No title

Graphs and Combinatorics (2009) 25:1–14 Digital Object Identifier (DOI) 10.1007/s00373-008-0826-4

Graphs and Combinatorics © Springer-Verlag 2009

On the Existence of Elementary Abelian Cycle Systems Anna Benini, Anita Pasotti Dipartimento di Matematica, Facolt`a di Ingegneria, Universit`a degli Studi di Brescia, Via Valotti 9, I-25133 Brescia, Italy. e-mail: [email protected], [email protected]

Abstract. We present some necessary and/or sufficient conditions for the existence of an elementary abelian cycle system of the complete graph. We propose a construction for some classes of perfect elementary abelian cycle systems. Finally we consider elementary abelian k-cycle systems of the complete multipartite graph. Key words. (regular) (i-perfect) cycle system, Partial i-differences. Mathematics Subject Classification (2000) 05B30.

1. Introduction Throughout this paper the standard notation of graph theory will be used, so K v , K m×s , Ck will denote the complete graph on v vertices, the complete multipartite graph with m parts of size s and the k-cycle, respectively. Given a set S, by K S we mean the complete graph whose vertex-set is S. Given two graphs K and , a (K , )-design is a set of graphs isomorphic to  whose edges partition the edge set of K (see [5]). A (K v , Ck )-design is also called k-cycle system of order v (see [6]). The problem of determining the spectrum of values of v for which there exists a (K v , Ck )-design for a given k attracted a large number of combinatorialists since the 60’s. One can easily see that necessary conditions for its existence are v odd and v(v − 1) divisible by 2k. Quite recently, it has been finally proved that these conditions are also sufficient. So, there exists a k-cycle system of order v if and only if k ≤ v, v is odd and v(v − 1) ≡ 0 (mod 2k). The if part of this theorem was solved by Alspach and Gavlas [2] in the case of k odd (for another ˜ recent proof see [8]) and by Sajna [22] and [23] in the case of k even. A k-cycle system of order v is i-perfect, where 1 ≤ i ≤  k2 , if for any edge [x, y] ∈ K v there is exactly one cycle of the system in which x and y have distance i. Given I ⊆ {1, 2, . . . ,  k2 }, by saying that a (K v , Ck )-design is I -perfect, we mean that it is i-perfect for all i ∈ I . If I = {1, 2, . . . ,  k2 }, an I -perfect (K v , Ck )-design is called a Steiner k-cycle system of order v and it is denoted by Sk S(v). A Kirkman k-cycle system of order v, denoted by K k S(v), is a Sk S(v) together with a partition of its cycles into 2-factors of K v . The major known existence results about i-perfect

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cycle systems and, in particular, about Steiner cycle systems can be found in [1], [3], [7], [12], [13], [15], [16], [17]. Here we only recall that a necessary condition for the existence of a Sk S(v) is that v and k be odd integers. Given an additive group G, a (K , )-design B is said to be regular under G (for short G-regular) if, up to isomorphisms, the vertices of K are the elements of G and whenever B is a cycle of B we also have B + g ∈ B for every g ∈ G. Of course by B + g we mean the graph defined by V (B + g) = V (B) + g

and

E(B + g) = {[x + g, y + g] | [x, y] ∈ E(B)}.

A design is said to be cyclic, abelian, non abelian, ... when it is regular under a group having the respective property. The existence problem of cycle systems for the cyclic case has generated a considerable amount of interest. Many authors have contributed to prove the following Theorem 1.1 which is, so far, the most important result about the existence of cyclic k-cycle systems (see [10], [11], [14], [18], [19], [20], [21], [24]). Theorem 1.1. For all v ≡ 1 or k (mod 2k) there exists a cyclic (K v , Ck )-design with the only definitive exceptions of (v, k) = (9, 3), (15, 15), ( p n , p n ) with p a prime and n > 1. In what follows we will deal with the existence problem of cycle systems for the elementary abelian case. We recall that an elementary abelian group is a finite group where every non-zero element has order a prime p. The only results we know about this case are the following: Theorem 1.2 (Bonisoli, Buratti, Mazzuoccolo [7]). There exists a 2-transitive elementary abelian Kirkman k-cycle system of order v if and only if (k, v) = ( p, p n ) for some odd prime p and some positive integer n. Theorem 1.3 (Granville, Moisiadis, Rees [13]). For every prime power q = 2ke + 1 with k odd, there exists an elementary abelian Steiner k-cycle system of order q. Proof. Let q be as in the statement and g be a primitive element of the finite field of order q, then an elementary abelian Steiner k-cycle system of order q is formed by the cycles (g i + α, g i+2e + α, g i+2·2e + α, . . . , g i+(k−1)·2e + α) for 0 ≤ i < e and for each field element α.



Remark 1.4. Note that the cycles of the previous construction are nothing but the blocks of the developments 1 of Ci = (g i , g i+2e , g i+2·2e , . . . , g i+(k−1)·2e ) where 0 ≤ i < e. 1 Let G be any finite additive group and D  = ∅ any subset of G. The incidence structure

dev D := {G, {D + x, x ∈ G}, ∈} is called the development of D (see [4]).

On the Existence of Elementary Abelian Cycle Systems

3

Also, in [25] A. Vietri proved the existence of an elementary abelian 3-cycle system of order (2d + 1)2 when 2d + 1 is a prime and d ≡ 0, 3, 8, 11(mod 12). In Section 3 some necessary and sufficient conditions for the existence of elementary abelian k-cycle systems (EA-k-cycle systems, for short) will be proved. It is convenient to give here some notation about finite fields, since we will use them for getting our constructions. Given a prime power q, the finite field of order q will be denoted by Fq , while Fq∗ will denote the multiplicative group of Fq . Also, we denote by C e the group of non-zero e-th powers in Fq , e being any divisor of q − 1. The cosets of C e in Fq∗ , also called cyclotomic classes of index e, are C e = g 0 C e , g 1 C e , g 2 C e ,. . . , g e−1 C e , where g is a primitive element of Fq . For simplicity, once g has been fixed, we set Cie := g i C e . 

When e = 2, we will write Fq and Fq instead of C 2 and C12 , respectively.

2. The Method of Partial i-differences to Construct Regular i-perfect (Kv , Ck )-designs From [12] we learn that any regular i-perfect k-cycle system can be characterized by the method of partial differences introduced by M. Buratti in [9]. Here we recall some definitions and the main result from [12], useful in the following. Definition 2.1. Let A = (a0 , a1 , . . . , ak−1 ) be a k-cycle with vertices in an abelian group G and let d be the order of the stabilizer of A under the natural action of G, that is d = |{g ∈ G : A + g = A}|. Given i ∈ {1, 2, . . . , k/2}, the multisets i A = {±(ah+i − ah ) | 0 ≤ h < k} ∂i A = {±(ah+i − ah ) | 0 ≤ h < k/d} where the subscripts are taken modulo k, are called the list of i-differences from A and the list of partial i-differences from A, respectively. Of course, d = 1 implies ∂i A = i A. More generally, given a set F of k-cycles with vertices in G, by i F and ∂i F one means the union (counting multiplicities) of all multisets i A and ∂i A respectively, where A ∈ F.   Theorem 2.2. Given I ⊆ 1, 2, . . . , k2 , a G-regular I -perfect (K v , Ck )-design is equivalent to a set F of k-cycles with vertices in G (called base cycles) such that ∂i F = G − {0} for all i ∈ I . As an immediate consequence we have: Corollary 2.3. A G-regular (K v , Ck )-design is equivalent to a set F of k-cycles with vertices in G such that ∂1 F = G − {0}. In Example 2.5 we will show how to construct an EA-(K 25 , C15 )-design just applying Corollary 2.3. For our purposes the following notation will be useful.

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Let a0 , a1 , . . . , ar −1 , x be elements of an additive group G, with x of order p. The closed trail represented by the concatenation of the sequences (a0 , a1 , . . . , ar −1 ) (a0 + x, a1 + x, . . . , ar −1 + x) (a0 + 2x, a1 + 2x, . . . , ar −1 + 2x) ··· (a0 + ( p − 1)x, a1 + ( p − 1)x, . . . , ar −1 + ( p − 1)x) will be denoted by [a0 , a1 , . . . , ar −1 ]x . Remark 2.4. Note that [a0 , a1 , . . . , ar −1 ]x is a ( pr )-cycle if and only if the elements ai , for i = 0, . . . , r − 1, belong to pairwise distinct cosets of the subgroup < x > in G. Also, if A = [a0 , a1 , . . . , ar −1 ]x is a ( pr )-cycle then ∂1 A = {±(ai − ai−1 ) | i = 1, . . . , r − 1} ∪ {±(a0 + x − ar −1 )}. Example 2.5. Let G = Z5 × Z5 and consider the following 15-cycles with vertices in G (with some abuse, let us write any pair (x, y) as x y): A = [00, 01, 13]10 = (00, 01, 13, 10, 11, 23, 20, 21, 33, 30, 31, 43, 40, 41, 03) B = [00, 13, 22]01 = (00, 13, 22, 01, 14, 23, 02, 10, 24, 03, 11, 20, 04, 12, 21) C = [00, 22, 32]02 = (00, 22, 32, 02, 24, 34, 04, 21, 31, 01, 23, 33, 03, 20, 30) D = [00, 11, 34]03 = (00, 11, 34, 03, 14, 32, 01, 12, 30, 04, 10, 33, 02, 13, 31). One can easily check that: ∂1 A = {±01, ±12} ∪ {±02} ∂1 C = {±22, ±10} ∪ {±20}

∂1 B = {±13, ±14} ∪ {±21} ∂1 D = {±11, ±23} ∪ {±24}

Setting F = {A, B, C, D} we have ∂1 F = G − {00}. Hence, by Corollary 2.3, {A, B, C, D} is a set of base cycles of an EA-15-cycle system of order 25. Explicitly, the required design is the following: {A + 0i | 0 ≤ i ≤ 4} ∪ {B + i0 | 0 ≤ i ≤ 4} ∪ {C + i0 | 0 ≤ i ≤ 4} ∪ {D + i0 | 0 ≤ i ≤ 4}.

