o

j = 2 p ( x ) - 3x + x 4 1 -p(x)

"

The result about l ( x ) follows in the same way. 9 From Theorem 13.5.3 one can obtain recurrences and asymptotic estimates of mn and ln.

Chapter 14 Pseudothreshold and Equistable Graphs 14.1

Introduction

In this chapter we study two classes of graphs containing the threshold graphs, namely the pseudothreshold graphs introduced by Chvs and Hammer [CH77] and the equistable graphs introduced by Payan [PayS0] and studied by Mahadev, Peled and Sun [MPS94] . Pseudothreshold graphs are graphs for which there exists a hyperplane such that the characteristic vectors (of sets of vertices) lying on one side of the hyperplane correspond to stable sets and the ones lying on the other side correspond to nonstable sets; the characteristic vectors lying on the hyperplane itself may correspond to stable or nonstable sets. These graphs are characterized in the next section. Equistable graphs are graphs for which there exists a hyperplane such that all the characteristic vectors lying on one side of the hyperplane correspond to stable sets, and in addition the characteristic vectors lying on the hyperplane itself correspond precisely to the maximal stable sets. As we see in this chapter, these two classes are very different despite the similarities in their definitions. In fact, neither of these two classes contains the other.

14.2

Pseudothreshold Graphs

In their key paper on threshold graphs, Chvs and Hammer [CH77] introduced the concept of pseudothreshold graphs as follows: 351

Pseudothreshold and Equistable Graphs

352

D e f i n i t i o n 14.2.1 A graph G - (V, E) is called a p s e u d o t h r e s h o l d g r a p h if there exist a mapping w" V --~ N (w ~ O) and a real t such that for every SC__V, S is a stable set ---> ~ x e s w(x) <_ t S isanonstable set ~ E~esw(x) >t. (14.1) Thus in a pseudothreshold graph, the hyperplane separating the characteristic vectors of stable and nonstable sets may contain vectors of both kinds of sets. In contrast, in a threshold graph the separation is strict, and the hyperplane does not contain characteristic vectors of nonstable sets (or equivalently of any sets). The same authors obtained the structure of pseudothreshold graphs. We present their results in this section. L e m m a 14.2.2 Let G - (V,E) be a pseudothreshold graph as in Definition 1~.2.1. Then the t in Equation (1~.1) must be positive if E 5r fg and may be assumed to be positive otherwise. P r o o f . If E - O, then t - 1 and w(x) <_ 0 will satisfy Equation (14.1). Assume that E 7(: ~. Since the empty set is stable, Equation (14.1) implies that t >_ 0. If t - 0, then since every singleton is a stable set, w(x) <_ 0 for each x C V, and since w ~ 0, there exists a z C V such that w(z) < O. Since E r O, there exist adjacent vertices x, y, not necessarily distinct from z. Then the set S - {x, y, z} violates Equation (14.1). Hence t > 0. tt From now on we assume that t > 0. For every graph G - (V, E), we define two subsets P0, Q0 of V as follows: No - {72 E V 9 ~721, ?22, u 3 E V, Uttl, ?2?./2, uu 3 E E , UlU2, ?22?23, ?_/3Ul @ E }

Qo - {v c v . 3Vl, y2, v3 E v, yyl, vv2, yv3, yll)3 ~ E , yly2, y2v3 ~ E } .

These definitions are illustrated in Figure 14.1. The next four lemmas are used in the characterization of pseudothreshold graphs. Lemma

14.2.3 Let G -

(V, E) be a pseudothreshold graph. Then u C Po ---> w(u) >_2t/3 u c Qo ~

w(u) <_ t/3.

r

IA

IV

V

O0

"1

~)

N

II

~3

D

C8

~

~

.

~

C8

.

r

C

.-,

0~ ~

~

dh

~

~ <

S

C

~,

.~

IV

IV

IA

4 - - t - + +

IA

IV

c~ 9

IV

IV

IV

+ 4 - + +

IA

4-

~

~

9

o

,O

m

m

w

oa

A w

~

I~I

o

O

(1)

0~

t~

r162

r162

O

O

t~

354

Pseudothreshold and Equistable Graphs

These implications follow easily by induction on m from Lemma 14.2.3.

9

L e m m a 14.2.5 If P* N Q* - ~, then every two vertices in P* are adjacent and no two vertices in Q* are adjacent. P r o o f . This follows from the definition of P* and Q*. L e m m a 14.2.6 No pseudothreshold graph contains an induced 3K2. P r o o f . Assume that a pseudothreshold graph contains a 3K2 with edges UiVi for i - 1, 2, 3. Then w(u~) + w ( u ~ ) + w ( u a )

_< t

~ ( ~ ) + ~ ( ~ ) + ~(v~)

_< t

~(~,) + ~ ( ~ )

>_ t

~ ( ~ ) + ~(v~)

_> t

~(~) + ~(~)

>_ t.

But these inequalities are clearly inconsistent with t > 0. The following theorem characterizes pseudothreshold graphs. T h e o r e m 14.2.7 For every graph G equivalent:

(V, E), the following conditions are

1. G is pseudothreshold; 2. P* n Q* - 0 and G is 3K2-free; 3. V can be partitioned into sets P, Q, R such that (a) every vertex in P is adjacent to every vertex in P U R, (b) no vertex in Q is adjacent to another vertex in Q u R, (c) there are no three mutually nonadjacent vertices in R. P r o o f . 3) =~ 1)" An appropriate assignment is

w(u)-

..

0, i f u c Q 1, i f u E R 2, if u C P

t-2.

14.2

Pseudothreshold Graphs

355

1) =~ 2): This follows from Lemma 14.2.4 and Lemma 14.2.6.

2) =~ 3): We prove this by means of a simple algorithm that terminates in O(rt 4) steps either by showing that Condition 2 does not hold or by constructing the partition described in Condition 3. The algorithm is as follows. First of all, find P* and Q*. (This can certainly be done in O(n 4) steps.) Then find out whether P * N Q* = ~. (If not, stop: Condition 2 does not hold.) Then set S = V - (P* U Q*); note that by the definition of P* and Q*, every vertex in S is adjacent to all the vertices in P* and to no vertex in Q*. Let So consist of all the vertices in S that are adjacent to no other vertex in S; define

P=P*,

Q=Q*USo,

R=S-So.

Find out whether there are three mutually nonadjacent vertices in R. If not, stop: by Lemma 14.2.5 P, Q and R have all the properties described in Condition 3. If, on the other hand, there are three mutually nonadjacent vertices Ul,U2,U3 C R, then each ui is adjacent to some vi E R. Using the fact that R N (P0 U Q0) = o , we may now easily verify that the set {Ul, u2, u3, Vl, v2, v3} induces a 3K2 in G. (To see this, note that the vj's are distinct, mutually nonadjacent and that each vj is adjacent to exactly one ui.) Hence Condition 2 does not hold. " The corollary below follows from the proof of the above theorem.

Corollary 14.2.8 If G is pseudothreshold, Equation (14.1) can be satisfied with t = 2 and w(u) C {0, 1,2} for all u.

"

We conclude this section with the following remarks.

Remarks 1. The degree sequences of pseudothreshold graphs on n vertices lie on the boundary of the convex hull of the degree sequences of all graphs on n vertices, because they satisfy the facet inequality

Z iEP

x ~ - ~ x~ _< IPl(n- 1 - I Q I )

(14.2)

iEQ

with equality, where P, Q, R is the partition of V as in Condition 3 of Theorem 14.2.7. This follows from Lemma 3.3.13.

356

Pseudothreshold and Equistable Graphs

2. The converse is not true. For instance, the graphs K1@K3,3 and K1@C6 have the same degree sequence (6, 4, 4, 4, 4, 4, 4) satisfying (14.2) with equality, where P is a singleton corresponding to the vertex of K1, Q is empty, and R consists of the other six vertices. But the first graph is not a pseudothreshold graph and the second one is. 3. The complement of a pseudothreshold graph need not be pseudothreshold. For instance, 3K2 is pseudothreshold. In view of the above remarks, the following questions are interesting.

Problem 1. Which degree sequences lie on the boundary? 2. Which degree sequences are potentially pseudothreshold?

14.3

Equistable Graphs

Payan [Pay80] introduced the equistable graphs, and later Mahadev, Peled and Sun IMPS94] obtained several interesting results on these graphs. We present the results of IMPS94] in this section. Definition 14.3.1 A graph G - (V,E) is said to be equistable if there exists a mapping ca" V --. N + such that for all S C_ V, S is a maximal stable set ~

c a ( S ) - ~ c a ( v ) - 1. vES

Payan [PayS0] showed that the threshold graphs and the domishold graphs are equistable, and announced that the P4-free graphs are equistable. In the next subsection we introduce a subclass of equistable graphs the strongly equistable graphs. We show that the strongly equistable graphs are closed under disjoint unions and joins, and therefore the P4-free graphs are strongly equistable. Later in this section, necessary conditions for equistability and sufficient conditions for strong equistability are obtained and used to characterize the equistability of some classes of perfect graphs, pseudothreshold, and outerplanar graphs. The structure of these equistable graphs is also given.

14.3

Equistable Graphs

357

Finally, we show that for a wide family of graphs, equistability implies strong equistability, and discuss the closure of various classes of strongly equistable graphs under graph substitution. An induced path on four vertices is denoted by P4(a, b, c, d) to indicate that its vertex set is {a, b, c, d} and its edges are ab, bc, cd.

14.3.1

Strongly Equistable Graphs

Let G = (1,1,E) be a graph. We denote by $ the set of maximal stable sets of vertices of G, and by 7- the set of all other nonempty sets of vertices of G. Put A - A(G) - {~v" V ~ R+'~v(S) - 1 for all S C $}. Thus A(G) is a set defined by a finite number of linear equations and nonnegativity inequalities, i.e., it is the intersection of an affine set with the first orthant. Since each vertex belongs to some maximal stable set, A ( G ) is bounded and therefore it is a polytope, possibly empty.

Definition 14.3.2 A graph G is said to be strongly equistable if for each T E 7- and each c <_ 1, there exists an w E .A such that w(T) 7~ c. L e m m a 14.3.3 Let G be strongly equistable, and assume that for each T E T , a forbidden value cr <_ 1 is given. Then there exists an co C .4 such that w(T) 7~ cT for all T E T . In particular, G is equistable, because we can take CT -- 1 for all T C 7-. P r o o f . For each T E 2r, the linear equations and inequalities defining .A(G) do not imply a)(T) = CT, since G is strongly equistable. In other words, the polytope A(G) is not contained in the hyperplane aJ(T) = CT. Therefore the intersection AT = A N {a~ : c0(T) = CT} constitutes a lower dimensional subpolytope of A. Consider the affine hull ~ of A, i.e., the set of all affine combinations of a finite number of elements of A. In the relative topology of 7{, ~4 has a positive volume whereas AT has zero volume. Since iT is finite, [J{AT : T C 7"} has zero volume and is therefore a proper subset of A. This means that A possesses an aJ which is outside every AT. 9

T h e o r e m 14.3.4 The strongly equistable graphs are closed under disjoint union.

358

P s e u d o t h r e s h o l d and E q u i s t a b l e G r a p h s

P r o o f . Let Gi -- (Vi, Ei), i - 1,2, be strongly equistable with V1 and 89 disjoint, and let G - (V,E), where V - 1/1 UV2, E E1 UE2. For each subset X of V, we use the notation X i - X N Vi. Let ,4, 7" etc. refer to G, and ~4i, T / e t c . refer to Gi. Then S C $ if and only if $1 E 81 and ~q2 C 32. To show that G is strongly equistable, let T C 7- and c _< 1 be given. Then T1 E T1 or T2 C ~ or T1 C 81, T2 - ~ or T1 - O, T2 C 32. Assume that T1 E T~. Then there exists an CVl in .A1 such that Cvl(T1) 7~ c. Define w" V ~ R0+ by (.U

w(v) -

(V)

0

if v E I/1

if v E V2.

Then w C A and aJ(T) - wl(T~) + 0 -r c. Assume that T1 E S1, T2 - ~. Consider arbitrary wl E ,41, aJ2 E A2. If c 7~ 0, define aJ" V ~ R + by

{

0 w2(v)

{

a21(V) if v E V1

i f v E V1 ifvEV2,

and if c = 0, define aJ by

0

i f v E V2.

In both cases w C ,4, a~(T) r c. The other cases are symmetric. Theorem

14.3.5

The strongly equistable graphs are closed u n d e r join.

P r o o f . Let Gi = (Vi, Ei), i = 1,2, be strongly equistable with V1 and V2 disjoint, and let G - (V, E) be their join, where V - 1/1 U V2, E - E1 U E2 U El2, and E12 is the set of all possible edges joining 1/1 and 89 We use the same notation for ,4, .di etc. as before. Then a subset S of V is in ,S' if and only if S1 E 81 and 5'2 - ~ , or the other way round. Based on aJi " V / ~ N +, we define the composition w of Wl and ~o2 to be w 9 V ~ N + given by

CO(?_)) _ S

/

v e yl

aJ2(v)

if v E V2.

If wi E .Ai for i - 1, 2, then the composition w is in .A. To show that G is strongly equistable, let T E 7" and c <_ 1 be given. Then T1 and T2 are not

14.3

Equistable Graphs

359

both empty. If T1 E 81 and T2 C 5'2, choose any wi C .Ai, i - 1, 2, and let be their composition. Then w E ~4 and aJ(T) - coI (rl)-t-co2(Z2) - 2 > c. In all other cases T1 C T1 or T2 C ~ , so without loss of generality, assume that T1 E T1. Choose anyw2 C ~42. Then c-cz2(T2) _< c_< 1, and since T1 E T1, there exists an col E r such that ~1 (T1) r c-~o2(T2). Let w be the composition of Wl and aJ2. Then w C ~4 and w(T) - wl(T1) + a32(T2) -r c . . . The following result is well-known. L e m m a 14.3.6 ([Sei74]) Let G - (V, E) be a graph. Then the following conditions are equivalent" 1. G is P4-free; 2. if G has more than one vertex, then G is the disjoint union or the join of smaller P4-free graphs.

Since the one-vertex graph is clearly strongly equistable, we obtain the following result from Theorems 14.3.4, 14.3.5 and Lemma 14.3.6, establishing the results of Payan mentioned above. C o r o l l a r y 14.3.7 The P4-free graphs are strongly equistable.

14.3.2

Some Strongly Equistable Perfect Graphs

We now present a useful necessary condition for a graph to be equistable. T h e o r e m 14.3.8 Let G = (V,E) be an equistable graph with an induced P4(vi, v2, v3, v4). Then every maximal stable set of G containing the two ends Vl, v4 must have a vertex adjacent to the two middle vertices v2, v3. P r o o f . Let {Vl, v4} 0 S be a maximal stable set ( 0 indicates disjoint union). Put P = {x e S : xv2 ~ E}, Q = {x e S : xv3 ~ E} and assume that, if possible, P U Q = S. Let us extend the stable sets {V2, V4} 0 P and {Vl, V3) 0 Q to maximal stable sets {v:, v4} @P @P' and {Vl, va} 0 Q 9 Q', respectively. Then P' N Q' = 0, because if v C P' N Q', then {Vl, v4, v} 0 S is a stable set properly containing the maximal stable set {Vl, v4} @ S. Now pick an w C fl, such that w ( X ) = 1 only if X is a maximal stable set of G. We have W(Vx) + w(v4) + w(S) - 1

Pseudothreshold and Equistable Graphs

360

02(V2) -~- 02(?24) -~- 02(P) --[-02(Pt) -- 1 02(Vl) + 02(?23) -~- 02(Q) -[--02(Qt) -- 1, and by adding the last two equations and subtracting the first, we obtain

~(v~) + ~ ( ~ ) + ~(P n Q) + ~(P') + ~(q') = ~ However, {v2, v3} 0 ( P n Q) 0 P' 0 Q' is a non-stable set (not multiset), contradicting the choice of ~. .. We call the graph shown in Figure 14.2 an "A". Figure 14.2: An "A" graph.

/k T h e o r e m 14.3.9 Let G be an A-free graph (for example a triangle-free graph, a bipartite graph etc.). Then the following conditions are equivalent:

1. G is equistable; 2. G is strongly equistable; 3. G is P4-free. P r o o f . This follows from Corollary 14.3.7 and Theorem 14.3.8. 9 We now present a useful sufficient condition for a graph to be strongly equistable. T h e o r e m 1 4 . 3 . 1 0 If S is a maximal stable set of a graph G = (V, E) such that for every pair u, v C V - S, (14.3)

then G is strongly equistable.

14.3

Equistable Graphs

361

P r o o f . The following four facts are used in the proof. F a c t 14.3.11 Let w" V ---, R+o . Then w E .A if and only if w(S)w(v)

1 -

w(Ns(v))

(14.4) for all v C V -

S.

(14.5)

For the "only if", (14.4) is trivial, and (14.5) is true because for every v E V - S , the set O - { v } U N s ( v ) i s maximal stable. Indeed, Q is clearly stable. Moreover, Q is maximal stable, because we cannot add any vertex of Ms(v) to Q, and we cannot add any vertex z E V - S to Q, since such x would satisfy N s ( z ) N N s ( v ) = e , and also N s ( x ) n N s ( v ) = e by (14.3), and so M s ( x ) = e , contradicting the maximality of S. For the "if" part, if co satisfies (14.4), (14.5) and X is any maximal stable set, then co(X) = co(X n S) + w(X - S). Now co(X - S') = ~ v e x _ s W ( N s ( v ) ) by (14.5). The neighborhoods Ms(v) for v E XS partition S " - X because, first, they are disjoint by (14.3), and, second, if there were a vertex y in ( S - X ) - U v e x - s N s ( v ) , then X U { y } would be stable, contrary to the maximality of X. Therefore 2 v e X - S W ( N s ( v ) ) = co(S - X) and w(X) = co(S) = 1. Hence w E A. F a c t 1 4 . 3 . 1 2 For every T C_ V (not necessarily T E T ) and w E ~4, w(T) -

Indeed, by Fact 14.3.11, w(T) - w(T N S) +

E

vCT-S

sES

where Xzns is the characteristic function of T N S. Using N r - s ( S ) -- NT(S), we see that XTAS(8) _qt_[NT-s(S)[ - INT[s]I . F a c t 1 4 . 3 . 1 3 If INT[s]I - 0 for all s E S, then T - 0 . Indeed, S N T - O and no vertex of S has a neighbor in T. Hence T - 0 , for otherwise S is not maximal stable. F a c t 1 4 . 3 . 1 4 If INT[s]I - 1 for all s C S, then T is a maximal stable set. Indeed, T is stable for the following reason: if two vertices of T - S are adjacent, then by (14.3) they have a common neighbor s E S, and so INT[S]I >_ 2; if x E T N S and y E T - S are adjacent, then INr[z]l >_ 2. r is maximal stable,

Pseudothreshold and Equistable Graphs

362

because we cannot add to T any vertex of S (such vertex is already in T or is a neighbor of some vertex of T), and we cannot add to T any vertex v E V - S (if {v} U (T - S) is stable, then by (14.3), N s ( v ) N N s ( T - S) - o , and so N s ( v ) C_ T N S by the condition of Fact 14.3.14, and moreover N s ( v ) 7/= rg by the maximality of S, so N s ( v ) meets T). To conclude the proof of Theorem 14.3.10, let T C T and CT _< 1 be given, and assume that, if possible, w(T) - Cr for all w C ~4. Then by Fact 14.3.12, for all w C ~4, -

(14.6)

sES

Therefore by Fact 14.3.11, Equation (14.6) is a consequence of (14.4) and (14.5), and hence a linear combination of them. In the matrix of coefficients of Equations (14.4) and (14.5), the columns of V - S contain an identity matrix in the rows of Equation (14.5) and 0 in the row of Equation (14.4). Since these columns also contain 0 in Equation (14.6), none of Equations (14.5) can contribute to the linear combination. Thus Equation (14.6) is a multiple of Equation (14.4), that is to say INr[s]l = cr for each s E S. Since Cr _< 1, it follows that Cr = 0 or Cr = 1. Therefore by Facts 14.3.13 and 14.3.14, T = O or T is a maximal stable set, contradicting T C T. .. By applying Theorem 14.3.8 and Theorem 14.3.10, we obtain necessary and sufficient conditions for some classes of graphs to be equistable.

Definition 14.3.15 A graph is called a b l o c k g r a p h if each of its blocks is a complete graph. T h e o r e m 14.3.16 Let G be a block graph. are equivalent:

Then the following conditions

1. G is equistable; 2. G is strongly equistable; 3. each block of G has a vertex that is not a cut vertex. P r o o f . 3) =~ 2): Let the blocks of G be B 1 , . . . , Bk, and let si be a non-cut vertex in Bi for i = 1 , . . . , k. Note that si belongs only to Bi, and N[si] = Bi. This implies that the set S' = { S l , . . . , sk} is maximal stable. Moreover, for u , v C V - S' we have: uv C E ~ u and v belong to the same Bi for some

14.3

Equistable Graphs

363

i 4---5, u and v have a common neighbor in S. Therefore Condition 2 follows from Theorem 14.3.10. 2) =~ 1)" This follows from Lemma 14.3.3. 1) ~ 3)" Assume that, if possible, G has a block B - {c~,..., ck} consisting entirely of cut vertices. Then k >_ 2. For each 1 _< i _< k, choose a neighbor di of ci such that di ~ B. If didj E E for some i r j, then there is a cycle dicicjdj meeting B, and hence contained in B, contradicting the choice of di. Therefore {all,..., dk} is a stable set. Extend it to a maximal stable set S. This S cannot have a common neighbor of Cl and c2, because S N B - 0. Thus the existence of P4(dl, Cl, c2, d2) contradicts Theorem 14.3.8. .. Recall that G(K, S) denotes a split graph G with split partition (K, S). Clearly every split graph can be written in the form G(K, S), where S is a maximal stable set (if it is not, we can move one vertex from K to S and make S maximal stable).

T h e o r e m 14.3.17 Let G(K, S) be a split graph with S a maximal stable set. Then the following conditions are equivalent:

1. G is equistable; 2. G is strongly equistable; 3. Every two vertices of K have a common neighbor in S. P r o o f . 3) =v 2): This follows from Theorem 14.3.10. 2) =v 1): This follows from Lemma 14.3.3. 1) =v 3): Let b and c be distinct vertices of K with no common neighbors in S. Vertices a C Ns(b) and d C Ns(c) exist by the maximality of S, and hence there is an induced P4(a, b, c,d). However, S has no vertex adjacent to both b and c, contradicting Theorem 14.3.8. 9

14.3.3

Pseudothreshold and Outerplanar Graphs

Recall the characterization of pseudothreshold graphs in Theorem 14.2.7. In that theorem we can require that GIRl has no dominating vertex, since such vertex can be put in P. We denote G with the partition as in that theorem by G(P, Q, R), always requiring that GIRl has no dominating vertex. L e m m a 14.3.18 A graph G is a P4-free graph without a dominating vertex

or stable set of size 3 i f and only if G -

W1 (~ "" | Wt for some t >_ 1,

364

Pseudothreshold and Equistable Graphs

where each W~ is of the form W~ - K~ U K[ and I(~, I([ are disjoint non-empty complete graphs. P r o o f . This follows easily from Lemma 14.3.6. 9 Note that G(P, Q, o) is a split graph, whose equistability is characterized by Theorem 14.3.17. The next theorem characterizes the equistability of the other pseudothreshold graphs. T h e o r e m 14.3.19 For a pseudothreshold graph G(P, Q, R) with R ~r 0, the following conditions are equivalent"

1. G is equistable; 2. G is strongly equistable; 3. G[R] is P4-free. P r o o f . 2) =~ 1): This follows from Lemma 14.3.3. 1) ~ 3): If G[R] contains an induced P4(a, b, c, d), then S = {a, d} U Q is a maximal stable set, since R has no stable set of size 3, and every vertex of P is adjacent to a. However, S has no vertex adjacent to both b and c, contradicting Theorem 14.3.8. 3) ~ 2): Lemma 14.3.18 describes the structure of G[R], and we adopt the same notation. We also put I ' ( i - {kil, ki2,...} and I ( [ - {k~l , k~2,...} for i = 1 , . . . , t. Clearly the maximal stable sets of G are the sets of the following two forms: {p} U NQ (p) with p C P, and { kij, k~t } U Q with kij E I(~, k~t E I'(~, i = 1 , . . . , t. Therefore aJ C .A if and only if aJ is non-negative and satisfies the following equations: m

w(p) + co(NQ (p) ) w(kij) + a3(k~t) + aJ(Q) -

1 1

Vp E P

(14.7)

Vi,j, 1.

(14.8)

It follows from (14.8) that aJ is constant on each / T i and on each K~. Therefore (14.7) and (14.8) are equivalent to the following system of linearly independent equations:

w(p) + w(-NQ(p))

-

1

Vp E P

32(~il) + (-U(]~I) -t- co(Q)

w(kij)--w(kil)

--

1

Vi

(14.10)

--

0

Vj _> 2, Vi

(14.11)

-

0

v j _> 2, vi.

( 4.12)

(14.9)

14.3

Equistable Graphs

365

To prove that G is strongly equistable, let T E T and c _< 1 be given, and assume that, if possible, w(T) - c for all w E ~4. Then the equation aJ(T) - c is a consequence of Equations (14.9)-(14.12), and therefore a linear combination of them. Therefore the equation w ( T ) - c takes the form t

E (~p[O,.)(p)-~-(.u(NQ(p))] Jr E ")/i[('u(~il ) + ("g(]r pEP t

-t- ~

+ (,M(Q)]

i=1 t

~_~ aij[w(kij) - co(kii)] + ~ ~ a'ij[w(k~j ) - w(k~l)]

i = 1 j_>2

(14.13)

i = 1 j_>2 t

pEP

i=1

' Since w(p) appears only once in the sysfor some multipliers 5p, ~/i, aij, aij. tern (14.9)-(14.12), and with a coefficient of 1, we have 5p = 1 if p E P N T, 5p=0ifpE P-T. Similarly for j _ > 2 , aij = 1 ifkij E K i n T , aij = 0 i f k~j E K~ - T, a~j - 1 if k~j C K[ N T, a}j - 0 if k~j E K~ - T. Therefore, in order for the coefficient of w(k~l) in the linear combination (14.13) to assume its correct value (1 if k~l E T, 0 if not), 3'~ must be equal to ]Is n T I. Similarly 7i must also be equal to IK[ N T I. Thus we can rewrite (14.13) in the following way: t

pEPnT

[~(p) + ~(xq(p))] + E ~ ( Q ) i=1 t

+ ~(/q n T) + ~(/(~

n T ) --

IP n TI + ' ~ ~,

(14.14)

i=1

where

~ - IIq n T I Note that

IK[ n TI.

(14.15)

t 1 ~/i - 71R n T I.

(14.16)

i=1

Therefore 1 IP N T] + ~]R n T] - c.

(14.17)

Further, in order for w(q), q E Q, to assume its correct coefficient in (14.14) (lifqEONT, 0ifqEO-T),wemusthave 1

INpnr(q)l + -~lR A T

]e 1 I - ~[ 0

for q E Q N T forqEQ-T.

(14.18)

Pseudothreshold and Equistable Graphs

366

From (14.15)-(14.17), c is an integer, and from c _< 1 we obtain c - 0 or C--1. C a s e 1" c - 0. Then by (14.17) P A T - R A T - 0 . Then the left-hand side of (14.18) is 0, which implies Q n T - O. Therefore T - O, contradicting TET. C a s e 2" c - 1. From (14.17), either P n T - {p} and R n T - O, or else P A T - O and R A T - {kij, k~l } (by (14.15)). In the first possibility, (14.18) can be rewritten as [I.N{p}(q).-

1 forqCQNT 0 for q E Q - T,

which means that Q N T - NQ(p). Thus T - { p } U N Q ( p ) , which is a m a x i m a l stable set, contradicting T C T. In the second possibility, the left-hand side of (14.18) is 1, which implies Q - T - o . Thus T - {kij, k~l} U Q, which is a m a x i m a l stable set, contradicting T E T. 9 D e f i n i t i o n 1 4 . 3 . 2 0 A planar graph is said to be o u t e r p l a n a r if it can be embedded in the plane so that every vertex lies on the boundary of the infinite region. Obviously, a graph is outerplanar if and only if each of its blocks is outerplanar. The following l e m m a is used to prove Theorem 14.3.23 below. L e m m a 1 4 . 3 . 2 1 Every 2-connected outerplanar graph with more than two vertices is a Hamiltonian graph. P r o o f . E m b e d the 2-connected outerplanar graph in the plane so t h a t all vertices lie on the b o u n d a r y B of the infinite region. Let C - V l , . . . , v k be ~ cycle all of whose edges lie on B. If C is not Hamiltonian, there is a vertex Ul ~ C adjacent to some vertex of C, say Vl. Since Ul is on B, Ul is in the exterior of C. Let v~ be any vertex of C, r :/: 1. By 2connectivity there is a cycle C' - V l t t l U 2 . . . containing the edge VlU 1 and the vertex v~. Let rn be the smallest index such that u,~ coincides with a vertex on C, say vj. Note that U l , . . . , u m - 1 are in the exterior of C and t h a t 2 _< j _< k. If j - 2, then V l , U l , . . . , u m , v 3 , . . . , v k is a cycle enclosing the interior of the edge VlV2, so that edge cannot lie on B, contradicting the choice of C. A similar contradiction occurs if j - k. If 2 < j < k, then either the cycle V l , U l , . . . , u m , v j + l , . . . , v k encloses v2 or else the cycle Vl, U l , . . . , Urn, V j - I , . . . , ?22 encloses vk, so either v2 or vk is not on B, a contradiction. 9

14.3

Equistable Graphs

367

D e f i n i t i o n 14.3.22 A flower Fn

-

Fn(?tl,...,?.tn;Yl,...,Vn)

is a graph consisting of an even cycle UlVlU2V2...UnVn whose chords are precisely the short chords UxU2, U2U3, . . . , U~Ul. Figure 1~,.3 illustrates the flowers F2 and F6.

Figure 14.3: Flowers.

V2

vx

U2

Y3 ..

U

V4 v

U

2

.Yl

6

,., V6

Ul

?,)2

~

~

U

2/ V5

F2

F6

T h e o r e m 1 4 . 3 . 2 3 Let G be an outerplanar graph. Then the following conditions are equivalent:

1. G is equistable; 2. G is strongly equistable; o

(a) each block of G is a flower, a square, or a complete graph on at most 3 vertices; (b) in a block of G that is a flower or a square, no vertex of degree 2 in the block is a cut vertex:

368

Pseudothreshold and Equistable Graphs

(c) in a block of G that is a complete graph, not all vertices are cut vertices.

P r o o f : 3) =~ 2)" By Theorem 14.3.4 we may assume that G is connected. If a block of G is a square, then by Condition 3b, this block is the entire G, and it is strongly equistable by being P4-free. If the block is a single vertex, then it is the entire G by connectivity and the same conclusion holds. We may therefore assume in addition that each block of G is a flower or a complete graph on two or three vertices. We construct a set S of vertices as follows" from each block that is a flower, put in 5' all the vertices of degree 2 in the block; from each other block put in S exactly one vertex that is not a cut vertex. Clearly the vertices of S in each block from a maximal stable set in the block, and further, by Condition 3b, S contains no cut vertex of G. Therefore S is a maximal stable set of G. It is easy to verify that S satisfies the condition of Theorem 14.3.10, and hence G is strongly equistable. 2) =~ 1)" This follows from Lemma 14.3.3. 1) =~ 3)" First we prove Condition 3a. Every block B of G is outerplanar. If B has at most four vertices, then B is clearly F2, a square, or a complete graph on at most three vertices. So assume that B has at least five vertices. By Lemma 14.3.21 B has a Hamiltonian cycle C - Zl . . . zk. We assume that G is embedded in the plane so that every vertex lies on the boundary of the infinite region. Therefore every chord of C lies in the interior of C. We now show the following two facts" F a c t 1 No f o u r consecutive vertices along C induce a P4. P r o o f . Assume without loss of generality that C has a P4(Xl, x2, z3, x4). By Theorem 14.3.8, G has a vertex v adjacent to both x2 and x3, but to neither Xl nor x4. Clearly v is in B, and hence v - xi for some i _> 6. Note that zi is the only vertex of G adjacent to both x2 and x3 by planarity. If x4xi-1 ~ E , extend { X l , X 4 ~ X i _ I } t o & maximal stable set S, and S will violate Theorem 14.3.8 with respect to P4(xl,x2, x3, x4). Therefore x 4 x i - , E E. See Figure 14.4. Now consider P4(x2, xi, xi-1, x4). The only vertex that can possibly be adjacent to both xi and xi-1 is x3, which is adjacent to x2, a contradiction to Theorem 14.3.8. This proves Fact 1. F a c t 2 Between every four consecutive vertices along C, there is exactly one chord, and this chord joins two vertices at distance 2 along C. P r o o f . Consider for example Xl,X2, X3, X4. By Fact 1, there is a chord between them, and it is enough to show that XlX4 is not a chord. If XlX 4 is & chord, then between x2, z3, x4, xs or between xk, x l , x 2 , x3 there is no chord,

14.3

Equistable Graphs

369

Figure 14.4: Illustrating the proof of Fact 1. Dashed lines indicate nonedges. X4 X3

9

X 2

Xl

Xi-1

contradicting Fact 1. This proves Fact 2. From Fact 2, it follows that k is even and B has a spanning flower fn(tll,

. . . , tln;

Yl,.

. . , Yn),

where n - k/2 >_ 3. It remains to show that the cycle U l . . . t t n is chordless in B. Assume for example that U l U i C E , where 2 < i < n. Then the P4(vl, Ul, u~, v~) and any maximal stable set S containing { V l , . . . , v~} contradict Theorem 14.3.8. This proves Condition 3a. To prove Condition 3b, consider first a block B that is a flower fn(tll,

. . . , tln;

Yl,

. . . ~ ~)n)

and assume for example that Vl is a cut vertex. Let x be any neighbor of t~1 outside B. Then the P4(x, vl, u2, v2) and any maximal stable set S containing {x, v2, v~} contradict Theorem 14.3.8. Similarly, consider a block B that is a square abcd, and assume for example that a is a cut vertex. Let x be any neighbor of x outside B. There is no vertex adjacent to both a and b, yet there is P4(x, a, b, c), again contradicting Theorem 14.3.8. The proof of Condition 3c is similar to the proof of Condition 3b. 9

370

14.3.4

Pseudothreshold and Equistable Graphs

Equistability, Strong Equistability, and Substitution

In Subsection 14.3.2 we saw among other things that for split graphs and block graphs, equistability is equivalent to strong equistability. Below we show that more generally, the same is true for all perfect graphs. Recall the definitions of .A, S, and T at the beginning of Subsection 14.3.1. Theorem

14.3.24

1. If G is a perfect graph and .A 7~ 0, then .A has an integer-valued weight function w. 2. A(G) has an integer-valued weight function if and only if G has a clique meeting all maximal stable sets. 3. If G is equistable, and if for every T C T there exists some co E A such that w(T) is not in the open interval (0,1), then G is strongly equistable. ~. In particular, if G is an equistable perfect graph, and more generally if G is an equistable graph with a clique meeting all maximal stable sets, then G is strongly equistable. P r o o f . 1" Consider the clique matrix M of the complementary graph G, whose rows are the incidence vectors of the maximal stable sets of G. By Ldvasz's Perfect Graph Theorem [Lov72], G is perfect. Hence by Chvs Theorem [Chv75], the extreme points of the polytope defined by Ma~ _< 1, aJ _> 0 are integer-valued. Note that A is a face of this polytope, and since by assumption A r 0 , .4 contains an extreme point of the polytope. 2: Let co be an integer-valued weight function in .4. Since every maximal stable set S satisfies aJ(S) - 1, S must contain exactly one vertex x with w(x) - 1. Let C be the set of all vertices x such that aJ(x) - 1. Clearly C meets all maximal stable sets. Moreover, C is a clique, otherwise some two vertices of C can be extended to a maximal stable set S with a~(g) >_ 2. Conversely, if C is a clique meeting all maximal stable sets, let aJ be the characteristic vector of C. Clearly w is an integer-valued weight function in

A. 3: Since G is equistable, there exists a non-negative weight function a / s u c h that J ( S ) - 1 if and only if S E $. To show that G is strongly equistable,

14.3

Equistable Graphs

371

let T E 7 " a n d c_< 1 be given. B o t h w a n d w ~ are in A. I f c < 0 there is nothing to prove. If c - 0, then co(T) r 0 since T r e (any vertex x C T can be extended to a set 5' C $; if w'(x) - O, then co'(S- x) - 1 even though S-x ~,5'). If0 < c < 1, then co(T) r c. If c - 1, t h e n a / ( T ) r csince

T s. If G is equistable, then ~4 -~ ~. If in addition G is perfect, then by Parts 1 and 2 above, G has a clique meeting all maximal stable sets. If G is equistable and has such a clique, then by Parts 2 and 3, G is strongly equistable. [] Note that the flower Fs is a non-perfect equistable graph having a clique meeting all maximal stable sets. The strongly perfect graphs [BC84] are the graphs in which every induced subgraph has a stable set meeting all maximal cliques. Their complements then have a clique meeting all maximal stable sets. Moreover, the strongly perfect graphs are perfect [BC84], and so are their complements. A clique of the form N[v] for some vertex v is said to be a simplicial clique. Clearly a simplicial clique meets all maximal stable sets. We now characterize the graphs satisfying the sufficient condition of Theorem 14.3.10 using simplicial cliques. 4:

T h e o r e m 14.3.25 For every graph G - (V, E), the following conditions are equivalent:

1. G has a maximal stable set S such that for every u, v C V - S, uv C E if and only if Ns(u) n N s ( v ) # ~; 2. every edge of G is in a simplicial clique. P r o o f : 1) ~ 2)" For every s C S, N[s] is a clique by the "if" part of Condition 1. These simplicial cliques cover E by the "only if" part of Condition 1. 2) =~ 1)" Let N[v~],... ,N[vk] be all the distinct simplicial cliques, and let S - {Vl,..., vk}. The set S' is stable, since if vivj C E, then N[vi] C_ N[vj] and N[vj] C_ N[vi], contradicting the distinctness of N[vi] and N[vj]. Further, S is maximal stable, because every vertex belongs to some simplicial clique by Condition 2. Again by Condition 2, if u, v C V - 5' and uv ~ E, then u and v cannot have a common neighbor vi, because N[vi] is a clique.

C o r o l l a r y 14.3.26 If G - (V, E) is a graph satisfying Condition 1 of Theorem 14.3.25, then for each x C V there exists some w C A with co(x) - 1.

~

~

~

C~

~

9 r bo

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Ch

~'~ ~ "

~"

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.'7-"

~

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9

C

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O ~ t O m" ~

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CbC]

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b.a

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('1)

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="

Fat]

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9

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~ ~ ~

---'

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=-.

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t,,a

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(1)

~.~.

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14.3

Equistable Graphs

373

P r o o f : For each COl E r 022 E r define co" V(H) ~ R + as in Lemma 14.3.28. Then ~o C A(H). Given any T C T ( H ) and c < 1, let T1 - T fl V(G1), T2 - T N V(G2). Then co(T) - col (T1) -t- col (x)co2(V2).