3. Existence of Elementary Abelian Cycle Systems We have already observed that necessary and sufficient conditions for the existence of a k-cycle system of order v are k ≤ v, v odd and v(v − 1) ≡ 0 (mod 2k). For the existence of an elementary abelian k-cycle system of order v a stricter condition is necessary. Proposition 3.1. If an EA-(K v , Ck )-design exists, then a) v = p n where p is an odd prime and n ≥ 1; b) k ≤ v; c) p( p n − 1) ≡ 0 (mod 2k).

On the Existence of Elementary Abelian Cycle Systems

5

Proof. Let B be the required system and let A be a k-cycle of B. Conditions a) and b) follow obviously. To prove condition c) one can easily see that the order of the stabilizer of A under the natural action of Znp is 1 or p and this implies that the order of the orbit of A is p n or p n−1 , respectively. Since the orbits partition the set n n n of blocks of B we have that p n−1 must be a divisor of p ( p2k −1) , that is p( p2k−1) ∈ Z.  The previous proposition suggests searching sufficient conditions distinguishing two different cases: the first one when p is a divisor of k and the second one in which p does not divide k. 3.1. Case: p Divides k Lemma 3.2. Let p be a prime such that p n ≡ 1(mod 2t). If there exists a sequence A = (a0 = 0, a1 , a2 , . . . , at ) of elements of F pn such that a) D := {ai − ai−1 | 1 ≤ i ≤ t} is a complete system of representatives for the cosets of C t in F∗pn b) {ai − a j | 0 ≤ i < j ≤ t − 1} ∩ {hat | 0 ≤ h ≤ p − 1} = ∅ then there exists an EA-( pt)-cycle system of order p n . Proof. Let B = [0, a1 , a2 , . . . , at−1 ]at . Condition b) together with Remark 2.4 make us sure that B is a ( pt)-cycle. Again by Remark 2.4, we obtain ∂1 B = −1 · D. Now, let S be a complete system of representatives for the cosets of −1 in C t and let F = {s · B | s ∈ S}. Since ∂1 F = S · ∂1 B = S · −1 · D = C t · D, by Condition a) it follows that ∂1 F = F∗pn , hence, by Corollary 2.3, we have that F is the set of base  cycles of an EA-( pt)-cycle system of order p n . Proposition 3.3. Let p be a prime. There exists an EA-( pt)-cycle system of order p n for all p n ≡ 1(mod 2t) such that t 2 (t − 1) pn − 1 ≥ . p−1 2

(1)

Proof. We consider the sequence A = (a0 = 0, a1 , . . . , at ) of elements of F pn so defined: for all i = 1, . . . , t − 1 we set t ai = ai−1 + xi−1 with xi−1 ∈ Bi−1 := Ci−1 − Ai−1

where Ai−1 = {a j − ak | 0 ≤ j < k ≤ i − 1}, and t at = at−1 + xt−1 with xt−1 ∈ Bt−1 := Ct−1 − Ah where Ah = {h(ai − a j ) − at−1 | 0 ≤ h ≤ p − 1, 0 ≤ i < j ≤ t − 1}.

Firstly, we prove that such a sequence exists, namely that Bi−1 = ∅ for all i = n t (t−1) t 1, . . . , t. By hypothesis p t−1 ≥ ( p − 1) t (t−1) 2 , which implies |C | ≥ ( p − 1) 2 . It

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is easy to check that |Ai−1 | = i(i−1) ≤ (t−1)(t−2) , so Bi−1 = ∅ for all i = 1, . . . , t −1. 2 2 t (t−1) Moreover, |Ah | ≤ ( p − 1) 2 + 1 and note that, for instance, 0 and −x0 are two t . Hence B elements of Ah which do not lie in Ct−1 t−1  = ∅ too. Now we have to prove that the sequence A defined as above satisfies Conditions a) and b) of Lemma 3.2. Since D := {ai − ai−1 | 1 ≤ i ≤ t} = {x0 , x1 , . . . , xt−1 }, where xi ∈ Cit for any i = 0, . . . , t − 1, Condition a) is satisfied. Now we deal with Condition b). Reasoning by contradiction, we suppose that there exist h, i, j with 0 ≤ h ≤ p − 1, 0 ≤ i < j ≤ t − 1 such that ai − a j = hat = h(at−1 + xt−1 ), that is hxt−1 = (ai − a j ) − hat−1 . If h = 0 then xt−1 = h −1 (ai − a j ) − at−1 , but xt−1 ∈ Bt−1 . A contradiction. If h = 0 then ai = a j implies  x j−1 = ai − a j−1 , again a contradiction, since x j−1 ∈ B j−1 . We observe that the previous Condition (1) is always satisfied when t < n. In this case, obviously under the necessary condition p n ≡ 1(mod 2t), we are able to give a concrete construction for an EA-cycle system. Construction 3.4. Case t < n. We consider the sequence A = (a0 = 0, a1 , a2 , . . . , at ) defined by the rule ai = ai−1 + g i−1 for i = 1, . . . , t, where g is a primitive element of F pn . We want to prove that B = [a0 , a1 , a2 , . . . , at−1 ]at is a ( pt)-cycle. By Remark 2.4, it is sufficient to show that the elements of {a0 , a1 , . . . , at−1 } belong to different cosets of the additive group at . Reasoning by contradiction, let i, j with 0 ≤ i < j ≤ t − 1 such that ai − a j = αat for a suitable α ∈ F p . Namely g j + ... + g i−1 = α(1 + g + ... + g t−1 ), but it cannot happen since g is primitive and t < n. It is easy to see that ∂1 B = −1·{1, g, g 2 , . . . , g t−1 }. Hence, by taking a complete system S of representatives for the cosets of −1 in C t , using Corollary 2.3 it follows that F = {s · B | s ∈ S} is the set of base cycles for an EA-(K pn , C pt )-design. Proposition 3.5. Let q = p dr be an odd prime power, where r > 1. There exists an EA-( pt)-cycle system of order q for every odd integer t dividing p d − 1. Proof. Let q − 1 = p n − 1 = 2et, with t as in the statement. Let g be a primitive element of Fq and set ε = g 2e . Starting from the construction of Theorem 1.3, now we are able to construct an EA-( pt)-cycle system of order q. We consider the closed trail Ci = [g i , g i ε, g i ε2 , . . . , g i εt−1 ]xi , where xi = g i (g − 1) (1 − εt−1 ) and i is taken modulo e. We want to prove that Ci is a ( pt)-cycle for all i ∈ Ze . By Remark 2.4, it is sufficient to show that the elements of {g i , g i ε, g i ε2 , . . . , g i εt−1 } belong to different cosets of the additive group < xi > for all i ∈ Ze . In other words, we have to prove that, for any λ ∈ F p and for any 0 ≤ j < k ≤ t − 1 g i (εk − ε j ) = λg i (g − 1)(1 − εt−1 )

∀i = 0, . . . , e − 2

(2)

and also g e−1 (εk − ε j ) = λ(1 − g e−1 )(1 − εt−1 ).

(3)

On the Existence of Elementary Abelian Cycle Systems

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Reasoning by contradiction, we assume that in (2) the equality holds for some λ ∈ F p and some 0 ≤ j < k ≤ t − 1. If λ = 0, then εk = ε j results in a contradiction. If εk − ε j λ = 0, we obtain g = λ−1 + 1, hence g ∈ F p (ε). Since g is a primitive ele1 − εt−1 n ment of Fq we obtain F p (ε) = Fq . On the other hand, by hypothesis, ppd −1 divides −1

2e and ppd −1 = 1. So g 2e = ε belongs to the proper subfield of Fq of order p d , that −1 is F p (ε) ⊂ Fq , again a contradiction. In a similar way we can show that (3) is also satisfied. Now, we consider the set F = {Ci | i ∈ Ze }. It is easy to check that ∂1 F = Fq∗ , thus, by Corollary 2.3, F is the set of base cycles of an EA-( pt)-cycle system of order q.  n

There exist parameters which satisfy Proposition 3.3 but not Proposition 3.5 (see Table 1) and also vice versa (see Table 2). On the other hand, Proposition 3.3 and Proposition 3.5 are not complementary. In fact there are EA-cycle systems whose existence is guaranteed by both these results (see Table 3) and others whose existence is guaranteed by neither Proposition 3.3 nor Proposition 3.5 (see Table 4). Of course Condition (1) of Proposition 3.3 is certainly satisfied when t = 1 (see also Theorem 1.2) and t = 2. Moreover, if t = 3 Proposition 3.3 gives the existence of an EA-3 p-cycle system of order p n for every prime p such that p n ≡ 1(mod 6) with ( p, n) = (5, 2), (7, 2). On the other hand, we have a direct construction for an EA-15-cycle systems of order 25 (see Example 2.5) and Table 1. p n t

3 6 7

5 5 11

7 4 5

11 6 45

11 6 63

13 4 15

13 8 238

13 8 510

17 4 15

17 8 290

11 2 5 1

11 4 15 2

17 8 870

23 4 5

23 6 91

23 6 143

Table 2. p n t d

3 6 13 3

5 6 31 3

7 2 3 1

7 6 57 3

7 6 171 3

7 10 2801 5

7 10 8403 5

19 2 9 1

19 4 45 2

19 6 381 3

19 6 3429 3

Table 3. p n t d

3 8 5 4

5 4 3 2

7 3 3 1

7 6 3 3

7 6 9 3

7 6 19 3

11 6 3 2

11 6 5 2

11 6 15 2

7 6 43

7 8 1201

19 4 3 2

19 4 5 2

19 4 9 2

19 4 15 2

Table 4. p n t

3 4 5

3 8 41

5 2 3

5 4 13

5 8 313

7 4 15

11 6 105

11 6 315

13 4 119

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Proposition 3.5 ensures the existence of an EA-21-cycle system of order 49. Besides, if we consider the case of t = 4, again Proposition 3.3 ensures the existence of an EA-4 p-cycle system of order p n for every prime p such that p n ≡ 1(mod 8) with ( p, n) = (5, 2), (7, 2), (11, 2), (13, 2), (17, 2), (19, 2). Anyway, we have directly checked the existence of such cycle systems. Hence we can conclude with the following: Proposition 3.6. There exists an EA-( pt)-cycle system of order p n for every prime p such that p n ≡ 1(mod 2t) with t = 1, 2, 3, 4 and t ≤ p n−1 . 3.2. Case: p Does Not Divide k In the following, by adapting a known construction for perfect cyclic cycle systems (see [12]), we are able to give a construction method for perfect elementary abelian k-cycle systems of order p n where p is not a divisor of k. Note that in this case condition c) of Proposition 3.1 becomes p n ≡ 1(mod 2k). Hence we consider only the case of k even, since the case of k odd has already been solved (see Theorem 1.3). Construction 3.7. Let q = p n ≡ 1 (mod 2k) be a prime power, where k = 2e t with e ≥ 1 and t odd. Let ε be a primitive t-th root of unity in Fq and let (x0 , x1 , . . . , x2e −1 ) be a 2e -tuple of elements of Fq∗ belonging to pairwise distinct cosets of < ε >. Let A = (a0 , a1 , . . . , ak−1 ) be the k-cycle defined by the rule a j = ε s xr

for j = 2e s + r, 0 ≤ r < 2e , 0 ≤ j ≤ k − 1

or, more explicitly, A = (x0 , x1 , . . . , x2e −1 , εx0 , εx1 , . . . , εx2e −1 , . . . , εt−1 x0 , εt−1 x1 , . . . , εt−1 x2e −1 ). Now, for any i = 1, . . . , k/2, we consider the multiset i A of all the i-differences of A and the list L i of the first 2e i-differences, that is L i = {ah+i − ah | 0 ≤ h ≤ 2e − 1}. It is easy to see that i A = −ε · L i , where −ε is the group of 2t-th roots of unity in Fq .   Let I be a subset of 1, 2, . . . , k2 containing 1. If L i results in a complete system e of representatives for the cosets of C 2 in Fq∗ for all i ∈ I then, taking a complete syse tem S of representatives for the cosets of −ε in C 2 and setting F = {s · A | s ∈ S}, e we have i F = S · i A = S · −ε · L i = C 2 · L i = Fq∗ . Hence, by Theorem 2.2 (note that in this case ∂i A = i A), F is the set of base cycles of an I -perfect EA-(K q , Ck )-design. Of course, applying Construction 3.7 is the more difficult the larger e and |I | are. We are able to obtain some results in the following two different cases: (e =  1, I = 1, 2, 4, 6, . . . , k2 − 1 ) and (e > 1, I = {1}).

On the Existence of Elementary Abelian Cycle Systems

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3.2.1. Case e = 1  Proposition 3.8. Let q = p n ≡ 1 (mod 2k) be a prime power and let I = 1, 2, 4,  6, . . . , k2 − 1 . For all k ≡ 2 (mod 4), with k = 2, there exists an I -perfect EA(K q , Ck )-design. Proof. Let k = 2t, t odd, and let ε be a primitive k/2-th root of unity in Fq . Following Construction 3.7, we have to find a pair of elements from two different cosets of ε ⊆ Fq∗ , say (x0 , x1 ), in such a way that x1 − x0 and εx0 − x1 belong to different 

cosets of C 2 = Fq ⊆ Fq∗ . We want to prove that there exists an a ∈ Fq such that   a+ε 1, a+1 is a suitable pair. Firstly, we observe belongs to ε and we can always find an x1 =  that x0 = 1   a+ε   q−1 a+ε k a+1  ∈ ε, because  a+1 | a ∈ Fq  = 2 > 2 = |ε|. Secondly, we note that 

ε−1 ε−1 and εx0 − x1 = a a+1 belong to different cosets of Fq , a being in Fq . x1 − x0 = a+1 At this point, Construction 3.7 makes us sure that

F = {s(1, x1 , ε, εx1 , ε2 , ε2 x1 , . . . , εk/2−1 , εk/2−1 x1 ) | s ∈ S} is the set of base cycles of an EA-(K q , Ck )-design, S being a complete system of representatives for the cosets of −ε in Fq . 

Finally, note that if x1 ∈ Fq then the design is I -perfect, where I = {1, 2, 4, 6, . . . , k/2 − 1}. In fact for any i ∈ I we have i F = S· < −ε > ·{x1 (εi/2 − 1), εi/2 − 1} = Fq · {x1 (εi/2 − 1), εi/2 − 1} = Fq∗ . Actually, we  such a choice is always possible. In fact, we know  can prove that  a+ε | a ∈ Fq and that A has size q−1 that x1 ∈ A = a+1 2 . By way of contradiction we assume that all the elements of A are squares, so A = Fq since they have the same size. But 1 is a square which does not lie in A. The assertion follows.  Example 3.9. {1, 2, 4}-perfect EA-(K 81 , C10 )-design As an explicit example, we apply the above proposition to get a {1, 2, 4}-perfect EA-(K 81 , C10 )-design. In F81 = Z3 [x]/x 4 +x+2 we choose x as a primitive element and ε = x 16 = 2x 3 + x+2 as a primitive 5-th root of unity in F81 . As in Proposition 3.8 we set: x0 = 1 ∈ ε,  a+ε a = 2x 3 + x 2 + x = x 75 ∈ F81 and, consequently, x1 = a+1 = x ∈ ε. Moreover, 

3 39 ∈ F . we note that x1 − x0 = x + 2 = x 44 ∈ F 81 81 while εx 0 − x 1 = 2x + 2 = x So, considering the cycle

C = (1, x, ε, εx, ε2 , ε2 x, ε3 , ε3 x, ε4 , ε4 x) and assuming the set S = {1, x 2 , x 4 , x 6 } as a complete system of representatives 2 4 6 for the cosets of −ε in F 81 , we can say that F = {C, x C, x C, x C} is a set of base cycles of an EA-(K 81 , C10 )-design B. Also, since x1 = x is a non-square, B is a {1, 2, 4}-perfect EA-10-cycle system of order 81.