If w2(T2) is not constant for all ~2 C ~4(G2), choose aJi - a~, and then aJ(T) is not constant for all ~o C ~4(H). If aJ2(T2) is a constant b > 1 for all co2 C ~4(G2), then again takew - w~, and then c0(T) - aJl(T1)+w2(T2) >_ b > 1 >_ c. In the remaining case aJ2(T2) is a constant b _< 1 for all w2 C .A(G2). Since G2 is strongly equistable, T2 - O or T2 C 8(G2). If T2 - 0 , then b - 0, T1 C T(G1), and w(T) - COl(T1); and since G1 is strongly equistable, we can choose col so that COl(T1) -r c. If T2 C 8(G2), then b - 1, T1 t2 {x} C T(G1), and aJ(T) - wl(T1 U {x}); again we can choose aJ1 so that aJl(T U {x}) -~ c, since G1 is strongly equistable. 9 C o r o l l a r y 14.3.30 Assume that GI satisfies the sufficient condition of Theorem 1~.3.10 and G2 is strongly equistable. Then Gl(x ~ G2) is strongly

equistable for each x C V(G1). P r o o f : This is a consequence of Corollary 14.3.26 and Theorems 14.3.10 and 14.3.29. C o r o l l a r y 14.3.31 Let C be the class of all strongly equistable graphs G

such that for each x C V(G), there exists a weight function co E .A(G) with co(x)- 1. Then C is closed under substitution. P r o o f : This follows from Theorem 14.3.29 by defining the weight function as in L e m m a 14.3.28. 9 The class C contains all the equistable graphs that we encountered in the last two subsections: in every one of these graphs, every vertex belongs to a clique meeting all the maximal stable sets. An interesting class of graphs from our point of view here is the class of co-Meyniel graphs, the complements of the Meyniel graphs. A Meyniel graph [Mey76] is a graph in which each odd cycle of length at least 5 has two or more chords. L e m m a 14.3.32

substitution.

The Meyniel and the co-Meyniel graphs are closed under

374

Pseudothreshold and Equistable Graphs

P r o o f i Let G1 and G2 be Meyniel, and assume that, if possible, H = Gl(x --+ G2) has an odd cycle C of length at least 5 with at most one chord. Then C has at least two vertices of G2, for otherwise G1 is not Meyniel. Also, C has at least two vertices of G1, for otherwise G2 is not Meyniel or C has a dominating vertex. Therefore the vertices of C alternate at least twice between G1 and G2, i.e., C has non-incident edges ab and cd with a, c in G1 and b, d in G2. It follows that ad and bc are edges of H. One of them, say bc, is an edge of C and the other, say ad, is a chord of C, for otherwise C has length 4 or C has two chords. The sequence of vertices along C from d to a in the direction abcd begins with the vertex d of G2 and ends with the vertex a of G1. This sequence has at least three vertices. Therefore it has adjacent vertices e in G1 and f in G2 such that at least one of them is distinct from a, b, c, d. If e is distinct from a, b, c, d, then eb is a second chord of C, and if f is distinct from a, b, c, d, then fc is a second chord of C, a contradiction. This proves that the Meyniel graphs are closed under substitution. The co-Meyniel graphs are also closed, as Gl(x --+ G2) = Gl(x ---+ G2). I The following is a result of Hokng [Hok87] (See also [Her90]). L e m m a 14.3.33 A graph is Meyniel if and only if, for each induced subgraph, each vertex belongs to a stable set meeting all maximal cliques. We can now prove the following about the co-Meyniel graphs: T h e o r e m 14.3.34 The equistable co-Meyniel graphs are strongly equistable, and are closed under substitution. P r o o f i By Lemma 14.3.33, if G is co-Meyniel, then each vertex of G belongs to a clique meeting all maximal stable sets. If in addition G is equistable, then G is strongly equistable by the "more generally" clause of Part 4 of Theorem 14.3.24. (In fact, the Meyniel graphs are known to be strongly perfect [Rav82] and thus perfect, and hence G is perfect. Therefore the strong equistability of G follows also from the more particular clause of Part 4 of Theorem 14.3.24.) Further, by the proof of Part 2 of Theorem 14.3.24, for each vertex x of G there exists an aJ E ,4 with a3(x) = 1. Therefore, if both G1 and G2 are co-Meyniel and equistable, they are strongly equistable, and GI(X ~ G2) is strongly equistable by Theorem 14.3.29 and co-Meyniel by Lemma 14.3.32. 9 We conclude this section with the following two c o n j e c t u r e s .

1. The strongly equistable graphs are closed under substitution. 2. All equistable graphs are strongly equistable.

Chapter 15 Threshold Weights and Measures 15.1

Introduction

In this chapter we consider three ways, in addition to the threshold dimension, to gauge by how much a given graph differs from being a threshold graph. The threshold weight of a graph was introduced by Wang and Williams [WW91] and also considered by Mahadev and Peled [MP91]. We assign a non-negative weight to each vertex and each edge of the graph, and define the score of a set of vertices as the sum of the weights of its vertices and of the edges induced by these vertices. We wish to choose the weights so that the score of each stable set is smaller than the score of each non-stable set, and the sum of the edge weights is minimal. Under a suitable normalization, the minimum value is the threshold weight of the graph, and by this definition it is zero if and only if the graph is a threshold graph. We discuss the threshold weight in Section 15.2. The threshold measure of a graph was introduced by Peled and Simeone [PS89a]. Here one assigns non-negative weights just to the vertices, as in the definition of a threshold graph, and considers that the score of a set of vertices is the sum of the weights of its vertices. In general we cannot make the score of each stable set smaller than the score of each non-stable set. Instead, we normalize the weights by requiring that the smallest score of a non-stable set be 1, and minimize the largest score of a stable set. The minimum value, which is smaller than 1 if and only if the graph is threshold, is the threshold measure of the graph. We discuss the threshold measure in Section 15.3. The threshold gap of a graph was 375

Threshold Weights and Measures

376

introduced by Hammer, Ibaraki and Simeone [HIS78, HIS81]. It is actually a property of the degree sequence of the graph, namely half of the smallest distance in the L 1 n o r m between the degree sequence and any threshold sequence of the same length. Thus it vanishes if and only if the given degree sequence is a threshold sequence. Arikati and Peled lAP94] introduced the majorization gap of the degree sequence as the minimum number of reverse unit transformations required to transform it into a threshold sequence. It turns out that the threshold gap is the same as the majorization gap. We discuss the threshold and majorization gaps in Section 15.4.

15.2

Threshold Weights

In this section, based on the work of Wang and Williams [WW91] and Mahadev and Peled [MP91], we define the threshold weight of a graph and discuss its properties. As we mentioned in the Introduction, to define the threshold weight of a graph G = (V, E), we assign a weight w~ >_ 0 to each v e r t e x i E V and a w e i g h t we >_ 0 to each edge e E E. The score o f a s e t S C_ V of vertices is computed as ~ i e v wi + ~eeE(S) we, where E(S) is the set of edges with both endpoints in S. We require that the score of each stable set be smaller than the score of each non-stable set. This requirement has the normalized form in which the score of each stable set is at most t - 1 and the score of each non-stable set is at least t for some threshold parameter t. By the non-negativity of the weights, it is sufficient to require this only for maximal stable sets and minimal non-stable sets (two adjacent vertices). We wish to minimize the sum of the edge weights, and its minimum value is defined as the threshold weight of G. We are thus led to the following formal definition.

Definition 15.2.1 The threshold weight W(G) of a graph G = (V, E) is the optimum value of the following linear programming problem. rain

~

w~

eEE

s.t.

~ wi ~ t - 1

for each maximal stable set S C_ V

iES

wi + wj + we >_ t for each edge e = ij E E

w>O.

(15.1)

15.2

Threshold Weights

377

An assignment (w; t) is called a feasible ( o p t i m a l ) t h r e s h o l d a s s i g n m e n t when it is feasible (optimal) for the linear programming problem (15.1). In general (15.1) involves an exponential number of constraints, since there are as many maximal stable sets. The graph G satisfies W(G) - 0 if and only if G is a threshold graph, since W(G) - 0 means that all the edge weights are zero. The assignment of zero vertex weights and unit edge weights is always feasible with t - 1, and this feasible threshold assignment has an objective function value equal to IEI, the number of edges. Therefore 0 < w ( a ) _< ]El for every graph G, where one extreme, 0, characterizes the threshold graphs. We call a graph G - (V, E) at the other extreme, i.e., one satisfying W ( G ) - IEI, ~ h~avy graph. In this sense, the heavy graphs are the most non-threshold graphs (of course, if E - O, then G is both a threshold graph and a heavy graph, but this is the only such case). A feasible threshold assignment remains feasible when restricted to an induced subgraph, and hence W ( H ) <_ W(G) for every induced subgraph H of G. Moreover, if C 1 , . . . , Ck are the connected components of G, then for the same reason ~ =k1 W(C~) <_ W(G). Recall that two edges are said to be in conflict if their endpoints induce an alternating 4-cycle. L e m m a 15.2.2 Let e and f be edges in conflict in G. Then every feasible threshold assignment (w; t) for G satisfies w~ + wf >_ 2. Consequently, if G is a non-threshold graph, then W(G) > 2. P r o o f . Let e - p q and f wp + wT _< t - 1;

rs, where pr and qs are nonedges. Then

wq+ws
wp + wq + w~ > t;

w~ + w~ + wf >_ t

and therefore

w~ + w f > 2 t - wp - w q - w ~ - w ~ > _ 2 .

D e f i n i t i o n 15.2.3 Let (w; t) be a feasible threshold assignment of a graph G. The edges e satisfying w~ < 1 are called the light edges of (w; t). Part (a) of Figure 15.1 illustrates a feasible threshold assignment whose light edges do not form a threshold graph, although the graph induced by their endpoints is threshold. Part (b) illustrates an optimal threshold assignment

378

Threshold Weights and Measures

(its optimality follows from Lemma 15.2.2) whose light edges form a threshold graph, but the graph induced by their endpoints is not threshold. However, the following lemma shows that the light edges of an optimal threshold assignment form a threshold subgraph (not necessarily an induced one). Figure 15.1" (a) a feasible threshold assignment whose light edges do not form a threshold graph. (b) an optimal threshold assignment the endpoints of whose light edges do not induce a threshold graph. t-4 t-5

1

1

0 A

0

4

2

0

A

A

3

w

./

2

x_

w

0

3

w

2

0

2

(b) L e m m a 15.2.4 Let ( w ; t ) be an optimal threshold assignment of a graph G - (V,E), and let L be the set of light edges of ( w ; t ) . Then (V,L) is a threshold graph. P r o o f . If (V, L) is not a threshold graph, it has an alternating 4-cycle with edges pq and rs and nonedges pr and qs. The nonedges pr and qs may be nonedges of G, or edges of G with weight at least 1. If pr is a nonedge of G, we have wp + w~ _< t - 1 by the constraints of (15.1), and if pr is an edge with wp~ _> 1, we have the same conclusion, otherwise wp~ could be decreased below 1, contradicting the optimality of the threshold assignment. Similarly wq + w~ _< t - 1. On the other hand, wp + wq + Wpq ~ t together with wpq < 1 imply that w p + w q > t - 1, and similarlyw~ + w ~ > t - 1. These four inequalities are inconsistent. ..

15.2

T h r e s h o l d Weights

379

The following result follows directly from the constraints (15.1). It is very useful in proving that certain graphs possessing symmetry properties are heavy. L e m m a 15.2.5 Let ab be an edge and cd a nonedge of a graph G. Every feasible threshold assignment (w; t) for G satisfying wc + Wd > Wa + Wb must satisfy Wab >_ 1. In particular, if e -- xy is an edge, zy is a none@e, and Wz > wx, then we > 1. The following corollary is an example of applying Lemma 15.2.5. C o r o l l a r y 15.2.6 All cycles of length 4 or more and their complements are heavy. Proof. Let G be the k-cycle Ck. Consider any optimal threshold assignment (w;t) of G. Each of the k cyclic permutations of the vertex weights of w around the cycle accompanied by a corresponding permutation of the edge weights is again an optimal threshold assignment. The average (w*; t) of these k optimal threshold assignments is yet another optimal threshold assignment (this follows from the convexity of the set of optimal threshold assignments). But (w*;t) enjoys the property that its vertex weights are constant. Let e be an arbitrary edge of G and let f be a nonedge adjacent to e, which exists since k > 4. By applying Lemma 15.2.5 to (w*; t), we see that w~ > 1. Thus G is heavy. The proof for the complements of cycles is similar. 9 The heavy graphs cannot be characterized by forbidden configurations, since they are not closed under taking induced subgraphs. In fact, every graph with at least one edge has an induced threshold subgraph with one edge, which is not heavy. The following theorem gives a convenient characterization of heavy graphs in terms of the optimum value of a linear programming problem having fewer variables and constraints than (15.1). T h e o r e m 15.2.7 A graph G [El

>

max

(V, E) is heavy if and only if

E dizi

iEv

s.t.

for each maximal stable set iES

Scv

z>_O,

(15.2)

T h r e s h o l d W e i g h t s and M e a s u r e s

380

where di is the degree of vertex i. Proof. The linear programming dual of (15.1) is max

~-~ ys S

s.t. S

e

Vi C V S ~i

(15.3)

e ~i

y~ _< 1

Ve C E

y>_O. The constraints of (15.3) imply that E s ys <_ IEI, equality holding if and only if y~ = 1 for all e C E. By the definition of heavy graphs and the duality theorem of linear programming, we thus have that G is heavy if and only if the following system of linear inequalities is solvable. ys -

IEI

S

~--~Ys >_ di

VicV

S ~i

y>_O. In other words, again using linear programming duality, if and only if IE]

_> min E Ys

=

max

s

s.t.

~ ys >_di Vi

E dizi iEv

s.t.

~ z i <_ 1 VS

S~i

iES

y_O

z_>O.

A direct proof of Theorem 15.2.7 not using LP duality can be found in [MP91]. We apply Theorem 15.2.7 to show that several types of graphs are heavy. L e m m a 15.2.8 Ps (the path on 5 vertices) is heavy.

15.2

Threshold Weights

381

P r o o f . Denote the vertices as 1 , . . . , 5 along the path. Let z be feasible for (15.2). Then Zl + z3 + zs _< 1, z2 + z4 _< 1, and z3 _< 1. After multiplying these inequalities by 1, 2, and 1, respectively, and adding, we obtain Zl + 2 ( z 2 + z3 + z4) + zs _< 4, which is the condition for a heavy graph in Theorem 15.2.7.

T h e o r e m 1 5 . 2 . 9 If G - ( V , E) is a regular graph and x(G) < G is heavy.

IVI/2,

th~

P r o o f . Let r be the degree of every vertex, and let S 1 , . . . , S X be a partition of V into X - x(G) stable sets. If z is feasible for (15.2), then x k=l iESk

iEV

which is the condition of Theorem 15.2.7 for G to be heavy. T h e o r e m 1 5 . 2 . 1 0 Let G = (V,E) have a proper vertex-coloring in which each color has at least two vertices and all the vertices of the same color have the same degree. Then G is heavy. In particular, the complete k-partite graph K ~ .....~nk is heavy if mj >_ 2 for each j, the perfect matching inK2 is heavy if m >_ 2, and so on.

P r o o f . Let '--q'l,-.., Sk be a partition of V into stable sets, and let Dj be the common degree of the vertices of Sj for each j - 1 , . . . , k. If z is feasible for (15.2), then k

iEV

k

Dy j=l

E Dj iCSj

j=l

d /2 -IEI, iEV

which is the condition of Theorem 15.2.7 for G to be heavy. 9 Recall that N ( x ) denotes the open neighborhood of a vertex x, namely the set of all neighbors of x. The next theorem has a construction that preserves heaviness. T h e o r e m 15.2.11 Let G be a heavy graph, let x be a vertex of G, and let H be obtained from G by adding a new vertex y such that N(y) C N ( x ) in H (y is not adjacent to x). Then H is heavy. In particular, if v is a non-isolated vertex of a heavy graph and we add a new vertex of degree 1 adjacent to v, the graph remains heavy.

Thre s hold Weights and Measures

382

P r o o f . Let di denote the degree of vertex i C V ( H ) in H. The degree of vertex i E V(G) in G is then given by

d'i

if i ~ N(y) _ f di, d i - 1, if i c N(y).

In particular, d2 - d~ >_ dy. Let z be a feasible solution of (15.2) for H. Then the following constitutes a feasible solution of (15.2) for G:

, {

zi -

zi, if/=fix z~ + zu, i f i - x .

By applying the constraints of (15.2) to z on H and Theorem 15.2.7 to z / on G, we obtain

E

di Zi --

iEV(H)

iq~N(y)U{x,y}

<<=

E

iq~N(y)U{x,y}

E

iq~N(y)W{x,y}

~

dizi nt- E dizi + iEN(y) ieN(y)

d',z~ + ~

d~z~ + dyzy

ieN(y)

d'~z, + IN(v)I + d'~"

iEn(y)

d'~z~+ IN(y)I + d'~z"

i~x,y

_< IE(o)I + IN(y)i- IE(H)I. This is the condition for H to be heavy by theorem 15.2.7. T h e o r e m 15.2.12 Every forest having at least two non-singleton connected components is heavy. P r o o f . The forest can be obtained from a matching inK2 with m _> 2, which is heavy by Theorem 15.2.10, by repeatedly adding vertices of degree 1, which preserves heaviness by Theorem 15.2.11. 9 Recall that a star is a tree with at most one vertex of degree larger than 1. It is a threshold graph, and so its threshold weight is zero. A bistar is as tree with exactly two vertices of degree larger than 1; in other words, it is obtained from two disjoint stars by joining their centers. In the following, we show that the triangle-free graphs are heavy, with the exception of stars and bistars plus isolated vertices.

15.2

Threshold Weights

Theorem

15.2.13 I f G -

383

(V, E) is a bistar, then W ( G ) - IE[ - 1.

P r o o f . Let a and b be the vertices of degrees rn + 1 and n + 1, respectively (i.e., a is adjacent to b and to rn additional vertices of degree 1, and similarly for b). To show that W ( G ) = IEI- 1 = rn + n, we exhibit feasible solutions of the dual linear programming problems (15.1) and (15.3) that achieve the same objective function value. The solution of (15.1) has t = 2, w~ = 1 for each edge e with the exception ofw~b = 0, and wi = 0 with the exception of w~ = Wb = 1. The edge constraints are clearly satisfied, and so are the maximal stable set constraints, as is easy to verify for all the maximal stable sets, namely S~ = N ( a ) , Sb = N(b) and So = V - {a, b}. The objective function value is rn + n. The solution of (15.3) has y~ = 1 for each edge e with the exception of W~b = 0, and Yso = n, YSb = rn, YSo = 0. Again it is easy to verify the constraints and the fact that the objective function value is rn + n. .. Isolated vertices are immaterial to the question whether a graph is heavy, so in the next theorem we assume that there are none. T h e o r e m 15.2.14 Let G be a triangle-free graph without isolated vertices, which is not a star or a bistar. Then G is heavy. P r o o f . Let (w; t) be an optimal threshold assignment for G - (V, E), and let L be the set of light edges for it. If L - O, then clearly G is heavy, so assume that L 7~ 0 . By Lemma 15.2.4, the edges of L form a triangle-free threshold graph T, which is necessarily a star. Assume first that L consists of a single edge e. Since G is triangle-free, has no isolated vertices and is not a star or a bistar, it has an edge f in conflict with e, and by Lemma 15.2.2 w~ + w/ _> 2. Every other edge has weight of at least 1, and so the sum of the edge weights is at least ]E I and G is heavy. Now assume that L has more than one edge, and thus T is a star with more than one edge. Let a be the center of T. Since G is triangle-free, one of the cases below must hold. C a s e 1" a is on some cycle C of length 4. Let a, b, c, d be the vertices along C (b and d may or may not be in T), let V l , . . . , vk be the vertices of T other than b, d that are adjacent to b, and let U l , . . . , ut be the other vertices of T. Then a, c, b, d, V l , . . . , vk induce a complete bipartite graph K2,k+2, which is heavy by Theorem 15.2.10. By adding to it U l , . . . , ul we obtain an induced subgraph H of G in which U l , . . . , ul have degree 1, and so H is heavy by Theorem 15.2.11. Since (w; t) remains a feasible threshold assignment when

384

Threshold Weights and Measures

restricted to the induced subgraph H, the sum of the edge weights of the edges of H is at least IE(H)]. Since g contains all the light edges, every other edge has weight at least 1, and therefore the sum of all the edge weights of G is at least IE(G)I, so G is heavy. C a s e 2" a is on some cycle C of length 5, but not on any cycle of length 4. Then C is a chordless cycle and so is heavy by Corollary 15.2.6. The vertices of T not already on C are not adjacent to any vertex of C except for a, since G is triangle-free and a is not on any cycle of length 4. As in Case 1, these vertices and the vertices of C induce a heavy subgraph H of G containing all the light edges, and we proceed as before. C a s e 3" a is not on any cycle of G of length smaller than 6, and there are paths a, p, q and a, r, s having only the vertex a in common. Then q, p, a, r, s induce a chordless path of length 5, which is heavy by Lemma 15.2.8, and the vertices of T not already on it are not adjacent to any vertex of it except for a. As before, these vertices and the vertices of the path induce a heavy subgraph H of G containing all the light edges, so G is heavy. C a s e 4" a is not on any cycle of G of length smaller than 6, and there are no paths as in Case 3. If G has an edge e in a different connected component than the one containing T, then e and any light edge induce a matching 2K2, which is heavy by Theorem 15.2.10. Hence by Theorem 15.2.11 T and e form a heavy induced subgraph H of G containing all the light edges, and we are done as before. Therefore we may assume that G is connected. Since G is not a star or a bistar and we are in Case 4, there is a chordless path of length 4 starting from a. The vertices of this path, together with any vertex of T not already on it, induce a chordless path of length 5, and the argument continues as in Case 3. .. a clique C c_ V of G - (V,E) is called a big clique when IE(C)I > ] E ( V - C)I, where E ( X ) denotes the set of edges with both ends in X C_ V. The following lemma characterizes graphs possessing a big clique. L e m m a 15.2.15 A graph G - (V,E) has a big clique if and only if the following inequalities have an integral solution.

dizi > [El

where di is the degree of i E V

iEV

for each maximal stable set S C V iES

z>O.

(15.4)

15.2

Threshold Weights

385

P r o o f . Let z be integral and satisfy (15.4). Then each zi is 0 or 1, and the set C = {i C V : zi = 1} is a clique. Therefore

[El < ~ & z ~ - ~ & - 21E(C)I + IBI, iEV

iEC

where B is the set of all edges joining C to V - C. Therefore IE(C)I > ] E ( V - C)I , so C is a big clique. Conversely, if C is a big clique of G, its characteristic vector z satisfies (15.4) by a similar argument, tt C o r o l l a r y 1 5 . 2 . 1 6 A graph with a big clique is not heavy. P r o o f . This is immediate from Theorem 15.2.7 and Lemma 15.2.15. 9 It was conjectured that the converse of Corollary 15.2.16 holds. Indeed it holds for triangle-free graphs by Theorem 15.2.14, since stars and bistars have a big clique. However, it fails even f o r / Q - f r e e and Ks-free graphs, as the following examples show. Let Gk,~ denote the graph obtained from the k-cycle Ck by adding r dominating vertices. Then G9,2 is not heavy, and yet has no big clique. To see this, denote the successive vertices of C9 by u o , . . . , us and the two dominating vertices by vo, Vl. Then every maximal clique of G9,2 = (V, E) has the form C = {v0, Vl, Uj, Uj+ 1} (addition mod 9), and ]E(C)[ = 6 = [E(V - C)], so C is not a big clique. To show that G9,2 is not heavy, set zi = 1 for i = v0, Vl and zi = 1/4 for i = u 0 , . . . , Us. It is easy to verify that z is a feasible solution of (15.2) with objective function value larger than IE]. By a similar argument Gk,T is not heavy and has no big clique for each r -2,3mod4

if k - 3 + (~+2). The above counterexample is Ks-free, but not K4-free. There are also K4-free counterexamples. Consider the 5-wheel G5,1 and attach a path of length 2 to one of its vertices. Call the resulting graph an umbrella if the path is attached to the center, and a kite if it is attached to another vertex. It is easy to see that both the umbrella and the kite have no big cliques. Both of them are also not heavy by Theorem 15.2.7, as can be seen from the (optimum) solution of (15.2): zi = 1 if i is the center of the wheel, zi = 1/2 if i is any other vertex of the wheel, and zi = 0 if i is one of the other two vertices. The conjecture does hold for perfect graphs, as shown by the next theorem. Theorem

1 5 . 2 . 1 7 Every non-heavy perfect graph has a big clique.

Thre s hold Weights and Measures

386

P r o o f . Let G = (V, E) be a non-heavy perfect graph. By Theorem 15.2.7, since G is not heavy, inequalities (15.4) have a solution. This means that the optimum solutions of the linear programming problem (15.2) have a value larger than ]El, and hence this linear programming problem has a basic feasible solution z with value larger than IEI. Since G is perfect, every basic feasible solution is integer by Chvgtal's theorem [Chv75], so z must be integer. Hence G has a big clique by Lemma 15.2.15. m To close this section, we mention the following open p r o b l e m s . 1. We do not know if the problem of recognizing a heavy graph belongs to NP, nor do we know if it belongs to co-NP. 2. The problem of recognizing a graph with a big clique is in NP, but we do not know if it is in P. 3. Within the class of non-heavy graphs, which graphs, in addition to perfect graphs, have an integer optimal solution to the linear program (15.2)?

15.3

Threshold Measures

This section is based on the work of Peled and Simeone [PS89a]. By the definition of threshold graphs, a graph is threshold if and only if we can assign a non-negative weight to each vertex so that the total weight of each stable set of vertices is smaller than the total weight of each non-stable set of vertices. By the non-negativity of the weights it is enough to require this separation just for maximal stable sets and minimal non-stable sets (pairs of adjacent vertices). In a suitable normalization of the weights, this means that a graph G = (V, E) is a threshold graph if and only if there exist vertex weights wv >_ 0, v C V and a threshold r < 1 such that ~ v e S Wv <_ r for each maximal stable set S and w~ + Wv _> 1 for each edge u v C E . In other words, the minimum of r is smaller than 1 when subject to the maximal stable set and the minimal non-stable set constraints and the non-negativity of w. We therefore regard the minimum of r subject to the above constraints as a certain measure of the non-thresholdness of G. This motivates the following definition.

15.3

387

Threshold Measures

1 5 . 3 . 1 The t h r e s h o l d m e a s u r e v ( G ) of a graph G - (V, E) is the optimal value of the following linear programming problem.

Definition

min

s.t.

r

~ wv <_ ~-

for each maximal stable set S c_ V

yES

(15.5)

wu + w, >_ l for each edge uv C E w>O.

To emphasize the dependence of the optimum w on G, we sometimes denote it by w(G). Thus the optimum value of 15.5 is smaller than 1 if and only if G is a threshold graph. In that case, the minimum can be found by the very efficient and elegant algorithm of Section 1.3. The optimum value is equal to 1 if and only if G is a pseudothreshold graph. We have seen in Section 14.2 an algorithm to recognize these graphs and a characterization of their structure, from which it follows that when the optimum of (15.5) is 1, there is always an optimum solution for which the components of w have values of 0, 1/2 and 1. Figure 15.2 illustrates optimal solutions of (15.5) for a threshold graph and a non-threshold graph. Like (15.1), the linear programming problem (15.5) has exponentially many constraints, since in general there are as many maximal stable sets. This indicates that finding T(G) may be hard. We show below that this is indeed the case. We begin our general discussion of the threshold measure by determining the threshold measure of a disjoint union of two or more graphs. 15.3.2 w(G1 [..J G2) --~ (w(G1),w(G2)). Consequently T(G1 U G2) -- 7"(G1) -k- T(G2). This generalizes immediately to the union of any number of graphs.

Theorem

P r o o f . For k = 1,2, it is convenient to denote by Ak and Bk the edge-vertex incidence matrix and the maximal-stable-set-vertex incidence matrix of Gk. In other words, Ak has a column for each vertex and a row for each edge of Gk, and the corresponding entry is 1 or 0 according as the vertex is or is not an endpoint of the edge. Similarly Bk has a column for each vertex and a row for each maximal stable set, and the corresponding entry is 1 or 0 according

Threshold Weights and Measures

388

Figure 15.2" Threshold measures of (a) a threshold graph and (b) a non-threshold graph. Tz

8 9

5 4

T-1 4

1

1

9

l

9

~ ~1

12 / '

(b) as the vertex belongs or does not belong to the maximal stable set. We also let e k and fk denote all-1 column vectors of appropriate dimensions. Then the linear programming problem (15.5) for Gk can be written as follows, where tt k and u k denote row vectors of dual variables corresponding to the vector constraints of (15.6)" min #k

s.t.

Vk

A k w k >_ ek - - B k w k + rk.f k > 0 Wk>_O.

(15.6)

The linear programming dual of (15.6) is the following" T(Gk) lVk

-

max

13,k e k

s.t.

ttkAk -- v k B k <_ 0

vkf

- 1

tt k >_ O,

(15.7)

~'k >_ O.

To formulate the corresponding linear programming problems for G1 U G2, we use the fact that its maximal stable sets are precisely the unions of maximal

15.3

Threshold Measures

389

stable sets of G1 and G2. If i and j index the maximal stable sets of G1 and G;, respectively, then the maximal independent set constraint of (15.5) for G1 tO G2 reads ( B l W l ) i -~- (B2w2) j ~ T. We can therefore write (15.5) for G1 [-J G2 as follows, where jttl, jM2, Pl, P2 and uij are the dual variables to the corresponding constraints of (15.8), and where ~1 and rl2 are auxiliary column vectors introduced for convenience: r(GIUG2)

--

rain T s.t.

/-t2 Pl P2 Pij

A l W l ~ el A2w2 k e2 - B l W l + T~I -- 0 - B 2 w 2 + r12 - 0

(15.8)

--(?]1)i- (?]2)j -t'- T ~ 0 W 1 ~ 0,

W2 ~ 0.

The linear programming dual of (15.8) is as follows" 7-(G1 [,.J G2) Wl w2 (?]1)i

-

max s.t.

I-I,l e l -~ ~2e2 ~t 1A1 - PlB1 < 0 t t 2 A 2 - p2B2 < 0 ( P l ) i - E j 17ij = 0 (m)j

T

-

aj

= 0

~ i j vii = 1 tt 1 > 0 , tt 2 > 0 ,

aj >_ O.

The theorem asserts that if wl, r~ and w~, r~ are optima of (15.6) for k - 1,2, respectively, then (w~, w~), r~ + r~ defines an optimum of (15.8) when the auxiliaries rl~, rl~ are determined from the equations of (15.8). To show this, note that the constraints of (15.8) are satisfied, and then consider optima t t ; , v ; of (15.7). They satisfy the following complementary slackness conditions: tt*k(Akw*k -- ek) = 0 (15.10) v*k(--Bkw*k + r[~f k ) = O.

(15.11)

We define variables for (15.9) as follows:

It is easy to see that these variables define a feasible solution of (15.9). They also satisfy all the required complementary slackness conditions: (15.10),

390

Threshold Weights and Measures

(15.11), and the condition .;

+

-(,;),-

- 0

By theorem 15.3.2 we may assume without loss of generality that G is connected. We now consider how the threshold measure of a join of two graphs is related to their threshold measures. Unlike the case of the union, we only give estimates, for a reason that will be shown soon. T h e o r e m 1 5 . 3 . 3 Let G1 and G2 be disjoint graphs with largest stable set sizes OL1 and a2, respectively. Then C~lt~2 ~1 -t'- C~2

~ T(G1 (~ G 2 ) ~

~1 max(o~ 1 , c~2 ) 9

P r o o f . The upper bound for T(G10 G2) follows from the fact that assigning to every vertex a weight of g1 and taking T -- 71 max(a1, a2) defines a feasible solution of (15.5) for G1 | G2. To prove the lower bound, let Sk be a stable set of size ak in Gk for k = 1, 2. For every feasible solution w, T of (15.5) for G1 @G2, Sk possesses a vertex vk satisfying Wvk <_T/ak. The non-stable set {Vl, V2} satisfies wvl + WV2 ~ 1, which gives the desired bound. ,, C o r o l l a r y 15.3.4 If ~ 1 - - OL2 phic), then T(Ga | G 2 ) - a/2.

--

OL

(in particular, if G1 and G2 are isomor-

An immediate consequence of Corollary 15.3.4 is that it is hard to compute the threshold measure in general. Theorem

15.3.5 It is NP-hard to compute the threshold measure of a graph.

P r o o f . This follows from a(G) = 2T(G | G). " Unlike T(G~U G2), T(G1 @ G 2 ) i s not a function of T(G1) and r(G2). Indeed, one can easily find graphs G, G' satisfying T(G) = T(G') but a(G) 5r a(G'), as illustrated in Figure 15.3. Now if there were a function f satisfying r(G1 | G2) = f(r(G1), T(G2)), we would have the absurdity

f(T(G), T(G)) -- T(G 9 G) - a(G)2 r a(G')2

=

a') -

15.3

391

Threshold Measures

Figure 15.3" Graphs with the same threshold measure but different largest stable set sizes. Optimal solutions of (15.5) are shown. r-1

c~-2

r-1

0

0

,iw

,w

1

0

0

,~,

1

1

c~-4 0

0

w

1

In the following we show that, in contrast to the situation described by Theorem 15.3.5 for general G, if G is bipartite, the linear programming problem (15.5), although still having exponentially-many essential constraints, is equivalent to a small, structured linear program with a symmetric coemcient matrix having 2's on the main diagonal and 0's and l's elsewhere. This small linear program can then be solved in polynomial time by the well-known methods of Khachian [Kha79] and Karmarkar [Kar84]. T h e o r e m 15.3.6 If G is a bipartite graph with an edge-vertex incidence matrix A, then

T(G)

-- min s.t.

eTpt A A r t t >_ e tt >_ O,

(15.12)

where e denotes an all-1 column vector of an appropriate dimension. Moreover, if It is an optimum of (15.12), then w - ATtt is an optimum of (15.5). Note that A A T is the edge adjacency matrix of G (or equivalently the adjacency matrix of the line-graph of G) with 2's along the main diagonal. P r o o f . By theorem 15.3.2 we may assume without loss of generality that G is connected, and in particular has no isolated vertices. Denote by B the maximal stable set vs. vertex incidence matrix of G and by f an all-1 column vector of an appropriate dimension. Then the definition (15.5) of r(G) can

392

Threshold

Weights

and Measures

be written as follows" T(G)

-

min

T

s.t.

Aw>e

(15.13)

Bw<_rf

w>O. m

The constraint B w <_ ~-f on w and ~- can be expressed as follows" r

>

max

wTb

s.t.

Ab < e

(15.14)

b binary, because A b <_ e and b binary imply that b is the characteristic vector of a stable set of vertices of G, so b is dominated by a row of B, and conversely. Since A has no zero columns, the constraint "b binary" of (15.14) can be relaxed to "b _> 0, integral". Since A is totally unimodular (because G is bipartite), the integrality constraint on b can be dropped. Thus by the duality theorem of linear programming, (15.14) is equivalent to the following: r

>

max

wTb

s.t.

A b <_ e

=

min

eTp,

s.t.

AT# > w

b_>O

(15.15)

#>_0.

Since A T >_ O, the inequality in (15.15) is equivalent to the existence of tt >_ 0 satisfying AT# _> w and r - eTpt. Therefore (15.13) is equivalent to the following: T(G)

--

min

7"

s.t.

Aw>e

(15.16)

ATIt >_ w eT# - r

#>o,

w>O.

The constraints involving w in (15.16), namely A w >_ e, ATtt >_ w and w >_ 0, can be replaced by A A T # >_ e, thus eliminating w. Indeed, this condition is clearly necessary, and conversely, if it holds and tt >_ 0, then w defined by w - A T # will satisfy its constraints in (15.16). This proves (15.12) and the moreover statement of the theorem. ..

15.3

Threshold Measures

393

There are several variations of Theorem 15.3.6. By the symmetry of A A T, the linear programming dual of (15.12) is given by "r(G)

-

max s.t.

err A A T v <_ e v>0.

(15.17)

Further, if G has edges, then w(G) > 0, which permits us to transform the variables/~ and r - e T # of (15.12) according to 1~-#

t T

1 T

and rewrite (15.12) as follows" =

max

t

=

s.t.

AAT)~ >_ te er,~ -- 1 ,~>0

max

min

( A A r,~)i

s.t.

eT~- 1 ,~>0.

(15.18)

Similarly, we can transform the variables v and r - e r u of (15.17) according to p=-- v S - - - -1 T

T

and rewrite (15.17) as follows" 1

=

min s.t.

s A A T p <_ se eTp--1 p>O

=

min

max

o

3

s.t.

"(AATp)j "---e Tp -1 p>_O.

(15.19)

The problem (15.19) is the linear programming dual of (15.18). The formulations (15.18) and (15.19) exhibit ~ 1 as the value of a two-person zero-sum game with payoff matrix A A T. The pure strategies of the players are the edges of G, and the payoff to the first player is the number of common endpoints of the two chosen edges. The optimal mixed strategies are the optimal ,~ and p.

:394

T h r e s h o l d W e i g h t s and M e a s u r e s

A vertex cover (or transversal) is a set of vertices meeting every edge. Using (15.17) and essentially reversing the steps in the proof of Theorem 15.3.6, we can obtain the following theorem, which together with Definition 15.3.1 gives a combinatorial rain-max relation for bipartite graphs:

T h e o r e m 15.3.7 If G -

( V , E ) is a bipartite graph with an edge-vertex incidence matrix A, then r ( G ) is the optimal value of the following linear programming problem:

max s.t.

r

~

zv >_ r

for each minimal vertex cover C c_ V

vec

(15.20)

zu+z~
for each edge uv 6 E

z>0.

Moreover, if v is an optimum of (15.17), then z of (15.20).

A T v is an o p t i m u m

P r o o f . By arguments similar to those of Theorem 15.3.6, we obtain the following from (15.17)" r(G) - m a x { r " A z < e, A T v < z, e T v -- r , z > O , v > 0}.

The existence of v > 0 satisfying A T v < z and e T v -- r is equivalent to T <_ m a x { e r r 9 A T v <_ z,~' > O} =

m i n { z T 6 9 A 6 >_ e, 6 >_ 0}

=

m i n { z T 6 9 A 6 > e , 6 binary}.

This means that with the vertex weights z, every vertex cover has a total weight of r or more, and establishes (15.20). The moreover part follows from the argument leading to the first equation of this proof. [] Figure 15.4 illustrates the optimal solutions of (15.12) and (15.17) for a tree. The optimal w and z are obtained from the illustrated tt and v, respectively, by summing them over the edges incident with each vertex. If a vector tt _> 0 satisfies A A T t t - e, then tt is an optimum of the linear programming problem (15.12) as well as of its dual (15.17). Stated differently, e v e r y / and j in (15.18) and (15.19) realizes the minimum or maximum there. This motivates the following definition.