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3.2.2. Case e > 1 Proposition 3.10. Let q ≡ 1(mod 2k) be a prime power, where k = 2e t with e > 1 and t odd. There exists an EA-k-cycle system of order q for any q > 22e−1 [(2e − 1)t − (2e − 3)] − 2e+1 + 1. Proof. Let g be a primitive element of Fq and let ε be a primitive t-th root of unity in Fq . Let (x¯1 , x¯2 , . . . , x¯2e −1 )c be the solution of the system ⎧ x2 − x1 = g(x1 − 1)c ⎪ ⎪ ⎪ ⎪ x3 − x2 = g 2 (x1 − 1)c ⎪ ⎨ .. . ⎪ e ⎪ ⎪ x e − x2e −2 = g 2 −2 (x1 − 1)c ⎪ ⎪ e ⎩ 2 −1 ε − x2e −1 = g 2 −1 (x1 − 1)c e

where c is a fixed element of C 2 . Note that such a solution exists if and only if

e −1 h D = 1 + c 2h=1 g = 0. In this case, for any i = 1, . . . , 2e − 1 we have

2e −1 h h ε + εc i−1 h=i g h=1 g + c x¯i = . D It is not hard to see that (x¯1 , x¯2 , . . . , x¯2e −1 )c1 = (x¯1 , x¯2 , . . . , x¯2e −1 )c2 if and only if e c1 = c2 . So, for any fixed c in C 2 provided D = 0, the set L 1 = {x¯1 − 1, x¯2 − x¯1 , . . . , x¯2e −1 − x¯2e −2 , ε − x¯2e −1 } results in a complete system of representatives for e the cosets of C 2 in Fq∗ . Now, let A be the closed trail A = (1, x¯1 , . . . , x¯2e −1 , ε, ε x¯1 , . . . , ε x¯2e −1 , ..., εt−1 , εt−1 x¯1 , . . . , εt−1 x¯2e −1 ). We want to prove that A is a k-cycle with vertices in Fq∗ , namely that there exists e an element c¯ ∈ C 2 such that D = 0, x¯ = 0 for any  = 1, . . . , 2e − 1, and / ε for any  = 1, . . . , 2e − 1 and 1 ≤ j < i ≤ 2e − 1. Note that, x¯ , x¯i · x¯ −1 j ∈ e for any fixed  = 1, . . . , 2e − 1, there is exactly one value of c ∈ C 2 giving x¯ = 0 e and there are at most t = |ε| values of c ∈ C 2 giving x¯ , x¯i · x¯ −1 j ∈ ε for each  = 1, . . . , 2e − 1 and for each pair (i, j) with 1 ≤ j < i ≤ 2e − 1. On the other hand, it is easy to see that x¯1 = 1 = x¯2 , x¯ = ε for any  = 1, . . . , 2e − 1 and e 2e x¯i · x¯ −1 j  = 1 for any 1 ≤ j < i ≤ 2 − 1. Thus, among the values of C we have to take off at most 2e + 2(t − 2) + (2e −3)(t −1) +

(2e −1)(2e −2) (t −1) = 2e−1 [(2e −1)t −(2e −3)] − 2 2

q −1 > 2e−1 [(2e − 1)t − (2e − 3)] − 2. 2e Hence the assertion follows from Construction 3.7.  e

elements. By the hypothesis on q, |C 2 | =

Of course, the bound determined in the previous proposition is as sharp as possible. Nonetheless, in the same way as in Proposition 3.10, we are able to construct an EA-(K 81 , C20 )-design, even if 81 > 105.

On the Existence of Elementary Abelian Cycle Systems

11

Example 3.11. EA-(K 81 , C20 )-design In F81 = Z3 [x]/x 4 +x+2 we choose x as a primitive element and ε = x 16 = 3 2x + x + 2 as a primitive 5-th root of unity in F81 . As in Proposition 3.10, we consider the system ⎧ ⎨ x2 − x1 = g(x1 − 1)c x3 − x2 = g 2 (x1 − 1)c ⎩ ε − x3 = g 3 (x1 − 1)c Fixing c = 1 we obtain the solution x1 =

x 2 + 2x + 2 = x 37 , x3 + x2 + x + 1

x3 =

x3 + x2 + 1 = x. x3 + x2 + x + 1

x2 =

2x 2 + x + 1 = x 77 , x3 + x2 + x + 1

It is easy to check that none of the x1 x2−1 = x 40 , x1 x3−1 = x 36 , x2 x3−1 = x 76 belongs to ε. So, assuming S = {1, x 4 } as a complete system of representatives for the cosets of −ε in C 4 , we can state that F = {s(1, x1 , x2 , x3 , ε, x1 ε, x2 ε, x3 ε, . . . , ε4 , x1 ε4 , x2 ε4 , x3 ε4 ) | s ∈ S} is the set of base cycles of an EA-(K 81 , C20 )-design. 4. Elementary Abelian Cycle Systems of Kms In this section we propose some results about elementary abelian cycle systems of the complete multipartite graph. As far as the authors are aware, at the moment the only known result about regular cycle systems of K m×s is the following (see [10] and [26]) Theorem 4.1. A cyclic k-cycle system of K m×k exists if and only if (m, k) = (3, 3) and

(m, k) ≡ (0, 1)(mod 2), (2, 2)(mod 4), (3, 2)(mod 4).

Also, it is obvious that removing all lines having a fixed direction from the affine plane of order q one gets a transversal design T D(q, q) (for the well known concept of a transversal design one can see, e.g., [4]) that equivalently can be seen as an EA-(K q×q , K q )-design. So we can state: Proposition 4.2. There exists an EA-(K pn × pn , K pn )-design for every prime p. We start determining a necessary condition for the existence of an EA-cycle system of K m×s .

12

A. Benini and A. Pasotti

Proposition 4.3. If an EA-(K m×s , Ck )-design exists, then a) ms = p n where p is a prime and n > 1; b) k ≤ ms; c) ps(m − 1) ≡ 0(mod 2k). Proof. Of course |V (K m×s )| must be a prime power, so let ms = p n where p is a prime and n > 1. Hence a) is satisfied. Obviously condition b) is true. To conclude, we prove c). For the same reason as in Proposition 3.1, p n−1 must be a divisor of n p n −s) of blocks of the required design. Hence ps(m−1) must be an the number p ( 2k 2k integer.  Example 4.4. EA-(K 2×4 , C4 )-design Let the vertices of K 2×4 be the elements of Z32 . We choose the following partition of the vertex-set {{000, 001, 010, 011}, {100, 101, 110, 111}}. Now, it is easy to see that the cycles C1 = [000, 101]010 ,

C2 = [101, 001]010 ,

C3 = [000, 110]010 ,

C4 = [001, 100]010

form an EA-(K 2×4 , C4 )-design. While it is trivial that a (K m×s , Ck )-design together with a (K s , Ck )-design gives a (K ms , Ck )-design, the following seems not. Proposition 4.5. If there exist an EA-(K m×s , Ck )-design and an EA-(K s , Ck )-design, then there also exists an EA-(K ms , Ck )-design. Proof. Let A and B be an EA-(K m×s , Ck )-design and an EA-(K s , Ck )-design, respectively, and let P1 , P2 , . . . , Pm be the parts of K m×s . For any positive integer n, let us denote by In the set {1, 2, . . . , n}. By hypothesis ms = p n for some odd prime p where n > 1, so m = p t and s = pr for suitable integers t and r . Since r A and B are elementary abelian, V (K m×s ) = Zt+r p and V (K s ) = Z p . Without loss of generality we can assume that the m parts of K m×s are the cosets of Zrp in Zt+r p . Namely, {g1 , g2 , . . . , gm } being a complete system of representatives for the cosets r of Zrp in Zt+r p , Pi = Z p + gi , for any i ∈ Im . Of course f i : x ∈ V (K s ) → x + gi ∈ V (K Pi ) results in an isomorphism between K s and K Pi for any i ∈ Im . Hence, for any k-cycle B ∈ B we can see that f i (B) results again in a k-cycle. Since Fi = { f i (B) | B ∈ B} is an EA-k-cycle system of K Pi for any i ∈ Im , it is not hard to see that A ∪ F1 ∪ F2 ∪ ... ∪ Fm is an EA-(K ms , Ck )-design.  Theorem 4.6. Let p be an odd prime. If there exist an EA-(K pt × pr , K pu )-design and an EA-(K pu , Ck )-design then there also exists an EA-(K pt × pr , Ck )-design. Proof. Let A be an EA-(K pt × pr , K pu )-design and let {K 1 , . . . , K α } be a set of representatives for the Zt+r p -orbits on A. As above, by In we denote the set {1, 2, . . . , n}.

On the Existence of Elementary Abelian Cycle Systems

13

By hypothesis there exists an EA-(K i , Ck )-design Bi , for every i ∈ Iα . For any arbitrary block  of A, there exist g ∈ Zt+r p and i ∈ Iα such that  = K i + g, hence g = {B + g | B ∈ Bi } is an EA-(, Ck )-design, say B . It is not hard to see that Bi +   B = ∈A B is an EA-(K pt × pr , Ck )-design. From Theorem 4.6 and Proposition 4.2, applying suitable propositions from Section 3, we can also state the following Proposition 4.7. There exists an EA-(K pn × pn , C pt )-design for every prime p such that p n ≡ 1(mod 2t) with t = 2, 3, 4 and t ≤ p n−1 . (cfr. Proposition 3.6). Proposition 4.8. There exists an EA-(K pn × pn , C pt )-design for any prime p such that p n ≡ 1(mod 2t) with

p n −1 p−1

≥

t 2 (t−1) 2

(cfr. Proposition 3.3).