15.3

395

Threshold Measures

3 Figure 15.4" Illustration of Theorem 15.3.6 for a tree. r - eTlu- eTv- 3"

oI

o

dk

A

0

1

lii1

3 A

1

5

0

A

1

0

3 A

0

A

1

7

D e f i n i t i o n 15.3.8 A bipartite graph with an edge-vertex incidence matrix A is called e q u i t a b l e if there exists a vector tt >_ 0 satisfying A A T t t - e. Clearly a graph is equitable if and only if all its connected components are equitable. Below we characterize the connected equitable bipartite graphs in terms of the maximum weight of a stable set of vertices under certain vertex-weights. The problem of finding a stable set with maximum weight in a bipartite graph has well-known network flow algorithms. We also give a characterization of equitable bipartite graphs in terms of a certain generalization of Hall's condition for the existence of a perfect matching in a bipartite graph. For trees the matrix A A T is non-singular, and we give very simple combinatorial algorithm for solving A A r t t = e for trees, leading to a characterization of equitable trees. We use the convention that the color classes of a bipartite graph are called red and blue. T h e o r e m 15.3.9 Let G be a connected bipartite graph having r >_ 1 red and b >_ 1 blue vertices. Assign a weight of ir to each red vertex and a weight of-g1

396

Threshold

Weights

and Measures

to each blue vertex. Then G is equitable if and only if the largest total weight of a stable set is 1.

P r o o f . Let i and j index the red and blue vertices of G - (V, E), respectively. We can write the linear programming problem (15.12) as follows, using the vertex variables w = ATtt, which by Theorem 15.3.6 constitute an o p t i m u m of (15.5): r(G)

-

min

~#~j i,j

s.t.

E

#ij -- Wi

J E

#ij -- Wj

(15.21)

i Wi nt- Wj ~ 1

ijcE

#~j >_ 0

ijEE

#~j - 0

ij~E.

If G is equitable, then (15.21) has an o p t i m u m satisfying wi + wj = 1 for all ij C E. Then by the connectivity of G all the wi are equal and all the wj b for all i are equal. Since ~ i wi - ~ i j pij - ~ j wj, it follows that wi - -;-4-g r wj = ~+b for all j, and r(G) - r+b" ~b Since w is a feasible solution of (15.5) every stable set S must satisfy ~ e s Wv < ;~b" This means that with the vertex weights given in the theorem, the total weight of every stable set is 1 or less. Since the stable set of all red vertices has a total weight of 1, the "only if" part of the theorem follows. Conversely, if G satisfies the condition of the theorem, then

1

~

max

s.t.

(

l~xi+

1 )

7" i

-b j

xi nt- xj ~ 1

~xj

ij E E

(15.22)

x binary. Since the constraint matrix of (15.22) is totally unimodular and there are no isolated vertices, the condition "x binary" can be relaxed to x >_ 0. Then

15.3

Threshold

397

Measures

by linear programming duality there exist

)~ij, ij

C E, satisfying

Aij

b

ijEE

1

~j >_ j:ijEE

for all i

r

(~5.23)

1

for all j

i:ijEE

A~j > 0

ij C E.

But every solution of (15.23) must satisfy all its inequalities except for the non-negativity constraints with equality, as can be seen by summing the /-inequalities or the j-inequalities. Therefore the vector A*=

rb A>O r+b -

satisfies AATA * - e, which proves the "if" part. 9 Next we characterize equitable bipartite graphs by a generalization of Hall's condition for the existence of a perfect matching in a bipartite graph. T h e o r e m 15.3.10 Let G be a connected bipartite graph having r >_ 1 red and b >_ 1 blue vertices. Then G is equitable if and only if every set X of red vertices satisfies 1

IN(X)l >_ -~ F ~h~

(15.24)

N ( X ) d ~ o t ~ th~ ~ t of ~ighbo~ of th~ w~tic~ of X .

P r o o f . Denote by R and B the sets of red and blue vertices, and assign weights of !~ and g1 to their vertices, respectively. Condition (15.24) states that the total weight of N ( X ) is at least the total weight of X, for each XcR. To prove the "if" part, let S be any stable set of vertices, and take X - S N R, so that S and N ( X ) are disjoint. Then B includes the set ( S N B ) U N ( X ) , whose total weight is at least the total weight of ( S N B ) U X = S by (15.24). Hence B is a stable set of maximum weight, and so G is equitable by Theorem 15.3.9. To prove the "only if" part, assume that G is equitable, and therefore the weight of every stable set is at most 1 by Theorem 15.3.9. For every set

Threshold Weights and Measures

398

-IXl+w(b-IN(X)l

1 X C R, the set X U ( B - N ( X ) ) i s stable, and hence 1 _> r giving (15.24). 9 From Theorem 15.3.10 and Hall's condition we immediately obtain the following.

Corollary 15.3.11 Let G be a bipartite graph with an equal number of red and black vertices. Then G is equitable if and only if G has a perfect matching. To characterize equitable trees, we need some notation. We continue the convention that i, i0 index red vertices and j, jo index blue vertices. For a tree T and any edge e = ij of T, deleting e from T produces two subtrees. We denote by Tij the subtree containing i and by Tji the subtree containing

j. P r o p o s i t i o n 15.3.12 Let T = (V, E) be a tree and let x be an assignment of vertex-weights, possibly negative, satisfying

E xi - E xj. i

(15.25)

j

Then there exist unique edge-weights Aij, ij E E satisfying ~ij

-

xi

for all i

-

xj

f o r aU j.

(15.26)

j:ijEE i:ijEE

An explicit formula for the ~iojo = E ieTio3o

xi-

,~ij i8 given by E

xj-

JET/0~ 0

E

xj-

jET30i 0

y~ xi.

(15.27)

iET~0,0

P r o o f . The existence and uniqueness of the Aij follow from an algorithm that solves (15.26) "from the leaves up". That is, choose a leaf (a vertex of degree 1), say i ~, and its unique neighbor j'. Set ~i,j, = xi, as dictated by (15.26), delete i' and the edge i'j' and replace x~ by x j , - x~,. The result is a smaller tree that still satisfies (15.25), to which we can apply induction. The basis of the induction is the one-edge tree, where (15.25) is utilized. To prove the explicit formula (15.27), we show below that the Aij defined by (15.27) satisfy

E j:iojEE

Xo.

(15.28)

15.3

399

Threshold Measures

Figure 15.5: Illustration of the proof of Proposition 15.3.12. i0

Tjxio

~/j2io

Tjkio

A similar proof holds for the other equation of (15.26). Let j l , . . . ,jk be all the blue neighbors of i0, as in Figure 15.5. Then

k

k(

E ~,oj- E ~,o~- E r=l

j:ioj6E

r=l

j

i

E x~j6Tj,.i o

Zx )

iETj,., o

xo)

proving (15.28). " We are now ready for a simple combinatorial characterization of equitable trees. T h e o r e m 1 5 . 3 . 1 3 Let T be a tree having r > 1 red and b >_ 1 blue vertices. For each edge e - iojo of T, define _ ~(~) r

r

b(~) b '

(15.29)

Tiojo containing io, respectively. Then T is equitable if and only if ~ >_ 0 for each edge e. In that case )~ is an optimum of the threshold measure problem (15.18).

Threshold Weights and Measures

400

P r o o f . Formula (15.29) is obtained from (15.27) by setting x~ xj __ ~. 1 Therefore by Proposition 15.3.12, )~ satisfies 1

A~j -

-

j:ijEE

all i

r

Aij

1 ~

-

1?- and

(15.30) all j,

i:ijEE

where T - (V, E). Thus if )~ _> 0, then )~ satisfies (15.23), and the argument of the "if" part of Theorem 15.3.9 shows that T is equitable. Conversely, if T is equitable, then as in the "only if" part of Theorem 15.3.9, the equations (15.30) admit a solution )~' _> 0. But by Proposition 15.3.12 these 1 equations have a unique solution given by (15.27), where xi - 1 and xj = -g, namely the ,k defined by (15.29). Therefore ,V - ,k and ,k _> 0 [] We conclude this section with two examples of equitable trees. T h e o r e m 15.3.14 Let T be a tree in which all red vertices have degree 2 (equivalently, T is obtained by subdividing each edge of a tree with a new vertex). Then T is equitable. Proof. To see vertex from j

T satisfies b - r + l . A l s o , each edge e - iojo of T satisfies r(e) - b(e). this, use the bijection mapping each red vertex i of Tiojo to the blue j of Tiojo such that i is the immediate successor of j along the path to i0. Therefore formula (15.29) gives A(e)-r(e)

(1 1) r

r + l

>0"

A complete k-ary tree is a rooted tree in which every internal vertex has exactly k children, and all the leaves have the same distance from the root. Theorem

15.3.15 Every complete k-ary tree is equitable.

P r o o f . Let T - (V, E) be a complete k-dry tree, and assume without loss of generality that its root is blue. Let h be the height of T (the distance from the root to each leaf) 9 The total number of vertices of T is IV] - k h]+~ l l l 9 The numbers r and b of red and blue vertices of T are functions of k and h. Specifically, if h is even, then b - P(k, h) "- 1 + k 2 +

k 2]Ch -

-

Q ( k , h) . -

kh + 2 - 1

k 4 --~-""" + ~h __

IVI - P ( k ,

1

h) ]~2

__1

~

1

15.4

401

Threshold and Majorization Gaps

and if h is odd, then k h+l -- 1 k:-i

b - R(k,h) "- g ( k , h - 1 ) -

k h+l - 1

- s(k, h) . - IVI - R(k, h) - k k~---=-V To compute the A~ of (15.29) for a given edge e, it is convenient to always use that subtree of T - e that contains the root, using either (15.29) or its analog with red and blue switched. Let 1 be the height of this subtree. If h is even, then

_

P(k,l) Q(k,1) V~(k~h) - Q(k, h)'

R(k,~)

S(k,1)

-d-(-i l h ) - P ( k, h ) '

for k even for k odd,

and if h is odd, then

~ -

P(k, 1) Q(k,l) 5 ( ; , ~ - R(k,h)' R(k,~) S(k,t)

for k even = 0,

for k odd.

In all cases, one obtains Ae _> 0.

15.4

Threshold

and

Majorization

Gaps

In this section we introduce two measures of non-thresholdness of a degree sequence: the threshold gap, based on the work of Hammer, Ibaraki and Simeone [HIS78, HISS1], and the majorizaion gap, based on the work of Arikati and Peled lAP94]. It turns out that these two measures coincide, and we investigate their properties and the relations between them. Recall from Section 3.1 that a vector d - ( d l , . . . , d~) with integer components is called a proper sequence if n - 1 >_ dn >_ " " >_ dl > 0. Recall also the definition of the corrected Ferrers diagram C(d) representing a proper sequence d (Definition 3.1.6), and of the corrected conjugate sequence d' and the corrected Durfee number m of d. Also recall (Theorem 3.2.2) that a proper sequence d is threshold if and only if C(d) is symmetric (i.e., d' - d), in which case C(d) is the adjacency matrix of the unique labeled realization of d. We define a distance between proper sequences as follows.

402

T h r e s h o l d Weights and M e a s u r e s

D e f i n i t i o n 15.4.1 The distance between proper sequences d and e of the same length n is defined as n

l i d - ell -

(15.31)

I & - e l. i=1

In other words, the distance is just 1/2 of the Ll-norm of the difference d - e . If ~'~'~in._=ldi and ~i~1 ei are even, as is the case when d and e are graphic sequences, then the distance lid- ~11 is an integer. The reason is that

9

di~ei

di>_ei

i

di<ei

Since 2in___l di and ~i~a ei are even, so are ~d,>~, di--~-~d,<e, di and ~d,<~, e i ~d,>~, ei, and therefore ~i~1 I d i - ei] is an even integer. We denote by 7~T,~ the set of proper threshold sequences of length n. The threshold gap of d is defined as follows. D e f i n i t i o n 15.4.2 If d is a proper sequence of length n, its t h r e s h o l d g a p is eE D"I'n

lid-

Thus the threshold gap of d is a measure of the non-thresholdness of d. In Subsection 15.4.1 we determine the threshold gap of a proper sequence and all the threshold sequences that realize it. Recall the definitions of the majorization ~ and strict majorization ~relations between sequences of length n (Definition 3.1.1). Recall also the definition of unit transformation (Definition 3.1.2). If a is obtained from b by a unit transformation, we say that b is obtained from a by a reverse unit transformation. By Condition 9 of Theorem 3.1.7, every degree sequence d is majorized by some threshold sequence, and by Condition 8 of Theorem 3.2.2 the majorization is strict if and only if d is not a threshold sequence. From this and from the Muirhead Lemma (Theorem 3.1.3) it follows that if d is any degree sequence, then some threshold sequence can be obtained from d by a finite number of successive reverse unit transformations. This motivates the following definition of the majorization gap of d as a measure of the non-thresholdness of d.

15.4

Threshold and Majorization Gaps

403

D e f i n i t i o n 15.4.3 The m a j o r i z a t i o n g a p R ( d ) of a graphical sequence d is the smallest number of reverse unit transformations required to transform d into a threshold sequence. Thus R(d) = 0 if and only if d is already a threshold sequence. In Subsection 15.4.2 we give a formula for R(d) and show that it is equal to the threshold gap of d. We also determine its m a x i m u m for a fixed number of edges or vertices, as well as exhibiting all the sequences that realize this maximum, which can be thought of as the most non-threshold in this sense. We also discuss the related questions of the difference gap and some others.

15.4.1

The Threshold Gap

Before we discuss the threshold gap, we introduce two binary operations on the proper sequences of length n. D e f i n i t i o n 15.4.4 If d and e are proper sequences of length n, their j o i n and m e e t are defined as the proper sequences d V e and d A e given by (dV e)i

-

max(di, ei)

i - 1,...,n

(dA e)i

-

min(di, ei)

i - 1,...,n.

It turns out that DT-~ forms a lattice under the operations of join and meet, having the largest element ( n - 1, n - 1 , . . . , n - 1) and the smallest element (0, 0 , . . . , 0). This follows from the following proposition. Proposition

15.4.5 DT~ is closed under joins and meets.

P r o o f . For integers k _> 1 and h >_ 0, we denote by S(k, h) the set of the smallest h positive integers excluding k, that is to say

S(k,h)-

{1,...,h}, {1, ,k-l,k+l,...,h+l},

ifhk.

If d is a proper threshold sequence whose unique labeled realization on the vertices 1 , . . . , n is G, then the neighborhood of each vertex i in G is Na(i) = S(i, di). Conversely, if a graph G on the vertices 1 , . . . , n has neighborhoods satisfying this condition, then G is a threshold graph with degree sequence d, so d is a threshold sequence.

404

Threshold

Weights and Measures

We prove the closure of 2)7",~ under joins; the results for meets can be proved similarly or deduced from the closure under complements. Let d, e, C 2)Tn, and let G and H be the threshold graphs on the vertices 1 , . . . , n in which vertex i has degree d~ and e~, respectively. Thus N o ( i ) = S(i, d~) and NH(i) = S(i, ei) for each i. Put f = dV e and F = G tO H. Then for each i we have

NF(i) -- S(i, d~) U S(i, e~) - S(i, max(d~, e~)) - S(i, f~). Hence f is a threshold sequence, i.e., f C ~)']"n. 9 We now define an easily-computed quantity t(d) associated with a proper sequence d, and show that it is equal to the threshold gap of d. D e f i n i t i o n 1 5 . 4 . 6 If d is a proper sequence of length n and corrected Durfee

number rn, we put 1

m

t(d) - -~ ~

!

1

d~[ -

Id~ -

7-51

n

E

!

Id - d l.

( 5.a2)

~=m+l

To justify the equality between the two expressions in the definition of t(d), consider the corrected Ferrers diagram C - C(d), and divide it into four submatrices M, A, B, 0 as illustrated in Figure 15.6. Figure 15.6: The corrected Ferrers diagram of a proper sequence. The submatrix M is full of l's except on its main diagonal; the submatrix O is full of O's. 1

m

m+l

n

M

B

A

O

m

m+l

Since each row and column sum of M is m - 1, d~ - di is equal to the difference between the i-th column sum of A and the i-th row sum of B. Since

15.4

Threshold and Majorization Gaps

405

the l's of A are at the top of the columns and the l's of B are at the left n ~" 1, if Cij r Cji of the rows, we have Id~- di I - ~j=~+~ u~j, where u~y - ~ O, otherwise Consequently m

m

i=1

n

i=1 j = m + l

A similar argument involving the rows of A and the columns of B shows that n

m

E Id:-d i=m+l

n

l-E E

Uij~

i=1 j = m + l

and therefore the equality in Definition 15.32 has been justified, and we also have an interpretation of t(d) as a measure of the deviation of C(d) from symmetry. Furthermore, we have ~i~=~ di - m ( m - 1) + a + b, where a and b are the numbers of l's in the submatrices A and B, respectively. Therefore if ~in=.l di is even, as is the case when d is graphic, a and b have the same parity, and it follows that t(d) is an integer. As a first step in proving that t(d) is the majorization gap of d, we associate with each proper sequence d of length n with corrected Durfee number m the sequences d and d defined by

d i - { d'i' f o r i - 1 , . . . , r n

di,

for i - m

+ 1,..., n

cli- { di, f o r / - 1 , . . . , r n d}, for i - m + l , . . . , n .

Referring to Figure 15.6, we see that C(d) is obtained from C(d) by replacing the submatrix B with A T, whereas C(d) is obtained by replacing A with B r. Therefore d and d are proper sequences and their Ferrers diagrams are symmetric, i.e., d, d C 7PTn. To find the corrected Durfee numbers r'n - re(d) and rh - re(d) of d and d, we notice that for C - C(d) we always have Cm+l,~n = 0 (for if Cm+l,,~ = 1 one would also have Cm,m+l = 1 by dm >_ din+l, and m could be increased by 1). In contrast, x - C~n,m+l can be either 0 or 1. From the definition of d it is clear that d - C(d) satisfies C',~+~,,~ = 0 and therefore rh - m. On the other hand, d' - C(d) s a t i s f i e s C m + l , m = x and Cm+2,m+l = 0, and therefore rh is m or m + 1 according as x is 0 or 1. We have therefore verified the following formula: ~-m

~_{ '

m, re+l,

if d i n - m - 1 if d i n > m - 1 .

(15.33)

Thre s hold Weights and Measures

406

It follows directly from (15.31) and (15.32) that for each proper sequence d, the proper threshold sequences d and d satisfy

lid- rill - l i d - rill -

t(d).

(15.34)

T h e o r e m 15.4.7 E v e r y proper sequence d of length n satisfies t(d) -

min l i d - cll.

cEDTn

P r o o f . Let g: C 77T~ be optimal, i.e., lid- ~11- mincev~rn lid- ell. Let d be the corrected Ferrers diagram of ~: and rh its corrected Durfee number. First we show that rn _< rh _< rh, where rh is the corrected Durfee number of d, given by (15.33). Assume that, if possible, rh < m. Put h = ~:,~+1 < rh, and define the sequence c by { ~i+1, fori-h+l,...,rh ci ~, for i - rh + 1 g:i, otherwise. Since C is symmetric and h < rh, c is a proper threshold sequence, and its symmetric corrected Ferrers diagram is obtained from ~' by enlarging its corrected Durfee,square from size rh to size rh + 1. We have C~+l - h < c~+1 - r h _< r n - 1 _< dm _< drh+l and therefore Ic~+l - d~+l I < 1~=~+1- d~+a I - ( ~ - h).

(15.35)

In addition, we have in any case that (i-

h+ 1,...,~).

(15.36)

By adding the inequalities (15.35) and (15.36), we obtain IIc-dll < II~-dll, contradicting the optimality of ~:. This proves that m < ~n, and a similar argument shows that rh < rh. We have shown that m < rh < rh, and therefore by (15.33) rh is either m or rh. We show below that l[~:- dll - l i d dll if r~ - m, and it can be similarly shown that I1~- dl[ - l i d - dll if ~ - ~ . Consequently, at least one of J and d is an optimal threshold sequence, and the required conclusion

that l i d - ~11-

t(d) will follow from (15.34).

As stated above, we consider the case ~ - m and have to show that I 1 ~ - d l l - lid-dll. Note that ~ ~nd d ~re proper threshold sequences with the

15.4

T h r e s h o l d a n d M a j o r i z a t i o n Gaps

407

same corrected Durfee n u m b e r m. If g: - d, we are done. Otherwise there must be some i - m + 1 , . . . , n such that g:i r a~i, because rows m + 1 , . . . , n of the s y m m e t r i c corrected Ferrers diagram determine it completely. Let k be the largest such i satisfying g:i > di, if any. Put h - g:k and define the sequence c' by

, ~ c.i-1, ci - ~ ci,

fori-k,h otherwise.

The sequence c' is proper, because by the maximality of k we have g:k > dk _> dk+l >_ ck+l, and by the s y m m e t r y of its corrected Ferrers diagram it is a threshold sequence. Clearly its corrected Durfee n u m b e r is again rn. Since c~ - g : k - 1 >_ dk - dk, we have I c ~ - d k [ - Ig:k--dkl-- 1. Hence I I c ' - d l l _< I]g;- dll because in any case I c ~ - dh] _< Ig:h- dhl + 1. It follows that c' is another optimal threshold sequence and we may replace g: with c'. R e p e a t e d application of this procedure eventually allows us to assume that ci <_ di for i - m + 1 , . . . , n. If we still have g: =/= d, there must be some i - rn + 1 , . . . , n such that ~i < di. Let k be the smallest such i. Put h - g:k and define the sequence c" by ,,

]' g : i + l ,

ci - ~ ci,

fori-k,h+l otherwise.

The sequence c" is again proper, because by the minimality of k we have g:k < dk _< dk-1 _< g:k-1, and by the s y m m e t r y of its corrected Ferrers diagram it is a threshold sequence. Its corrected Durfee n u m b e r is again rn. Since c~ - g:k + 1 _< dk - dk, we have [[c"-d][ _< IIg:- dll as before, so c" is optimal and we may replace g: with c". R e p e a t e d application of this procedure allows us to assume finally that ci - di for i - m + 1 , . . . , n. This implies that g: - d and completes the proof. " For a given proper sequence d of length n, we denote by 2)Tn(d) the set of proper threshold sequences of length n at m i n i m u m distance from d, that is to say, by T h e o r e m 15.4.7, lPT~(d) = {c C 2P7"~ : [ I c - all = t(d)}. We determine this set below. For this purpose we put

d+-dvd,

d--dAd.

Since d, d E 2)Tn, we have d +, d- E 2)Tn by Proposition 15.4.5. It is easy to see that d + - max(d~, d'~) and d~- - min(d~, d'~). To justify (15.32), we have argued that t(d) is equal to one half of the n u m b e r of positions where the submatrix A in Figure 15.6 differs from B T.

408

T h r e s h o l d Weights and M e a s u r e s

It is easy to see that both lid + - d l l quantity, which means that

and lid- - d l l

are equal to the same

d +, d- C DT'n(d).

(15.37)

We need the following l e m m a to determine D'Y~(d). Lemma

1 5 . 4 . 8 Let d be a proper sequence and h, k integers satisfying d-~ <

h <_ d +. Then the following implications hold:

d; < & G+ > d k

G+ > & ---->, d ; < & .

P r o o f . We prove only the first implication for k - rn + 1 , . . . , n; the other cases are similar. The hypothesis d~- < dk implies t h a t d~ < dk and hence d~- - d~ and d + - dk. Therefore the condition d~- < h _< d + of the l e m m a reads d~ < h <_ dk. Since k _> rn + 1, the condition dk _> h is equivalent to d~ _> k - 1 and the condition d~ < h is equivalent to dh < k - 1. Therefore dh < d~, which means dh < d +. 9 We are now ready to characterize ~D'Yn(d). Theorem

1 5 . 4 . 9 Let d be a proper sequence of length n. Then -

{c

Z

7. . d - <

< d + }.

In other words, I)T~(d) is the sublattice of 7?T~ with the largest element d + and the smallest element d-. P r o o f . First we prove that if c C Z)Tn and d- _< c _< d +, then c C Z)'Yn(d). Observe that since d + and d- have the same corrected Durfee n u m b e r rn and c lies between them, c too has the same corrected Durfee n u m b e r m. We m a y assume t h a t c -r d - , for otherwise the result follows from (15.37). Let k be the largest subscript i satisfying d~- < ci. This k must be larger t h a n m, for otherwise ci - d~ for each i > m, which would imply that the threshold sequences c and d- coincide. We assume that dk -- d +, i.e., that d;- < dk; the case dk - d~- is similar. P u t h - ck. Since rn < k, we have h < rn and therefore Ch _> k - 1. Since d~- < h _< d + and d~- < dk, L e m m a 15.4.8 gives us t h a t d + > dh - d~. If D - - C ( d - ) is the corrected Ferrers diagram of d - , then from h - ck > d~- we have D-~h -- O, hence by the s y m m e t r y of D -

15.4

T h r e s h o l d a n d M a j o r i z a t i o n Gaps

we have D~k - 0, and hence d~ < k given by

1

~

409

1 _< Ch. Consider now the sequence c 1 fori-h,k otherwise.

ci--1,

ci - ~ Ci

This sequence is proper because of the s y m m e t r y of C(c) and the fact that if ck - ck+l, then Ck+l > d~- >_ d~-+l , contradicting the maximality of k. By the s y m m e t r y of C(c) and the definition of h it is also clear that C(c 1) is symmetric, so c 1 C DT-n. By Ch > d~ and ck > d~- we also have d- <_ c I _< d + and IIc I - d-II < I I c - d-II. Furthermore, from ck _< d + - dk we have Iclk--dkl -- Ick--dkl+ 1, and from Ch > d-~ - dh we have Iclh--dhl -- Ich--dhl--1; consequently IIc1 - a l l - I I c - all. By repeating this argument, we obtain a finite sequence c - cO,..., c s - d- of proper threshold sequences all at the same distance from d. Therefore c e DT-n(d) by (15.37). Conversely, given that c C D T n ( d ) , we show that c _< d+; the argument for d- _< c is similar. Assume otherwise, and let k be any subscript satisfying ck > d +. Let G and H be the threshold graphs on 1 , . . . ,n satisfying dega(i ) - ci and degH(i ) -- d + for all i. Let h be the largest subscript such that kh is an edge of G. Since degH(k ) < dega(k), kh is not an edge of H. Therefore degH(h ) < dega(h), that is to say d + < Ch. Consider the sequence c' given by , ( ci-1, fori-k,h ci - ~ ci, otherwise. By our familiar argument, c' C DT-n, and by ck > d + > dk and Ch > d + > dh, we have ] [ c ' - dll < I I c - dll , contradicting c e "D'-f'n(d). 9

15.4.2

The Majorization Gap

In order to find an explicit expression for the majorization gap R ( d ) , we use the following notations. For a sequence d - ( d l , . . . , dn), let S k ( d ) denote the k-th partial sum of d, i.e 9, Sk(d) - 2~=1d~. k Note that Sn(d) - Sn(d'). For a proper sequence d - ( d l , . . . , dn), put 6i(d) - (d~i - di) +,

i - 1 , . . . , n,

where x + - max(x, 0), and n -

i=1

Threshold Weights and Measures

410

Our first main result is a formula for the majorization gap, to be proved below"

T h e o r e m 1 5 . 4 . 1 0 For every proper degree sequence d,

R(d)

6(d)

-

E x a m p l e . Let d = (2,2,2,2,2). Then d ' = ( 4 , 4 , 2 , 0 , 0 ) and 5(d) = 4. The theorem asserts that R(d) = 2. A reverse unit transformation from 5 to 1 transforms d into f = (3, 2,2,2, 1), and a reverse unit transformation from 4 to 1 transforms f into e = ( 4 , 2 , 2 , 1 , 1 ) . Since e = e', e is a threshold sequence. Before we prove Theorem 15.4.10, we discuss some preliminary results. For any two sequences a = ( a l , . . . , an) and b = ( b l , . . . , bn), we define

i-1,...,n

5i(a, b) - (ai - bi) +, and n

6(a,

-

6 (a, i=1

If a and b are integer sequences, a ~ b, a l _> . " _> a~ and bl > " ' " > bn, we define U ( a , b) to be the m i n i m u m number of unit transformations required to transform a into b. Observe that under these conditions the following are equivalent: (i) a = b; (ii) 5(a, b ) = 0; (iii) U(a, b ) = O. Lemma

15.4.11 Let a -

(al,...,an)

quences such that a ~ b, aa >_ .'. U(a, b) - 5(a, b).

and b - ( b l , . . . , b n ) >_ an and bl >_ ""

be integer se-

> bn.

Then

P r o o f . It is easy to see that if a r b and c is obtained from a by a unit transformation, then 5(a, b)+ 1 _> 5(c, b) _> 5(a, b ) - 1. Thus U(a, b) >_ 5(a, b). Also, as given on page 135 of [MO79], if a r b, we can always perform a unit transformation on a to obtain a c such that c ~ b and 5(c, b) = 5(a, b ) - 1. Thus U(a, b)<_ 5(a, b). ..

15.4

T h r e s h o l d and Majorization Gaps

411

It is easy to check that for integer x, (x+l)+_x+_

{ 1, i f x _ > 0 0, otherwise, 2, if x_>0 1, if x - - 1 0, otherwise,

x +-

(x + 2) +

x+_(x_l)+_

x+-(x-2)

{ 1, i f x > _ l 0, otherwise,

+-

[ 2, if x>_2 / 1, if x - 1 0, otherwise. (15.38)

We use these facts to prove the following lemma.

L e m m a 15.4.12 Let

d - (dl,...,dn) be a proper sequence and a s s u m e that e is obtained f r o m d by a reverse unit t r a n s f o r m a t i o n . Then

6(d)- 2 < 6(~) < ~(d)+ 2. Proof. Represent d by its corrected Ferrers diagram and assume that e d - up + U~. Let or-

{ d~+l, d~+2,

ifd~<~r-I ifd~_>Tr-1,

7-

{ do, dp+l,

ifdo<

p

ifdp>p.

Thus the transfer is from row p, column T of C ( d ) to row 7r, column or, and consequently 7r 5r cr and p 5r r. See Figure 15.7. We have four cases. Case 1" 7r 7~ r and p -/ a. Since e - d - U p + u~,, for 1 <_ i <_ n, we have e i - - d i except for e~= d~+l, e o d R - 1 , and e i - - d i except for D

l

l

l

l

~-d~+l,~=d~ .

6~(~)- (~"

i. T h ~ n. 6 ~ ( ~ l - ( ~. l

!

d~

~ ) + - ( .d "

e~) + - ( d ~ - d ~ + 1) +. Hence 5~(e) - 5~(d) -

-1, ifd'-d~> 0, otherwise,

5o(e) - 5o(d) -

1, i f d ' - d ~ >_0 0, otherwise.

5~

1, i f d ' p - d p > O 0, otherwise,

Similarly, - 5~

-

-1, ~ ( e ) - 5~(d) -

ifd'~-d~>l O, otherwise.

I

1

l)+,~nd

412

Threshold

Weights and Measures

Figure 15.7: Illustrating a transfer from p to 7r. Solid lines represent l's and possibly . ' s .

T

~T

I

1 ....

I

C a s e 2: 7r 7~ T and p - or. Here, ~ r ( e ) - ~ ( d ) and 8 ; ( e ) - ~ ( d ) Case 1. Also 6 p ( e ) - ( e ' p - ep) + - ( d ' p - d p + 2) +. Hence

8 p ( e ) - 6p(d) -

2, if d'p - dp > 0 1, if d'p - - d p - - - - 1 O, otherwise.

C a s e 3" 7 r - T and p5r a. H e r e S p ( e ) - S p ( d ) )+ - ( d rI - d ~ - 2 ) Case 1. S i n c e 6 r ( e ) - ( G - e/r --2,

6r(e)-6r(d)

-

are as in

-1, O,

and 5o(e)-5~,(d) + we obtain

are as in

if d'~ - dr > 2 if d" - d~ - 1 otherwise.

C a s e 4" 7r - ~- and p - ~r. We now h a v e S r ( e ) - S r ( d ) as in Case 3, and 5 p ( e ) - 6p(d) as in Case 2. To complete the proof, note that 5i(e) - 5i(d) except for i E {~r, p, ~, ~-}. For future reference, note the following from the proof of L e m m a 15.4.12. Necessary and sufficient conditions for 6(e) - 5 ( d ) - 2 in Cases 1, 2, 3 are: 1. if 7r -~ T, p -~ or, then (a) d ' -

d~> 1,

15.4

Threshold and Majorization Gaps

(b)

r

413

0,

(c) f ~ - d ~ < O , (d) s

d~>_ 1.

2. if 7r 7~ r, p - (r, then (a) d ' -

(b)

d~>_ 1,

l

(c) d ' -

d~_> 1.

3. if 7c - r, p -r a, then

d'-

2,

(b) d ' p - d p < O ,

d'-

O.

L e m m a 1 5 . 4 . 1 3 Let d be a proper sequence. For 1 ( i , j (_ n, if di < j - 1, ! then dj < i. P r o o f . The proof is simple. L e m m a 1 5 . 4 . 1 4 Let d C D~ and let p be the largest index i such that di < d~. Then p < n. Further, 1. A s s u m e d'p (_ p -

!

1 and let q - d;. Then dq - p -

!

1 and dq ( p -

2.

2. A s s u m e dpt >_ p and let q - dpt + l. Then dq - p and dql <_ p - 1 . P r o o f . Since Sn_l(d) < Sn_l(d') and Sn(d) - Sn(d'), we have d= >_ d~, so p < n by definition of p. The fact that p < n justifies our mentioning of column p + 1 (of C) below. In both cases q is the position of the last 1 in I ! column p, so Cqp - 1. The assumption on p implies dp+ 1 _< dp+l _< dp < dp. ' 1 < dp, ' implying that Cq,p_t_ 1 r 1. We prove the two parts as Hence dp+ follows" !

1. See Figure 15.8. If d; _< p - 1 , then p > q and Cq,p+l r "k. It follows that Cq,p_t. 1 -- 0, and therefore (since Cqq - * and Cqp - 1) dq - p - 1. Also dp < d; - q gives Cpq - 0, and again C(q, q) - * implies d; _< p - 2.

T h r e s h o l d Weights and M e a s u r e s

414

Figure 15.8: Illustrating Part 1 of Lemma 15.4.14. q

P

~ * ~ 1 0 0

*

!

2. In this case, apply L e m m a 15.4.13 with i - p and j - 1 - dp. T h e n ! j - dip + 1 - q and dq < p, as required. In our case p < q. See Figure 15.9. We have seen t h a t Cq,;+l r 1, b u t Cqp - 1, so dq _> p. If q > p + 1, t h e n Cq,p+l = 0 and dq - p, as required. If q - p + 1, t h e n Cqp - Cp+l,p = 1 and Cqq - , . On the other h a n d , Cq,q+l m u s t be 0 if ! it exists, for otherwise Cp;;+l would be 1, c o n t r a d i c t i n g dq < p. T h u s again dq - p.

Figure 15.9: Illustrating Part 2 of Lemma 15.4.14.

Lemma 1,...,p-

p

q

,

0

1

*

Let d C Dn and let p be as in L e m m a 15.~.1~. FOr r 1, dr >_ p implies dr < dlr.

15.4.15

-

-

15.4

Threshold and Majorization Gaps

415

Figure 15.10" Illustrating the proof of Lemma 15.4.15. r

p

t

~ * ~ I 0

0

*

P r o o f . A s s u m e to the c o n t r a r y t h a t d~ > d'~. Let t - d~ + 1 > p (since d~ > r, t is the position of the last 1 in row r. See Figure 15.10). By c o n s t r u c t i o n C~t - 1, and by a s s u m p t i o n Ct~ - O, so d't ~_ r and dt < r - 1. B u t t h e n dlt > dt c o n t r a d i c t i n g the definition of p. 9 Lemma

15.4.16

Let d be a proper sequence. If there exists a k such that

d'k < d'k+l, then 1. d'k - k - l , d ' k +

~-k;

2. k is unique. P r o o f . 1" If d~ < k - 1 or d~ > k, then d~+ 1 < d~ since the l's of C(d) are left-justified. Hence d~ - k - 1. T h e n again by this reason and the a s s u m p t i o n t h a t d~ < d~+l, we obtain d~+ 1 - k. 2" T h e s a m e p r o p e r t y of C(d) implies t h a t the conditions d~ - i - 1 and d~+ 1 - i are possible for at most one i. ,, 1 5 . 4 . 1 7 (1) Let d C Dn have corrected Durfee number m, and let C(e) be obtained from C(d) by deleting the last 1 in row i, where 1 < i < rn and (a) if i < m, then di > di+l; (b) if i - r n , then dm >_ m. Then e C D , . Further, if d} > d,, then 5(e) > 5(d). (2) Let d G Dn, and let C(e) be obtained from C(d) by adding a 1 to the end of column i, where 1 < i < rn and (a) if i > 1, then d~ < d~_l;

Lemma

416

Threshold Weights and Measures

(b) if i = m, then dm >_ m. Then e e D , . Further, if d~ >_ d~, then 5(e) > 5(d). P r o o f . 1: The a s s u m p t i o n s on i g u a r a n t e e t h a t e is a proper sequence. The c o l u m n of the 1 t h a t is removed is j = 1 + di > m. This implies t h a t e E Dn. Further, if d~ > di, then Cji - 1, and thus dj ~_ i. But the a s s u m p t i o n s on i i m p l y d~ - i. Therefore by equation (15.38),

5(e) - 5(d) - (d'i - di + 1) + - (d'i - di) + + (d~ - dj - 1) + - (d} - dj) + - 1 + O. 2: This is similar to Part 1.

..

Let d c Dn with d~ > 0 and m - 2. Then S(d) <_ n with equality if and only if dl - d2.

2,

Lemma15.4.18

P r o o f . Note t h a t the condition dn > 0 implies n _> 2. Using d2 < dl _< n - 1, d~ - 1, d2 _> 1, and di - 1, d~ - 2, for i - 3 , . . . , d2 + 1, we obtain n

~(d)

-

~ (d'~ - d~)+ i=1

=

(n-

_< ( n -

1-dl)+

-~-(d2- 1)(2-

1)

1 -d2) + + d2- 1

=

n-l-d2+d2-1

"~

rt ~2.

We need three more l e m m a s to prove T h e o r e m 15.4.10. Lemma

1 5 . 4 . 1 9 Let d C D~. A s s u m e that dn > O, d~ < d'a, and let p be the

largest i such that di < d~. Let q be such that dq > dqt and either (a) q < n or (b) q - n , dn > 2. Then Sk(d) < X k ( d ' ) - 1,

for k -

p,...,q-

1.

Since d C Dn, Sk(d) <_ Sk(d'). A s s u m e t h a t , if possible, Sk(d) >_ Sk(d') - 1 for some k satisfying p _< k _< q - 1. W h e n q < n we have

Proof.

X~(d) > S~(d')d~ > d~,

( a s s u m p t i o n on p) ( a s s u m p t i o n on q) ( a s s u m p t i o n on p)

I dq >_ dq-+-i

(d~ < dl ~ d'n - 0 , d. > 0)

d. > d ' + l .

d~ > d~,

i-k+l,...,q-1 i-q+l,...,n-1

15.4

417

Threshold and Majorization Gaps

Adding all these inequalities, we obtain a contradiction, S~(d) > S ~ ( d ' ) + 1. W h e n q - n, the inequality for dq drops, but by assumption the inequality for dn can be strengthened by 1, and the same contradiction is obtained.

1 5 . 4 . 2 0 Let d C Dn, a s s u m e that dl < d'1, dn > O, and let p be as in Lemma 15.~. 19. Then

Lemma

Sk(d) <_ S k ( d ' ) - 1,

k - p,...,n-

1.