Proposition 4.9. Let p n = p dr be an odd prime power, where r > 1. There exists an EA-(K pn × pn , C pt )-design for every odd integer t dividing p d − 1 (cfr. Proposition 3.5). Proposition 4.10. There exists an EA-(K pn × pn , Ck )-design for any prime p such that p n ≡ 1(mod 2k) and any k ≡ 2(mod 4), with k = 2 (cfr. Proposition 3.8). Proposition 4.11. Let q ≡ 1(mod 2k) be a prime power, where k = 2e t with e > 1 and t odd. There exists an EA-(K q×q , Ck )-design for any q > 22e−1 [(2e − 1)t − (2e − 3)] − 2e+1 + 1 (cfr. Proposition 3.10). Acknowledgements. The authors are grateful to Prof. Marco Buratti for helpful suggestions on the item of this paper and to the referee for providing valuable remarks.

References 1. Adams, P., Billington, E.J.: Completing some spectra for 2-perfect cycle systems. Austral. J. Combin. 7, 175–187 (1993) 2. Alspach, B., Gavlas, H.: Cycle decompositions of K n and K n − I . J. Combin. Theory Ser. B 81, 77–99 (2001) 3. Bennett, F.E., Ge, G., Zhu, L.: Existence of Steiner seven-cycle systems. Discrete Math. 252, 1–16 (2002) 4. Beth, T., Jungnickel, D., Lenz, H.: Design Theory. Cambridge University Press, (1999) 5. Bryant, D., El-Zanati, S.: Graph decompositions. In: Handbook of Combinatorial Designs, Second Edition. Colbourn, C.J., Dinitz, J.H., (eds.), Chapman & Hall/CRC, Boca Raton, FL, pp. 477–486 (2006) 6. Bryant, D., Rodger, C.A.: Cycle decompositions. In: Handbook of Combinatorial Designs, Second Edition. Colbourn, C.J., Dinitz, J.H., (eds.), Chapman & Hall/CRC, Boca Raton, FL, pp. 373–382 (2006) 7. Bonisoli, A., Buratti, M., Mazzuoccolo, G.: Doubly transitive 2-factorizations. J. Combin. Designs 15, 120–132 (2007) 8. Buratti, M.: Rotational k-cycle systems of order v < 3h; another proof of the existence of odd cycle systems. J. Combin. Designs 11, 433–441 (2003)

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9. Buratti, M.: Cycle decompositions with a sharply vertex transitive automorphism group. Le Matematiche (Catania) 59, 91–105 (2004) 10. Buratti, M., Del Fra, A.: Existence of cyclic k-cycle systems of the complete graph. Discrete Math. 261, 113–125 (2003) 11. Buratti, M., Del Fra, A.: Cyclic hamiltonian cycle systems of the complete graph. Discrete Math. 279, 107–119 (2004) 12. Buratti, M., Rania, F., Zuanni, F.: Some constructions for cyclic perfect cycle systems. Discrete Math. 299, 33–48 (2005) 13. Granville, A., Moisiadis, A., Rees, R.: Nested Steiner n-cycle systems and perpendicular array. J. Combin. Math. Combin. Comput. 3, 163–167 (1988) 14. Kotzig, A.: Decomposition of a complete graph into 4k-gons. Mat. Fyz. Casopis Sloven. Akad. Vied 15, 229-233 (1965) (in Russian) 15. Lindner, C.C., Rodger, C.A.: 2-perfect m-cycle systems. Discrete Math. 104, 83–90 (1992) 16. Lindner, C.C., Stinson, D.R.: Steiner pentagon systems. Discrete Math. 52, 67–74 (1984) 17. Manduchi, E.: Steiner heptagon systems. Ars Combin. 31, 76–85 (1991) 18. Peltesohn, R.: Eine losung der beiden heffterschen differenzenprobleme. Compos. Math. 6, 251–257 (1938) 19. Rosa, A.: On cyclic decompositions of the complete graph into (4m + 2)-gons. Mat. Fyz. Casopis Sloven. Akad. Vied 16, 349–352 (1966) 20. Rosa, A.: On cyclic decompositions of the complete graph into polygons with odd number of edges (Slovak). Casopis Pest. Mat. 91, 53–63 (1966) 21. Rosa, A.: On decompositions of a complete graph into 4n-gons. Mat. Casopis Sloven. Akad. Vied 17, 242–246 (1967) (in Russian) ˜ 22. Sajna, M.: Cycle decompositions of K n and K n − I . Ph.D. Thesis, Simon Fraser University, July (1999) ˜ 23. Sajna, M.: Cycle decompositions III: complete graphs and fixed length cycles. J. Combin. Designs 10, 27–78 (2002) 24. Vietri, A.: Cyclic k-cycle systems of order 2kn + k; a solution of the last open cases. J. Combin Designs 12, 299–310 (2004) 25. Vietri, A.: Difference families in Z2d+1 ⊕Z2d+1 and infinite translation designs in Z⊕Z. Graphs and Comb. 23, 111–121 (2007) 26. Wu, S.L., Fu, H.L.: A note on cyclic m-cycle systems of K r (m) . Graphs and Combin. 22, 427–432 (2006) Received: February 12, 2008 Final version received: October 25, 2008

Graphs and Combinatorics (2009) 25:15–25 Digital Object Identifier (DOI) 10.1007/s00373-008-0822-8

Graphs and Combinatorics © Springer-Verlag 2009

Perpendicular Rectangular Latin Arrays Robert Brier, Darryn Bryant Department of Mathematics, University of Queensland, QLD 4072, Australia.

Abstract. A set {A1 , A2 , . . . , At } of rectangular arrays, each defined on a symbol set X , is said to be t-perpendicular if each t-element subset of X occurs precisely once when the arrays are superimposed. We investigate the existence of sets of r by s rectangular arrays which are row-Latin, column-Latin and t-perpendicular. For example, we show that for all odd n, there exists a pair of row- and column-Latin 2-perpendicular r by s arrays with symbol set X of  size n if and only if r s = n2 and r, s ≤ n.

1. Introduction In this article we investigate a variant of the notion of orthogonality for Latin squares. See [1] for a survey of results on orthogonal Latin squares. An r by s array A of symbols is said to be row-Latin if (a)

A(i, j) and A(i, j  ) are distinct for 1 ≤ i ≤ r and 1 ≤ j < j  ≤ s,

and column-Latin if (b)

A(i, j) and A(i  , j) are distinct for 1 ≤ i < i  ≤ r and 1 ≤ j ≤ s.

A Latin square of order n is an n by n row- and column-Latin array of symbols chosen from an n-set. A pair of Latin squares is orthogonal if each ordered pair of elements of their underlying set occurs exactly once when the squares are superimposed. More precisely, a pair {A, B} of Latin squares, each based on the same symbol set X with |X | = n, and with rows and columns indexed by {1, 2, . . . , n} is orthogonal if {(A(i, j), B(i, j)) : 1 ≤ i ≤ n, 1 ≤ j ≤ n} = X × X. In this paper we will be interested in covering all unordered, rather than ordered, pairs of elements and we shall use the term “perpendicular” to describe this property. The usage of this term is consistent with its existing usage in the literature, such   as in perpendicular arrays. A perpendicular array PA(t, k, n) is a k by nt array with entries chosen from an n-set such that each column has k distinct entries and each set of t rows contains each set of t distinct entries as a column precisely once. See [2] for a survey of results on perpendicular arrays.   Of course, the number of unordered pairs of an n-set is n2 , and so some unordered pairs will occur more than once if two Latin squares are superimposed. Hence

16

R. Brier, D. Bryant

we instead consider r by s rectangular arrays with r, s ≤ n and which are both row- and column-Latin. We call such an array an r by s rectangular Latin array of order n, when its symbols are chosen from an n-set. Note that an n by s rectangular Latin array of order n is equivalent to a Latin rectangle. We will also be interested in the more general problem of covering, by superimposing a set of t rectangular Latin arrays, all t-subsets of the underlying set of the arrays. We give the following formal definition. Definition 1.1. Let A1 , A2 , . . . , At be r by s rectangular Latin arrays, each defined on a symbol set X . Then A1 , A2 , . . . , At are said to be t-perpendicular if each t-element subset of X occurs precisely once in the r by s array having the t-set {A1 (i, j), A2 (i, j), . . . , At (i, j)} in row i and column j. When t = 2, we may write just perpendicular rather than 2-perpendicular. Example 1.1. A set of 4-perpendicular 3 by 5 rectangular Latin arrays of order 6. 1 2 3

5 6 1

2 3 4

4 5 6

3 4 5

2 3 4

6 1 2

1 2 3

5 6 1

4 5 6

4 5 6

1 2 3

3 4 5

2 3 4

6 1 2

3 4 5

4 5 6

5 6 1

6 1 2

1 2 3

Another question of interest is the existence of sets of k > t, rather than just t, r by s rectangular Latin arrays which have the property that any t of them are t-perpendicular. We examine this question in Section 4. 2. Existence of t-perpendicular Rectangular Latin Arrays In this section we examine the following question. Question 1. For which values of t, r , s and n does there exist a set of t-perpendicular r by s rectangular Latin arrays of order n ?   Clearly we require r s = nt and r, s ≤ n. The sensible values of t to consider are in the range 1 ≤ t ≤ n. The cases t = 1, t = n − 1 and t = n are trivial. There exists a 1-perpendicular r by s rectangular Latin array of order n if and only if r s = n (a single r by s array containing n distinct symbols once each). There exists a set of (n − 1)-perpendicular r by s rectangular Latin arrays of order n if and only if r s = n (the arrays A1 , A2 , . . . , An−1 where A1 is any r by s array containing each element of Zn once each and A x (i, j) = A1 (i, j)+(x −1) (mod t) for 1 ≤ x ≤ n −1, 1 ≤ i ≤ r and 1 ≤ j ≤ s). There exists a set of n-perpendicular r by s rectangular Latin arrays of order n if and only if r = s = 1 (the arrays A1 , A2 , . . . , An where the single entry in Ai is i).

Perpendicular Rectangular Latin Arrays

17

Thus we may restrict our attention to values of t in the range 2 ≤ t ≤ n − 2. It is straightforward   to verify that for 3 ≤ t ≤ n − 3, the only values of t, r , s and n satisfying r s = nt and r, s ≤ n are (t, r, s, n) ∈ {(3, 4, 5, 6), (3, 5, 7, 7), (4, 5, 7, 7), (3, 7, 8, 8), (5, 7, 8, 8)} where without loss of generality we assume r ≤ s. We can’t have n ≥ 9 since then n  2 ≥ r s for 3 ≤ t ≤ n − 3. The following examples show that the arrays exist > n t in each of these cases. Example 2.1. A set of 3-perpendicular 4 by 5 rectangular Latin arrays of order 6. 1 2 6 3

2 3 1 4

3 4 2 1

4 5 3 6

5 6 4 2

2 1 3 5

3 5 4 2

4 3 5 6

5 4 6 1

6 2 1 4

3 4 5 1

4 2 6 5

5 6 1 2

6 1 2 3

1 5 3 6

Example 2.2. A set of 3-perpendicular 5 by 7 rectangular Latin arrays of order 7. 3 1 2 6 5

4 2 3 7 6

5 3 4 1 7

6 4 5 2 1

7 5 6 3 2

1 6 7 4 3

2 7 1 5 4

1 2 5 3 7

2 3 6 4 1

3 4 7 5 2

4 5 1 6 3

2 4 1 5 3

3 5 2 6 4

4 6 3 7 5

5 6 2 7 4

6 7 3 1 5

7 1 4 2 6

5 7 4 1 6

6 1 5 2 7

7 2 6 3 1

1 3 7 4 2

Example 2.3. A set of 4-perpendicular 5 by 7 rectangular Latin arrays of order 7. 4 1 2 7 3

5 2 3 1 4

6 3 4 2 5

7 4 5 3 6

1 5 6 4 7

2 6 7 5 1

3 7 1 6 2

3 5 1 2 4

4 6 2 3 5

5 7 3 4 6

6 1 4 5 7

7 2 5 6 1

1 3 6 7 2

2 4 7 1 3

2 3 5 4 1

3 4 6 5 2

4 5 7 6 3

5 6 1 7 4

6 7 2 1 5

7 1 3 2 6

1 2 4 3 7

1 2 4 3 6

2 3 5 4 7

3 4 6 5 1

4 5 7 6 2

5 6 1 7 3

6 7 2 1 4

7 1 3 2 5

Example 2.4. A set of 3-perpendicular 7 by 8 rectangular Latin arrays of order 8.

18

R. Brier, D. Bryant

1 2 3 5 8 7 6

2 3 4 6 1 8 7

3 4 5 7 2 1 8

4 5 6 8 3 2 1

5 6 7 1 4 3 2

6 7 8 2 5 4 3 3 5 7 1 4 8 2

7 8 1 3 6 5 4 4 6 8 2 5 1 3

8 1 2 4 7 6 5 5 7 1 3 6 2 4

2 3 4 6 7 5 8 6 8 2 4 7 3 5

7 1 3 5 8 4 6

3 4 5 7 8 6 1 8 2 4 6 1 5 7

4 5 6 8 1 7 2 1 3 5 7 2 6 8

5 6 7 1 2 8 3

6 7 8 2 3 1 4

7 8 1 3 4 2 5

8 1 2 4 5 3 6

1 2 3 5 6 4 7

2 4 6 8 3 7 1

Example 2.5. A set of 5-perpendicular 7 by 8 rectangular Latin arrays of order 8. 8 2 4 6 7 5 3

1 3 5 7 8 6 4

2 4 6 8 1 7 5

3 5 7 1 2 8 6

4 6 8 2 3 1 7

5 7 1 3 4 2 8

6 8 2 4 5 3 1

7 1 3 5 6 4 2

7 1 3 5 4 8 6

8 2 4 6 5 1 7

1 3 5 7 6 2 8

2 4 6 8 7 3 1

3 5 7 1 8 4 2

4 6 8 2 1 5 3

5 7 1 3 2 6 4

6 8 2 4 3 7 5

6 8 2 3 5 7 1

7 1 3 4 6 8 2

8 2 4 5 7 1 3

1 3 5 6 8 2 4

2 4 6 7 1 3 5

3 5 7 8 2 4 6

4 6 8 1 3 5 7

5 7 1 2 4 6 8

5 7 8 2 3 1 4

6 8 1 3 4 2 5

7 1 2 4 5 3 6

8 2 3 5 6 4 7

1 3 4 6 7 5 8

2 4 5 7 8 6 1

3 5 6 8 1 7 2

4 6 7 1 2 8 3

4 5 7 1 2 3 8

5 6 8 2 3 4 1

6 7 1 3 4 5 2

7 8 2 4 5 6 3

8 1 3 5 6 7 4

1 2 4 6 7 8 5

2 3 5 7 8 1 6

3 4 6 8 1 2 7

This leaves only the cases t = 2 and t = n − 2. For t = n − 2, we will see in Section 4 that there is a set of t-perpendicular n−1 2 by n rectangular Latin arrays of order n whenever n is a power of an odd prime, see Theorem 4.1. The smallest case not covered by this result is a set of 4-perpendicular 3 by 5 rectangular Latin arrays of order 6, and such a set is given in Example 1.1. All other cases of Question 1 remain open for t = n − 2.

Perpendicular Rectangular Latin Arrays

19

For t = 2 and n odd we completely settle Question 1 by showing that there is a pair of perpendicular r by s rectangular Latin arrays of order n whenever r s = nt and r, s ≤ n, see Section 3. For t = 2 and n even we have a complete solution to Question 1 in the case r = n2 and s = n − 1. The following theorem says that in this case the arrays exist. Theorem 2.6. Let n be an even integer. There exists a pair of perpendicular rectangular Latin arrays of order n.

n 2

by (n −1)

Proof. The following two arrays consisting of elements from Zn suffice to prove the theorem. Note that for d = 1, 2, . . . , n2 − 1, pairs of elements which are d apart in the cycle (0, 1, 2, . . . , n − 1) are obtained when the d-th or (n − d)-th columns are superimposed, and that pairs of elements which are n2 apart in the cycle (0, 1, 2, . . . , n − 1) are obtained when the n-th columns are superimposed (where the columns are labeled 1 to n).

n 2

0 1 .. .

1 2 .. .

−1 1 2 .. . n 2

n 2

n−1 0 .. . n 2

−2

n−1 0 .. . n 2

n 2

2 3 .. .

−2

n 2

2 3 .. .

n−2 n−1 .. .

+1

n 2

+1

−3

n−2 n−1 .. . n 2

n 2

3 4 .. .

−3

n 2

3 4 .. .

n−3 n−1 .. .

+2

n 2

+2

−4

n−3 n−1 .. . n 2

n 2

··· ··· .. .

−4

···

4 5 .. .

··· ··· .. .

+3

···

n 2 n 2

n 2

+1 +2 .. .

n 2

n 2

+1 .. .

0

n−1

n 2

0 1 .. .