P r o o f . We have Sk(d) < Sk(d') since d E Dn. If Sk(d) - Sk(d'), then there exists an i, k < i < n such t h a t d~ < d~, since d~ > d'n and S n ( d ) - Sn(d'). But then i > p, contradicting the definition of p. tt Lemma

15.4.21

Under the assumptions of Lemma 15.~.20:

1. if d'p < p - 1, then Sk(d) < S k ( d ' ) - 1 for k - 1 , . . . , dpI - 1 ; !

2. if dp >_ p, then Sk(d) <_ Sk(d') - 1 for k - 1 , . . . , p -

1.

P r o o f . 1" Let q - d p I < p - 1 . By L e m m a 15.4.14, dq - p - 1 and dq/ < p - 2 , so dq > d'q. I f d i < d} for i - 2 , . . . , q 1, then the assumption dl < d~ implies Sk(d) < S k ( d ' ) - 1 for k - 1 , . . . , q - 1, as required. So assume t h a t there exists an i, 2 < i < q - 1, such that di > d}, and let r be the smallest such i. Since r < q, d~ k dq - p - 1 . But the fact that & > d'~ and L e m m a 15.4.15 imply d~ < p - 1. Therefore d~ - p - 1. We assert t h a t di > d} for i - r , . . . , q . Indeed, for such i, p - 1 d~ > di k dq - p - 1, so di - p - 1. This implies, by the construction of C, t h a t for i - r , . . . , q 1, d'~ > q - 1 > i, and therefore d} > d'~+1 by I I l L e m m a 15.4.16. Hence d~ >_ d~+ 1 >_ ..- >~ dq_ 1 ~_ dq. Thus for i - r , . . . , q, d~ <_ d~ < d~ - p - 1 - d i , proving the assertion. Now for k - 1 , . . . , r 1, the facts dl < d~ and di <_ d~ for i - 2 , . . . , k imply t h a t Sk(d) <_ Sk(d') - 1. Also, for k - r , . . . , q - 1, if Sk(d) - Sk(d'), then Sk+l(d) > Sk+l(d') as dk+l > d~+ 1 by the assertion. But this contradicts d C D~. Hence again Sk(d) <_ S k ( d ' ) - 1. ! 2" Let q - d p + 1 _> p + 1. From L e m m a 15.4.14, dq - p. S i n c e p < q,

Threshold Weights

418

and

Measures

dp _> dq - p. Thus d l > . . . > dp. Using L e m m a 15.4.15, we obtain di _< d~ for i - 1 , . . . , p - 1. This and the assumption d l < d~ complete the proof. 9 P r o o f of T h e o r e m 15.4.10. From L e m m a 15.4.12 we have R(d) >_ [ ~ 1 , as a reverse unit transformation can decrease 5(d) by at most 2. We show that if 5(d) > 0, we can always construct a proper sequence e such that 1. e is obtained from d by a reverse unit transformation, 2. e is a degree sequence, and 3.

= e(d)-

2.

It then follows that 5(d)is even and R(d) - 6(d) 2 " To show that e is a degree sequence we use the Berge Condition of Theorem 3.1.7, i.e., & ( e ) is even and Sk(e)_< Sk(e')for k = 1 , . . . , n . To show that 5(e) = 5 ( d ) - 2 we use the observation made after the proof of L e m m a 15.4.12. Without loss of generality, we may assume that dn > 0, for if dn = 0, then we work with c - ( d l , . . . ,d~-l). We may also assume that d 1 < d~, for if d 1 - d i ( - 7 " t - 1), then we work with c - ( d 2 - 1 , . . . , d ~ - 1). We distinguish two cases. C a s e 1" There exists an i > 1 such that di < d~. Let p be the largest such i. The basic idea is to transfer the 1 at the end of column p to the end of row 1. Let s = dl + 2 be the destination column for the moving 1. We have two subcases now. ! ! C a s e 1.1" dp _< p - 1 . Then q - dp is the source row for the moving 1. Also dq = p - 1 by L e m m a 15.4.14 and hence p is the source column for the moving 1. Define e = d - uq + Ul. Then for i = 1 , . . . , n , (a) ei = di except for e l - - d l q- 1 and eq - - d q - 1, and (b) e Ii - - d Ii except for epI - d Ip - 1 and Cs,

1.

We first prove that e is a degree sequence. Since 5's(e') = S'n(e), it sumces to prove S'k(e) _< S'k(e'), for k = 1 , . . . , s - 1. This is done as follows: Sk(e)=Sk(d)+l_< Sk(d') = S k ( e ' ) for k = 1 , . . . , q - 1 (by L e m m a 15.4.21), Sk(e) = Sk(d) < Sk(d') = Sk(e') for k = q , . . . , p - 1,

15.4

Threshold and Majorization Gaps

Sk(e)=Sk(d)<_Sk(d')-i

419

=Sk(e')

fork=p,...,s-1

(by L e m m a 15.4.20). To show that 5(e) = 5 ( d ) - 2, note that (p, r, 7r, a) = (q, p, 1, s), where p, T, 7r, a are as defined in the proof of L e m m a 15.4.12. By assumption p -r 1. ! Also 0 < dn _< dp < dp i m p l i e s p _ < s - l , and q < p. H e n c e q < s - l , giving q =fi s. Now the necessary and sufficient conditions for 5(e) = 5 ( d ) - 2 are verified as follows" (a) d ~ - d 1 ~ 1 by assumption; (b) d ' q - dq < 0 as dq - p - 1 and d'q _< p - 2 by L e m m a 15.4.14; (c) d'~ - d~ < 0 as d~ _> d~ _> 1 and d'~ - 0 ; and (d) d ' p - dp >_ 1 by definition of p. ! C a s e 1.2" dp > p. From L e m m a s 15.4.20 and 15.4.21 we have Sk(d) < Sk(d'),

k-

1,..., n-

1.

(15.39)

Let q - dp! + 1 be the source row for the moving 1, and define e as in Case 1.1. We first show that Sk(e) _< Sk(e') for k = 1 , . . . , s - 1. It follows from <_ d ~ <_ L e m m a 15.4.14 that dq - p and dq' <_ p - 1 . Note t h a t s - 1 - d l + l n - 1. We have: Sk(e) = Sk(d) + 1 <_ Sk(d') = Sk(e') for k = 1 , . . . , p - 1 (by (15.39)), -

_<

+

-

(since ep - dp _< dpI _ _ 1 _ _ epl ) , Sk(e) = S k ( d ) + 1 <_ S k ( d ' ) - 1 = Sk(e') for k = p + 1 , . . . , q 1 (by L e m m a 15.4.19), Sk(e)=Sk(d)<_Sk(d')-i =Sk(e') fork=q,...,s-1 (by (15.39)). To show that 5(e) = 5 ( d ) - 2, observe that again (p, T, 7r, a) = (q,p, 1,s). ! If q r s, then (a), (c), and (d) are as in Case 1.1, and (b) d q - dq < 0 by L e m m a 1 5 . 4 . 1 4 . If q - s, t h e n d ~ - d l >_ 1 a n d d ' p - d , _ > 1 as before, and I / ! dq - dq ~ - 2 since dq - p ~ 2 and dq - d s - O. C a s e 2: i - 1 is the only index with the property di < d}. Then dn - 1, for dn >_ 2 implies d~ - n - 1 - d~ > dl >_ d2, contradicting the assumption of the case. We assert that d~ >_ dl -t- 2. Indeed, we show that d~ - dl ~- 1 implies that S~(d) is odd, contradicting the assumption that d is a degree sequence. We have d~ - d 1 + 1 , d~ <_ di, i-2,...,n-1, d" - d ~ - l - 0 .

420

Threshold Weights and Measures

By adding we obtain ~n(d') < Sn(d) = Sn(d'). Thus all the inequalities above are in fact equalities, i.e., d~ - di for i - 2 , . . . , n - 1. Therefore the sequence c = ( d l , . . . , dn-1,0) has a symmetric corrected Ferrers diagram. This implies that S~(c) is even, which in turn implies that S n ( d ) is odd. This proves the assertion. Let s = d l q- 2 and define e = d - un + U l. Then for i = 1 , . . . , n, ei = di except for el - dl -t- 1 and en - d n - 1 - 0, and e Ii - d Ii except for e II - d II - 1 and e ts - 1. To show that e is a degree sequence, set q = s (_< n - 1), p = 1 and apply L e m m a 15.4.19. Then for k = 1 , . . . , s - 1, Sk(e) = S k ( d ) + 1 <_ S k ( d ' ) - 1 = Sk(e'). Clearly for k = s , . . . , n, Sk(e) <_ Sk(e'). It remains to verify the necessary and sufficient conditions for 5(e) = 5 ( d ) - 2. Note that here (p, T, 7r, ~r) = (n, 1, 1, s). Observe that s ~ n by the assertion. Now (a) d~ - d 1 ~ 2 b y the assertion; (b) d~ - d n = - 1 < 0; and

(~) d'~- d ~ - - d ~ < - d ~ < 0. We now study the connection between the majorization gap R ( d ) and the threshold gap t(d). Specifically we show that these gaps are equal and every threshold sequence achieving the majorization gap also achieves the threshold gap. Theorem

15.4.22 For every p r o p e r degree sequence d,

t(d)

-

R(d).

P r o o f . Let A - {i " 1 < i < m , d ~ - d i > 0}, A' - {i " 1 < i < m , d } - d ~ < 0}, and B - {i 9 m + l <_ i <_ n , d ' ~ - d i >_ 0}. Define a+ - E~eAd'~--d~, a _ -- ~ i e A , d ~ - di, /3+ - ~ i e B d } di. It follows from the equality (15.4.6) that a _ - - / 3 + . Therefore 1

t(d)

=

~

=

, 1 1 [di - d~l - ~(a+ - a _ ) - ~(~+ + 3+)

~

-~ ~ AK

~1 ~ iEAuB

(d'~ - d~) - ~1 ~~( d , -,

d~)+ _ 15(d) - R(d),

i=1

where the last equality follows by Theorem 15.4.10. 9 The following result states that all threshold sequences achieving the majorization gap R ( d ) also achieve the threshold gap t(d). Recall that if a ~ b, U ( a , b) denotes the m i n i m u m number of unit transformations required to transform a into b.

15.4

Threshold and Majorization Gaps

421

T h e o r e m 15.4.23 Let d be a proper degree sequence and e a proper threshold sequence such that e ~ d and U(e, d ) - R(d). Then lie- dll- t(d) P r o o f . It suffices to prove lie- dll - R(d) by virtue of Theorem 15.4.22. Since e > d, we have S n ( e ) - Sn(d), and so

( e i - di) ei >di

~

( d i - ei).

(15.40)

ei < di

We then have, where n is the length of d and e, n

lie- d[I -- ~ ~ le~- &l i=1

=

1 ~ ~

1 (~, - <) + ~ ~

ei >di

-

~

(< - ~)

ei
(ei-di)

(by (15.40))

ei>di n

=

E(~,-

<)+

i=1

=

~(~, d)

=

U(e,d)

=

R(d).

(by Lemma 15.4.11)

We remark that the assumptions that e is threshold, l i e - dll - t(d), and S~(d) - Sn(e) do not imply that e > d. For example consider d (6,6,4,4,3,3,1,1) and e - ( 7 , 6 , 4 , 3 , 3 , 2 , 2 , 1 ) . We now consider the problem of characterizing the degree sequences with m a x i m u m majorization. We show that for a fixed number of edges and a variable number of vertices, R(d) is maximized precisely when d is the degree sequence of a matching plus isolated vertices. For a fixed number n of vertices and a variable number of edges, we exhibit all the degree sequences that maximize R(d), and they turn out to be almost ~-regular. As is customary, we denote the constant sequence (a, a , . . . , a) of length n by a n, the sequence (a, b , . . . , b) by ab n-l, etc. T h e o r e m 15.4.24 Let d be a proper degree sequence with a fixed sum 2q > O. Then R(d) <_ q - 1.

Furthermore, equality holds if and only if d is of the form d - 12qO~.

Threshold Weights and Measures

422

P r o o f . We have

~(d)

-

~ ( d ' , - d~)+ i>l

=

( d '1 -

dl)+

-t-

E ( d t i - di) + i>2

_< d'l -d~ + ~; d'~

(~n~ dl _> d~ ~na d~, d~ _> 0).

i>2

If dl > 2, then 5(d) < Ei>l d ~ - 2 - 2q - 2. If dl - 1, then d - 12q0r, and so 5(d) - 2 q - 2. Conversely, let d satisfy ~ > l ( d } - d ~ ) + - 2 q - 2 . Assume that, if possible, dl>

1. T h e n

(d i - d l ) + 2 (d~-d~)+

dl-dl _< dl _< d~,

--

i>_2.

Adding these inequalities, we obtain 2q - 2q, hence all the above inequalities hold as equalities. Therefore for i >_ 2, if d} > 0, then di - O. But this fails for i - 2, as d~ >_ dl - 2. [] Let us now consider the majorization gap of the degree sequences of a fixed length n. We make use of the following functions:

f(n)

f(n)-l, -

L~J

[~],

g(n) -

f(n),

ifn-3mod4 otherwise.

We reproduce the definition of Dn for convenience. For positive integers n, Dn - { d -

( d l , . . . , dn)" d is proper, Sk(d) < Sk(d') for k - 1 , . . . , n}.

proper degree sequence of length 5(d) <_ g(n). Further, for n >_ 5, equality holds if and only if

Theorem

15.4.25

Let d be a

for n ~ 3 mod 4

d-

F~I

~

o~

d-l~J

n.

Then

~

n-3 2 '

for n - 3 mod 4 or

To prove Theorem 15.4.25, we find it convenient to prove the following theorem:

15.4

Threshold and Majorization Gaps

Theorem

15.4.26

423

For d C D~, 5(d) <_ f(n). Further, for n > 5, equality

holds if and only if d -

[~-~1 n or d -

We need several results to prove these theorems. about the functions f ( n ) and g(n).

1. f ( n -

1) < f ( n ) for n > 1, f ( n -

2. f ( n ) > n 3. g ( n - 1 )

First, some simple facts

1) < f ( n ) for n > 3;

2 for n _> 2, with strict inequality for n > 5;

< g(n) for n > 4, g(n) > n + l f o r n > 7 .

Next, three lemmas about corrected Ferrers diagrams. 1 5 . 4 . 2 7 Let 0 r d C D n . Put s - dl + 1, p - d~s, q - d~p + 1, r - ds. Thus s > _ r e > p > 1 and q > m > r > 1. Assume that the following hold (See Figure 15.11): Lemma

1. s > m + 1, and if equality holds, then p < m; 2. q < s ; 3. r < p - r . Then: 1. there exists a sequence e C

Dn

such that

e I

--

dl-1 and 5(e) > 5(d)+2;

2. there exists a sequence f c D~ such that (a) (i) f l

-

dl -

1 or (ii) f l

-

dl

and f'~ - 1;

(b) 5 ( f ) > 5(d) + 1; (c) S n ( f ) and Sn(d) have the same parity. P r o o f . We begin by proving s t a t e m e n t 1. First note that p _< m _< q and r < p. Secondly, we m a y assume that the first r columns of C(d) are full, i.e., d~i - n - 1, i - 1,...,r, (15.41) for otherwise we can fill column 1, then 2, and so on up to r (by adding l's at the end of these columns) without going out of D~, and increase 5(d) at each step, by L e m m a 15.4.17.

424

Threshold Weights and Measures

Figure 15.11: Illustrating L e m m a 15.4.27. r

p

m

q

s

*

0

m

0

Counting in two ways the number of l's in rows 1 , . . . , s and columns r + 1 , . . . , p of C(d), we obtain p

~

d:-(p-r)(q-1)+ i=r+l

(15.42)

(di-r).

i=q+l

Since d e Dn, ~p(d) ~ ~p(d'). This implies, by (15.41) and (15.42), p(s - 1 )

p


~

d'i - r ( n - 1 )

+ (p- r)(q-1)

i=r+l

•

(di - r).

i=q+l

Therefore p(s - q) <_ r(n - 1 - q) + r + EiLq+l (di ~--q+l ( d ~ - r), and it follows that (p-

+

r)(s - q) < r(n - l - s) + p +

~

F) < r ( r t - -

(di - r).

1 - q) + p +

(15.43)

i=q+l !

!

' for if d~ < d~+1 for some By L e m m a 15.4.16, dr+ 1 _> dr+ 2 >_ 9.. > dp, i < p _< q, a contradiction. Also r < i < p, then q - 1 _< d~ < d~+1 -

15.4

425

Threshold and Majorization Gaps

di-s-1

I a <_s-2. andd~+

fori-r+l,...,p,

d~ > d'~,

Hence

i - r + 1,...,p.

(15.44)

I I l l Again by Lemma 15.4.16, dlq+l ~ dq+ 2 ~ . . . ~ ds, f o r if d i < di+ 1 f o r some q < i < s, then d~+1 - i > q >_ m, a contradiction. Therefore for i - q + 1 , . . . , s , di <_ dq+l _< p - 1 and d~ >_ d'~ - p . Hence

d'~ > d,,

i-

q + 1,...,s.

(15.45)

Using (15.41), (15.44) and (15.45), we have q

5(d) - r(n - s) + ~

(d'i - di) + + ~

i=p+l

(d'i - di).

(15.46)

i=q+l

Define a sequence e such that C(e) is obtained from C(d) by making the first p columns full and deleting the s-th column, i.e., , ei-

f n-l, / 0, d~

fori-l,...,p fori-s otherwise

It is easy to check (using s > m + 1) that e C Dn by Lemma 15.4.17. Further, s--1

~(~) -

~(~,,-

~)+

i=1 q

= p(n-s+l)+

s-1

y~ (d'i - d i ) + +

y~ (d'i - p )

i=p+l

=

i=q+l q

~(~ - ~) + ( p - ~)(~ - ~) + p + ~

(d'~ - d~) + +

i=p+l

(d'~ - p)

(~s d' - p)

i=q+l q

=

r(n-s)+(p-r)(n-s)+p+

Z

(d',- d,) + +

i----p+l

(d'i - di) i--q+l

=

~

( p - di)

i-q+l

~(d) + ( p - ~)(~ - ~) + p -

~ i=q+l

( p - d~)

(by (15.46))

Threshold Weights and Measures

426

=

5(d)+(p-r)(n-s)+p-

)_2 ( P - r ) + i=q+l

=

~(d) + (p - ~)(~ - ~) + p -

~_~ ( d i - r ) i=q+l

(~ - q ) ( p - ~) + ~

(d~-

~)

i=q+l

> 6(d)+ ( p - r)(n - s ) - r(n - 1 - s) = ~(d) + ( p - ~)(~ - ~ ) - r(~ - ~ ) + >

~ ( d ) + ~.

( ~ ~ < p-

(by (15.43))

~)

So 5(e) >_ 5(d)+ 2. This proves Statement 1. We now prove Statement 2. If S'~(e) has the same parity as S~(d) (before the achievement of (15.41)), then f = e has the required properties. If the parities differ, take f = e + u l . In this case C ( f ) can be obtained from C(d) by making the first p columns full and deleting all the l's except the first 1 in the s-th column. So f C Dn by Lemma 15.4.17. The other required properties of e follow from S~(I) = S~(e)+ 1, 5~(f) = 5 1 ( e ) - 1, and 5~(f) = 5~(e) for i = 2,...,n. ,, L e m m a 15.4.28 relaxes one of the assumptions of L e m m a 15.4.27. Lemma

15.4.28

Under the conditions of Lemma 15.4.27 except r <_ p - r"

1. there exists a sequence e G Dn such that

e I

--

d l - 1 and 5(e) >_ 5(d) + 1;

2. there exists a sequence f C D~ such that (a) (i) f l -- dl - 1 or (ii) f a - dl and f~ - 1; (b) in case (i) above, 5(f) > 5(d) + 1; in case (ii) above, 5(f) >_ 5(d); (c) S~(f) and con(d) have the same parity. P r o o f . If r _< p - r, then the results follow from Lemm~ 15.4.27, so assume r > p - r. As in the proof of Lemma 15.4.27, we may assume that (15.41) holds. We also deduce Equation (15.46) as before. Define a sequence e such that C(e) is obtained from C(d) by deleting the s-th column. Then c E D~ by L e m m a 15.4.17 and

~(~)

s--1

-

E(~'~-

~)+

i=1 q

=

r(n+l-s)+

~ i=p+l

s-1

(d'i - d i ) ++ ~ i=q+l

(d'i - d i )

15.4

427

Threshold and Majorization Gaps

=

~(~ - ~) + ~ +

Z

(d~ - d~) + +

i=p+l

=

5(d)+r-p+r

>

~(d).

( d ' , - d,) - d: + d~ i=q+l

(by (15.46) and the definition o f p a n d r )

( ~ ~ > p - ~)

This proves Statement 1. To prove Statement 2, define f from e as in the proof of L e m m a 15.4.27. 9 L e m m a 15.4.29 is a variation of L e m m a 15.4.27. L e m m a 1 5 . 4 . 2 9 Let 0 ~ d C Dn. Put s - da + 1, p - dis, r - ds. A s s u m e that s - m + 1 and p - m (see Figure 15.12). Figure 15.12" Illustrating Lemma 15.4.29.

r

m8

.

m 8

0

-k

0

O.

Then 1. if m > 3, then there exists a sequence and 5(e) > 5(d) + 1;

e C Dn such that el - d l - 1

2. if m > 4, then there exists a sequence f c Dn such that (a) (i) f l = d l -

1 or (ii) f l - - d ,

and f~ = 1;

(b) 5 ( f ) > 5(d)-4- 1; (c) S ~ ( f ) and Sn(d) have the same parity. P r o o f . We begin by proving Statement 1. First, observe that m < n - 1, because m = n - 1 and p = m would imply Sin(d) > Sm(d'), contradicting d E Dn. Secondly, assume without loss of generality that (15.41) holds as in

428

Threshold Weights and Measures

t h e p r o o f of L e m m a 15.4.27. Define a s e q u e n c e e such t h a t C ( e ) is o b t a i n e d f r o m C(d) by m a k i n g t h e first m - 1 c o l u m n s full, if necessary, a n d d e l e t i n g t h e c o l u m n s - rn + 1, i.e.,

ei! m

rtml 07

d~, Then e E

Dn

~ for i - 1 , . . . , r n for i - m + 1 otherwise.

1

by L e m m a 15.4.17. Using t h e facts d} - n -

di

m, d~ - m - 1 for i - r + 1 , . . . , definition of m a n d r), we o b t a i n

5(d) - r ( n -

1 - m)+ m-

m,

d i 'n +

r < (m-

1 -

m,

1 for i - 1 , . . . , r, < 1 (by _ m -

dm+l - r

1)(n - 1 - m ) +

m - r.

Also, 5(e) - ( r n -

1)(n - 1 - ( r n

- 1)) - (rn - 1)(n - 1 - r n ) + m - 1.

It is now clear t h a t if r > 1, t h e n 5(e) > 5(d). If r = 1, t h e s a m e c o n c l u s i o n holds, since m - 1 > 1 a n d n - 1 - m > 0. T h i s proves S t a t e m e n t 1. W e now prove S t a t e m e n t 2. If Sn(d) before t h e a c h i e v e m e n t of (15.41) a n d S~(e) h a v e t h e s a m e parity, t h e n f = e has t h e r e q u i r e d p r o p e r t i e s . O t h e r w i s e t a k e f = e + Um+l. O n c e again, C ( f ) can be o b t a i n e d f r o m C(d) by m a k i n g t h e first m - 1 c o l u m n s full a n d d e l e t i n g all t h e l ' s e x c e p t t h e first 1 in c o l u m n m + 1. T h e r e f o r e f C Dn by L e m m a 15.4.17. F u r t h e r , (~(f) = ( ~ ( e ) - 1, as (~l(f) = ( ~ l ( e ) - 1, a n d 5~(f) = 5~(e) for i = 2 , . . . , r n . Hence 5 ( f ) = (rn - 1)(n - 1 - rn) + m - 2. N o w it is clear t h a t if r > 2, t h e n 5 ( f ) > 5(d). If r - 2, t h e s a m e c o n c l u s i o n holds since m - 1 > r ( b e c a u s e m >_ 4) a n d n - 1 - m > 0. Finally, if r - 1, t h e n t h e conclusion holds again, since

5(f)

-(m-2)(n-l-rn)+n-l-m+m-2 >_

(m-2)(n-l-rn)+m-1

>

n-l-rn+m-1

(asn-2>rn)

(asm>_4andn-l-m>O)

=

W e are now r e a d y to prove T h e o r e m s 15.4.25 a n d 15.4.26.

15.4

Threshold and Majorization Gaps

429

P r o o f o f T h e o r e m 1 5 . 4 . 2 6 . T h e s t a t e m e n t is true for n - 1, so a s s u m e n > 2. If d n - 0, t h e n by induction on n, 8(d) < f ( n 1) < f ( n ) , and f ( n - 1) < f ( n ) for n > 3. We m a y therefore assume t h a t d~ > 0, and c o n s e q u e n t l y m > 2. If m 2, then 5(d) < n - 2 by L e m m a 15.4.18, and hence 8(d) _< f(n). E q u a l i t y holds if and only if d~ - d2 and n - 2 - f ( n ) , which implies n < 4. Hence we m a y assume m _> 3. P u t s - dl + 1, p - d~s, q - dp~ + 1, r - ds, and observe as before t h a t p _< m < q, and m < s Proposition. If s > m + 1, then there exists a sequence e C Dn such that el

-

-

dl

-

1 and 5(e) > 5(d). Consequently we may assume that s - m.

To prove the proposition, first assume t h a t q > s. Define a sequence e such t h a t C(e) is o b t a i n e d from C(d) by m a k i n g the first p columns full, and deleting the s-th column. T h e n e C D~ and 5(e) > 5(d) by L e m m a 15.4.17. Now a s s u m e q < s. T h e n r d~ < dq+l < p - l , and h e n c e r < p. I f s > m + l or p < m, t h e n the required sequence e exists by L e m m a 15.4.28. If s - m + 1 and p - m, t h e n the required sequence e exists by L e m m a 15.4.29. This proves the proposition, and we m a y assume t h a t s - m. F u r t h e r , we m a y a s s u m e t h a t the first m - 1 columns of d are full, for otherwise we m a k e t h e m full w i t h o u t leaving D~, t h e r e b y increasing 8(d) by L e m m a 15.4.17. Then d - ( m - 1) n ,

5(d)- (m-

1)(n-

1 -(m-

1)),

and 5(d) reaches a m a x i m u m when n--1

m-

2 ' n ~ 7-1,7,

1 -

nodd neven.

(15.47)

Therefore 5(d)_<

__tl )2, ~(~-1),

n odd neven.

This m e a n s t h a t 5(d) <_ f(n). Further, for n _> 5, equality holds if and only if d - ( m - 1) n, where m - 1 is given by E q u a t i o n (15.47), i.e., if and only if d-[~-~l

Proof of Theorem

n

or

d - L - ~ - ! j n.

1 5 . 4 . 2 5 . We use the n o t a t i o n E~ - {d E D ~ - S~(d) even},

Threshold Weights and Measures

430

i.e., En(d) is the set of all proper degree sequences of length n. Since E~ C_ Dn, we have 5(d) _< f ( n ) by Theorem 15.4.26. For n ~ 3 mod 4, f ( n ) - g(n) and hence 5(d) _< g(n). The cases of equality for d C Dn are when d - [ _ ~ ] n or d - [~-~]~. Since these d belong to E~ when n ~ 3 m o d 4 , all the conclusions of Theorem 15.4.25 are established in this case. However, for n = n-l~2 is odd, whereas (~(d)is even by Theorem 15.4.10, since 3 rood 4, f ( n ) - (--5-~ d is a degree sequence. Therefore (~(d) < f ( n ) - 1 - g(n) for n - 3 mod 4. It remains to track down the cases of equality for n - 3 mod 4. Thus from now on, we assume that n - 3 rood 4, n _> 7, and ~i(d) - g(n). If d n - - 0 , then ~(d) <_ g ( n - 1) < g(n), contradicting our assumption, so we may assume that dn > 0, and therefore m > 2. Also m - 2 implies, by L e m m a 15.4.18, that ~i(d) _< n - 2 < g(n), again contradicting our assumption, so we assume that m > 3 . P u t s - d l + l . Proposition. If s > m + 2, then there exists a sequence f E En such that

(~(f) > ~(d), contradicting ~(d) - g(n). Consequently we may assume that s-m ors-m+ l. We prove the proposition by showing that, when s > m + 2, 1. if d's - 1, then there exists an f E En such that 5(f) > 5(d); 2. if d'~ >_ 2, then there exists an f E E~ such that 6(f) >_ 5(d), and if equality holds, then fl - d~ (so that f and d have the same s) and f;-1. C a s e 1" d's - 1. The basic idea is to work with the sequence ( d l 1, d 2 , . . . , d~) and introduce a 1 at the end of the first row if the parity becomes odd. Put t - s - l , p - d ~ t , q d tp + l . T h e n p _ < m_< q. Assume that q >_ t. Define a sequence f such that C ( f ) is obtained from C(d) by deleting the last 1 in column s - 1 and the 1 in column s. Then f C E~ and 5(f) > 5(d) by L e m m a 15.4.17. Therefore we may assume q < t. Put r = dr. See Figure 15.13. C a s e 1.1: t > m + 1 or p < m. Define a sequence c such that C(c) is obtained from C(d) by deleting the last 1 in the first row. Note that c E D~ by L e m m a 15.4.17 and, further, c satisfies all the hypotheses of L e m m a 15.4.28. By the latter, there exists a sequence g E Dn such that 5(g) >_ 5(c) + 1. The required sequence f is defined by f

_ J" g, (g~ + 1 , 9 2 , . . . , g ~ ) ,

\

Sn(g) even S~(g) odd.

15.4

Threshold and Majorization Gaps

431

Figure 15.13" Illustrating Case 1 of the Proposition in the proof of Theorem 15.4.25. r

m

p

m

ts

.jrm

0

To see this, recall that in the proof of L e m m a 15.4.28, C(g) is obtained from C(c) by making the first p columns full and deleting the t-th column (if r _< p - r ) , or by making the first r columns full and deleting the t-th column (if r > p - r). Note that when S~(g) is odd, C(f) could be obtained from C(c) by making the appropriate columns full and then deleting all but the first 1 in column t. Therefore f C Dn in this case by L e m m a 15.4.17. Clearly f E Dn also holds when S~(g) is even. Further, Sn(f) is even in both cases. Thus f C E~. From the construction of c and f it is clear that 5(c) - (5(d)+ 1 and (5(f) _> ( 5 ( g ) - 1. Therefore 5(f) >_ 5 ( g ) - 1 _> (5(c) - 5 ( d ) + 1. 1.2" t - m + 1 and p - m. This is similar to Case 1.1, except that here we use L e m m a 15.4.29 instead of L e m m a 15.4.28. Case

C a s e 2" 2 _< d', _< rn. Put p - d'~ and q - d'p + 1. If q _> s, then we may remove the last two l's in the s-th column without leaving En and thereby increase (5(d) by L e m m a 15.4.17. We therefore assume that q < s. Then the required f exists by L e m m a 15.4.28.

This completes the proof of the proposition and hence we may assume that s - m or s - m + 1.

Threshold Weights

432

and

Measures

It is convenient here to define a new function. Let a _> 6 be a fixed integer such t h a t a - 2 m o d 4. For 0 _< k _< a, define k(a

h~(k) -

k(a-

-- k),

k)-

k

l,

even

k odd,

i.e.,

h~(k)- 2 L"~-"~ 2 J. Note t h a t the m a x i m u m of ha(k) occurs at k -

~2

1, 7, 7 + 1~, ~

and the

m a x i m u m is ~2-4 4 " Proposition. if and only if

If s - m or s - m + 1, then 5(d) _< h n - l ( m -

d

-

(m-l)

d

-

(m -

~

1), with equality

formodd

1) n - 1 ( m -

2)

or

d - m(m

-

1) n - 1

f o r m even.

To prove the proposition, consider first the case t h a t s - m. If columns 1 , . . . , m - 1 are m a d e full in this order, then d stays in Dn and 5(d) increases at each step by L e m m a 15.4.17. For odd m, the resulting d - ( m - 1) n also belongs to E~, and so it is the only sequence of E~ satisfying s - m t h a t m a x i m i z e s 5(d). T h e m a x i m u m in this case equals ( m - 1 ) ( n - 1 - ( m 1)) hn-1 (m1 ) . For even m, the above d does not belong to E~, and so the only sequence of E~ satisfying s - m t h a t maximizes 5(d) is the previous sequence in the filling-up process, n a m e l y d - ( m - 1 ) ~ - l ( m - 2). T h e m a x i m u m in this case equals ( m - 1)(n - 1 - ( m 1 ) ) - 1 - h~-i (m - 1). Now consider the case t h a t s - m + 1. We assert t h a t if d'~ _> 2, t h e n there exists f C E~ such t h a t 5 ( f ) > 5(d), and consequently we m a y a s s u m e t h a t d'~- 1. To prove the assertion we distinguish two cases. ! C a s e 1" 2 _< d~ < m. P u t p - d~, r - d~, q - dp + 1. We m a y a s s u m e t h a t q < s, for otherwise the last two l's in the s-th column of d m a y be r e m o v e d w i t h o u t leaving E~, t h e r e b y increasing 5(d) by L e m m a 15.4.17. T h u s q - m. If r _< p - r , then the required f exists by L e m m a 15.4.27, so we a s s u m e t h a t r > p - r. Using dli > rn and di - rn for i -- 1 , . . . , r, d~ - m - 1 and di - rn ' for i - r + 1 , . . . , p, d~ - di - r n - 1 for i - p + 1 , . . . , m , din+ 1 - - p, d m + l - r and d} - 0 for i - rn + 2 , . . . , n , we obtain 5(d) - ~ ( d ' i - di) + p i=1

r.

(15.48)

15.4

Threshold

and Majorization

Gaps

433

Let f be a sequence such that C(f) is obtained from C(d) by (a) deleting the s-th column and (b) adding a 1 at the end of the (r + 1)-st column if p is odd. Then f C En by L e m m a 15.4.17 and 7-

5(f)

>

~-~(d'~- ( d ~ - 1)) i=1 r

=

r

i=1

>

5(d).

(by (15.48) a n d r > p - r )

C a s e 2" d~, - m. If m > 4, then the required f exists by L e m m a 15.4.29. Now assume the special case m - 3. Then d - 32d'2-21n-d'2-1 with d~ > 2 (see Figure 15.14).

Figure 15.14: Illustrating a special case in the proof of Theorem 15.4.25. m

* 1

1 1

1 1 * 1 1 0

1

1

m s

:

d;-t-1

*

:

1 1 1

n

8

1 1

0

1

If d~ - 2, then Sn(d) - n + 6 is odd, since n - 3 mod 4, contradicting d C En. Therefore d~ >_ 3. Hence 5(d) - ( n - 4 ) + + ( d ; - 3 ) + + ( 2 3) + + ( 3 - 2 ) + - n+d~-6 _< 2 n - 7 , and writing n - 4 k + 3 , we have 5(d) <_ 8 k - 1 < 4k 2 + 4 k - ( 2 k + 1) 2 - 1 - g(n), contradicting the assumption This completes the proof of the assertion, and consequently we may assume that d ~, - 1. If m is odd, then make the first m - 1 columns of C(d) full and delete the last 1 in the first row without leaving En, thereby increasing 5(d) by L e m m a 15.4.17. Therefore m may be assumed to be even. Then it is easy to

Threshold Weights and Measures

434

check as before that d = m ( m - 1)n-1 is the only sequence in En satisfying s =rn + 1 and d~ = 1 that maximizes 5(d), and 5 ( d ) = hn-1( r n - 1). This proves the proposition, and so 5(d) _< h ~ _ l ( r n - 1). The function h ~ _ l ( m - 1) reaches its m a x i m u m when rn-

1 - n-1 2

1'

or

rn-

1 --

n -21

'

or

m--l--

_~ +1.

The m a x i m u m i s ( _ ~ ) 2 _ 1 - g(n). By the previous proposition it is achieved only by the following sequences d: when r n -

1 - n-x 2

when r n -

1 - -~

1 (rn odd)" (rn even)"

d-

(

or d when r n -

)nl ~

2 ---fn + l (~__A1 ) ~-1

1 - - ~ + 1 (rn Odd)"

See [AP94] for related results about the difference gap and the fact that every degree sequence d such that R(d) - 1 has a realization that is both bithreshold and cobithreshold.

Chapter 16 Threshold Graphs and Order Relations 16.1

Introduction

In a type of test frequently used in social sciences, the subjects in a set A are presented with all the dichotomous items in a set B. For instance, suppose A is a set of children and B is a set of skills. Do all the children learn the skills in the same order? At what age do children learn a given skill? To answer these questions, define a relation R from A to B such that aRb if and only if child a has learned skill b. The data can be interpreted by the assumption that easier skills are learnt first. This idea, which is due to G u t t m a n [Gut44], amounts to the existence of two scales f : A --. IR and g : B --. N, such that

aRb if and only if f(a) > g(b). In this interpretation, f is a scale measuring the subjects ability, and g is a scale measuring the difficulty of the skill. Ducamp and Falmagne [DF69] have shown that for finite sets A and B, such scales f, g exist if and only if the following condition holds for R:

(aRb and cRd) ~

(aRd or cRb).

We have already seen essentially this result in Section 4.5. A relation R satisfying this condition is called biorder. Doignon et al. [DDF84] have investigated biorders of arbitrary cardinality. The results in the next two sections 435

T h r e s h o l d Graphs and Order R e l a t i o n s

436

are based on their work. In the finite case, many of their results were obtained independently by Bouchet [Bou71] and Cogis [Cog79]. Observe that if G is a bipartite graph with color classes A, B and edge set {ab : aRb}, then G is a difference graph if and only if R is a biorder. As we see in the next two sections, some of the results on difference graphs and threshold dimension were obtained independently by these authors using the terminology of biorders and bidimensions. We define below various order relations that are used in this chapter. First recall that a relation R C A x A is said to be reflexive if aRa for each a E A; irreflexive if -,aRa for each a E A; symmetric if aRb ---5, bRa for all a, b E A; asymmetric if aRb m ~ -~bRa for all a, b E A; antisymmetric if for all a 5r b, aRb ----5, -,bRa; transitive if (aRb, bRc) ~ a r c for all a, b, c E A; negatively transitive if the complement of R (i.e., (A x A ) - R) is transitive; complete if for all distinct a, b E A, aRb or bRa; strongly complete if for all a, b E A, aRb or bRa; and countable if A is a countable set. D e f i n i t i o n 16.1.1 Let R be a binary relation on A, i.e., R C A • A. Then

9 R is a q u a s i o r d e r if it is reflexive and transitive; 9 R is a w e a k o r d e r if it is transitive and strongly complete; is a s i m p l e o r d e r if it is transitive, antisymmetric and strongly complete;

9 R

9 R is a s t r i c t s i m p l e o r d e r if it is transitive, asymmetric and complete; 9 R is a s t r i c t w e a k o r d e r if it is asymmetric and negatively transitive; 9 R is a p a r t i a l o r d e r if it is reflexive, transitive and antisymmetric; 9 R is a s t r i c t p a r t i a l o r d e r if it is transitive and asymmetric; 9 R is an e q u i v a l e n c e r e l a t i o n if it is reflexive, symmetric and transitive.

16.2

Biorders

437

A partially ordered set is also called a poset. A complete partial order is also called total or linear. If R is a partial order on A, then a least element of a subset B of A is an element a C B such that aRb for every b C B. Note that a least element need not exist. A well-ordered set is a poset in which every nonempty subset has a least element. In particular, a well-ordered set is a chain (i.e., a set on which the partial order is total). The Well-Order Principle, equivalent to the Axiom of Choice, states that every set can be well-ordered.

16.2

Biorders

Recall that a biorder is a relation R from A to D such that for all a, b, c, d,

(aRb and cRd) ----5, (aRd or cRb).

(16.1)

We characterize biorders in this section. All the results are due to [DDF84]. We first need some notations and definitions. An ordered pair (a, d) is often denoted by ad. Thus aRd ~ ad C R. The converse relation R -1 is the set of all ordered pairs da such that ad C R. The complement of R is denoted by R. The relative product R S of two relations R and S is defined by

R S - {a f " aRd, dS f for some d}. It is easy to see that a binary relation R is a biorder if and only if

~/~-1/~ C /~.

(16.2)

Observe that if we associate a digraph G with the relation R the natural way (i.e., A U D as the vertex set and {ad: aRd} as the arc set), then G is a Ferrers digraph if and only if R is a biorder. Hence the graphical representation of Condition (16.1) is given in Figure 2.1 or Figure 2.2. Considering the forbidden subgraphs of Figure 2.2, it is easy to prove the following proposition. P r o p o s i t i o n 16.2.1 Let R be a biorder on a set E. If 1~ is reflexive on E, then R is strongly complete and negatively transitive on E. If R is irreflexive on E, then R is asymmetric and transitive on E.

T h r e s h o l d Graphs and Order R e l a t i o n s

438

Definition 16.2.2 Let R be a relation from A to D. For a C A and d C D we put a R - {y e D " aRy} and R d - {x e A" xRd}. Furthermore, we define a relation RA on A and a relation RD on D by aRAb

if and only if

aR 2 bR

dRDe

if and only if

Rd C_ Re

and for all a, b E A and d, e C D.

Some useful facts concerning RA and RD are collected below. P r o p o s i t i o n 16.2.3 For every relation R from A to D we have 1. RA and RD are quasi orders;

2. RA - -RR -1 and RD -- R-l-R; 3. the following conditions are equivalent: (a) R is a biorder; (b) RA is a weak order on A; (c) RD is a weak order on D.

We now turn to the problem of embedding a given relation R in the reals, in the following sense. Definition 16.2.4 A relation R from A to D is realizable w i t h r e s p e c t to < (respectively <) if there exist two mappings f " A ~ N and g" D ---+N such that aRd

if and only if

f ( a ) < g(d)

(resp. f ( a ) < g(d))

for all a C A and d C D.

It is useful to notice the following. P r o p o s i t i o n 16.2.5 (f,g) is a realization of R with respect to < if and only if (g, f ) is a realization of R -1 with respect to <.

16.2

Biorders

439

It is also easy to verify that if R is realizable with respect to _< or with respect to <, then R is a biorder. The converse is true if R is countable as stated in the following theorem and proved after Proposition 16.2.10. T h e o r e m 16.2.6 Let R be a countable relation from A to D. following conditions are equivalent:

Then the

1. R is a biorder; 2. R is realizable with respect to <_; 3. R is realizable with respect to <. As a first step in proving Theorem 16.2.6, we see that there is no loss of generality in assuming that A and D are disjoint.

D e f i n i t i o n 16.2.7 A (disjoint) d u p l i c a t i o n of a relation R from A to D is a relation R' from A' to D' defined as follows:

1. A' is a copy of A; i.e., there exists a one-to-one correspondence c~ from A' to A. 2. D' is a copy of D; i.e., there exists a one-to-one correspondence 5 from D' to D. 3. A' and D' are disjoint. ~. a'R'd' Proposition

if and only if

c~(a')RJ(d').

16.2.8 Let R' be a duplication of R.

1. R is a biorder if and only if R' is a biorder. 2. R is realizable with respect to <_ (resp. <) if and only if R' is realizable with respect to <_ (resp. <). The following is an important result that is used later. P r o p o s i t i o n 16.2.9 Let R C_ A x D for two disjoint sets A and D. Among all the relations Q on A U D such that

1. Q is a quasi order,

440

Threshold Graphs and Order Relations

2. Q N ( A • D ) -

R,

there exists one Qn that is maximal in the sense that it contains all the others. It is given by the formula QR -- R U RA U RD U R - 1 R R -1. P r o o f . Let Q be any relation on A U D satisfying Conditions 1 and 2. Then for every a, b E A and c, d C D,

9 aQb ~

aRAb since R C Q and Q is transitive,

9 cQd ~

cRDd since R C Q and Q is transitive,

9 aQc~aRcsinceQN(AxD)-R, 9 dQb ~ arc.

-~(dR-I-RR-lb) since R C_ Q, Q is transitive, and aQc ------>

Thus Q c_ Qn. It is straightforward to verify that Qn satisfies Conditions 1 and 2. 9 P r o p o s i t i o n 16.2.10 Let R C_ A x D for two disjoint sets A and D. Then the following conditions are equivalent:

1. R is a biorder; 2. Q R is a weak order on A U D. P r o o f . Assume first that R is a biorder. Then from Part 3 of Proposition 16.2.3, it follows that RA, RD are weak orders. In addition, for a E A and d E D, a-Rd ---> --,(dR-l~/~-la) ~ dQna. Hence it follows that for every a, b G A U D, aQnb or bQna. Therefore Qn is a weak order. Conversely, assume that Qn is a weak order. For every a, b E A and c, d E D such that arc, bRd and -~aRd, we have cR-I-RR -1 b. Hence -~cQRb. But then, Qn being a weak order, we must have bQRc and therefore bRc. Thus R is a biorder. " We now prove Theorem 16.2.6, which characterizes biorders among the countable relations. P r o o f of T h e o r e m 16.2.6. We only prove 1) -----5, 2). Then the rest of the implications easily follow.

16.2

Biorders

441

1) =~ 2): By Proposition 16.2.8 we may assume that A and D are disjoint. Let QR be the maximal extension of R as in Proposition 16.2.9. QR is a weak order on A U D by Proposition 16.2.10. Since A U D is countable there exists a function h defined on A U D such that for all p, q C A U D

pQnq

if and only if

h(p)<_ h(q).

(We can construct h by enumerating A U D and defining h inductively.) Thus if f, g are the restrictions of h to A, D respectively, then (f, g) is a realization of R with respect to _<. " One can derive from Theorems 16.2.14 and 16.2.16 below that the relation <_ on R is realizable with respect to _< but not realizable with respect to <. Similarly, the relation < on R is realizable with respect to < but not realizable with respect to <_. This shows that Theorem 16.2.6 does not hold in general for uncountable sets. We give other characterizations of biorders below. First we need some definitions and notations. D e f i n i t i o n 16.2.11 Let R be a relation from A to D. A subset M of A U D is said to be w i d e l y d e n s e if for all a E A and d C D, aRd implies the

existence of an element rn E M such that n

m

either (m C A, mRd, mRAa) or (m C D, dRDrn, arm).

(16.3)

A subset M of A U D is said to be o r d e r - d e n s e if for every p, q E A U D, qQnp implies the existence of an element m C M such that pQnm and mQnq, where as usual Qn is the maximal quasi order extension of R in the sense of Proposition 16.2.9. P r o p o s i t i o n 1 6 . 2 . 1 2 Let R C_ A x D be a biorder with A M D - 0 . every widely dense subset is order-dense.

Then

P r o o f . Let M be order-dense and let qQnp. We must show that pQRm and mQnq for some m E M. C a s e 1" p,q E A. Since qQnp, we have qRAp. Hence there exists some d C D such that pRd and qRd. Since M is order-dense, there exists some m C M such that either (a) q-Rrn and dRDrn or (b) m--Rd and mRaq. If (~), then from pRd, dRDm, it follows that pQRm. Since QR is a weak order and q-Rm, it follows that mQRq. If (b), then from pRd and mRd and from the

442

Threshold Graphs and Order Relations

fact that Qn is a weak order it follows that pQndQnm and hence pQRm. Also from mRAq it follows that mQRq. Case 2" p, q C D. This case is similar to Case 1. Case 3" p C A, q C D. From qQnp it follows that there exist a C A, d E D such that pRd, aRd and aRq. From aRd it follows that there exists some m C M such that either (a) aRm and dRDm or (b) mRd and mRAa. If (a), then from pRd and dRDm it follows that pQnm. Also since Qn is a weak order and aRm, aRq, it follows that mQnq. If (b), then since pRd, mRd, we have pQnm. Also from mRAa and aRq, we have mQRq. Case 4" p E D, q E A. From qQRp, we have qRp. Hence there exists some m E M such that either (a) qRm and pRDm or (b) mRp and mRAq. If (a), then from qRm we have mQRq. also from pRDrn, we have pQRm by the definition of Qn. If (b), then from mRp, we have pQnm. Also from mRAq we have mQnq by the definition of Qn. 9

L e m m a 16.2.13 ( B i r k h o f f - M i l g r a m T h e o r e m [Rob79, p. 112])

For every relation R on A, the following conditions are equivalent: .

There exists a function h" A ---, N satisfying aRb if and only if h(a) < h(b);

2. R is a weak order with a countable order-dense subset. T h e o r e m 16.2.14 For every relation R C_ A x D, R is realizable with respect

to < if and only if R is a biorder with a countable widely dense subset. P r o o f . Observe that the concepts of biorder, realizable, widely dense set remain invariant under duplication. See Proposition 16.2.8. Hence we may assume that A gl D - O. "If"" Let M be a countable widely dense subset for the biorder R. By Proposition 16.2.12, M is an order-dense subset, and by Proposition 16.2.10, Qn is a weak order. Hence by Lemma 16.2.13 there exists a function h 9 AUD --. IR such that aQnb if and only if h(a) <_ h(b). The restrictions of h to A and D constitute a realization of R with respect to <_ by Proposition 16.2.9. " O n l y if"" Assume that ( f , g ) is a realization of R with respect to _<. We have already observed that R must be a biorder. It will be shown that A has a countable subset A* that is widely dense. By a similar argument, one could alternatively show that D has a countable subset D* that is widely dense.

16.2

Biorders

443

Consider all the real numbers c~i in f ( A ) for which there exists some real number 5i in g(D) satisfying 5i < c~i and (5i, c~i)n f ( A ) -

~.

(16.4)

With every such c~i let us associate a particular 5i for which (16.4) holds. Notice that the open intervals (Si, c~i) are nonempty and disjoint. Indeed, consider two intervals (5i, c~i), (Sj, c~j) with c~i < c~j. We have c~i <_ 5j since otherwise c~i E (Sj,c~j)in contradiction with (16.4). As a consequence, the collection of c~i is countable. We can thus find a countable subset A 1 of A such that f(A~) contains all the c~. Furthermore, let us select a subset A~ of A such that for every pair (p < q) of rationals the following is true: (p, q)N I ( A ) 7/= rg implies the existence of an element m* C A~ such that p < f(m*) < q. Clearly, A~ can be taken to be countable. It remains to show that the countable set A* - A~ U A~ is widely dense. Let a C A and d C D be such that aRd. We therefore have f(a) > g(d). Let us consider two cases: (a) (g(d), f ( a ) ) n I ( A ) - ;g. In this case, f(a) is one of the c~. Let m* be the element in A~ such that f(m*) - f(a). One readily sees that m*Rd and m*RAa and Condition (16.3) is fulfilled. (b) (g(d), f ( a ) ) n I(A) :/: ;g. Let a' be an element in A such that g(d) < f(a') < f(a). Furthermore let p and q be rational numbers such that

g(d) < p < f(a') < q < f(a). Finally, let m* be an element in A~ such that p < f(m*) < q. Therefore we have g(d) < f(m*) < f ( a ) , which implies m*Rd and m*RAa and Condition (16.3) is again fulfilled. 9 According to Proposition 16.2.5, R is realizable with respect to < if and only if R -1 is realizable with respect to <. We are thus led to define below a strictly dense condition on R as a restatement of a widely dense condition o n R -1 . D e f i n i t i o n 16.2.15 Let R C_ A x D. A subset M of A U D is said to be s t r i c t l y d e n s e if for all a C A, d E D, aRd implies the existence of an

element m E M such that either ( m e D, aRm, mRDd) or (m e A, aRAm, mRd).

(16.5)

444

Threshold Graphs and Order Relations

Theorem 16.2.16 now follows as a corollary to Theorem 16.2.14. T h e o r e m 16.2.16 For every relation R from A to D, R is realizable with respect to < if and only if R is a biorder with a countable strictly dense subset. The following results on the realizability of interval orders follow easily from Theorems 16.2.14 and 16.2.16. T h e o r e m 16.2.17 For every relation R on a set E, the following conditions are equivalent:

1. R is an interval order (i.e., an irreflexive biorder) with a countable widely dense subset; 2. there exist mappings u 9 E ~ R and r 9 E ~ N + such that, for all p, q C E, pRq if and only if u(p) + r(p) <_ u(q). T h e o r e m 16.2.18 For every relation R on a set E, the following conditions are equivalent:

1. R is an interval order with a countable strictly dense subset; 2. there exist mappings u 9 E ~ R and r 9 E ~ R + such that, for all p, q C E, pRq if and only if u(p) + r(p) < u(q).

16.3

Bidimensions

We now consider the problem of the representation of an arbitrary relation by the intersection of a collection of biorders. We remark that if R has cardinality 1, then both R and R are biorders. Hence every relation R C_ A x D is a union (intersection) of biorders from A to D. We use intersection of biorders to define bidimension. D e f i n i t i o n 16.3.1 The b i d i m e n s i o n dB(R) of a relation R is the smallest cardinal a such that there exists a family of biorders {Bi " i C I} with ] I ] - c~ and

R- A i

(16.6)

16.3

Bidimensions

445

The bidimension is also called the i - b i d i m e n s i o n . Similarly we define the u - b i d i m e n s i o n using union instead of intersection.

In the finite case, the bidimension is the same as the Ferrers dimension of Section 2.3, and it is interesting to compare that section with our results below. The following is an example of a bidimension being an arbitrary cardinal. m

D e f i n i t i o n 16.3.2 A relation R C_ A x D is called a c r o w n if R is a oneto-one correspondence from A to D. The common cardinality a - I A I - IDI is called the size of the crown. P r o p o s i t i o n 16.3.3 The bidimension of a crown R of size a > 2 is equal to a. We leave the proof as an exercise. The following theorem represents the bidimension by an embedding in a product of linear orders. T h e o r e m 16.3.4 The i-bidimension (resp. u-bidimension) of a relation R C_ A x D is the smallest cardinal a such that there exist a family of linear orders {(Wi,_
p r i ( f ( a ) )
where pri denotes the canonical projection of W on Wi.

P r o o f . We only consider the case of i-bidimension since u-bidimension is similar. The proof follows from Lemma 16.3.5 below, which generalizes (but is easier than) Theorem 16.2.14. 9 L e m m a 16.3.5 For R C A x D, A n D - 0 , the following conditions are equivalent: 1. R is a biorder; 2. there exists a linear order (W, <) and two mappings f " A ~ W , and g" D --~ W such that aRd if and only if f ( a ) <_ g(d).

T h r e s h o l d Graphs and Order Relations

446

P r o o f . 2) =~ 1)" This follows from the definition of biorders. 1) =~ 2)" By Proposition 16.2.10, QR is a weak order. Define the equivalence relation E on A U D by x E y if and only if x Q R y and yQRx. Then the quotient W - Q / E is a linear order. Let h be the canonical homomorphism from A UD to W defined by h ( x ) - { y ' x E y } . Then since Q n N ( A x D) - R, the restrictions f, g of h to A, D respectively satisfy the required property. 9 Note that if R is countable, then we can use Theorem 16.2.6 in place of Lemma 16.3.5 to derive the following proposition in place of Theorem 16.3.4. P r o p o s i t i o n 16.3.6 Let R be a countable relation f r o m A to D. Then the bidimension of R is the smallest cardinal a for which there exist two mappings f " A ~ IR ~, and g " D ~ IR ~ such that for all a C A and d E D aRd

if and only if

f ( a ) < g(d),

where < denotes the product order on R ~.

Recall that the dimension of a quasi order Q (resp. strict partial order P) on a set E is the smallest cardinal a for which there exists a family of weak orders (resp. strict weak orders) on E of cardinal a whose intersection is Q (resp. P). It is denoted by d ( Q ) ( r e s p , d(P)). Thus from Proposition 16.2.3 the following three conditions are equivalent" d s ( R ) - 1, d ( R A ) - 1, d ( R D ) - 1. Also if R is a crown of size > 2, then we have 2 - d ( R n ) - d(RD) < We may have d(RA) - 2 and d(RD) arbitrarily large as the next example shows. Let S be any set of cardinal a > 2. D e f i n e A t o b e S a n d D to be the set of all one-element subsets of S together with all their complementary subsets in S. Define R by a R d if and only if a C_ d. One can check that 2 - d(RA) < d(RD) - ds (R) - a. We show in this section that for every quasi order Q, dB(Q) -- d(Q). D e f i n i t i o n 16.3.7 A n n - a l t e r n a t i n g cycle of a relation R is a finite sequence al, d l , . . . , an, d~ such that (ai, di) C R

for i - 1, . . . , n,

(di, ai+l ) r R -1 for i - 1 , . . . , n, where an+l - al. We say that this cycle i n c l u d e s the pairs (ai, di) C R and the pairs (di, ai+l) C /~.

16.3

447

Bidimensions

Figure 16.1" A 4-alternating cycle. al

a2 /

~"

a3 Z

/

/

dl

a4

/

d2

d3

d4

See Figure 16.1 for an example of an alternating cycle. L e m m a 16.3.8 A relation is a biorder if and only if it has no alternating cycle. P r o o f . The "if" part is clear since the forbidden subgraph is a 2-alternating cycle. For the "only if" part we prove by induction on n > 2 that a biorder R has no n-alternating cycle. The case n - 2 follows since R is a biorder. Assume the condition for n - k, and consider an alternating cycle al , dl , . . . , ak+l , dk+l . From a k R d k , ak+lRdk+l and ak+lRdk we infer akRdk+l since R is a biorder. Hence the sequence al , dl , . . . , ak-1, dk-1, ak, dk+l

is a k-alternating cycle of R, yielding a contradiction. D e f i n i t i o n 16.3.9 A n n-alternating cycle al, d l , . . . , an, dn is called b a r e if f o r all i , j - 1 , . . . , n a i R d j ----'z i - j .

See Figure 16.2 for a bare 3-alternating cycle. Clearly every alternating cycle contains a bare alternating cycle. D e f i n i t i o n 16.3.10 Let R be any relation. The h y p e r g r a p h a s s o c i a t e d w i t h R is the hypergraph H ( R ) whose vertices are the ordered pairs in R and such that e is an edge of H ( R ) if and only if e is the collection of

448

Threshold Graphs and Order Relations

Figure 16.2" A bare 3-alternating cycle. a-

a2

IL q'- ...

/

a3

t \

f \

• /

f

) ( X

1

f \

I"#"

f \

/ /

x

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x..x

I

dl

d2

d3

all pairs in R included in some alternating cycle. The b a r e h y p e r g r a p h w i t h R is the hypergraph Hb(R) obtained from H ( R ) by deleting every edge that strictly contains another edge and every vertex not contained in the remaining edges.

associated

Figure 16.3 shows some examples of these hypergraphs and bare hypergraphs. L e m m a 16.3.11 1. A set of pairs in R forms an edge of Hb(R) if and only if it consists of all the non-arcs included in some bare alternating cycle of R.

2. A pair in R is a vertex of Hb(R) if and only if it is included in some 2-alternating cycle of R. 3. If R is a quasi order, then the vertices of Hb(I~) are precisely the incomparable pairs in R, that is to say the pairs (re, n) such that (m, n), (n, m) e R. In that case {(m, n), (n, m)} is an edge of Hb(R). P r o o f . Part 1 follows from the definitions. Part 2 follows from Part 1. Part 3 follows from Part 2. " Recall that the chromatic number x(H) of a hypergraph H is the least cardinal a such that the vertices of H can be partitioned into a classes, each one containing no edge of H having cardinality 2 or more, that is to say, no such edge is monochromatic. In the rest of this section we prove the equalities dB(R) - x(Hb(R)) - X(Hb(QR)) -- dB(QR) - d(Qn). We need a number of lemmas for this purpose.

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450

Lemma

Threshold Graphs and Order Relations

16.3.12 For every relation R, we have

1. d , ( R ) -

x(H(R)) - X(Hb(R));

2. if ds(R) - c~, then R is the intersection of c~ biorders included in RU RR-1R. P r o o f . 1" To show x(H(R)) <_dB(R), consider any family of biorders {Bi" i C I} such that R - ('1~B~. Each B~ is a set of vertices of H(R) that cannot contain an edge of H(R), for otherwise Bi would have an alternating cycle, contrary to L e m m a 16.3.8. Furthermore, we have Ui B i - - R. Now take a strict simple order < on I and define

c,- BZ- U j
The sets Ci partition the vertices of H(R) and none of them contains an edge of H(R). We thus have x(H(R)) <_dB(R). To prove the reverse inequality, let us assume that {Di 9 i C I} is a partition of R such that n o Di contains an edge of H(R). For each i C I, let C~ be such that (i) D~ C_ C~ C_ R, (ii) C~ does not contain any edge of H(R), and (iii) C~ is maximal for these properties. Clearly we have R C_ ('1~C~ = Ui Ci c_ Ui D i - R, and so

R - fl . i

We will have ds(R) <_x(H(R)) if we show that B~ - C~ is a biorder for each

icI. If this is not true, we have aBid, bBie, aBie, and bBid for some a, b E A and d,e C D, where R C_ A x D. We show below that this implies that Ci - Bi contains an edge of H(R), a contradiction. We have four cases to consider. C a s e 1" (a,d) e R and (b,e) E R. In this case {(a, e), (b,d)} is an edge of H(R) contained in B-T. C a s e 2" (a,d) e R and (b,e)C B ~ - R. By the maximality of C ~ - B~, there must be an edge of H(R) contained in Ci U {(b,e)} and not contained in Ci. That is, we can find b2,..., bq C A, e 2 , . . . , eq E D such that {(b, e), (b2, e 2 ) , . . . , (bq, %)} is an edge in H(R) with (bj, ej) E Ci for j - 2 , . . . , q . The set {(b,d), (b2,e2),..., (bq, eq), (a,e)} is then an edge of H(R) contained in C~ since by hypothesis (b, d) C C~ and (a, e) E C~. C a s e 3: (a, d) C B i - R and (b, e) E R. This case is analogous to Case 2.

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16.3

Bidimensions

453

-- { ( a l , d l ) , . . . , ( a k , dk)} is monochromatic in is an edge of Hb(R), to obtain a contradiction. either ai = mi or aiRmi, and similarly either ni therefore from mi+lQnni and the transitivity of taken cyclically. Therefore S is indeed an edge of

Hb(R). We show Observe that for = di or niRdi. It Qn that ai+lRdi, Hb(R).

that S each i, follows indices ..

L e m m a 1 6 . 3 . 1 6 If B is a biorder extending a relation R, then there exists a minimal biorder B m such that R C B m C t3. m

P r o o f . Consider the family/3 of biorders B' satisfying R C B' C_ B, partially ordered by inclusion. By Hausdorff Maximal Principle, equivalent to the Axiom of Choice, the poset /3 has a maximal chain C. The intersection C - N{B " B C C} of all the biorders in C is also a biorder. To see this, assume that a, b C A and d, e C D are such that aCd, bCe, aCe, bCd. We must have able and bB2d for some B1,/32 C C. Since C is a chain, we have either Ba C_ B2 or B2 C B1. Assuming the former, we have aBld, bBle, able, bBld in contradiction to the fact that B1 is a biorder. Thus C is a lower bound for C, and by the maximality of C, C must be a minimal element of B. ,, L e m m a 1 6 . 3 . 1 7 Let R C_ A • D and let B be a minimal biorder extending R. Then,

1. B A R 2. B -

B-

RBD;

B B - 1 B C R.

P r o o f . 1" BA being reflexive, one has R C BAR and BAR C BAB C_ B. Hence, by the minimality of B, the equality BAR - B will be established if we prove that BAR is a biorder. Let a, b C A and d, e C D be such that aBaRd and bBARe. Since BA is a weak order we may assume that we have aBAb. Since BA is transitive (i.e., BABA C_ BA), this implies aBARe and proves that BAR is a biorder. The equality B - RBD can be proved in a similar way. 2" It is sufficient to notice that for every biorder B, if (a, d) G B - B B - 1 B , then B - {(a, d)} is also a biorder (otherwise abe, bBd, and bBe for some b C A and e C D, yielding (a, d) C B B - 1 B ) . " The following lemma states a much deeper property of minimal biorders.

454

Threshold Graphs and Order Relations

L e m m a 16.3.18 Let B be a minimal biorder extending a relation R. If R is transitive, then B is transitive. Furthermore, B has the same loops as R. P r o o f . The transitivity of B is a result of the following computation based on Part 1 of Lemma 16.3.17:

B 2 -(BAR)(RBD)-

BA(RR)BD C_ BARBD - BBD C_ B.

As for the loops of B, let us remark that, B being transitive, a B B 1Bd implies a ~ d. According to Part 2 of Lemma 16.3.17, this implies that all the loops in B are loops in R. 9 L e m m a 16.3.19 For every quasi order Q, we have dB(Q) - d(Q). P r o o f . Since every weak order is a biorder, we have dB(Q) <_ d(Q). In order to prove that dB(Q) >_ d(Q), let us consider a family of biorders {B~" i C I} such that Q - Ni B~ and Ill- dB(Q). According to Lemma 16.3.16, for each i E I there exists a minimal biorder B ~ such that Q c_ B ~ C_ Bi. We have thus Q -~B?. (16.7) i

Using Lemma 16.3.18 we see that each B ~ is transitive and reflexive. B ~ is thus a weak order (see Proposition 16.2.1). The desired inequality follows from the fact that Q is, according to (16.7), the intersection of Ill weak orders. [] Theorem

16.3.20 For every relation R C A x D with A N D - 0, we have d . ( R )

-

-

-

P r o o f . Applying Part 1 of Lemma 16.3.12 to R and QR, we obtain the first and third equalities. The second equality follows from Lemma 16.3.15. The last equality follows from Lemma 16.3.19 and Proposition 16.2.9. .. T h e o r e m 16.3.21 For every strict partial order P, d s ( P ) - di(P) (i.e., the interval dimension of P). P r o o f . This is similar to Lemma 16.3.19 (remember that the interval orders are the irreflexive biorders). []

16.4

Relations of Bidimension 2

16.4

455

R e l a t i o n s of B i d i m e n s i o n 2

In this section we study the relations of bidimension 2. We characterize them and the uniqueness of their representation as the intersection of two biorders in terms of the graphs defined as follows. 16.4.1 Let R be a relation. The g r a p h a s s o c i a t e d w i t h R is the graph G(R) whose vertices are the ordered pairs in R and whose edges are the sets {(a, e), (b, d)} such that aRd, are, bRe, bRd. The b a r e g r a p h a s s o c i a t e d w i t h R is the graph Gb(R) obtained from G(R) by deleting every vertex not contained in an edge. Definition

It is easy to see that every coloring of H(R) is a coloring of G(R), hence x(G(R)) < x(H(R)). For some relations R we may have strict inequality, as shown by the example of Cozzens and Leibowitz in Section 6.1. Remark.

F a c t 16.4.2

1. The edges of Gb(R) are the edges of cardinality 2 of Hb(R); 2. Gb(R) and Hb(R) have the same vertices;

P r o o f . Part 1 follows from Part 1 of Lemma 16.3.11. Part 2 follows from Part 2 of Lemma 16.3.11. Part 3 follows from Parts 1 and 2. Lemma

16.4.3 For every relation R C_ A • D with A N D - 0,

P r o o f . This is similar to Lemma 16.3.15. 9 Let C(P), I(P) denote the comparability and the incomparability graphs of a poset P. This means that P is the vertex set of both C(P) and I(P), and xy is an edge of C(P) (resp. I(P)) if and only if xPy V yPx (resp.

-~(xPyVyPx)). A poset P has dimension at most 2 if and only if I(P) is a comparability graph.

Lemma

16.4.4 ( D u s h n i k - M i l l e r T h e o r e m )

456

T h r e s h o l d Graphs and Order R e l a t i o n s

P r o o f . " O n l y if"" Let P - L1 VI L2 where L1 and L2 are linear orders. Then L11 VI L2 is a transitive orientation of I(P). "If"" Let P* be a transitive orientation of I(P). Define L1 - P U P* and L2 - P U ( p . ) - l . Then it can be checked that L1, L2 are linear orders and L1 N L2 - P. L e m m a 16.4.5 ([GH64]) A graph G - (V, E) is a comparability graph if and only if G has no sequence v0,...,v2t of vertices with vivi+l E E and v~v~+2 ~ E for i - 0 , . . . , 2 1 (indices mod 21 + 1). The following theorem gives a convenient characterization of relations of bidimension 2. Theorem

16.4.6 For every relation R C_ A x D with A N D - 0, dB(R) <_ 2

if and only if X(Gb(R)) < 2. P r o o f . dB(R) - X(Hb(R)) >_ X(Gb(R)), by Theorem 16.3.20 and Part 3 of Fact 16.4.2. Thus if dB(R) _< 2, then X(Gb(R)) _< 2. Conversely, we show that if X(Gb(R))<_ 2, then dB(R)<_ 2. Observe that

dB(R) - d ( Q R ) - d(P),

X(Gb(R)) - x(Gb(QR)) - x(Gb(P)),

where P is the poset defined as the quotient P - Q n / E and xEy if and only if xQny A yQnx: the first chain of equalities follows from Theorem 16.3.20 and definitions; the second from Lemma 16.4.3 and definitions. We finish the proof by showing that for every poset P, if ~(Gb(P)) _< 2, then d(P) _< 2, or, equivalently, that if d(P) > 2, then Gb(P) has an odd cycle. By Lemma 16.4.4, since d(P) > 2, I(P) is not a comparability graph and hence, by Lemma 16.4.5, there exists a sequence m 0 , . . . , m21 of elements of P such that {mi, mi+l} is an incomparable pair and {m~, m~+2} is a comparable pair of P for each i, with indices taken mod 21. Since P is reflexive and miPmi+l and mi+lPmi, the pairs (mi, mi+l) and (mi+l, mi) are adjacent vertices of Gb(P). Consider the cyclic sequence of the 21 + 1 vertices (m~, m~+l) of Gb(P). By virtue of rni+lPmi+l, the consecutive vertices (rn~, rni+l) and (m~+~, mi+2) are adjacent if and only if m~Pm~+2. If they are not adjacent, then mi+2Pmi and hence (mi+l, mi) and (mi+2, mi+l) are adjacent, in which case we insert them between (mi, mi+l) and (m~+l,m~+2), obtaining a path of length 3 between (m~, m~+l) and (m~+l, m~+2). Having done all the appropriate insertions, we obtain an odd cycle in Gb(P). []

16.4

457

Relations of Bidimension 2

D e f i n i t i o n 16.4.7 A graph G - (V, E) is u n i q u e l y o r d e r a b l e when IVI 1 or there exist exactly two (opposite) partial orders on V whose comparability graph is G. D e f i n i t i o n 16.4.8 Given a graph G, we say that two of its edges are dir e c t l y e q u i v a l e n t if they are of the form ran, rap with np not an edge. The transitive closure of this reflexive symmetric relation is an equivalence relation on the set of edges of G, whose equivalence classes are called the i m p l i c a t i o n c l a s s e s of G. Fact 16.4.9 Let P be a poset of dimension 2. Then P is uniquely representable as an intersection of two linear orders if and only if I(P) is uniquely orderable. P r o o f . This follows from the proof of Lemma 16.4.4. Fact 16.4.10 (see [Gol80]) A graph is uniquely orderable if and only if it is a comparability graph with only one implication class. L e m m a 16.4.11 A partial order P of dimension 2 has a unique representation as an intersection of two linear orders if and only if the graph Gb(P) is connected. P r o o f . From Lemma 16.4.4 and Facts 16.4.9 and 16.4.10, it is enough to show that I(P) has only one implication class if and only if Gb(P) is connected. Associate with each edge mn of I(P) the two adjacent vertices (m, n), (n, rn) of Gb(P). If two edges rnn and rnp of I(P) are directly equivalent, then { (n, m), (m, p) } or {(p,m), (re, n)} are edges of Gb(P). It follows that if I(P) has only one implication class, then Gb(P) is connected. On the other hand, if {(re, n), (p,q)} is an edge Gb(P), then rnn and pq are in the same implication class of I(P). Indeed, rnq and np are nonedges of I(P) by definition of a 2-alternating cycle, whereas rnn, pq and nq are edges of I(P) by the transitivity of P, and hence mn is directly equivalent to nq, which is directly equivalent to pq. It follows that if Gb(P) is connected, then I(P) has only one implication class. .' L e m m a 16.4.12 For every relation R C_ A x D with A N D -

1. if Gb(QR) is connected then Gb(R) is connected;

r

we have

458

T h r e s h o l d Graphs and Order R e l a t i o n s

2. if (re, n) and (a,d) are as in Definition 16.3.13, then mn and ad are edges of I(QR) belonging to the same implication class. P r o o f . 1" Let (a, d) and (a', d') be any two vertices of Gb(R). Since they are also vertices of Gb(QR),there exists a path (a, d) -- (?Ttl, rt 1), (m2, rt2),... , (?T/k, ~tk) --

(a', d')

in Gb(Qn). Replacing each (mi, ni) by the associated (ai, di) as in Definition 16.3.13, we obtain a path from (a,d) to (a',d')in Gb(R) by Part 2 of Lemma 16.3.14. 2: Since ( m , n ) i s a vertex of Hb(QR), we have mQnn and Qn has a 2alternating cycle including the pairs (m, n) and (rn', n') of Qn. By the transitivity of Qn we have nQnm. Therefore mn is an edge of I(QR). By Part 3 of Lemma 16.3.14, (a, d) is a vertex of Hb(R), and hence aRd and aQnd. By Part 2 of Lemma 16.3.14, R has a 2-alternating cycle including the pairs (a, d) and (a', d') of R, and hence by Proposition 16.2.9, we have dQna. Therefore ad is an edge of I(Qn). To show that mn and ad are in the same implication class of I(Qn), we consider separately the four cases of Definition 16.3.13. In Case a), ad - mn and there is nothing to prove. In Case b), we have nRd, hence nQnd, and so nd is a nonedge of I(Qn), and a d - md and rnn are directly equivalent. Case c) is similar to Case b). In Case d), we have mQnd by the transitivity of Qn and dQnm by Proposition 16.2.9. Therefore md is an edge of I(Qn). Moreover, am and nd are nonedges of I(Qn) by virtue of a r m and nRd. Hence rnd is directly equivalent to both mn and ad. 9 T h e o r e m 16.4.13 Let R C A x D with A N D - O and dB(R) - 2. Then

R is uniquely representable as an intersection of two biorders contained in R U R R - 1 R if and only if the graph Gb(R) is connected. P r o o f . Notice that by Part 2 of Lemma 16.3.12, R is indeed representable as the intersection of two biorders included in R t_JRR -1R. The only issue is the uniqueness of the representation. " I f " : Let R = B1 N B2, where each Bi is a biorder contained in R U R R -1R. Every vertex (a,d) of Gb(R) belongs to an edge {(a,d), (a',d')} of Gb(R). The biorder B~ must contain one of (a,d) and (a',d'), say (a,d). Then (a, d) C B~ - B2 and (a', d') E B 2 - B~. Thus every vertex of Gb(R) belongs to one of B1 - R and B 2 - R, and no edge meets both of them. In other words,

16.4

459

R e l a t i o n s of B i d i m e n s i o n 2

(B1 - R, B2 - R) is a proper coloring of Gb(R). Since Gb(R) is connected, the coloring is uniquely determined and with it B1 and B2. " O n l y if"" Assume that Gb(R) is not connected. Then by Part 1 of Lemma 16.4.12, Gb(QR) is not connected. Hence Gb(P) is not connected, where P is the poset defined by P - Q n / E and xEy if and only if zQny A yQnx. Hence by Lemma 16.4.11, P has two representations P - L1 N L2 = L i N L~, where Li and L~i are linear orders and L1, L i agree on one implication class of I(P) and disagree on another. It follows that Qn - W1 N W2 = W; N W~, where W~, W[ are weak orders and W~, W~ agree on one implication class of I(QR) and disagree on another. By Part 2 of Lemma 16.4.12, each implication class of I(Qn) has an edge ad with (a, d) e A x D. Therefore by Lemma 16.2.10, Wi N (A x D) and W[N (A x D) define two representations of R, as required. ,, The following theorem exhibits many ways of representing a relation R of bidimension 2 as the intersection of two biorders, under a certain condition. Theorem

16.4.14

Let R C_ A x D, A N D - 0 ,

and assume that R admits

no 3-alternating cycles. Then 1. d . ( n )

-

2. if d , ( R ) <_ 2, then for every proper coloring R U R1 and R U R2 are biorders.

(R1,R2) of Gb(t~), both

P r o o f . 1" It is easy to see that R admits no bare k-alternating cycles for any k >__3. Hence Hb(R) - Gb(R) and the result follows from Theorem 16.3.20. 2: Let B - R U R1. It is enough to show that B is a biorder. If not, then B contains a 2-alternating cycle a, d, a', d' with aBd, a'Bd ~, a B f , a'Bd. There are four cases. (i) If aRd, a'Rd', then (a,d') and (a',d) are adjacent vertices of Gb(R) and hence one of them is in B, a contradiction. (ii) If aRd, a'-Rd', then (a', d')is in R1 and hence is a vertex of Gb(R). Let (a", d") be adjacent to ( a ' , )d ~ . Then a, d, a', d", a", d ~ is a 3-alternating cycle in R, a contradiction. (iii) If aRd, a'Rd', then we argue as in (ii). (iv) If a-Rd, a'-Rd', then both (a, d) and (a', d') are vertices of Gb(R). Let (b, e) and (b', e') be vertices adjacent to (a, d) and (a', d'), respectively. Then b, d, a t, e ~, b', d ~, a, e is a 4-alternating cycle. Hence R admits a 3-alternating cycle, a contradiction. 9

Threshold Graphs and Order Relations

460

16.5

Multiple Semiorders

A special kind of biorders is the relation called semiorder, defined below. D e f i n i t i o n 16.5.1 A binary relation P on a finite set X is said to be a s e m i o r d e r if there exist a function f 9 X ~ 11{ and a nonnegative real number t such that for all x, y E X ,

xPy ~

f ( x ) > f(y)-4-t.

We then say that (f; t) is a c o n s t a n t t h r e s h o l d r e p r e s e n t a t i o n of P. Notice that every semiorder is indeed a biorder, but the converse is not true, as we see in Corollary 16.5.6. The above definition is generalized to multiple semiorders as follows. D e f i n i t i o n 16.5.2 Let 7) = ( P 1 , . . . , Pro) be an m-tuple of binary relations on the same finite set X . A c o n s t a n t t h r e s h o l d r e p r e s e n t a t i o n for 7) consists of a real-valued function f on X and real numbers t l , . . . , tm such that for all j = 1 , . . . , m and x , y C X ,

xPjy ~

f ( x ) > f ( y ) + tj.