+1 .. .

n−1

n 2

−1 

Other than the cases covered by  the above theorem, the smallest even n for which there exist r and s such that r s = n2 with r ≤ s ≤ n is n = 16; where we have r = 10 and s = 12. A pair of perpendicular 10 by 12 rectangular Latin arrays, P and Q say, of order 16 is shown below. The underlying set for the arrays consists of the elements of the group Z4 × Z4 and this group was used in its construction. These arrays are partitioned into fifteen 2 by 4 subarrays, and the differences Pi, j − Q i, j are constant within each of these subarrays. (0,0) (1,0) (2,0) (3,0) (2,2) (3,2) (0,2) (1,2) (3,1) (0,1)

(0,1) (1,1) (2,1) (3,1) (2,3) (3,3) (0,3) (1,3) (3,2) (0,2)

(0,2) (1,2) (2,2) (3,2) (2,0) (3,0) (0,0) (1,0) (3,3) (0,3)

(0,3) (1,3) (2,3) (3,3) (2,1) (3,1) (0,1) (1,1) (3,0) (0,0)

(2,0) (3,0) (1,1) (2,1) (0,0) (1,0) (3,1) (0,1) (0,2) (1,2)

(2,1) (3,1) (1,2) (2,2) (0,1) (1,1) (3,2) (0,2) (0,3) (1,3)

(2,2) (3,2) (1,3) (2,3) (0,2) (1,2) (3,3) (0,3) (0,0) (1,0)

(2,3) (3,3) (1,0) (2,0) (0,3) (1,3) (3,0) (0,0) (0,1) (1,1)

(3,0) (0,0) (3,3) (0,3) (1,1) (2,1) (1,0) (2,0) (1,2) (2,2)

(3,1) (0,1) (3,0) (0,0) (1,2) (2,2) (1,1) (2,1) (1,3) (2,3)

(3,2) (0,2) (3,1) (0,1) (1,3) (2,3) (1,2) (2,2) (1,0) (2,0)

(3,3) (0,3) (3,2) (0,2) (1,0) (2,0) (1,3) (2,3) (1,1) (2,1)

20

R. Brier, D. Bryant

(0,1) (1,1) (2,3) (3,3) (2,0) (3,0) (1,2) (2,2) (2,1) (3,1)

(0,2) (1,2) (2,0) (3,0) (2,1) (3,1) (1,3) (2,3) (2,2) (3,2)

(0,3) (1,3) (2,1) (3,1) (2,2) (3,2) (1,0) (2,0) (2,3) (3,3)

(0,0) (1,0) (2,2) (3,2) (2,3) (3,3) (1,1) (2,1) (2,0) (3,0)

(3,1) (0,1) (0,0) (1,0) (1,2) (2,2) (2,3) (3,3) (1,1) (2,1)

(3,2) (0,2) (0,1) (1,1) (1,3) (2,3) (2,0) (3,0) (1,2) (2,2)

(3,3) (0,3) (0,2) (1,2) (1,0) (2,0) (2,1) (3,1) (1,3) (2,3)

(3,0) (0,0) (0,3) (1,3) (1,2) (2,1) (2,2) (3,2) (1,0) (2,0)

(2,1) (3,1) (1,3) (2,3) (3,2) (0,2) (3,3) (0,3) (3,0) (0,0)

(2,2) (3,2) (1,0) (2,0) (3,3) (0,3) (3,0) (0,0) (3,1) (0,1)

(2,3) (3,3) (1,1) (2,1) (3,1) (0,0) (3,1) (0,1) (3,2) (0,2)

(2,0) (3,0) (1,2) (2,2) (3,2) (0,1) (3,2) (0,2) (3,3) (0,3)

3. Pairs of Odd Order Perpendicular Rectangular Latin Arrays In this section we construct, for all odd values of n and all r and s satisfying the obvious necessary conditions, a pair of perpendicular r by s rectangular Latin arrays of order n. A simple construction gives us the following theorem which settles the case r = n−1 2 and s = n, for n odd. Theorem 3.1. Let n be an odd integer. There exists a pair of perpendicular rectangular Latin arrays of order n.

n−1 2

by n

Proof. The following two arrays consisting of elements from Zn suffice to prove the theorem. Note that for d = 1, 2, . . . , n−1 2 , pairs of elements which are d apart in the cycle (0, 1, 2, . . . , n − 1) are obtained when the d-th rows are superimposed (where the rows are labeled 1 to n). 0 1 .. .

1 2 .. .

2 3 .. .

··· ··· .. .

n−1 0 .. .

n−3 2

n−1 2

n+1 2

···

n−5 2

1 3 .. .

n−2

2 4 .. .

n−1

3 5 .. . 0

··· ··· .. . ···

0 2 .. .

n−3 

To construct pairs of odd order r by s perpendicular rectangular Latin arrays when r, s < n, we will work in Za × Zb for some a and b where n = ab. For convenience, we first make the following definition. Definition 3.1. Given any p by q array A whose entries are from the set Za × Zb , define π1 (A) and π2 (A) to be the two p by q arrays where π1 (A)’s entries are from Za , π2 (A)’s entries are from Zb , and A(i, j) = (π1 (A)(i, j), π2 (A)(i, j)) for 1 ≤ i ≤ p and 1 ≤ j ≤ q. Our arrays will be generated in Za × Zb from starter arrays which we define as follows. The subsequent lemma describes how starter arrays can be used to generate pairs of odd order perpendicular rectangular Latin arrays.

Perpendicular Rectangular Latin Arrays

21

Definition 3.2. A pair of (a, b, c, d)-starter arrays is a pair A and B of d by c arrays with entries from the group Za × Zb such that (i) {A(i, j) − B(i, j) : 1 ≤ i ≤ d, 1 ≤ j ≤ c} ∪ {B(i, j) − A(i, j) : 1 ≤ i ≤ d, 1 ≤ j ≤ c} = Za × Zb − {(0, 0)}, (ii) π1 (A) and π1 (B) are row-Latin, and (iii) π2 (A) and π2 (B) are column-Latin. Lemma 3.1. Let n be an odd integer and let r and s be integers satisfying (i) r, s ≤nand (ii) r s = n2 . If there exists a pair of (a, b, c, d)-starter arrays such that n = ab, r = ad, s = bc and n−1 2 = cd, then there exists a pair of r by s perpendicular rectangular Latin arrays of order n. Proof. Let A and B be a pair of (a, b, c, d)-starter arrays satisfying the conditions of the lemma. We construct the required pair of r by s arrays as follows. Let A∗ be the r by s array with rows and columns indexed by {1, . . . a} × {1, . . . , d} and {1, . . . , b} × {1, . . . , c} respectively and A∗ ((i, j), (k, l)) = (i, k) + A( j, l) for 1 ≤ i ≤ a, 1 ≤ j ≤ d, 1 ≤ k ≤ b and 1 ≤ l ≤ c. Similarly, let B ∗ be the r by s array with rows and columns indexed by {1, . . . a} × {1, . . . , d} and {1, . . . , b} × {1, . . . , c} respectively and B ∗ ((i, j), (k, l)) = (i, k) + B( j, l) for 1 ≤ i ≤ a, 1 ≤ j ≤ d, 1 ≤ k ≤ b and 1 ≤ l ≤ c. We now show that A∗ and B ∗ have the required properties. First we show that each is both row- and column-Latin. To show that A∗ is column-Latin, suppose that A∗ ((i 1 , j1 ), (k, l)) = A∗ ((i 2 , j2 ), (k, l)). That is, (i 1 , k) + A( j1 , l) = (i 2 , k) + A( j2 , l). Since π2 (A) is column-Latin, this implies that j1 = j2 , from which it follows that i 1 = i 2 . Thus, A∗ is indeed column-Latin, and similar arguments show that A∗ is row-Latin and that B ∗ is row- and column-Latin. We now show that A∗ and B ∗ are perpendicular. It is sufficient to show that any pair of elements of Za ×Zb occurs when A∗ and B ∗ are superimposed. Let (x, y) and (u, v) be distinct elements of Za × Zb . Since A and B are a pair of (a, b, c, d)-starter arrays, we can assume without loss of generality that (x, y)−(u, v) = A( j, l)−B( j, l) for some j ∈ {1, . . . , d} and l ∈ {1, . . . , c}. Thus, letting (i, k) = (x, y) − A( j, l) we see that A∗ ((i, j), (k, l)) = (x, y) and B ∗ ((i, j), (k, l)) = (i, k) + B( j, l) = (x, y) − A( j, l) + (A( j, l) − (x, y) + (u, v)) = (u, v). So we see that the pair {(x, y), (u, v)} does indeed occur when A∗ and B ∗ are superimposed and we conclude that A∗ and B ∗ are perpendicular.  In order to construct the necessary starter arrays required to prove our main result for this section, we will need the following result which is due to Hall [3].

22

R. Brier, D. Bryant

Theorem 3.2. [3] Let G be an additive abelian group of order n and let y1 , . . . , yn be (not necessarilly distinct) elements of G such that y1 + . . . + yn = 0. Then there exist two permutations x1 , . . . , xn and z 1 , . . . , z n of the elements of G such that xi + yi = z i for i ∈ {1, . . . , n}. Theorem 3.3. Let n be an odd integer and let r, s be integers satisfying (i) r, s ≤n, and (ii) r s = n2 . Then there exists a pair of r by s perpendicular rectangular Latin arrays of order n. Proof. The case r = n−1 2 and s = n is settled in Theorem 3.1, so we can assume r ≤ s < n. We define a, b, c and d as follows   n−1 n−1 ,s , n = ab, = cd. a = gcd(n, r ), c = gcd 2 2 We also define β and δ by r = aβ

and

s = cδ.

Note that abcd =

n(n − 1) = aβcδ 2

and

bd = βδ.

Now, since a = gcd(n, r ), n = ab and r = aβ we have gcd(b, β) = 1. Similarly we also have gcd(d, δ) = 1. Since bd = βδ, this means that b = δ and d = β. So in fact r = ad

and

s = bc

and we can dispense with β and δ. We also note that since r, s < n, b>d

and

a > c.