(16.8)

We denote this representation by (f; t l , . . . , tin). When such a mapping f exists, we call P a m u l t i p l e s e m i o r d e r and ( t l , . . . ,tin) its c o n s t a n t t h r e s h old v e c t o r . Clearly, if 7~ is a multiple semiorder, then Pj must be irreflexive if tj >_ 0 and reflexive if tj < 0. In psychology, Pj may capture one level of the preferences that a person expresses among the elements of X, and the existence of a representation of the form (16.8) is a very strong consistency requirement for such preferences. We report in this section Doignon's results [Doi77] characterizing multiple semiorders. For more details on applications of such families of relations in psychological measurement and in the theory of "probabilistic consistency", refer to [Doi77]. We need the following two notations. 1. For a binary relation R, its dual R' is the converse of its complement, i.e., aR'b if and only if bRa. 2. We associate with the m-tuple 7~ - ( P 1 , . . . , P,~) of binary relations on X a digraph G(7 )) - (X, A), where A is the multiset P1 U ... U P,~ U P ; u . . . uP' .

16.5

Multiple Semiorders

461

D e f i n i t i o n 16.5.3 Let P = ( P 1 , . . . , Pro) be an m-tuple of binary relations on X , A c y c l e f r o m 79 is a directed cycle in G(T)). We associate with this ! cycle the numbers pj,pj of arcs that belong to Pj, P~ for each j - 1 , . . . , m, ! and set qj - pj - pj. We assume in the definition of a cycle that pj > 0 for at least one j. A k - c y c l o n e f r o m P is a nonempty union of at most k cycles from 79. ! We define the numbers pj, pj, qj for a cyclone as for a cycle, and say that the cyclone is b a l a n c e d if qj = 0 for each j. We are now ready to characterize semiorders. L e m m a 16.5.4 Let G = ( V , A , w ) be a weighted multiple digraph, where w : A ~ N. The weight of a subset of A is obtained by summing w over it. Then the following two conditions are equivalent:

1. there exists a mapping f : V ~ R such that for all ( x , y ) C A, f ( x ) >_ f ( y ) + w(x, y); 2. no (directed) cycle of G has a strictly positive weight. Clearly, one may replace "cycle" by "simple cycle" in Condition 2. P r o o f . Condition 1 implies Condition 2 by summing the inequalities over the arcs of the cycle. The converse follows from the Farkas L e m m a (see [Sch86].

L e m m a 1 6 . 5 . 5 Let 7) = ( P 1 , . . . , P m ) be an m-tuple of binary relations on X . The m-tuple ( t l , . . . , t m ) of real numbers is a constant threshold vector for 7-9 if and only if each cycle from ~P satisfies

qltl nt-''" -~- qmtm > O. P r o o f . Assume that (f; t l , . . . , t i n ) iS a constant threshold representation for T ~. Since there is only a finite number of strictly positive values f ( x ) - f ( y ) tj, we can find an e > 0 that is a lower bound for all these values. We then have

xPjy ---.. f(x) >_ f(y) + tj + and also

xPjy - .

f(x) >_ f(y) - tj

Threshold Graphs and Order Relations

462

It follows that f and the following weights for G(T') satisfy Condition 1 of Lemma 16.5.4: w(x~ y) -- I tj--tj,-4- e, ifif xPjy xP~y. Therefore every cycle of G has a non-positive weight. Thus m

m

E pj(tj + ~)+ E p}(-tj) < O, j-1

j=l

which implies m

E qjtj > O. j=l

Conversely, since there is only a finite number of simple cycles from 7', we can find an e > 0 such that each cycle satisfies rn

m

E qjtj - e E p j >_ O. j=l

j=l

Using the same weights on G and taking the argument in reverse order, one derives the existence of a mapping f such that (f; t ~ , . . . , tin) is a constant threshold representation for 7'. 9

binary relation P is irreflexive and for all a, b, c, d C X,

Corollary 16.5.6 A

a

semiorder if and only if P is

(aPb A cPd) ~

(aPd V cPb),

(aPb A bPc) ~

(aPd V dPc).

Figure 16.~ shows the configurations forbidden by these conditions. P r o o f . The "Only if" part follows from the definition. To prove the "if" part, we assume that P does not have the configurations of Figure 16.4, and show that the 1-tuple (1) is a constant threshold vector for P. By Lemma 16.5.5, what we need to show is that q > 0 for every simple cycle from P. We show this by induction on the length of the cycle. The statement is true for cycles of length 1, 2 or 3 since the configurations G1, G2 and G3 of Figure 16.4 are forbidden. Let C be a simple cycle of length >_ 4 with q _< 0, if that were possible. If all arcs of C are in P, then we must have G3 or G2, a

16.5

Multiple Semiorders

463

Figure 16.4" The forbidden configurations for a semiorder.

G1

G2

G3

G4

G5

contradiction. Hence C has arcs in P'. Assume that C has two consecutive arcs in P. Then it has consecutive vertices a, b,c, d with aPbPcP'd, and hence aPd holds since Gs is forbidden. We can use the arc (a, d) to shortcut C and obtain a shorter cycle with the same q _< 0, contradicting the induction hypothesis. The only remaining possibility is that C has no consecutive arcs in P, and since q _< 0, its arcs actually alternate between P and P' and q = 0. Consider consecutive vertices a, b, c, d with aPbP'cPd. Then aPd holds since G4 is forbidden, and we can use the arc (a, d) to shortcut C and obtain a shorter cycle with q = 0, again contradicting the induction hypothesis. 9 The following theorem characterizes multiple semiorders. T h e o r e m 16.5.7 Let T' - ( P 1 , . . . , P r o ) be an m-tuple of binary relations on X , with m > 1. There is a constant threshold representation for ~P ( P 1 , . . . , Pro) if and only if no m-cyclone from 7:) is balanced. P r o o f . " O n l y if"" Assume that P admits a constant threshold representation ( f ; t l , . . . , t m ) . Then

xpjy ~

f ( x ) - f(y) > tj,

xpjy ~

f ( x ) - f(y) >_ -tj.

Threshold Graphs and Order Relations

464

If we consider any k-cyclone, write these implications for all its arcs and sum the resulting inequalities, we obtain m

m

m

o > E pJJ + E p}(-tJ)- E(pJ- p})tj. j=l

j= l

j=l

Moreover, the inequality is strict since at least one arc from some Pj is used by the cyclone. This clearly implies that pj r p} for at least one j. Hence the cyclone is not balanced. "If"" Assume that no cyclone from P is balanced. Observe that each Pj is either reflexive or irreflexive (otherwise we construct a balanced 2-cyclone by taking one loop in Pj and another one in P~). Consider first the case m - 1. If P1 is irreflexive, we show that the 1-tuple (1) is a constant threshold vector for P1. By Lemma 16.5.5, it is enough to show that q > 0 for each cycle. If C were a cycle with q _< 0, we could add Iq[ loops from P~ to obtain a balanced 1-cyclone, a contradiction. Similarly, If P1 is reflexive, then ( - 1 ) is a constant threshold vector for P1. Now assume that m >_ 2. Since there is only a finite number of values f ( x ) - f(y), we may always look for a nonzero threshold tl, and by an appropriate change of scale, even assume that tl - +1. We treat the case that P1 is irreflexive, setting tl - 1 (the other case is similar). By Lemma 16.5.5, we have to show the existence of a common solution ( t 2 , . . . ,try) to all the inequalities

q2t2 + ' " + qmtm > --ql associated with simple cycles from P. By Helly's Theorem (see [Chv83, p. 266]) applied to the convex sets defined by these inequalities in the Euclidean space of all ( m - 1)-tuples (t2, t 3 , . . . , tin), it is sufficient to show that every m of these inequalities have a common solution, say

qi2t2 + " " + qimtm > -qil

(16.9)

with i - 1 , . . . , m. By the Farkas Lemma, this is equivalent to the following assertion" given real numbers 11,..., lm with li >_ 0 for each i, and li > 0 for at least one i, m

-

i=1

0

j -

2,...,

(16.10)

16.5

Multiple Semiorders

465

implies m

liqil

> 0.

(16.11)

i=1 !

Since qij -- Pij - - P i j is an integer, we only have to verify the assertion for rational numbers l~ (because the real tuples ( / 1 , . . . , lm) satisfying l~ >_ 0 and (16.10) form a polyhedron with rational extreme points and directions). It follows that we need only consider li's that are natural numbers. Now each of the equations in (16.9), for fixed i, comes from a cycle Ci from 7). Assuming that li is a natural number, we consider the cycle C[ obtained by traversing the cycle Ci li times, and then the union U of these C~. Then U is an m-cyclone that uses ~i~__1lip}j arcs from P~ and ~i~=1 lipij arcs from Pj. Equation (16.10) tells us that for j - 2 , . . . , m , these two quantities are equal. Since by assumption U is not balanced, we deduce m

liqil 7~ O. i=1

In order to establish (16.11), and thus complete the proof, it remains to show that (16.10) together with m

liqil < 0

(16.12)

i=1

lead to a contradiction. If (16.12) were true, we would of course have one of the qil strictly negative (because all li are nonnegative), and thus the mcyclone U would use at least one arc (x,y) from P1. Now we form a new m-cyclone U' by adding m

I ~ liqil I i=1

times the arc (x, x) to U. Since P~ is irreflexive, (x, x ) i s in P~, so the new arcs increase the left-hand side of (16.12) to zero. Thus U' is a balanced m-cyclone, a contradiction. 9 Let us make the following remark. The proof above used only a restricted version of Helly's Theorem, namely the convex sets were open half spaces in I~ m-1 (a similar result for closed half spaces can be found in [Chv83, p. 146]). This restricted version follows easily from the Farkas L e m m a as follows. Assume that A x > b is unsolvable, where A has m - 1 columns. Then Am > tb, t > 0 is also unsolvable. Hence by the Farkas L e m m a there exist y > O, Yo > O, (y, yo) r (0, 0), satisfying y T A -- O , - - y r b + Yo -- O, and,

466

moreover, we m a y find such (y, eliminating yo, we are left with c o m p o n e n t s and satisfying y r A most m inequalities among Ax

Threshold Graphs and Order Relations

Y0) with at most m positive components. By a non-zero y >_ 0 having at most rn positive -- 0, yTb >_ O. Therefore a subsystem of at > b is unsolvable.

Chapter 17 Enumeration 17.1

Introduction

In this chapter we look at enumeration results of some families of graphs. For convenience we assume that all our graphs have the vertex set N = {0,...,n1}. Recall that two graphs G = (N,E) and H = (N,F) are the same unlabeled graph when they are isomorphic, i.e., when there is a permutation 7r of N such that ij C E if and only if ~r(i)Tr(j) e F. We say that G and H are the same labeled graph when E = F. Many enumeration results are expressed in terms of ordinary and exponential generating functions. The ordinary generating function (ogf) of a sequence a0, al, a 2 , . . . is ~ > 0 a~ x~ and its exponential generating function (egf) is ~ > 0 anxn/n!" The elementary properties of these functions are discussed in most introductory combinatorics books. In this chapter we have occasion to use the Burnside Lemma and other basic enumeration results, which can be found in the same books. We have already seen in Section 13.4 and 13.5 that decomposition techniques can be used to find the egf of the number of unlabeled box-threshold, matroidal and matrogenic graphs. See also Section 12.4 for the egf of the number of unlabeled domishold and hereditary pseudodomishold graphs. In Section 17.2 we discuss enumeration of threshold graphs, both unlabeled and labeled. It turns out that the enumeration of unlabeled threshold graphs can be used to derive the q-binomial theorem, and the enumeration of labeled threshold graphs can be used to derive a theorem of Frobenius connecting the Eulerian polynomial with Stirling numbers of the second kind. In 467

468

Enumeration

Section 17.3 we enumerate difference graphs, both unlabeled and labeled. Recall that difference graphs are bipartite. We refer to a bipartite graph G = (N, E) together with a fixed partition of N into two stable sets as bipartitioned. Section 17.3 discusses labeled and unlabeled difference graphs, either bipartitioned or not.

17.2

E n u m e r a t i o n of Threshold Graphs

We begin with the enumeration of unlabeled threshold graphs, based on Peled [Pel80]. Denote by r,~mk the number of unlabeled threshold graphs having n vertices, m edges, and m a x i m u m clique size k, and consider the generating function

Tn(x, q) - ~ 7"nmkqmxk. m,k We evaluate theorem. Lemma

T~(x, q) in two ways, and compare them to obtain the q-binomial

17.2.1

Tn(x, q) = x(1 + qx)(1 + q2x)... (1

+

qn-lx).

(17.1)

P r o o f . For each binary vector u = ( u 0 , . . . , Itn-1) ~ {0, 1}" with u0 = 1, we construct a graph G(u) - ( N , E ) on the vertex set N - { 0 , . . . , n 1} where E is such that for all i < j we have ij E E if and only if uj = 1. By Condition 4 of Theorem 1.2.4, G(u) is a threshold graph for each u, and every threshold graph can be obtained in this way. Moreover, G(u) is isomorphic to G(v) if and only if u - v. Indeed, the "if" part is obvious. Conversely, assume that u r v yet G(u) is isomorphic to G(v), and let j be the largest subscript such that uj --fi vj. Then G(u') is isomorphic to G(v'), where u ' = ( u 0 , . . . , uj) and v ' = ( v 0 , . . . , vj), since removing an isolated or a dominating vertex does not affect isomorphism. But this is impossible, since one of G(u') and G(v') has an isolated vertex and the other one does not. From this we already see that the number of unlabeled threshold graphs on n vertices is 2 "-1. The number of edges of G(u) is ~2j=0 n-1 jUj. Further, the set K - {j 9 uj 1} is a m a x i m u m clique of G(u). Indeed, K is clearly a clique containing 0, and consider any other clique K ~ % K. Then K ~ contains some element i > 0

17.2

E n u m e r a t i o n of T h r e s h o l d Graphs

469

with ui = 0, and so i is adjacent only to elements j C K such that i < j, i.e., to fewer than IKI elements. Hence IK'I _< IKI. We conclude that the unlabeled graphs enumerated by Tnmk are in oneto-one correspondence with the binary sequences u such that n-1

u0-1,

n-1

m-Eju

j,

k-Eu

j =o

j,

j =o

and (17.1) follows. 9 We remark that by putting q = i in (17.1) and extracting the coefficient of x k, we find that the number of unlabeled threshold graphs having n vertices n--1 and maximum clique size k is ~m r~,~k - (k-l)" This result can also be obtained by counting symmetric corrected Ferrers diagrams having n rows and corrected Durfee number k (Definition 3.1.6). The diagram is determined by the sequence of its first k row sums, which are non-increasing, do not exceed n - 1, and the last of them is equal to k - 1. The number of such [ n-1 \ sequences is [k-l), as can be seen by standard techniques. The next lemma makes use of the Gaussian polynomials [n.]_

[3J

(q)~

,

where(q)n-(1-q)(1-q2).-.(1-qn).

(q)j(q)n-j

L e m m a 17.2.2 T~(x,q) -

.

J

1

"

P r o o f . If G is a split graph having a split partition (K, S) with K a clique and S stable, and if K is a maximal clique, then K is in fact a maximum clique, since no vertex of S can be adjacent to all the vertices of K. If in addition G is a threshold graph, then by Condition 3 of Theorem 1.2.4, G is determined up to isomorphism by n, IK[, and the degrees of the vertices of S. Hence the unlabeled threshold graphs enumerated by rnmk are determined by the partitions of the integer m - (k2) (the number of edges outside IKI)into at most n - k positive parts (the positive degrees in S) not larger than k - 1. It is well-known [And76, p. 33] that the ogf for the number of partitions of integers into at most A positive integers not larger than B is [A+B]. Therefore rn~nk is the coefficient of qm-(~) in In-l], k-1 which is equal to the coefficient of nl q m x k in t h e s u m ~ j =

[ j-1 n-l] q (~)xj 9

This proves 17 92 9

"

Enumeration

470

By comparing (17.1) with (17.2), changing the index of summation and increasing n by 1, we obtain the following. C o r o l l a r y 17.2.3 (The q - B i n o m i a l T h e o r e m )

(1 + x)(1 +

qz)...

(1 +

q~-ax) - j~o

q(~)xJ"

(17.3)

Identity (17.3) goes back to Cauchy [Cau93] and probably to Euler. Many important identities are easily derivable from (17.3) and vice versa. For example, by letting n --+ cx~, we can obtain the following formal-power-series identity of Euler" (1 - x)(1

- qx)(1

-

q2x) . . . .

q('~)(-x)J

j~0 (1 - q)(1 - q~ :- (1 -

qJ)"

(17.4)

From (17.4), again using (17.3), we obtain the companion identity of Euler:

1 (1 - x)(1

- qx)(1

oo -

q2x)...

xj

= j~0 (1 - q)(1 - q2)...

(1 -

qi)"

(17.5)

From (17.4) and (17.5) we can get, again using (17.3), the following identity of Heine: (1 - ax)(1 - qax)(1 - q 2 a x ) . . . = ~-,~ (1 - a)(i - qa~_2..!l_. - qJ-la)x j (1-x)(1-q/)(1-q2x)... ~o ( 1 - q ) ( 1 - u --). ( 1 - q J ) (17.6) Andrews [And76] derives many series-product identities from (17.6), including the q-binomial theorem itself. Knuth [Knu71] discusses an interesting connection between the fact that [~] is the number of k-dimensional subspaces of GF(q) ~ and the fact that it is the egf for the number of partitions of integers into at most k parts not exceeding n - k. The connection is via matrices in row-echelon form, whose shape can be interpreted as a threshold graph. We now present the work of Beissinger and Peled [BP87] on enumerating labeled threshold graphs and the identity of Frobenius. Let tnjk denote the number of labeled threshold graphs with n vertices, j isolated vertices, and exactly k distinct vertex-degrees. For convenience we set tojk = ~jo~ko,

(17.7)

17.2

E n u m e r a t i o n of T h r e s h o l d G r a p h s

471

where ~ is the Kronecker symbol. We denote by

tn - ~ tnjk

(17.8)

j,k

the total number of labeled threshold graphs on n vertices. We determine the generating function n

F(x,y,z)- ~ t~jk~.yJz k.

(17.9)

n,j,k

Theorem

17.2.4

The generatingfunction defined by (17.9) is given by 1/z+e~Y-1 F(x,y,z)- (1 - xz)1/z - e~ + 1" (17.10)

P r o o f . It is clear that for n >_ 2 one has t~jk - 0 unless 0 _< j _< n and 1 _< k <_ n - 1 (k cannot equal n, because at least two degrees must be equal, as the graph cannot have both isolated and dominating vertices). We list below some other properties of tnjk.

tnjk--(~)tn-j,O,k-1 forj>_l

(17.11)

since any number of isolated vertices can be added to or removed from a threshold graph without affecting its thresholdness, as follows from Condition 4 of Theorem 1.2.4.

tnok- ~tnjk

for n >__2

(17.12)

j>__l

because, by Theorem 1.2.4, the complement of a threshold graph with n >_ 2 vertices, no isolated vertices, and k distinct vertex-degrees is a threshold graph with n vertices, one or more isolated vertices, and k distinct vertexdegrees. Equations (17.11) and (17.12) can be combined to obtain the recurrence

The boundary conditions (17.14) and (17.15) below apply to all classes of graphs, not just to threshold graphs. t~,~_,,k - 0

(17.14)

472

Enumeration

tnnk -- 5kl

(17.15)

for n > 1.

As an aid in determining F(x, y, z), consider the generating function

XnZk

yj(x, z) - E E t~j~ ~!

(17.16)

n>0 k>0

Then r 0 ( x , z) Xn

E E t~o~-fzk n)>0 k > 0

Xn

= 1 + o + E E t~o~., Z k

by (17.7),(17.14)

n>2 k>0

Xn

= 1+ E E Et~J~-~z k

by (17.12)

n>_2k>_Oj>_l

X~zk

= 1 + ~ ~ ~ ~njk n! j>_l n>_2 kkO

= 1 + ~-~ ( F j ( x ' z ) - O>_k~ t ~

k>_O ~-'~tljkxzk

-- l-Jr-j~>l ( F j ( x ' z ) - - E ~ j Ok>_Oe ~ k o Z k - -

by (17.16)

k>_O

by (17.7),(17.14),(17.15). Thus

Fo(x, z)

-

~

j>l

Fj(x, z)+ 1 - xz.

(17.17)

It follows from (17.11) and (17.16)that

xj Fj(x, z) - z~Fo(x, z) for j _> 1.

(17.18)

By summing (17.18) for j > 1 and using (17.17), we obtain

Fo(x,z)- 1 + xz - z(e ~ - 1)F0(x,z) and therefore F 0 ( x , ~) -

1-xz 1 - ( e ~ - 1)z

x-l/z e~ - (1 + l/z)"

(17.19)

17.2

E n u m e r a t i o n of T h r e s h o l d Graphs

473

By (17.18) we have

F(x,y,z)-

~ Fj(x,z)y j - Fo(x,z)(1 + z(e ~ y - 1)) j>o

and (17.10) follows. 9 To expand functions with denominators like that of (17.10), we use the Eulerian polynomials An(u), defined as follows (cf. Comtet [Com74] or Riordan [Rio58]): u-

1

U -- e v(u-1)

1

v~

U n>l

n'

= 1 + - ~ An(u)--r,

A o ( u ) - 1.

(17.20)

One combinatorial interpretation of the Eulerian polynomials is that for n _> 1, the coefficient of u m in An(u) is the number of permutations 7rl,..., 7r~ of 1 , . . . , n having m - 1 ascents, that is to say, positions j such that 7rj < 71"j+ 1 [GKP89, Equation (7.56)]. By substituting u - 2 in (17.20), we obtain 1

Xn

-

2

~

(17.21)

gn

n>O

where gn - A~(2)/2 for n >_ 1 and go - 1. The well-known numbers g~ were studied by Cayley in the 19th century and several references to them can be found in Knuth [Knu73] and Good [Goo75]. gn is the number of ordered partitions of the set { 0 , . . . , n - 1}, that is to say, the number of ways to partition the set into non-empty blocks and to arrange the blocks in a sequence. Equivalently, g~ is the number of those n-letter words over the alphabet { 1 , 2 , . . . } that for some k contain precisely the letters { 1 , . . . ,k}, repetitions allowed. To see this interpretation of gn, we only have to write 1

2-e ~

=

1

1 - ( e ~ - 1)

=

~-~(eX

-

1) k

k>0

and note that (e ~ - 1)k is the egf for the number of words over the alphabet { 1 , . . . , k} with each letter appearing at least once. Because of this interpretation of g~, we have

g n - ~ kk=0 !

nk '

(17.22)

474

Enumeration

where (~} denotes the Stirling number of the second kind, i.e., the number of partitions of { 0 , . . . , n - 1} into k unordered non-empty blocks. This agrees with the well-known expansion (see Comtet [Com74] or Riordan [Rio58])

1)

E k>0

"

k

"

Knuth [Knu73, Ex. 5.3.1.4] presents the following asymptotic for gn: g--5~-1 ( 1 ) nl9 -- 2(ln2)n+ 1 + k>, ~ Re (ln2 + 27rik) ~+1

n > - 1"

(17.23)

It is not hard to use (17.23) to obtain that

1

(1)

We can now enumerate the labeled threshold graphs. T h e o r e m 1'/'.2.5 The number of labeled threshold graphs on n vertices is given by 1, if n - 0,1 t~ 2(g~ - - ngn-1), if n _> 2. (17.24) P r o o f . According to (17.9) and (17.8), F(x, 1, 1) = E~>0 tnxn/n!" evaluate F(x, 1, 1)using (17.10), we obtain (1 - x)e ~ (2 2-e ~ -(l-x)2-e

) ~

1

If we

x~

-l-at-x-~-2E(gn-ngn_l)----(, n>2

~"

and the result follows. .. We now see a combinatorial argument that gives another proof of Theorem 17.2.5, as well as of the theorem of Frobenius. We denote by Tnk the set of labeled threshold graphs with the vertices { 0 , . . . , n - 1}, no isolated vertices, and exactly k distinct vertex-degrees, so that t~0k = IT~kl.

(17.25)

We also denote by Dnk the set of ordered partitions ( B 1 , . . . , Bk) of { 0 , . . . , n 1} into non-empty blocks B~ such that IB~] >_ 2. It is easy to see that

,D~k, - k ' { nk} - n(k - 1)'{k " - 1 1} "

(17.26)

17.2

475

E n u m e r a t i o n of Threshold Graphs

L e m m a 17.2.6 There exists a bijection between Dnk and Tnk. P r o o f . Both sets are empty if n - 1, so we may assume that n > 2. Given a partition d - ( B 1 , . . . , Bk) E Dnk, we construct a labeled graph G on the vertices 0 , . . . , n - 1 by joining each vertex of Bi to all the other vertices of B1 U - . . U Bi if i + k is odd, and to none of them if i -+- k is even. Then G is a threshold graph by Condition 4 of Theorem 1.2.4, and it has no isolated vertices, since the vertices of Bk are dominating. We show that G has exactly k distinct vertex-degrees by showing that two vertices v C Bi and w C Bj have the same degree if and only if i - j. The "if" part is obvious. Conversely, assume that i < j. If i + j is even, then i -t- 1 < j and every vertex of Bi+l is adjacent to exactly one of v and w, hence v and w have different degrees. If k + i is even and k + j is odd, then v has some neighbor in B1U---UB~ (because IB~I > 2), whereas w has no such neighbor; therefore v and w again have different degrees. A similar argument shows the same conclusion if k -4- 1 is odd and k -4-j is even. Therefore G C Tnk. Given G C Tnk, we construct an ordered partition d as follows. Let G1 - G. If Gi has been constructed as a threshold graph with two or more vertices, then it has either isolated vertices or dominating vertices, but not both, and we remove all these vertices to obtain Gi+l. By an argument similar to the one in the previous paragraph, we can show that two vertices of G have the same degree if and only if they are removed at the same time. Since G has exactly k distinct vertex-degrees, the last non-empty graph is Gk, and it has at least two vertices. We put the vertices of Gi - Gi+~ into the block Bk+~-i for i -- 1 , . . . , k and obtain an ordered partition d - ( B 1 , . . . , Bk) C Dnk. It is easy to see that the two mappings defined above between Dnk and Tnk a r e inverses of each other, and the proof is complete. 9 The second proof of Theorem 17.2.5 can be seen as follows. From (17.25), (17.26) and Lemma 17.2.6, we obtain " k

"

1} 1

(17.27)

for n >_ 1, and this equation also holds for n - 0 by our convention that took - ~k0. If we sum (17.27) over k, the left-hand side becomes t~/2 for n >_ 2 by (17.12), the right-hand side becomes g n - ng~_~ by (17.22), and (17.24) results. The reason we presented the first proof of Theorem 17.2.5 is that it enables us to prove the theorem of Frobenius [Frol0] (see also Comtet [Com74]),

IV

0

~.

I

....

I

._

~ ~,

~

I

I

~

~

IV

I

I

I

I

I--t,

~

~

I ~

oc--F-'

0

~

"~

i,~

~

~

9

~ ~

~

,~ o

o

~

I

CD

0

~

~.~

t'~

IV ~

I

9

~

.-

I

~

~

I

'ov

o

~

I I

;._,

o

~ ~

I

"--,.1

-~

0

~

O

c--c-

0

I

"

O0

-q

9

II

'--'

~1~

+

II

I

~ I

I

I

II

I

c~

I

~-~

I

o~

9

II

I

II

~

~

~

:~

~ 9

~ 9

~o~

IV

~

'ov

o ~ "~

i.-~~

~

0

~

~

o-'

~'~

0

~-

~

~

~

~

l:r'

0

eo

~

O~ ~

I

~

~

~. .

to

~

~

0 ~ o0

0~

0

""

~<

~'0~

l~-~

~

i.,J~

=

17.3

Enumeration of Difference Graphs

17.3

477

E n u m e r a t i o n of Difference Graphs

This section is based on Peled and Sun [PS94a]. We begin our discussion with enumerating labeled difference graphs, and express the results in terms of the numbers gn defined in (17.21) or (17.22). 17.3.1 Let dn be the number of labeled bipartitioned difference graphs on n vertices. Then

Theorem

ifn-O 2gn, i f n > _ l .

d n _ _ { 1,

P r o o f . By Theorem 2.4.4, dn is equal to the number of ordered partitions of { 0 , . . . , n - 1 } of the form (X0, X 1 , . . . , Xk, Yk,..., Y1, Y0), where only the blocks Xo and Y0 may be empty. We define a mapping f from the set of all such partitions P into the set of ordered partitions of { 0 , . . . , n - 1 } with nonempty blocks. We obtain f ( P ) by removing the empty blocks of P, if any. Then for n >_ 1, f is precisely two-to-one. To see this, let Q - ( B ~ , . . . , Bk) be an ordered partition with non-empty blocks. If k is even, then

f - l ( Q ) _ {@,(~,B1,...,Bk,~)}, and if k is odd, then

f - l ( Q ) _ {(O, B1,...,Bk),(B1,...,Bk, O)}. The same result can also be obtained by writing the egf for the d,. The egf for the number of ordered partitions of { 0 , . . . , n - 1 } into 2 k + 2 blocks, where only the first and last blocks may be empty, is e~(e ~ - 1)2ke ~. Therefore the egf for the d, is

k>0

e2~(e ~ - 1)2k _ e2~ e~ 2 - 1-(e ~-1) 2 = 2-e ~ = 2-e ~

1,

so d 0 - 2 9 0 - 1 - 1 anddn-2gnforn>_ 1. 9 One can introduce other parameters into the the generating function. For example, suppose that we have a set of p + q elements and want to construct an ordered partition the first p elements into k + 1 ordered blocks where only the first block may be empty, and an ordered partition of the last q elements into k + 1 blocks where only the last block may be empty. The number of

478

Enumeration

xP xq

possible ways to do this is the coefficient of ~ 7

in e~(e ~ - 1)k(e y - 1)ke y.

9

The coefficmnt of (p+q),, namely times the previous coefficient, is the number of ordered partitions of p + q distinct elements into 2k + 2 blocks, where only the first and last blocks may be empty. By summing over all k > 0 we find that

ex+Y

dpq xPYq

(p + q)!

1 -(e

~ -

1)(ey -

1)'

where dpq is the number of labeled bipartitioned difference graphs with color classes of cardinality p and q. By manipulating the egf for the dn we can derive identities involving Stirling numbers of the second kind. For example, using the fact that e = ( e ~ 1)k= d (e ~ - 1) k + l dx k + l

_

1 d x~~-,-d{ +1)1{ r k + 1 dx ~>0 , . ( k " k+l

}

-

"

k + l

'

we find

xs

1 + E ~2Ej! s>_l

" j_>0

{;} -

E k>0

([e~( e ~ - 1)]k) ~ - E k_>0

~.k!

{r + k -~

and therefore 2

j>0 J!

{ ~ } _ ~ > 0 ( : ) k ~ > ok?2 { r + l } { s - r + l } k+ 1 k+ 1 _ _

T h e o r e m 17.3.2

s>l.

The number of labeled difference graphs on n vertices is

(l+gn)/2. Proof. By Theorem 2.4.4 every difference graph consists of zero or more isolated vertices and a connected difference graph, and conversely. As in Theorem 17.3.1, the egf for the number of labeled non-empty bipartitioned connected difference graphs is ~k>l(e ~ - 1) 2k. Hence the egf for the number of labeled non-empty connected difference graphs is 71 E k ~ l ( ex - 1) 2 k where 1 the factor g comes from exchanging the two color classes. By adding 1 we also include the empty graph, and by then multiplying by e~ we allow for the

17.3

E n u m e r a t i o n of Difference G r a p h s

479

isolated vertices. Thus the egf for the number of labeled difference graphs is given by

(

e~ l + a . y -k>l ~(e~_l)2k 1

= z e~

(1+

) 1(

)

_ 2 e~ 1 + k>0 ~(eX-

1 )l(e~ 1 _ (e~_ 1)2 - ~

-+

1) 2k

1 ) 2_e~

and the number of labeled difference graphs on n vertices is the coefficient of W., namely (1 + gn)/2. " In the rest of this section we consider the enumeration of unlabeled difference graphs. We denote by | the set of unlabeled threshold graphs on n vertices, and by A, the set of unlabeled bipartitioned difference graphs on rz vertices. By (17.1) we have I O n l - 2n-l, as already noticed. We use the following connection between IZX~Iand IO~l. X n

L e m m a 17.3.3 [An[- IOnl + IAn_ll

7/

>_ 1.

Proof. Consider the mapping f 9 An -+ | given by f ( D ) - T, where D - (X,Y; E) C A~ is a bipartitioned difference graph with bipartition (X, Y), T - ( X U Y, E U E x ) , and E x is the edge-set of the complete graph on X. By Theorem 2.1.9 f is onto | but it is not one-to-one. To find f - x ( f ( D ) ) , we note that T and D are determined by their degree partitions according to Theorem 1.2.4 and Theorem 2.4.4. In T, the degree of every vertex of Y/ is [Xk U . . - U Xk-i+x[ and the degree of every vertex of Xi is IYk tO... U Yk-i+x [q-IX0 U . . . LI Xk[- 1. Since Xi and Y/are non-empty for i >_ 1, it follows that the degree partition of T is simply (Y0,..., Yk) unless one of the following conditions holds in D" 1. x 0 - o , IYkl - 1, in which case the degree partition of T is ( Yo , . . . , rk-1,

Yk

I,.,J X l

, X2 , . . . , Xk

);

2. I X 0 [ - 1, in which case the degree partition of T is ( ro , . . . , rk - l , Yk

[,-J X o

, X l , . . . , X k).

It follows that f - l ( f ( D ) ) - {D} unless Condition 1 or 2 holds for D, in which case f - l ( f ( D ) ) - {D,D*}, where D* is obtained from D by moving

480

Enumeration

the singleton Yk to X in Condition 1 and the singleton X0 to Y in Condition 2. Note that if Condition 1 holds for D, Condition 2 holds for D* and conversely. It follows that IOnl - Izx f- Izx'~l, where A" denotes the set of unlabeled bipartitioned difference graphs satisfying Condition 1 or 2. We complete the proof by indicating a bijection from A~ to An-l- It is given by deleting the singleton Yk in Condition 1 and the singleton X0 in Condition 2. The resulting bipartitioned difference graph D' has isolated vertices in X in Condition 1 and does not have them in Condition 2. Therefore we can uniquely reconstruct D from D' by adding a singleton Yk to Y if D 1 has isolated vertices in X and adding a singleton X0 to X if D' has none. This shows that indeed we have a bijection. 9 Theorem

17.3.4

The number of unlabeled bipartitioned difference graphs

on n vertices is 2 n. P r o o f . By Lemma 17.3.3 and the fact that IO~1- 2 n-l, we have [ A n [ - 2 n-1 q- 2 n-2 + " "

q- 20 + [ A o [ - 2n - 1 + 1 - 2 ~.

Another way to see this result is to modify the bijection between | and the set of binary words of length n - 1 into a bijection between A~ and the set of binary words of length n. We omit the details. 9 17.3.5 The number of unlabeled difference graphs on n >_ 1 vertices is 2~-2 + 2[~J -1.

Theorem

P r o o f . Again, according to Theorem 2.4.4, a difference graph consists of zero or more isolated vertices and a connected difference graph. Hence the required number is ~ - - 0 c~, where c~ is the number of unlabeled difference graphs with no isolated vertices on s vertices. According to Theorem 2.4.4, a bipartitioned difference graph with no isolated vertices is determined up to isomorphism by the cardinalities of the blocks X 1 , . . . , Xk and Y1,..., Yk, so the number of unlabeled bipartitioned difference graphs with no isolated vertices on s vertices is equal to the number of compositions of s into an even number of parts, i.e., sequences ( x l , . . . , x k , Y l , . . . , Y k ) of positive integers whose sum is s. For s _> 2, there a r e (;[21) compositions of s into 2k parts, and therefore a total of ~k (~k-~l) - 2s-2 compositions of s into an even number of parts. Since we are counting non-bipartitioned difference graphs

17.3

481

E n u m e r a t i o n of Difference Graphs

with no isolated vertices, Cs equals the number of inequivalent compositions of s into an even number of parts, where the composition ( x l , . . . , xk, yl,. 9 9 yk) is considered equivalent to itself and to ( y l , . . . ,xk, zl,... ,xk). We find c~ by the Burnside Lemma. All 2 ~-2 compositions into an even number of parts are invariant under the identity permutation. A composition into an even number of parts is invariant under the permutation 7r that switches color classes if and only if it has the form (if:l,...,3g.k, X l , . . . , X k ) , and ( x a , . . . , x k ) must then be a composition of s/2. Hence there are (~s even)compositions of s into 2k parts that are invariant under 7r, where the Iversonian (s even) stands for 1 if s is even and for 0 otherwise. By summing over all k >_ 1, we find that there are 2~/2-1(s even) compositions of s into an even number of parts that are invariant under 7r. Hence by the Burnside L e m m a c~ = 89(2 ~-2 + 2~/2-1(s even)) for s >_ 2. Clearly Co = 1 and cl - 0. therefore the number of unlabeled difference graphs on n >_ 2 vertices is given by n

~ - ~ Cs s~O n

1+ 0+ ~

(2 s-3 +

2s/2-2(s

even))

s~-2 n

= 2+ ~

(2 s-3 + 242-2(s even))

s----3

=2+2~-2-1+

(1+2+4+...2[~J

-2)

- 1 + 2 '~-2 + 2[~J -~ - 1 - 2 ~-2 + 2[~J -~. This value is also correct for n - 1.

9

Chapter 18 Extremal Problems 18.1

Introduction

Consider the following problems. How many connected threshold subgraphs can a graph have? What graphs with n vertices and rn edges have the largest number of connected threshold subgraphs? These problems still remain open. Paul ErdSs conjectured that for a fixed number of vertices, the complement of a matching contains the largest number of threshold subgraphs (private communication). Erd6s et al. [EGOZ89] consider the following problems. Given a graph G with n vertices and m edges, what is the largest number of edges that a chordal subgraph of G can have? What about an interval subgraph or a threshold subgraph? We report the results of these authors on interval subgraphs and threshold subgraphs in Section 18.2. Boesch et al. [BBB+90] considered the following problem. Which graphs on n vertices and m edges maximize the sum of squares of the degrees? It turns out that these graphs must be threshold graphs. However, not every threshold graph maximizes the sum of squares of the degrees. Only partial results are known in characterizing these threshold graphs. We report these results in Section 18.3. 483

484

Extremal Problems

18.2

Large Interval and Threshold Subgraphs of Dense Graphs

Given a graph with n vertices and m edges, what is the largest number of edges a chordal subgraph can have? What about an interval subgraph or a threshold subgraph? ErdSs et al. [EGOZ89] studied this problem for m >_ n2/4 + 1. They gave a complete answer for chordal subgraphs, and partial answers for interval and threshold subgraphs. We report their results for interval and threshold subgraphs. All results in this section are from [EGOZ89]. Let G be the class of graphs with n vertices and m edges, and 7 / b e a class of graphs closed under vertex deletion. Then there is a number f such that every member of ~ contains a member of 7-/with at least f edges, and there is a member of G containing no member of 7t with more than f edges. In practice, f may be hard to find, and it may be possible to bound it by finding a lower bound fl (there must be a member of 7-t with at least fl edges) and an upper bound f2 (there need not be a member of 7-t with more than f2 edges). The questions above are only interesting if the graphs contain sufficiently many edges. The complete bipartite graph Kn/2,n/2, for example, has no triangles and thus contains no interval graphs larger than a path ( n - 1 edges) and no threshold graph larger than a star (n/2 edges). We only consider graphs with n vertices and at least n2/4 + 1 edges (sufficient to guarantee the existence of a triangle), and refer to them as dense graphs. We define the size of a graph as the number of edges in the graph. The results in this section can be summarized as follows (all graphs are assumed to be dense graphs)" 9 Every graph has an interval subgraph of size > (1 + c)n. 9 There is a graph with no interval subgraph of size > 3n/2 - 1. 9 Every graph has a threshold subgraph of size > (1 + c)n/2. 9 There is a graph with no threshold subgraph of size > (1 + 0.5)n/2. The two constants c remain to be determined. By the star of a vertex a we mean the graph consisting of all vertices in the closed neighborhood of a, along with the edges joining a to all its

18.2

Large Interval and Threshold Subgraphs of Dense Graphs

485

neighbors. By the star of an edge ab we mean the union of the star of a and the star of b. We call a graph G an edge-star or an e-star if G has an edge ab such that G is the star in G on the edge ab. We need the following Lemmas. L e m m a 18.2.1 Let G be a threshold graph with no K4. Then G has an edge ab (called the r o o t edge) such that every edge of G is either incident to a (the r o o t v e r t e x ) , or is of the form bc for some c such that abc is a triangle of G. Proof. Let a, b be the first two vertices in the non-increasing order of degrees in the threshold graph G. " It is a known principle that if a graph has average degree A, it has a vertex of degree considerably larger than A or almost no vertices of degree much below A. The next lemma makes this more formal. L e m m a 18.2.2 For every 1 > e > 0 there is a d > 0 (depending only on e) such that every graph with n vertices and average degree A has a vertex of degree exceeding (1 + d)A or fewer than en vertices of degree at most (1 - e)A. Proof. Choose d < e2/(1 - e ) . Suppose the conclusion is false. Then the en smallest degrees do not exceed ( 1 - e)A each, for a total not exceeding e ( 1 - e)An, and the remaining ( 1 - e)n degrees do not exceed (1 + d)A each, for a total not exceeding (1 - e ) ( 1 + d)An. Thus the grand total degree for the whole graph is at most (1 + d - e 2 - ed)An < An, contradicting the fact that the total degree is An. 9 Another commonly-used fact is that in the neighborhood of the largestdegree vertex there must be a vertex of "not too small" degree. The next lemma states this more formally. L e m m a 18.2.3 Let G have average degree exceeding n/2. Let the largestdegree vertex of G have degree (1 + c)n/2. Then in its neighborhood there must be a vertex of degree exceeding

1-c+

2c 2 ) n i + c 2"

Proof. If not, then each of the (1 + c)n/2 vertices in the neighborhood has degree not exceeding that, while each of the (1 - c ) n / 2 vertices not in that

Extremal Problems

486

neighborhood has degree not exceeding (1 + c)n/2. Again, the total degree of all the vertices is no more than n2/2, a contradiction. .. Yet another general result is that if not too many vertices of small degree are deleted from a dense graph, the resulting graph is still dense. The next l e m m a formalizes this. L e m m a 18.2.4 Let 1 > e > O. If G is a dense graph on n vertices, and at most en vertices of degree at most ( 1 - e ) n / 2 are deleted from G, the resulting graph is still dense. P r o o f . The number of deleted vertices is k < en and the total number of edges deleted is at most k(1 - e ) n / 2 . The number of remaining edges is therefore at least

~ -~-+ 1 - k(1-

n

( ~ - k) ~

e)~ -

4

k(2~+ 1+

4

k) > ( ~ - k) ~ -

4

+ 1.

L e m m a 18.2.5 ([Edw]) If G has n vertices and at least n2/4 + 1 edges, then G has an edge that is on at least n/6 triangles. T h e o r e m 18.2.6 There is a constant c > 0 such that for sufficiently large n, every dense graph on n vertices has an interval subgraph with at least (1 + c)n edges. P r o o f . Let e - 0.1 and d - 0.01. If there is a vertex with at least (1 + d ) n / 2 neighbors, then the largest-degree vertex a has degree (1 + f ) n / 2 , where d _< f < 1. Then by Lemma 18.2.3, a has a neighbor b with degree exceeding (1 - f + 2_]_!_~ ~ The e-star on edge ab is an interval graph with more than l+fJ~" d~ < -l+f ~ this (1 + -l +~f ) n 1 edges. For sufficiently large n ~ and for c < V -e-star has at least (1 + c)n edges. Otherwise, by L e m m a 18.2.2, there are e2 fewer than en vertices of degree at most (1 - e)n/2 (because d < i-~-~, as required in the proof of L e m m a 18.2.2). If all these vertices are deleted, the resulting graph G ~ is still dense by Lemma 18.2.4. By L e m m a 18.2.5, G' has an edge ab lying on at least ( n - en)/6 - 0.15n triangles. Let abc be one of these triangles. The set of edges consisting of these triangles, the star on a, and the star on c form a graph H with at least n(0.45 + 0.45 + 0.15) - 2 edges. This is at least (1 + c)n for sufficiently large n and c < 0.05. Further, it is easy to construct an interval model for H. 9

18.2

Large Interval and Threshold Subgraphs of Dense Graphs

487

T h e o r e m 18.2.7 There is a constant c > 0 such that every dense graph on n vertices has a threshold subgraph with at least (1 + c)n/2 edges. P r o o f . Let 0 < e < 1/4 and 0 < d < e2/(1 - e) be two constants. If there is a vertex a of degree at least (1 + d)n/2, then the star on a is a threshold graph with at least (1 + c)n/2) edges for each c, 0 < c _< d. Otherwise, by Lemma 18.2.2, the number of vertices of degree at most (1 - e)n/2 is k < en. Delete all these vertices to obtain G ~. By Lemma 18.2.4, G ~ is dense. Hence, by Theorem 18.2.5, G' has an edge ab lying on at least ( n - k)/6 triangles. The union of these triangles and the star on a is a threshold graph whose number of edges is at least

n ~(1-r provided c _< ( 1 -

n-k 6

n >~(1-r

n-en 6

2 = 5 ~(1 - r

n ->(l+c)g'

4e)/3.

F a c t 18.2.8 There is a dense graph on n vertices whose largest interval subgraph has exactly (3/2)n - 1 edges. P r o o f . Consider the dense graph G obtained by adding an edge ab to the complete bipartite graph Kn/2,n/2. Let H be any interval subgraph of G. Delete one vertex from H, making sure it is a or b if H contains one of them. The resulting induced subgraph of H is still an interval subgraph of a bipartite graph on n - 1 vertices, and hence is a forest and has at most n - 2 edges. Adding back the star on the deleted vertex adds at most (n/2) + 1 edges, showing that H has a total of at most (3/2)n - 1 edges. On the other hand, let abc be any triangle in G. It is easy to see that the union of the stars of a, b and c is an interval graph with exactly (3/2)n - 1 edges. .. F a c t 18.2.9 For n sufficiently large and any fixed e > O, there is a dense graph on n vertices having no threshold subgraph with at least (1.5 + e)n/2 edges. (Hence the c in Theorem 18.2.7 cannot exceed 1/2.) P r o o f . Let A and B be vertex sets of size (1 + c)n/2 and ( 1 - c)n/2 respectively and add all ( 1 - c 2 ) n 2 / 4 edges from A to B, where c = 1/3. Now divide A into two equal parts A1 and A2 and add (cn)2/4 + 1 edges between those two parts, distributed as regularly as possible. The degree of a vertex in B is simply (1 + c)n/2, while each vertex in A1 has (1 - c)n/2 edges leading to

488

Extremal Problems

B and at most just over ((cn)2/4)/((1 + c)n/4) = nc2/(1 + c) edges leading to A2; so no star can be a big enough threshold graph. Consider a threshold subgraph containing a triangle. The graph is tripartite so has no K4; hence by Lemma 18.2.1 it has a root vertex and a root edge. We distinguish three cases: (i) root edge from A1 to A2, root vertex in either set; (ii) root edge from B to A1, root vertex in B; (iii) root edge from B to A1, root vertex in A1. All other cases are equivalent to these. (i) The root edge is from A1 to A2 and lies on at most ( 1 - c)n/2 triangles (the other vertex of the triangle must be in B); the root vertex has in addition about nc2/(1 +c) edges from A1 to A2. The total size of this threshold graph is about (1 - c)n + nc2/(1 + c), which equals (1.5)n/2 (since c = 1/3). (ii) The root edge joins B to A1. It lies on about c2n/(1 + c) triangles, and the degree of the vertex at B is at most (1 + c)n/2, of which c2n/(1 + c) edges have already been counted. The total size of this graph is thus about (1 + c)n/2 + c2n/(1 + c), which again equals (1.5)(n/2). (iii) The root edge again joins A1 to B, but we consider also the edges from the vertex in A1 to B, of which there are (1 - c)n/2. The total size of the threshold graph is about 2nc2/(1 + c) + (1 - c)n/2 = n/2. ..

18.3

M a x i m i z i n g the S u m of Squares of Degrees

Recall that a graph having n vertices and e edges is called an (n, e)-graph. In this section we consider the following problem. Given n and e, which (n, e)graphs have the largest sum of squares of the degrees? We refer to these graphs as a-optimal. Boesch et al. [BBB+90] consider this problem and show that these graphs must be threshold. However, not every threshold graph is a-optimal. For every valid choice of n and e, the authors define two threshold (n, e)graphs, which always exist and are unique, and prove that at least one of them is a-optimal. This enables us to compute the optimal value for any valid choice of n and e. We give examples showing that not both need be a-optimal, as well as a-optimal graphs that are not of this form. This leaves open the problem of characterizing the threshold graphs that are ~r-optimal. In this section we report the results of [BBB+90]. If r is a function from graphs to integers, then a graph G is called r

18.3

M a x i m i z i n g t h e S u m of Squares of Degrees

489

optimal if r >_ r for all graphs G' with the same number of vertices and edges as G. One such function of interest here is a(G) - ~ n d~, where ( d l , . . . , dn) is the degree sequence of G. Throughout this section we assume that n and e denote, respectively, the number of vertices and edges of a graph. Thus 0 <_e _< (2). As usual, Ki, Sj and Kl,k denote the complete graph on i vertices, the edgeless graph on j vertices and a star on k + 1 vertices, respectively. Lemma

18.3.1 Every a-optimal graph is a threshold graph.

P r o o f . If G is a graph with vertices a, b, c such that ac is an edge, bc is a nonedge and deg(b) > deg(a), consider the graph G' obtained from G by replacing edge ac with bc. It is easy to see that a(G') > a(G). Thus if a graph G is a-optimal, then for every two vertices a, b, deg(a) >_ deg(b) if and only if N[a] D_ N(b). Therefore G is a threshold graph by Condition 5 of Theorem 1.2.4. 9 Lemma

18.3.2 A graph G is a-optimal if and only if its complement G is

a-optimal. P r o o f . The result is immediate, since a(G) - ~"~in__l[(rt -- 1) - d~]2 or(G), where k(n, e)is a constant depending only on n and e. Lemma

18.3.3 Let G -

+

K1 0 H be an (n, e)-graph. Then

1. If G is a-optimal, then H is a-optimal. 2. Conversely, if H is a-optimal and if there is a connected a-optimal (n, e)-graph, then G is a-optimal. P r o o f . Part 1 follows from the fact that a(IQ | H) = k(n, e)+ a(H), where k is a constant depending only on n and e. To prove Part 2, let G t be a connected a-optimal graph. Then G ~ is a threshold graph of the form K1 | H', and both H and H' are ( n - 1, e - n + 1)-graphs. By part 1, H' is ~r-optimal, hence a(H) = a(H'), a(G) = cr(G'), and G is a-optimal. 9 D e f i n i t i o n 18.3.4 For fixed n and e, G1 and G2 are the following threshold

(n, e)-graphs (Figure 18.1)" (~l(p, q, r) --

I~p (~ ( ~q [,_JKl,r),

490

Extremal Problems

where n - p + q + r + l, e - ( ~ ) + p ( q + r + l ) + r ,

G2(p, q, r) - Sp I.J (I(q (]~ KI,~),

Figure 18.1: The (7, 6)-graphs of the forms G1 and G2.

GI(0, 0, 6) L e m m a 18.3.5

G2(2, 0,4)

1. G~ (p, q, r) is the complement of G2(p, q, r).

2. For every valid n and e, there exist (n, e)-graphs of the form G1 and G2.

~. If G1 (p, q, r) and G1 (pt qt rt) are (n, e)-graphs, they are isomorphic, and similarly for G2.

P r o o f . 1" This is obvious from the definition. 2" By part 1, it is enough to show that G2(p, q, r) exists whenever 0 < e < (2). Let k be the largest integer satisfying (k2) < e. Then k < n. If k - n~ then e -

(2) and we can t a k e q -

n-l,p-r-

0. I f k _< n - l ,

take

qe - ( k 2 ) _> 0, r - k - q > 0 (for otherwise e > ( ~ ) + k - (k+l), contradicting the maximality of k), and p - n - k - 1 > 0. 3: Let G1 (p, q, r) and GI(p', q', r') be (n, e)-graphs. If p - p', then clearly q _ ql and r - r' and we are done. Now assume that p' - p + 5 and 5 > 0. When p is increased by 1, (~) + p ( n - p)increases by n - p 1. Therefore

p,+,n

p 1,+,n

2,+

p

18.3

Maximizing the Sum of Squares of Degrees

491

Hence e - r' >_ (e - r) + (n - p - 1) - e + q, i.e., q + r' _< O, and we conclude that q - r' - 0 as well as 6 - 1. Therefore G1 (p, q, r) - G1 (p, 0, r) - Kp+a | S~_p_a,

G~ (p', q', r') - G~ (p', q', O) - Kp, | Sq, - Kp+, | Sn-p-1. We frequently use the following operation in the proof of Theorem 18.3.7 below (see Figure 18.2). Let G be a graph, a and b two vertices of G of degrees a and/3, respectively, T1 a set of m neighbors of a, all different from b and having the same degree tl, and T2 a set of m non-neighbors of b, all different from a and having the same degree t2. Then G ( a , b, Tx, T2) is obtained from G b y d e l e t i n g all edges from a to T1 and adding all edges from b to T2. A simple calculation shows that

a(G(a, b, T1, T2)) - ~r(G) - 2m(/3 - a + t2

-

tl

-~- m

-t- 1)

(18.1)

Figure 18.2: An operation used in the proof of Theorem 18.3.7.

G

T~

a

T~

T2

b

T2

G(a,b, T1,T2)

F a c t 1 8 . 3 . 6 If e < 3 and n is arbitrary, then both G1 and G2 are or-optimal

(n, e) -graphs. P r o o f . This follows from L e m m a 18.3.1, since if e _< 3, all the threshold (n, e)-graphs have the same ~r. 9 Theorem

1 8 . 3 . 7 For every fixed n and e, at least one of the (n, e)-graphs

G l a n d G2 is a-optimal.

492

Extremal Problems

P r o o f . We prove the result by induction on n. Obviously, the result holds for n - 1. Assume that for every e' <_ ( ~ 1 ) , one of the a-optimal ( n - 1, e')graphs is of the form G1 or G2. We may also assume that some connected (n,e)-graph is a-optimal. (If not, the complement G of every a-optimal (n, e)-graph G is connected and a-optimal by Lemma 18.3.2. If we can show that one of these G is of the form G1 or G2, then by L e m m a 18.3.5 G is also of the form G1 or G2.) By Lemma 18.3.1, a connected a-optimal (n, e)-graph contains a dominating vertex, deleting which we get an ( n - l , e - n + l ) - g r a p h . By the induction hypothesis there exists a a-optimal (n - 1, e - n + 1)-graph H of the form G1 or G2. By Lemma 18.3.3, G - K I | is a a-optimal (n, e)-graph. If H is of the form G1, then G is also of the form G1 and the proof is complete. So we assume that H is of the form G2(p, q, r), and there is no a-optimal (n - 1, e - n + 1)-graph of the form G1. We may also assume that p > O, (18.2) for otherwise G is of the form G2 and we are done. We have q + r >_ 3,

(18.3)

for otherwise H has at most 3 edges and hence by Fact 18.3.6, there is also a a-optimal (n - 1, e - n + 1)-graph of the form G1, contrary to our assumption. The proof is completed by obtaining an (n, e)-graph G' of the form G1 or G2 such that a(G') _> ~r(G). Observe that in G, every vertex of Sp has degree 1, every vertex of Kq has degree q + r + 1, every vertex of Kr has degree q + r, there is a vertex x of degree q + 1 adjacent to every vertex of Kq, and there is a vertex y of degree n - 1. See Figure 18.3. C a s e 1: q >_ 1. We first show that

p<_q+r-1.

(18.4)

Suppose p > q + r - 1. We show that H cannot be or-optimal, a contradiction. Starting with H, let 9 a be the vertex x of degree q, 9 b be a vertex in Kq of degree q + r, 9

9 T2 be a set of q - 1 vertices in S'p (which exist by p > q + r - 1), and

18.3

493

Maximizing t h e S u m of Squares of Degrees

Figure 18.3: The graph G used in the proof of Theorem 18.3.7.

s~

Y

G -

9 H'-

Is

@

G2(p, q, r)

H(a,b,T~,T2).

T h e n c r ( H ' ) - a ( H ) - 0 by (lS.1). Note t h a t H ' has the form ((/s I,.JS;)@ K~) U Sp_q+l, where K ' and S' are cliques and stable sets of the i n d i c a t e d cardinalities, all positive. P u t m - min(q + r - 2 , p - q + 1), and observe t h a t m > 0 by (18.3) and our a s s u m p t i o n . In H', let 9 b be the s a m e v e r t e x b as before, now of degree 2q + r - 1, 9 c be a v e r t e x in K~+r_ 1 of degree q + r - 1, 9 T; be a set of m vertices

*' in I(q+r_ 1 other

9 T~ be a set of m vertices

in Slp_q_t_l of

9 H"-

H'(c,b,T~,T~).

t h a n c, of degree q + r -

degree 0,

1

494

Extremal Problems

Then a( H") - (r( H') = 2 m ( m + 2 - r ) by (18.1). I f m = q + r - 2 , then ( r ( H " ) - a ( H ' ) = 2mq > 0 since we are in Case 1. I f m = p - q + l , then ~r(H")-a(H') = 2m(p+3-q-r) > 0 because p > q + r 1. This p roves (18.4). We now show t h a t r = 0. For otherwise, let m = min(p, r) > 0 by (18.2), and in G let 9 a be the vertex y of degree n - 1, 9 b be the vertex x of degree q + 1, 9 T1 be a set of m vertices from Sp of degree 1, 9 T2 be a set of m vertices from Kr of degree q + r, and

9 G ' - G(a,b, T1,T2). Then ~r(G')-a(G) = 2re(re+q-p) by (18.1). If m = p, then a(G')-cr(G) = 2mq > 0, and if m = r, then a ( G ' ) - a ( G ) = 2m(r + q - p ) > 0 by (18.4), so in both cases we have a contradiction to the a - o p t i m a l i t y of G. This proves t h a t r = 0. Thus H is of the form Kq+l Ig Sp. Now if p = 1, then G is in the form G2, as required. Otherwise p _> 2 by (18.2), and we let 9 a be the vertex y of degree n - 1, 9 b be a vertex of Sp of degree 1, 9 T1 - Sp - {b}, a set of vertices with degree 1, 9 T2 be a set of p - 1 vertices in Kq (which exist by (18.4) and r - 0) of degree q + 1, and

9 G ' - G(a,b, T1,T2). Then c r ( G ' ) - a ( G ) - 0 by (18.1). Thus G' is also a-optimal. Clearly, G' is an (n, e)-graph of the form G ~ ( p - 1, p, q - p + 1). C a s e 2" q - 0. Then H is of the form K~ t2 Sp+l, which is isomorphic to a2(p + 1, 1, 0). By (18.3) and q - 0 we certainly have r - 1 >_ 1, and we are again in Case 1. 9 R e m a r k . For (7, 6)-graphs, G2 is not a-optimal, since a ( G l ) > or(G2) (see Figure 18.1). By considering the complements of these two graphs as well,

18.3

Maximizing the Sum of Squares of Degrees

495

we conclude that neither G1 nor G2 need always be a-optimal. Moreover, the (6, 7)-graph G of Figure 18.4 is not of the form G1 or G2, yet is a-optimal along with G1 and G2, since all these graphs have the same cr value. Figure 18.4: Examples of it-optimal (6, 7)-graphs.

GI(1, 2,2)

G2(1, 1,3)

G

Chapter 19 Other Extensions 19.1

Introduction

In this chapter we discuss three unrelated concepts on threshold graphs. In Section 19.2 we look at a generalization of threshold graphs based on geometrical embeddings of graphs. The threshold graphs can be thought of as the graphs for which there exist a non-negative real weight w~ associated with each vertex u and a positive real threshold t such that u v is an edge if and only if w u w v > t . If we allow the weights and threshold to be any reals, we obtain the generalized threshold graphs characterized by Reiterman et al. in [RRS89]. If the weights are replaced by vectors and the product is replaced by the scalar product, we obtain a further generalization of the threshold dimension. In Section 19.3 we introduce a new class of graphs called C-tolerance intersection graphs, due to Jacobson et al. [JMM91]. Golumbic et al. [GM82, GMT84] defined the tolerance graphs by assigning to each vertex u an interval I~ and a non-negative tolerance t~, such that u v is an edge if and only if IIu n Iv I _> min(t~, t~). These graphs clearly generalize the interval graphs. Monma et al. [MRT88] defined the threshold tolerance graphs by assigning to each vertex u a non-negative weight w~ and a tolerance t~ such that u v is an edge if and only if w~ + w~ > min(t~, t~). Both these classes are special cases of C-tolerance intersection graphs. In Section 19.4 we study the universal graphs for the class T~ of threshold graphs on n vertices. A d-universal graph for a class C of graphs is a graph G such that every graph in d is an induced subgraph of G. Hammer and 497

498

Other Extensions

Kelmans [HK94] completely characterize those T~-universal graphs with the least number of vertices that are themselves threshold graphs.

19.2

Geometric Embeddings of Graphs

Given a simple graph G, a representation of G is a function that assigns to each vertex x of G a vector ~ C R d, such that two vertices x and y are adjacent in G if and only if x and y satisfy some specified geometrical condition. Representations of graphs of this type have been considered by numerous authors. Reiterman et al. [RRS89] consider the representation of a graph G = (V, E) such that for some real threshold t, xyCE

if and only if

~y_t.

The scalar product dimension dp(G) is the minimum number d such that G admits such a representation in R d. The authors show that dp(G) <_ t(G), where t is the threshold dimension. They call graphs for which dp(G) = 1 generalized threshold graphs and characterize them. We report their results here. Recall from Theorem 1.2.4 that G = (V, E) is a threshold graph if and only if there exist a function w : V --, N and a threshold number t C R such that xyCE if and only if w ~ + w v > t , or equivalently, using w' - e ~, t' xy E E

-

e t,

if and only if

w~wy~ ~ >_ t',

wherew t:V--+N +,t ~EN +,andN +={xCN:x>0}. Relaxing f r o m N + to N, we have the following generalization of threshold graphs. D e f i n i t i o n 19.2.1 A graph G = (V, E) is a g e n e r a l i z e d t h r e s h o l d g r a p h when there are a threshold t C N and a label x C N 1 associated with each vertex x, such that for all distinct x, y C V, xy C E

if and only if

x y >_ t.

(19.1)

Thus the threshold graphs are precisely those generalized threshold graphs that can be represented by a positive labeling and a positive threshold. Generalized threshold graphs are not necessarily threshold. For instance, the

19.2

Geometric Embeddings of Graphs

499

graph 2K2 is a generalized threshold graph, as shown in Proposition 19.2.2. However, if in Definition 19.2.1 the labeling and the threshold are nonnegative, then G = (V, E) is in fact a threshold graph, as is easy to see. We can always ensure, by slightly decreasing t if necessary, that the inequality in (19.1) never holds with equality. P r o p o s i t i o n 19.2.2 A graph is a generalized threshold graph if and only if it is the disjoint union of two threshold graphs or the complement of a difference graph. P r o o f . Suppose G - (V, E) is the disjoint union of threshold graphs G1 = (V1, EI) and G2 - ( 89 E2), and let x ~ x C R (x e V), ti > 0 represent G~ for i - 1, 2, where x >_ 0 (x E V). By a simple rescaling we may assume that t 1 t 2 t . Then the labeling x ~ x ~ defined by x'-

~" x, ( -x,

for x E V~ for x C 1/2

together with the threshold t represent G. Suppose the complement G is a difference graph with color classes V1 and V2. By Theorem 2.4.3, if we add to G all the edges joining vertices of V1, it becomes a threshold graph T. Let x H x be a positive labeling and t > 0 a threshold representing T, such that x y =/= t for all distinct x, y. Then G is represented by x ~ x ~ and t ~, where x'-

~ x, [ -x,

for x C V1 for x E

t ! --

--t.

Conversely, let G be a generalized threshold graph represented by x H x and a threshold t such that the inequality in (19.1) never holds with equality. Put

{xc

<0}.

First, assume t > 0. Then there is no edge between V1 and 1/2. The induced subgraph G1 - (VI, El) is a threshold graph as the labeling is non-negative on V1. As for the induced subgraph G2 - (1/2, E2), we use the non-negative labeling x ~ - x and the same threshold t to see that G2 is also a threshold graph. Next, assume t _< 0. Then V1 and 1/2 are cliques and the non-negative labeling x ~ [x[ together with the threshold Itl represent a threshold graph

500

Other Extensions

(V, E") such that for all x C 1/1 and y C 1/2, we have xy E E if and only if xy ~ E". Thus by Theorem 2.4.3 G is the complement of a difference graph.

C o r o l l a r y 19.2.3 The generalized threshold graphs have Dilworth number at most 2. Proof. This follows from Propositions 19.2.2 and 2.4.2. 9 The following theorem gives a characterization of generalized threshold graphs by forbidden induced subgraphs. The proof can be found in [RRS89]. T h e o r e m 19.2.4 ([RRSS9]) A graph is generalized threshold if and only if it does not contain any of the graphs G1,..., G10 of Figure 19.1 as induced subgraphs.

Figure 19.1: The forbidden induced subgraphs for generalized threshold graphs.

": 9 i i i I t ! G1

G2

Ga

G4

G5

G6

G7

G8

G9

GlO

We now consider the dimension of generalized threshold graphs. Recall that the threshold dimension t(G) of a graph G = (1/, E) is the minimum number t of threshold spanning subgraphs G1,..., Gt of G whose union is E; in other words, two vertices form a nonedge in G if and only if they form a nonedge in each of G1,..., Gt. By taking exponents, we can say equivalently

19.2

Geometric Embeddings of Graphs

501

that the threshold dimension t(G) of G is the minimum number t with the property that there are a function f " V ~ IRt with f ( x ) > 0 (x E V) and a vector c > 0 in IRt, such that for distinct x, y C V,

xy~E

if and only if

f(x),J'(y)
where 9 denotes the coordinatewise product and a < b means ai < bi for all i. Again, by a simple rescaling we may require that each component of e is the same number t. We now consider replacing a , y by the usual scalar product x y , admitting also negative values.

Definition 19.2.5 A representation of a graph G - (V, E) in R d is a labeling x ~ x (x C V) with values in ]I~ d together with a threshold t C N, such that for all distinct x, y C V we have

xy C E

if and only if

x y > t.

The s c a l a r product d i m e n s i o n dp(G) of a graph G is the minimum number d such that G admits a representation in N d. The assignment x ~ x, where x is the row of the vertex-edge incidence matrix of G corresponding to x, together with the threshold t - 1, provide the simplest example of a representation of G. This shows that dp(G) is well-defined and gives the trivial upper bound dp(G) _< n, where n is the number of vertices (since the vectors a generate a subspace of dimension _< n over N). A better upper bound for dp(G) is established in the following theorem, which shows that the scalar product representation is more efficient than the threshold one, in the sense that it represents more graphs using small dimensions.

T h e o r e m 19.2.6 Every graph G satisfies dR(G) < t(G). P r o o f . Let G - (V,E), t(G) - s, and E - U~=IEk , where the (V, Ek) are threshold graphs. Let V - Dok U ... U D mkk be the degree partition of (V, Ek), k - 1 , . . . , s, as in Definition 1.2.3. Recall that for every distinct x , y C V with x C D~, y C D~,

xyCEk

if and only if

i+j>_mk+l.

502

Other Extensions

We construct a labeling x ~ x with values in IR~ by p u t t i n g 1

(x)k - M i- ~'~k f o r k - 1 , . . - , s ,

ifxcD/k,

where M > s is arbitrary. If xy C E, then xy C Ek for some k and x C D/k, y E D) for some i , j satisfying i + j _> mk + 1. Hence 1

1

x y > xkyk -- M i - ~ m k M j-~mk - M i+j-mk >_ M.

If xy ~ E for distinct x, y, then for every k - 1 , . . . , s we have xy ~ Ek. Let x E D~, y E D). Then i + j - i n k < 1. Thus xkyk -- M ~+j-mk _< 1, and hence xy-

~-~xkyk < S < M. k=l

This proves t h a t our labeling together with the threshold t - M represent G in N s, and hence dp(G) <_ s. 9 R e m a r k . t(G) can be much larger than dp(G). For instance, dp(C~) - 2 ( T h e o r e m 19.2.8), whereas t(Cn) >_ n/2 for n _> 4. W h e n G is the disjoint union of two or more graphs each of which has at least one edge, then again dp(G) is much smaller than t(G), as can be seen from the proposition below. Proposition

1 9 . 2 . 7 If G is the disjoint union of graphs G I , . . . , Gn, then

dp(G) <_ m a x ( d p ( G 1 ) , . . . , d p ( G ~ ) ) + 1. P r o o f . Let m - m a x ( d p ( G 1 ) , . . . , d p ( G n ) ) - ~ - 1. For each of the graphs G~ - (V~, E~), consider a fixed representation x ~ x (x C V~) in N TM with a threshold ti such that the last coordinate of every ~ is 0. Consider the m x m matrices 1 0 01 0 0 Ai _ 0 0

0 0

cosi~n - s i n 'z n

siniZn cos 'z n

and vectors d i - ( 0 , . . . , O, v / t - ti) E R m, where t > m a x / t i is for the mom e n t unspecified. We show that if t is sufficiently large, then the assignment

19.2

503

Geometric Embeddings of Graphs

x ~ x', where ~ e ' - Ai(x + d i) for x C V/, i - 1 , . . . , n, represents G in IRm with the threshold t. Indeed, if x E !Vi and y C Vj, then

x'y' - A i ( x + d i ) A J ( y + d j) - ( x + d i ) T A iT A J ( y + d j) - ( x + d i ) T A J - i ( y + d j) m-2 (j -i)Tr ~(j = ~ xkyk + Xm-lY~n-1 COS ~ + ~ / t - tjxm-1 sin k=l

-

T/

v/t - t~ sin ~ (jy -m -i)Tr 1

i)Tr ?~

+ ~ / ( t - t i ) ( t - tj)cos (j - i)Tr

n

n

Now if i - j, then x'y' - x y + t - ti, and hence x y > t if and only if x y > ti, which happens for x :fi y if and only if xy is an edge of Gi. If i -r j, then obviously lira x y -- cos (j - i)Tr < 1. t ---*oo

t

r/

Hence x y < t for all x in Gi, y in Gj if t is sufficiently large. This verifies that we have defined a correct representation for G. 9 R e m a r k . In general, the upper bound of Proposition 19.2.7 cannot be improved. For instance, d e ( K 2 ) = de(2K2) = 1, but dp(3K2) > 1. On the other hand, it is not known whether de(V, UE~) <_ ~ de(V, E~). We conclude this section with some examples. Theorem

19.2.8

1. dp(C~) - 2 for every n > 4. 2. d p ( P n ) - 2

>__ 5

3. dp(nK2) - 2 for every n >_ 3. ~. dp(Kn,~) - n for every n >_ 1. P r o o f . 1" Let Xl , . . . , X n be vertices of Cn along the cycle. For i - 1 , . . . , n, the vectors 27ri 27ri) R2 cos~,sin-C n

n

represent Cn in IR2 with the threshold t - cos 2_~. Hence dp(Cn) <_ 2. The reverse inequality follows by Proposition 19.2.2. 2" It follows from Part 1 that dp(P~) _< 2, since P~ is an induced subgraph of Cn+l. The reverse inequality holds for n >_ 5 by Proposition 19.2.2.

504

O t h e r Extensions

3: This follows from Propositions 19.2.7 and 19.2.2. 4: Let the two color classes of Kn,~ be { X l , . . . , X n } and {Yl,...,Yn}. We prove that if Kn,n can be represented in R d, then I(n_l,n_ 1 c a n be represented in R d-~. From this and from the fact that d p ( K 2 , 2 ) = d p ( C 4 ) = 2, it follows that d p ( K n , ~ ) >_ n. Let X l , . . . ,Xn, Y l , . . . ,Yn together with a threshold t represent K~,n in R d.

We have x i x j < t,

fori,j=l,...,n,i-Cj,

yiyj < t

for i , j - 1 , . . . , n .

xiYj >_ t

Set Xn U

-

-- Yn

-

Ilxn -- Y~II

(notice that the denominator is nonzero). Let us consider the projections xi,t Yi! of the vectors xi, yi, i - 1 , . . . , n, onto the hyperplane u x - O. We have x 'i -

xi -- ( u x i ) ~ ,

Yi' -

Yi - ( u y ~ ) u .

We show that these projections represent Kn-l,n-1 (on X l , . . . ,Xn-1 and y l , . . . , y~-l) with the same threshold t. Indeed, for i < n we have Xn UXi

-

IlXn Xn

~Y~ -

!

!

!

!

!

!

XiXj

YiYj

Yn

XnXi

w

--

Yn

--

ynXi

IlXn - y~ll < o, x n y i -- YnYi > O,

IlXn -- Y~II y~ --

and hence for i , j - 1 , . . . , n xiYj

--

-- Ynll xi

IlXn

-- Ynll

1

-

(x~- (,,x,),,)(yj -(uyj)u)

=

x~yj- (ux~)(uyj)>

t,

-

x~xj-

t,

-

y ~ y j - ( u y ~ ) ( ~ y j ) < t.

(~x~)(uxj)<

As the hyperplane u x - 0 is a ( d - 1)-dimensional subspace, the above may be transformed into a representation in R d-1. We have proved d p ( K n , n ) >_ n. The reverse inequality is obtained by checking that the vectors xi

-

( O , . . . , O , n + 1 , 0 , . . . , 0 ) C R ~ (with n + 1 at the i th coordinate),

yi

_

(1,1,...,1) CR n

represent

Kn,n in

R n with t - n + 1.

19.3

19.3

Tolerance Intersection Graphs

505

Tolerance Intersection Graphs

Recall the definitions of interval graphs, comparability graphs and permutation graphs (Definition 1.4.1). Golumbic and Monma [GM82] introduced a generalization of interval graphs using tolerances, whose graphs were called tolerance graphs or interval tolerance graphs in [GM82, GMT84]. For these graphs one can associate with each vertex v an interval Iv on the real line and a positive real number tv, its tolerance, such that two vertices u and v are adjacent if and only if [I~ n I,[ >_ min(t~, t.), where Ill denotes the length of the interval I. Two significant special cases are given by the following two propositions. P r o p o s i t i o n 19.3.1 ( [ G M S 2 ] ) A graph G is an interval tolerance graph with equal tolerances if and only if G is an interval graph. P r o o f . Every interval graph G has an interval model in which the non-empty intersections of intervals have positive lengths. If we take the smallest of these lengths as our tolerance, we obtain an interval tolerance model with equal tolerances for G. Conversely, let G have an interval tolerance model with constant tolerance t. Replace each interval I~ = [a, b] with the interval

i~v _ ( [a + t/2, b - t/2], [ 0,

ifb-a>_t otherwise.

Then I'~ NIIv -~ ~ if and only if tin N / , I _> t, and hence G is an interval graph.

Proposition

19.3.2 ( [ G M 8 2 ] ) A graph G is an interval tolerance graph with tv = [Iv[ for each vertex v if and only if G is a permutation graph.

P r o o f . The condition of the 'only if' part amounts to the fact that vertices u and v are adjacent if and only if Iu C_ Iv or Iv C_ In. In that case, consider a fixed order < on the vertex set of G, and orient every edge uv of G from u t o y when either Iu C Iv, or I~ = I~ and u < v. This gives a t r a n s i t i v e orientation of G. Also, orienting every nonedge uv of G from u to v when m i n l u < minI~ gives a transitive orientation of the complement G. As we observed after Definition 1.4.1, this means that G is a permutation graph.

506

Other Extensions

Conversely, assume that G is a permutation graph, which means that G and G are comparability graphs. By Lemma 16.4.4, every transitive orientation of G is a poset of dimension at most 2. Therefore we can represent each vertex of G by a point (Vl, v2) E (R+) 2 so that vertices u and v are adjacent in G if and only if (Ul,U2) _< (Va,V2) or (Vl,V2) _< (Ul,U2). We can then define the intervals I~ - I-v2, vl] for all vertices v, and obtain the result that vertices u and v are adjacent in G if and only if I~ C_ I~ or Iv C_ I~. 9 Monma et al. [MRT88] call a graph a threshold tolerance graph if there are weights Wv and tolerances tv for each vertex v such that uv is an edge if and only if wu + Wv >_ min(tu, tv). By Condition 8 of Theorem 1.2.4, if all the tolerances are equal, we obtain precisely the threshold graphs. The complements of threshold tolerance graphs are referred to as coTT graphs. Jacobson et al. [JMM91] use the term C-tolerance (S, #)-graphs to refer to a broader family of classes of graphs, extending the interval tolerance graphs and the coTT graphs. These graphs, also called the C-tolerance intersection graphs, provide a unified approach for investigating and characterizing several interesting classes of graphs. To define these graphs, consider the definition of tolerance graphs and replace the intervals {Iv} with a family S - {Sv}vev of subsets of a set Z on which a measure # is defined. Replace 'min' with a positive function r of two real arguments. As before, {tv}~ev is a set of positive real numbers. Now replace the condition for adjacency with uv@E

if and only if

#(SurlSv)>_r

tv),

and we have the definition of C-tolerance (S, #)-graphs. For fixed S, #, r the tolerance intersection graphs are closed under induced subgraphs. Thus a characterization by forbidden induced subgraphs may be possible. Also, the two special subclasses when all the tolerances are equal and when tv = #(S v) can be studied separately. Some of the interesting possibilities for r and S that Jacobson et al. consider are as follows: 1. r b) could be min(a, b), max(a, b) or a + b. The corresponding classes are called rain-tolerance, max-tolerance, or sum-tolerance intersection graphs, respectively. 2. 5' could be a family of intervals on the real-line, finite non-empty sets, or subgraphs, in which case tt would be the length of the intervals, the size of the sets, or the number of vertices in the subgraphs, respectively. In particular, if S is a family of nested sets, i.e. S - { S 1 , . . . , S~} with

19.3

Tolerance Intersection Graphs

$1 C_ . . .

507

C_ Sn, the associated tolerance interval graphs are called

~-tolerance chain graphs. Below we list some of the results of [JMM91]. The proofs are simple. 19.3.3 1. Every graph is a rain-tolerance intersection graph of substars of a star for some constant tolerance.

Theorem

2. A graph is a rain-tolerance intersection graph of substars of a star with tolerances equal to the number of vertices in substars if and only if it is a comparability graph. 3. A graph is a C-tolerance chain graph with constant tolerances if and only if it consists of a clique and isolated vertices. ~,. A graph is a rain-tolerance chain graph with tolerances equal to the set-sizes if and only if it is a complete graph. 5. A graph is a max-tolerance chain graph with tolerances equal to set-sizes if and only if it is a disjoint union of cliques. 6. A graph is a sum-tolerance chain graph with tolerances equal to set-sizes if and only if is an edgeless graph. The following results do not assume that the tolerances are restricted to the special cases where all tolerances are equal or where tv - #(Sv). 19.3.4 ( [ J M M 9 1 ] ) 1. A graph is a rain-tolerance chain graph if and only if it is a threshold graph.

Theorem

2. A graph is a max-tolerance chain graph if and only if it is an interval graph. 3. A graph is a sum-tolerance chain graph if and only if it is a co T T graph. Proof. 1" Let G - (V, E) be a rain-tolerance chain graph with S1 C_ . . . C_ S~ as the family of nested set and t a , . . . , t~ as the corresponding tolerances. Thus we have for u < v,

uvE E

if and only if

] S u ] - [ S u N I v ] >__min(t~,tv).

508

Other Extensions

In particular, if t~ _< IS~ l, then vertex 1 is dominating. If t 1 ) ISll and vertex 1 is not isolated, let u be adjacent to 1. Then tu <_ ISll and hence u is dominating. In any case, G contains an isolated or a dominating vertex. So does every induced subgraph of G by the heredity of being a min-tolerance chain graph. Hence G must be a threshold graph by Condition 3 of Theorem 1.2.4. Conversely, let G be a threshold graph, and consider its degree partition D o , . . . , Dm as in Definition 1.2.3. For each vertex v C Dy, define the set S~ and the tolerance tv by S,-

{1,...,j+l}, {1, , m + 1},

t~- { re+l, 2+m-j,

if j < L~J ifj >

Observe that the Sv are nested sets. i _< j _< [~], then ISuNSvl-i+l

if j < [~J ifj >

Further, for u E Di and v C Dj, if

<m+l-min(tu,

tv),

so u and v are not adjacent. If i _> j > [~J, then

n Xvl-

+ 1 _> min(t~, t~),

so u and v are adjacent. Finally, if/_< [~J < j, then ISunSvl-i+l_>rn-j+2

if and only if

i+j>m.

Since by Condition 6 of Theorem 1.2.4, u and v are adjacent if and only if i + j > m, G is a min-tolerance chain graph with the assigned nested sets and tolerances. 2" Let G be a max-tolerance chain graph on the vertices V l , . . . , v ~ . We may assume without loss of generality that the associated nested sets are { 1 , . . . , r i } for i - 1 , . . . , n , where 1 _< r~ _< ... _< r~. Let the corresponding tolerance of vertex vi be ti. We may also assume that ti _< ri for all i, for if ti > ri, then vi is isolated and we can disregard vi. Consequently, we can associate the interval Ii - [ti, ri] with each vertex vi. Now vertices vi and vj are adjacent if and only if min(ri, rj) >_ m a x ( t i , t j ) . This holds if and only if Ii N Ij 5r 0 , and thus G is an interval graph. Conversely, if G is an interval graph with associated intervals [ti, ri], i - 1 , . . . ,n, we may assume without loss of generality that the ti and ri are positive integers. Then the sets { 1 , . . . ,ri} along with the tolerances ti represent G as a max-tolerance chain graph.

19.4

509

Universal Threshold Graphs

3: Recall that the complements of threshold tolerance graphs (coTT graphs) have the adjacency defined by

ij C E

if and only if

wi + wj < min(ti, tj).

Again we may assume that the ti are positive integers. Hence the coTT graphs are the same as the sum-tolerance chain graphs with nested sets { 1 , . . . , t i } and tolerances wi + 1/2. 9 R e m a r k . There is no known characterization of threshold tolerance graphs by forbidden subgraphs. However, a polynomial-time algorithm for recognizing this class of graphs is given in [MRT88].

19.4

Universal Threshold Graphs

In [HK94], Hammer and Kelmans studied the universal graphs for threshold graphs. Given a class C of graphs, a graph G is called a C-universal graph, or simply a C-UG, if every graph in C is isomorphic to an induced subgraph of G. If in addition G has the least number of vertices among all C-UGs, then G is called a minimum C-UG, or simply a C-MUG. Observe that if d is closed under complements, then G is a C-UG whenever its complement is, and similarly for C-MUGs. Likewise, if C* is the class of all induced subgraphs of the graphs in C, then every C-UG is a C*-UG, and similarly for MUGs. Let T~ denote the class of all threshold graphs on n vertices. We characterize all the :m-MUGs that are themselves threshold (i.e., the threshold T~-MUGs), a result originally proved by Hammer and Kelmans [HK94]. N o t a t i o n . We denote by (G, xy) the graph obtained from a graph G by adding two adjacent new vertices x, y, and joining y to all vertices of G. In other words, (O, xy) = (O U x) @ y. Similarly, we denote by (G,g-~y) the graph obtained by adding two nonadjacent new vertices x, y, and joining y to all vertices of G. In other words, (a,

= (G 9 y) u x.

Note that (G, xy) = (G, yx). Theorem

19.4.1 For every graph G, the following conditions are equivalent:

1. G is a (threshold)2/-~-UG;

510

Other E x t e n s i o n s

2. (G, xy) is a (threshold)~n+I-UG; 3. (G,-~y) is a (threshold)~n+I-UG. P r o o f . By Condition 4 of Theorem 1.2.4, if one of the three graphs is threshold, so are the other two. 1) =~ 2): Assume that G is a ~ - U G . Let T be any threshold graph on n + 1 vertices. If T contains an isolated vertex x, then T - {x} is isomorphic to an induced subgraph of G, and hence T is isomorphic to an induced subgraph of (G, xy). Otherwise T contains a dominating vertex y, and the proof is similar. 2) ~ 1): Assume that (G, xy) is a Tn+I-UG. Let T be any threshold graph on n vertices. The graph T' = T U {a} is isomorphic to an induced subgraph of (G, xy) on some vertex set X. Clearly, X does not contain y, since T' contains an isolated vertex. We may assume without loss of generality that x is the isolated vertex. It follows that T is isomorphic to the subgraph induced by X - {x} in G. The fact that 1) ,z---->, 3) follows from 1) ~ 2) by a simple complementarity argument. 9 Notation. Let a ~ = ( a l , . . . , a t ) b e a 0-1 sequence. Let H = G(a ~) be the graph constructed from G as described below: 1. s e t H : = G ; 2. for i = 1 , . . . , r , do H "-

(H, xiyi), (H, x-~),

if ai = 1, if ai = O.

1. Observe that G(a ~) is isomorphic to G(b ~) if and only if a ~ = b ~, and that, as with threshold graphs, there is a 1-to-1 correspondence between the set of 0-1 sequences a ~ and the set of graphs of the

F a c t 19.4.2

form 2. From Theorem 19.~. 1, it follows that G(a n) is a T~+k- UG if and only if G is a Tk- UG. 3. Since K1 is a threshold ~ - U G , K l ( a ~-a) is a threshold Tn-UG for each 0-1 sequence a n-1. Thus there are at least 2 n - 1 threshold T~-UGs, each with exactly 2 n - 1 vertices.

19.4

Universal Threshold Graphs

511

In fact, the 2 n-1 threshold Tn-UGs K l ( a n - l ) are all the threshold T~-MUGs, as we show below. T h e o r e m 1 9 . 4 . 3 If G is a Tn-MUG, then

1. the vertices of G can be partitioned into a stable set { S l , . . . , s n } and a clique { C l , . . . , Cn-1}. Hence G contains exactly 2n - 1 vertices. 2. If the si are indexed so that deg(sl) <_ . . . _< deg(sn), then deg(sn) n - 1 and k - l <_ deg(sk) <_ k for k - l , . . . , n - 1 . P r o o f . By Part 3 of Fact 19.4.2, G contains at most 2 n - 1 vertices. Now Part 1 follows since both K~ and its complement are induced subgraphs of G. To prove Part 2, first consider the threshold graphs of the form I~ | Kn_~ for r - 1 , . . . , n. The graph G contains t h e m all as induced subgraphs. Now setting r - 1, we obtain deg(s~) - n - 1. Then setting r - 2, we obtain deg(sn_~) _> n - 2. Then setting r - 3, we obtain deg(s~_2) >_ n - 3. Continuing in this way, we obtain deg(sk) >_ k - 1 for k - 1 , . . . , n - 1. Similarly, G contains as induced subgraphs all threshold graphs of the form I~UKn_~ for r - 1 , . . . , n - 1 . Now setting r - 1 we obtain deg(s~) _< 1. Then setting r - 2 we obtain deg(s2) _< 2. Continuing in this way, we obtain deg(sk) <_ k for k - 1 , . . . , n 1. Hence the result. 9 We can now replace 'UG' with ' M U G ' in Theorem 19.4.1, as stated below. Theorem

19.4.4

1. For every graph G, the following conditions are equivalent: (a) G is a (threshold) ~n-MUG; (b) (G, xy) is a (threshold) 7-n+I-MUG; (c) (G,-2--yy) is a (threshold) ~ + I - M U G . 2. G(a n) is a 7-~+k-MUG if and only if G is a Tk-MUG. P r o o f . By T h e o r e m 19.4.3, a ~ - M U G contains exactly 2 k - 1 vertices. Hence the result follows from Theorem 19.4.1 and Part 2 of Fact 19.4.2.

Theorem 19.4.5

the form K1 (a n-l).

The threshold 7-n-MUGs are precisely the

2 n-1

graphs of

512

Other Extensions

P r o o f . If G is a :m-MUG, then it is of the form described in Theorem 19.4.3. If in addition G is threshold, then N ( s l ) C_ . . . C_ N(sn). It follows that if the c~ are indexed so that deg(c~) < . . . _< deg(cn_l), then N(sk) is either { c l , . . . , ck-1} or { C l , . . . , ck}. Hence there are at most 2 ~-1 threshold T~-MUGs. But there are at least 2 n-1 by Part 3 of Fact 19.4.2. .. We close this section with the following three facts. F a c t 1 9 . 4 . 6 It follows from Theorem 19.~.3 that 1(1(0n-l) has the least number of edges and K1 (1 n-1 ) the most number of edges among all 7-~-MUGs. (Here, as usual, 0 n-1 - ( 0 , . . . , 0 ) and I n-1 - ( 1 , . . . , 1).) Thus the extreme values for the number of edges for 7-~-MUGs are achieved by threshold graphs. F a c t 1 9 . 4 . 7 The A graph illustrated in Figure 1~.2 is a non-threshold 7"3MUG, and n - 3 is the smallest value of n for which a non-threshold 7-~MUG exists. By Part 2 of Theorem 19.~.~, we can construct from the A graph non-threshold 7-n-MUGs for every n >_ 3. F a c t 1 9 . 4 . 8 The graph illustrated in Figure 19.2 satisfies the necessary conditions of Theorem 19.~.3, but is not a 7-4-UG, since it has no induced subgraph isomorphic to K1,3. It is the smallest such example in terms of the number of vertices or edges. By Part 2 of Theorem 19.~.~, we can construct from it similar examples for every n >_ 4.

Figure 19.2" A non-'Y4-UG satisfying the conditions of Theorem 19.4.3. 81

82

@

Cl

83

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List o f N o t a t i o n s "~ (graph isomorphism), 3 > (strict domination), 241 L (domination), 241 (vicinal preorder), 3 ~ c (domination with respect to G), 341 ~>t,- (domination in I•), 111 --. (domination equivalence), 241 >~d (dominal preorder), 314 II (incomparable), 1 (strict majorization), 60 (majorization), 60 U (graph union), 4 @ (graph join), 4 @ • (graph composition), 324 (3 (split g r a p h composition), 304 A (symmetric difference), 79 IR+ (positive reals), 2 IR+ (nonnegative reals), 2 {~} (Stirling n u m b e r of the second kind), 474 [3] (Gaussian polynomial), 469 II II (proper sequence distance), 402 #v (storage bits), 103

B(D) (bipartite representation), 35 b(G) (boxicity), 148 C(d) (corrected Ferrers diagram), 62 c(G) (cubicity), 149 ch(B) (difference dimension), 141 x(G) (chromatic number), 4 x(H) (hypergraph chromatic number), 5 C. (cycle on n vertices), 4

D(G) (Dilworth number), l l l , 241 d(Q) (dimension), 446 D(X) (Dilworth n u m b e r of a set), 111

dB(R) (bidimension), 444 dE(D) (Ferrers dimension), 47 dz(P) (interval dimension), 147 D,~ (polytope of degree sequences), 76 do(P) (partial o r d e r dimension), 47 de(G) (scalar product dimension), 501 Dp,q (polytope of bipartite degree sequences), 94 d' (corrected conjugate sequence), 62 ds(G) (split dimension), 163 DS,, (set of degree sequences), 75

(A, B; E ) (bipartite graph), 4 a(G) (stability number), 4 (c~,/~) (polar graphs), 168, 328 ALTCa(H) (alternating cycle of H rel. to G), 21 528

529

List of Notations

19Sp,q (set of bipartite degree se-

No(v) (neighborhood), 3 Nc(x) (neighbors of x in C), 341 Na[v] (closed neighborhood), 3

quences), 94 d* (conjugate sequence), 62 19T,~ (set of threshold sequences), 402

w(G) (clique number), 4

E(W) (induced edges), 3

P(D) (dimensional order), 50

f (Durfee number), 62

G(R) (graph of a relation), 455 G(x -+ .H) (graph substitution), 134 G* (conflict graph), 124 G[W] (induced subgraph), 3 "y(G) (domination number), 242, a~a F+(v) (out-neighborhood), 5 I'-(v) (in-neighborhood), 5 C~(R) (bare graph of a relation), 455

H(R) hypergraph of a relation), 447

H~(R) (bare hypergraph of a relation), 448

I(P) (incomparability graph), 140 idimp(G) (i-dimension), 128 J2k (cocktail party graph), 314 n(G) (clique cover number), 4 Kn (complete graph), 5 Kv, .....p~ (completemultipartitegraph), 5 m (corrected Durfee number), 63 inK2 (perfect matching), 5, 282

NA(B) (neighbors of/3 in A), 341 N(v) (non neighborhood), :3

P,~ (path on n vertices), 4

R(d) (majorization gap), 403 p(G) (vertex cover number), 4 Sn (edgeless graph), 5

t(d) (threshold gap), 404 t(G) (threshold dimension), 98 r(G) (threshold measure), 387 T,~ (n-vertex threshold graphs), 509

U(D) (underlying graph), 35 ndimp(G) (u-dimension), 128

Author Index Cogis, O., 8, 33, 36, 38, 39, 46, 48, 50, 176, 213, 436 Comtet, L., 473-475 Cozzens, M. B., 109, 124, 129, 134136, 455

Alavi, Y., 27-29 Andrews, G. E., 470 Arikati, S., 33, 376, 401,434, 468, 469 Ariyoshi, H., 165 Aspvall, B., 22

de Werra, D., 53, 176-178, 241,245, 246, 250, 252 Dilworth, R. P., 241 Doignon, J.-P., 7, 176, 213, 435, 437, 460 Ducamp, A., 7, 33, 176, 213, 435, 437 Dushnik, B., 46, 47, 140 Dutton, R., 483, 488

Behzad, M., 27-29 Beissinger, J. S., 470 Benzaken, C., 53, 241, 245, 246, 250, 252, 313, 314 Berge, C., 17, 62, 93, 176, 189, 371 Boesch, F., 483, 488 Bondy, J. A., 5 Bouchet, A., 33, 47, 436 Brandstgdt, A., 17 Brigham, R., 483, 488 Burkard, R. E., 116 Burr, S., 483, 488

Ecker, K., 7, 10, 134 Edmonds, J., 90 Edwards, C. S., 486 Entringer, R. C., 258 ErdSs, P., 119, 124, 126-128, 483, 484 Euler, L., 470 Even, S., 18, 140

Cauchy, A., 470 Cayley, A., 473 Chernyak, A. A., 3, 111, 164-166, 168, 169, 258,304, 313,323, 327, 328, 335, 336,339,341 Chernyak, Z. A., 3, 111, 164-166, 168, 169,313, 323,327, 339 Chvs V., 5, 7, 8, 10, 17, 22, 25, 29, 97, 98, 123, 124, 175178, 295, 351,370, 371,386, 464, 465

Falmagne, J.-C., 7, 33, 176, 213, 435, 437 Fishburn, P. C., 36, 37 FSldes, S., 111, 112, 241-243, 245, 252, 272, 296 Ford, L. R., 62, 93 530

Author Index

Frobenius, G., 475 Fulkerson, D. R., 62, 72, 93 Gale, D., 65, 93 Gallai, T., 119 Garey, M. R., 5, 22, 116, 141, 151, 152, 165, 169 Giles, R., 90 Gilmore, P. C., 37, 131,456 Golumbic, M. C., vii, 10, 17-19, 21, 22, 25, 27, 130, 242, 255, 457, 497, 505 Good, I. J., 473 Graham, R. L., 473 Gr5tschel, M., 17, 18 Gutman, I., 13, 75 Guttman, L., 435 Gys163 A., 483, 484 Hakimi, S. L., 71 Halsey, M. D., 129 Hammer, P. L., 7, 8, 10, 21, 22, 25, 29, 33, 52, 53, 71, 83, 97, 98, 111, 112, 116, 117, 123, 124, 175-179, 188, 192, 195, 241-243,245,246,250,252, 272,295, 296,313,314, 351, 376, 401,498, 509 Harary, F., 22, 179 Hardy, G. H., 59, 60 Havel, V., 71 Hayward, R., 255 Henderson, P. B., 7, 10, 15, 101 Hertz, A., 374 Ho~ng, C. T., 176-178, 241, 253, 374 Hoffman, A. J., 37, 72, 131,456 Hoogeveen, H:, 63

531

Ibaraki, T., 21, 22, 71, 83, 176, 192, 195, 213, 376, 401 Ide, M., 165 Jacobson, M. S., 497, 506, 507 Johnson, D. S., 5, 22, 116, 141,151, 152, 165, 169 Kaplan, H., 21 Karmarkar, N., 391 Kelmans, A. K., 498, 509 Khachian, L. G., 391 Kimbel, R., 145 Knuth, D. E., 470, 473, 474 Koop, G. J., 7, 105 Koren, M., 8, 75, 77, 91 Leibowitz, R., 109, 124, 134-136, 455 Lempel, A., 18, 140 Lesniak-Foster, L. M., 27-29 Littlewood, L. E., 59, 60 Lovs L., 17, 18, 253, 370 Ma, T.-H., 176, 213, 223, 224, 229 Mahadev, N. V. R., 22, 176-179, 188,241,242,253,255,351, 356, 375, 376, 380 Marchioro, P., 272, 303 Margot, F., 150, 158, 237, 258 Marshall, A. W., 59, 60, 93, 410 McAndrew, M. H., 72 McMorris, F. R., 497, 506, 507 Meyniel, H., 373 Miller, E., 46, 47, 140 Minty, G., 22, 25 Monjardet, B., 33 Monma, C. L., 497, 505, 506, 509 Morgana, A., 272, 303

532

Muirhead, R. F., 60 Mulder, M., 497, 506, 507 Muroga, S., 9 Murty, U. S. R., 5 Nordhaus, E. A., 27-29 Olkin, I., 59, 60, 93, 410 Ordman, E. T., 7, 15, 99, 101-103, 105, 124, 126, 128, 483,484 Ore, O., 33, 46, 47, 50 Orlin, J., 14 Ozaki, H., 165 Patashnik, O., 473 Payan, C., 90, 241, 253, 313, 314, 320, 351,356 Peled, U. N., 21, 22, 29, 33, 52, 63, 75, 94, 176, 177, 188, 192, 195, 213,258,272,336, 351, 356,375,376,380,386,401, 434, 468-470, 477 Petreschi, R., 188, 272, 303 Plass, M. F., 22 Pnueli, A., 18, 140 Poljak, S., 124 Pdlya, G., 59, 60 Rao, S. B., 3 Raschle, T., 21, 176, 179, 190 Ravin&a, G., 374 Rawlinson, K. T., 258 Reed, B., 497, 506, 509 Reiterman, J., 497, 498, 500 Riguet, J., 33 Riordan, J., 473, 474 Roberts, F. S., 129, 149, 442 R6dl, V., 497, 498, 500 Ruth, E., 13, 75

Author Index

Ryser, H. J., 65, 93, 121 Schrijver, A., 17, 18, 90, 461 Seinsche, D., 359 Shamir, R., 21 Sierksma, G., 63 Simeone, B., 29, 71, 83, 117, 258, 272, 303, 375,376, 386,401 Simon, K., 21, 176, 179, 190 Sifiajovs E., 497, 498, 500 Spinrad, a. P., 176, 213 Srinivasan, M. K., 33, 63, 75, 94 Stanley, R. P., 92 Sterbini, A., 188, 227 Stockmeyer, L., 141 Stoer, a., 5 Sun, F., 33, 351,356, 477 Sun, X., 33, 52 Tarjan, R. E., 22 Tindell, R., 483, 488 Trotter, W. T., 242, 255, 497, 505, 506, 509 Tsukiyma, S. I., 165 Tyshkevich, R. I., 3, 258, 272, 304, 327, 328,334-336, 341,348, 349 Wang, C., 375, 376 Williams, A., 375, 376 Witzgall, C., 5 Yannakakis, M., 8, 53, 139, 154 Zaks, S., 7, 10, 134 Zalcstein, Y., 7, 10, 15, 101, 124, 126, 128, 483, 484

Index 2-connected graph, 4 2-satisfiability, 22 2-summability graph, s e e conflict graph 2-threshold graph, 177, 178 complement is perfectly orderable, 178 is perfectly ordrable, 177 is weakly triangulated, 177 recognition, 191,224 strict, 188 characterization, 186, 188 is comparability graph, 188 recognition, 188

two, 213 big clique, 384 existence, 384 implies graph not heavy, 385 partial converse, 385 biorder, 7, 437 characterization, 447 bipartite graph, 4, 328 bipartitioned graph, 468 Birkhoff-Milgram Theorem, 442 bisimplicial edge, 255 bistar, 382, 383 bithreshold graph, 179 characterization, 185 block, 4 block graph, 362 Bollobgs Criterion, 64, 73 Boolean function, 2 biregular, 179 graphic, 179 regular, 179 threshold, 9 box, 257, 261,262 box-threshold, see BT graph boxicity, 244 is at most Dilworth number, 244 brittle graph, 253 BT graph, 257, 259 characterization, 261 Burnside Lemma, 481

adjacency matrix, 2 adjacent edges, 2 adjacent vertices, 2 aggregation of inequalities, 98 alternating 4-anticircuit, 33 alternating 4-cycle, 9, 10, 12, 72 alternating cycle, 21,190, 446 bare, 447 alternating cycle in S, 191 antichain, 1 arc of a digraph, 5 Axiom of Choice, 437 Berge Criterion, 64, 73 bidimension, 444, 454, 459 characterization, 445, 446 533

534

C4-free graph, 318 caterpillar, 242 center of a star, 5 chain, 1 chain graph, s e e difference graph characteristic vector, 2 chord of a cycle, 130 chordal bipartite graph, 255 chordal graph, s e e triangulated graph chromatic number, 4, 459 hypergraph, 5,454 circle graph, s e e overlap graph circuit, 271 circular-arc graph, 130 representation, 130 claw, 18 clique, 4, 17 k-, 4 clique cover, 18 threshold, 19, 20 clique cover number, 4 clique number, 4 clique partition, 4 cobithreshold graph, 180 cocktail-party graph, 314 cocomparability graph, 18, 112 codomishold graph, 337 cointerval graph, 18, 37 characterization, 37 color classes, 4 color-regular, s e e CR coloring, 4 communications network, 258 comparability and weakly triangulated graph is domination graph, 255

Index

comparability graph, 18, 19, 115, 140, 188, 250, 255 characterization, 189, 456 comparable, 1,140, 241 complement of a graph, 4 complete k-ary tree, 400 complete graph, 4 complete multipartite graph, 5 completely connected, 260 completely disconnected, 261 composition, 304, 324, 329, 480 configuration .T, 272 conflict, 175 conflict graph, 124, 175,180, 186, 191 conflict, edges, 124, 186 conjugate sequence, 61 corrected, 62 connected component, 4 connected graph, 4 consumer-priority flow, 263 core, 300 cotriangulated graph, 18 coTT graph, 506 recognition, 509 couple, 271 CR (color-regular), 260 CR connected, 260 crown, 445 cycle, 4 cyclic scheduling, 7, 105, 106 decomposition at the level (a,/3), 329 BT graph characterization, 344 canonical at the level (a,/3), 330 in the class (a,/3), 330

Index

uniqueness, 333 domishold graphs, 338 graph isomorphism, 334 hierarchy, 333 in the class (c~,~), 329 matrogenic graph, 348 indecomposable, 349 matroidal graph, 348 indecomposable, 349 non-polar uniqueness, 333 split graph isomorphism, 334 threshold graph, 348 unigraph, 348 uniqueness, 330 degree, 3 degree of an edge, 3 degree partition, 9, 11, 15, 24, 25, 55, 79, 84, 508 degree sequence, 3, 8, 13, 64 bipartite, 93 characterization, 93 BT characterization, 264, 307 is forcible, 259 characterization, 92 difference, 93 characterization, 94, 95 recognition, 96 domishold is edge-unigraphic, 339 forcibly P-graphic, 3 forcibly domishold, 340 is unigraphic, 340 matrogenic, 303 is unigraphic, 303 split graph characterization, 310

535 polytope, 76 bipartite, 94, 95 boundary, 355 dimension, 91 edges, 80, 83, 84 facets, 91 linear description, 87 potentially P-graphic, 3 realization, 3, 59 split, 92, 120 characterization, 92 is forcible, 120 strongly unigraphic, 3 threshold, 72, 73 characterization, 92 is strictly unigraphic, 72 unigraphic, 3 dense graph, 484 large interval subgraph, 486,487 large threshold subgraph, 487 diameter, 5 is at most D(G) + 1,242 difference graph, 7, 8, 33, 37, 38, 53-57 characterization, 54, 55 digraph, 5 bipartite representation, 35 underlying graph, 35 Dilworth number, 53,241,242,243 < 2, 53, 112, 250, 500 < 3 is brittle, 253 < 4 is perfect, 253 computing, 242 of a vertex set, 111 Dilworth Theorem, 1,241,250 dimension, 446, 454 biorder, see bidimension boxicity, 148

536

Index

cubicity, 149 difference, 141,147-149 two, 213 Ferrers, 33, 47, 50, 52 is poset dimension, 50 two, 213

characterization, 315 domistable graph, 314, 318 Durfee number, 62 corrected, 63, 120, 404, 469 Dushnik-Miller Theorem, 455

i-, 1 2 8

edge, 2 edge clique cover number, 101,102 edge-degree sequence, 3 edge-unigraph, 3 edge-vertex incidence matrix, 2 edgeless graph, 4 ellipsoid method, 18 embedding, 46 end of an edge, 2 enumeration BT graphs, 344 domishold graphs, 326 HP graphs, 324 labeled difference graphs, 477, 478 labeled threshold graphs, 471, 474 asymptotics, 474 matrogenic graphs, 349 matroidal graphs, 349 threshold graphs, 336, 468,469 unlabeled difference graphs, 480 equidominating graph, 314, 320 equistable graph, 356 A-free is strongly equistable, 360 block graph is strongly equistable, 362 co-Meyniel is strongly equistable, 374 necessary condition, 359 outerplanar is strongly equistable, 367

difference, 213 interval, 147 partial order, 8, 47, 50, 52, 140 as difference dimension, 221 two, 221 split, 163 <_ 2, 164 threshold, 98, 101, 123, 135, 149 complete multipartite, 134 is at most D(G), 243 matrogenic, 302 matroidal, 293, 295 restricted, 136, 138 triangle-free, 123, 124 u-, 1 2 8

difference, 213 dimensional order, 50 directed path, 5 disjoint union, 4 domigraphic inequality, 319 characterization, 319 dominating set, 242, 313 dominating vertex, 3 domination, 111,241 domination graph, 242,254, 255 domination number, 242,313 is at most D(G), 243 domishold and equidominating graph characterization, 320 domishold graph, 314, 318

537

Index

perfect is strongly equistable, 370 pseudothreshold is strongly equistable, 364 split is strongly equistable, 363 substitution, 372 equitable graph, 395 characterization, 395, 397 equitable tree characterization, 399 example, 400 ErdSs-GMlai Criterion, 64, 73, 92, 119, 120 Eulerian polynomial, 473 exclusion graph, 100,101,103, 104 Farkas Lemma, 461,464, 465 Ferrers diagram, 35 corrected, 62, 63, 262, 469 Ferrers digraph, 8, 33, 35-37, 4749 bipartite representation, 38 characterization, 39 forbidden configurations, 34 is interval order, 37 underlying graph, 37 frame, 266 BT characterization, 266, 267 Frobenius Theorem, 476 Fulkerson- Hoffman- McAndrew Criterion, 64 Gale-Ryser Theorem, 65, 93 Gaussian polynomial, 469 generating function exponential, 467 ordinary, 467 graph, 2

graph join, 4 graph property, 3 graph substitution, 134, 372 graph union, 4 graphical sequence, see degree sequence Grfinbaum Criterion, 64, 73 Guttman scale, 109 extended, 110 Hgsselbarth Criterion, 64, 73 Hall's Theorem, 4, 25, 26 Hamiltonian cycle, 5 Hamiltonian graph, 25 Hasse diagram, 1 Hausdorff Maximal Principle, 453 Havel-Hakimi Theorem, 71 head, 5 heavy graph, 3 7 7 characterization, 379 construction, 381 example, 379-383 Helly's Theorem, 464, 465 hereditary property, 3 hereditary pseudodomishold, see HP graph H-free graph, 4 HP graph, 323 characterization, 324 hypergraph, 5 T'-~ 5

Hamilton cycle, 150 edge of, 5 ideal, 140 implication class, 457 in-degree, 5 in-degree partition, 38

538 in-isolated, 5 in-neighborhood, 5 incomparability graph, 140 incomparable, 1, 140, 241 independence system, 271 intersection graph, 98 interval dimension, 454 interval graph, 18, 37, 115, 129, 130, 148, 252, 318 is circular arc, 130 is tolerance graph, 505 unit, 18, 129, 130 interval order, 37, 147, 444 characterization, 37 interval tolerance graph, 505 isolated vertex, 3 isomorphism, 3 labeled graph same, 467 labeling, 3 LCL (Largest Cycle Length), 22, 24, 27 length of a path, 5 light edges, 377 linear extension, 140 corresponds to maximal difference subgraph, 216 linear order, 46, 47, 140, 437 loop of a digraph, 5 loop of a graph, 2 majorization, 8, 60, 61, 65 strict, 60 majorization gap, 403 bound, 421,422 equals threshold gap, 420 formula, 410

Index

matching, 4 perfect, 4 special, 23, 25 largest, 24 matching with loops, 266, 267 matrogenic graph, 259, 272 characterization, 296 Dilworth number, 302 structure, 302 threshold dimension, 302 matroid, 271 matroidal graph, 271 characterization, 272 is triangulated, 295 strict, 284 structure, 284 structure, 288, 291 threshold dimension, 293, 295 maximal element, 1 maximum clique, 18 2-threshold graph, 179 threshold, 19, 20 weighted, 19 maximum stable set, 18 2-threshold graph, 179 threshold, 19, 20 weighted, 19 maximum total matching, 28 Meyniel graph, 373 characterization, 374 substitution, 373, 374 minimal element, 1 minimally imperfect graph, 252 minimum coloring, 18 2-threshold graph, 179 threshold, 19, 20 minimum number of resource units, 102

539

Index

minimum total cover, 27 missed by a matching, 4 MUG, s e e universal graph, minimum Muirhead Lemma, 60, 402 multiple semiorder, 460 characterization, 463 n-cycle, 5 (n, e)-graph, 2 neighborhood, 3 nested sets, 1 net, 276 net-complement, 276 NP-complete r-cyclic scheduling, 161,162 boxicity, 149 clique, 151 cubicity, 149 difference dimension, 145 < 3, 145 Hamiltonicity, 22 split graph, 116 HTrH, 162 interval dimension height 1,148 largest threshold subgraph, 153 longest cycle, 22 MON3PART, 158 NMTS, 152 one-in-three 3-SAT, 152 partial order dimension, 8 < 3, 144 height 1,145 partition into cliques, 165 POLAR, 171 POLAR(~), 169

smallest threshold supergraph, 153 split dimension, 165 bipartite, 166 threshold dimension, 124 split, 150 vertex cover, 165 weighted 2-threshold cover, 156 weighted 2-threshold partition, 156 on/off-weekend, 105 order-dense set, 441 out-degree, 5 out-degree partition, 38 out-dominating, 5 out-neighborhood, 5 outerplanar graph, 366 overlap graph, 129, 130 representation, 129 P completion, 21 P4-free graph characterization, 359 is strongly equistable, 359 pancyclic graph, 27 parallel processing, 7, 99 partial order, 46, 47, 49, 50, 139 height, 145 height 1,147 Kimbel's result, 145 path, 4 Peled-Srinivasan Criterion, 65, 73 pentagon, 5 perfect graph, 17 Perfect Graph Conjecture, 17 Perfect Graph Theorem, 17, 370 perfect matching, 282

i40 ~erfect order, 176 )erfect orientation, 253 ~erfectly orderable, 176, 177, 178, 253 ~ermutation graph, 18 is tolerance graph, 505 ~olar equivalence, 329 ~olar graph, 111,168, 328 hierarchy, 328 recognition, 169 ~olar isomorphism, 328 ~olar partition, 168, 328 ~oset, s e e partial order ?reorder, 1, 46, 49 dominal, 315 vicinal, 10, 12, 14, 55, 241 principal clique, 135 principal stable set, 135 problem r-cyclic scheduling, 151 chromatic number < 3, 141 clique, 151 cyclic scheduling with ratio constraint, 105, 106 HTrH, 151 largest threshold subgraph, 150 MON3PART, 151 NMTS, 152 one-in-three 3-SAT, 151 partial order dimension < 3, 141 partition into cliques, 165 POLAR, 169 POLAR(~), 169 sandwich, 21 split, 21, 22 threshold, 21 smallest threshold supergraph,

Index 151 vertex cover, 165 weighted 2-threshold cover, 151 weighted 2-threshold partition, 151 proper sequence, 62 property i-dimensional, 128 u-dimensional, 128 pseudodomishold graph, 323 pseudothreshold graph, 352 characterization, 354 has half-integer weights, 355 PV-chunk, 100, 104 PV-chunk definable graph, 7, 101 q-Binomial Theorem, 470 realizer, 213 regular graph, 5 relation antisymmetric, 436 asymmetric, 436 bare graph of, 455,459 bare hypergraph of, 448 complete, 436 countable, 436 dual of, 460 equivalence, 436 graph of, 455 hypergraph of, 447 irreflexive, 436 negatively transitive, 436 partial order, 436 quasi order, 436, 454 realizable, 438 characterization, 439,442,444 reflexive, 436

Index

simple order, 436 strict partial order, 436 strict simple order, 436 strict weak order, 436 strongly complete, 436 symmetric, 436 transitive, 436 weak order, 436 relation ~d is preorder, 314 relative product, 437 representation of a graph, 501 resource allocation, 7, 101-104 rigid circuit graph, see triangulated graph Ryser Criterion, 64, 73 saturated by a matching, 4 scalar product dimension, 501 example, 503 is at most t(G), 501 of disjoint union, 502 semiorder, 460 characterization, 462 separator, 9, 14, 100 set-packing problem, 7, 97 a-optimal graph, 488 example, 489, 491 is threshold, 489 signed graph, 179 balanced, 179 simplicial clique, 371 simplicial vertex, 253 sink, 5 source, 5 split graph, 10, 23, 111, 150, 318, 328 characterization, 112, 120

541 chromatic number, 121 clique number, 121 comparability graph characterization, 115 corrected Durfee number, 121 Dilworth number _< 2, 115 Hamiltonian, 116 interval graph characterization, 112 matrogenic graph characterization, 304 recognition, 163 split isomorphism, 333 split partition, 111,150 splittance, 117, 119 computing, 120 square, 5 stability number, 4 stable set, 4, 17 star, 5 Stirling number of the second kind, 474 strictly dense set, 443 strong unigraph, 3 strongly equistable graph, 357 is closed under disjoint union, 357 is closed under join, 358 is equistable, 357 substitution, 372, 373 sufficient condition, 360 strongly perfect graph, 371 subgraph, 3 edge-induced, 3 induced, 3 spanning, 3 supplier-priority flow, 262 support, 2

542

Index

tail, 5 test-and-set, 100, 103, 104 threshold completion, 21 threshold function, 2 threshold gap, 402 equals majorization gap, 420 formula, 406 realizations, 408 threshold graph, 7, 8, 35-37, 130, 318 add/delete edge, 79 characterization, 10 generalized, 498, 500 characterization, 499 forbidden subgraphs, 500 has D ( G ) <_ 2, 500 Hamiltonian, 22, 25-27, 106,163 is interval, 130 is overlap, 130 maximum matching, 32 maximum stable set, 31 Orlin weights, 14, 15, 101,103, 104 perfect matching, 32 recognition, 19, 20 total covering number, 30 total matching number, 30 threshold graph with loops, 266, 267 threshold hypergraph, 7 r-, 1 5 0

threshold measure, 387 bipartite graph is polynomial, 391 rain-max relation, 394 is NP-hard, 390 of disjoint union, 387 of join, 390

threshold number, 98 threshold partition number, 124, 126 threshold signed graph, see TS graph threshold tolerance graph, 5 0 6 threshold weight, 376 T~-MUG construction, 511 non-threshold, 512 threshold characterization, 511

T.-UG construction, 510 tolerance chain graph

r

507

max-

interval, 507 minis threshold, 507 is

sum-

is coTT, 507 tolerance graph, see tolerance intersection graph tolerance intersection graph r 506 max-, 5 0 6 rain-, 5 0 6 sum-, 5 0 6 total cover, 27 total covering number, 27 total dual integrality, 90 total matching, 27 total matching number, 28 total order, 1,437 transfer, 72,260 transitive orientation, 18 transitively orientable graph, see comparability graph

Index

transportation network, 262,263 tree, 5 triangle, 5 triangle-free graph, 123 triangulated graph, 17, 18, 112, 130 trivially perfect graph, 27 TS graph, 53, 245 characterization, 250 is comparability graph, 250 split graph characterization, 252 twins, 135 UG, s e e universal graph unigraph, 3 uniquely orderable graph, 457 characterization, 457 unit interval graph, 149 unit transformation, 60, 61 reverse, 402 universal graph, 509 minimum, 509 unlabeled graph same, 467 vertex, 2 vertex cover, 4 vertex cover number, 4 vicinal preorder, 3 weakly triangulated graph, 176, 177, 254, 255 Well-Order Principle, 437 well-ordered set, 437 widely dense set, 441

543