Since n is odd, there is a set S of n−1 2 elements of Za × Zb such that S ∪ −S = Za × Zb − {0}. Choose any such set of n−1 2 elements and let C be a d by c array containing each of them exactly once. For i = 1, 2, . . . , d let yi1 , . . . , yic be row i of π1 (C) and choose yi(c+1) , . . . , yia so that yi1 + . . . yia = 0 (mod a) (recall that c < a and so this is clearly possible). By Theorem 3.2 there exist two permutations xi1 , . . . , xia and z i1 , . . . , z ia of Za , such that xi j + yi j = z i j for j = 1, 2, . . . , a. Let A1 and B1 be the two arrays defined by letting, for i = 1, 2, . . . , d, row i of A1 be xi1 , . . . , xic and row i of B1 be z i1 , . . . , z ic . Note since xi1 , . . . , xia and z i1 , . . . , z ia are permutations, A1 and B1 are both row-Latin.  For j = 1, 2, . . . , c let y1 j , . . . , yd j be column j of π2 (C) and choose y(d+1) j, . . . ,    ybj so that y1 j +. . . ybj = 0 (mod b) (recall that d < b and so this is clearly possible).  and z  , . . . , z  of Z , By Theorem 3.2 there exist two permutations x1 j , . . . , xbj b bj 1j    such that xi j + yi j = z i j for i = 1, 2, . . . , b. Let A2 and B2 be the two arrays defined

Perpendicular Rectangular Latin Arrays

23

by letting, for j = 1, 2, . . . , c, column j of A2 be x1 j , . . . , xd j and column j of B2  and z  , . . . , z  are permutations, A and be z 1 j , . . . , z d j . Note since x1 j , . . . , xbj 2 bj 1j B2 are both column-Latin. We can now construct a pair A and B of (a, b, c, d)-starter arrays from A1 , A2 , B1 and B2 by defining A and B to be the arrays such that π1 (A) = A1 , π2 (A) = A2 , π1 (B) = B1 and π2 (B) = B2 . The fact that A and B are indeed a pair of (a, b, c, d)starter arrays follows immediately from the fact that A1 and B1 are both row-Latin, A2 and B2 are both column-Latin, and {B(i, j) − A(i, j) : 1 ≤ i ≤ d, 1 ≤ j ≤ c} = {(z i j , z i j ) − (xi j , xi j ) : 1 ≤ i ≤ d, 1 ≤ j ≤ c} = {(yi j , yi j ) : 1 ≤ i ≤ d, 1 ≤ j ≤ c} = {C(i, j) : 1 ≤ i ≤ d, 1 ≤ j ≤ c}. Hence, by Lemma 3.1 there exists a pair of r by s perpendicular rectangular Latin arrays of order n.  4. k-sets of Rectangular Latin Arrays in which Any t Are t-perpendicular We begin this section with some necessary conditions for the existence of a k-set of r by s rectangular Latin arrays of order n in which any t are t-perpendicular, and then we prove an existence result. Lemma 4.1. A necessary condition for the existence of a k-set of rectangular Latin arrays of order n such that any t are t-perpendicular, is that a perpendicular array PA(t, k, n) exists. Proof. Let A1 , . . . , Ak be a k-set of r by s rectangular Latin arraysof order n such that any t are t-perpendicular. Then clearly the array whose r s = nt columns are  A1 (i, j), A2 (i, j), . . . , Ak (i, j) for 1 ≤ i ≤ r and 1 ≤ j ≤ s is a PA(t, k, n). Lemma 4.2. When n is even there is no 3-set of pairwise perpendicular rectangular Latin arrays of order n. Proof. Suppose there exists a 3-set of pairwise perpendicular rectangular Latin arrays of order n. Then by Lemma 4.1 there exists a PA(2, 3, n). Now, in any two rows of a PA(2, 3, n), the number of occurrences of each symbol is n − 1. Thus, if a symbol occurs say a times in row 1, then it occurs n − 1 − a times in row 2 and n − 1 − a times in row 3. That is, the symbol occurs 2(n − 1 − a) times in rows 2 and 3 and we thus have 2(n − 1 − a) = n − 1. We conclude that n is odd.  When n = 3 it is easy to construct an n-set of pairwise perpendicular rectangular Latin arrays of order n, just take the 1 by 3 arrays (1, 2, 3), (2, 3, 1) (3, 1, 2). However, the following lemma shows that this is impossible for n > 3.

24

R. Brier, D. Bryant

Lemma 4.3. For n > 3 there is no n-set of pairwise perpendicular rectangular Latin arrays of order n. Proof. For even n the result follows immediately from Lemma 4.2. So let n ≥ 5 be odd and suppose there are n pairwise perpendicular arrays A1 . . . An . Without loss of generality we can assume that A x (1, 1) = x for x = 1, 2, . . . , n. Using the same argument as in the proof of Lemma 4.2 we see that each symbol occurs n−1 2 times in each array. For x = 1, 2, . . . , n, define i x,2 , i x,3 , . . . , i x, n−1 and jx,2 , jx,3 , . . . , jx, n−1 by 2 2 letting (1, 1), (i x,2 , jx,2 ), . . . , (i x, n−1 , jx, n−1 ) 2

2

be the cells in the array A x where the symbol x occurs. Now, (i x,u , jx,u ) is distinct from (i y,v , j y,v ) for 1 ≤ x < y ≤ n and 2 ≤ u, v ≤ n−1 2 ; otherwise the pair {x, y} occurs twice when A x and A y are superimposed. Thus we have        (i x,u , jx,u ) : 1 ≤ x ≤ n, 2 ≤ u ≤ n − 1  = n n − 1 − 1 = n(n − 3) .   2 2 2 Moreover, for x = 1, 2, . . . , n, since A x is row- and column-Latin, we have i x,u = 1 and jx,u = 1 for u = 2, 3, . . . , n−1 2 . Thus we have      (i x,u , jx,u ) : 1 ≤ x ≤ n, 2 ≤ u ≤ n − 1  ≤r s − r − s + 1 = n(n − 1) − r − s + 1.   2 2 So we have n(n − 3) n(n − 1) ≤ −r −s +1 2 2 from which it follows that r + s ≤ n + 1. But we also have rs =

n(n − 1) , 2

and it is not to hard to check that this is impossible for n > 3.



We now prove that the upper bound of n − 1 on the size of a set of pairwise perpendicular r by s rectangular Latin arrays of order n (given in Lemma 4.3) can be achieved when r = n−1 2 , s = n, and n is an odd prime power. Theorem 4.1. When n is an odd prime power there exists an (n − 1)-set of rectangular Latin arrays of order n such that • any pair are perpendicular and • any n − 2 are (n − 2)-perpendicular.

n−1 2

by n

Perpendicular Rectangular Latin Arrays

25

Proof. Let G F(n) be the finite field of order n and let S be a subset of G F(n) such that S ∪ (−S) = G F(n)\{0} and S ∩ (−S) = ∅. For each k ∈ G F(n) define Ak to be the n−1 2 by n array with rows indexed by S, columns indexed by G F(n), and with Ak (i, j) = ik + j for each i ∈ S and j ∈ G F(n). We will verify that {Ak : k ∈ G F(n)\{0}} is the required set of arrays. Checking that these arrays are row- and column-Latin is trivial (A0 is not row-Latin, but it is used in the proof). We now check that any pair of arrays chosen from {Ak : k ∈ G F(n)} is perpendicular. Suppose that {Ak (i 1 , j1 ), Al (i 1 , j1 )} = {Ak (i 2 , j2 ), Al (i 2 , j2 )} for some k = l. That is, we have either i 1 k + j1 = i 2 k + j2

and

i 1l + j1 = i 2 l + j2

i 1 k + j1 = i 2 l + j2

and

i 1l + j1 = i 2 k + j2

or

In the first case it follows immediately that i 1 (k − l) = i 2 (k − l) and hence that i 1 = i 2 and j1 = j2 . In the second case it follows that i 1 (k − l) = i 2 (l − k) and hence that i 1 = −i 2 , which contradicts our definition of S. So no pair occurs more than once when Ak and Al are superimposed and so since Ak (i, j) = Al (i, j) implies k = l, we can conclude that any pair of the arrays is perpendicular. The fact that any n − 2 arrays chosen from {Ak : k ∈ G F(n)\{0}} are (n − 2)-perpendicular follows immediately from the fact that the array not chosen and A0 are perpendicular.  Acknowledgement. This research was supported by the Australian Research Council.

References 1. R. J. Abel, C. J. Colbourn and J. H. Dinitz, Mutually Orthogonal Latin Squares (MOLS), in The CRC Handbook of Combinatorial Designs, 2nd edition (Eds. C. J. Colbourn, J. H. Dinitz), CRC Press, Boca Raton (2007), 160–193. 2. J. Bierbrauer, Ordered Designs, Perpendicular Arrays, and Permutation Sets, in The CRC Handbook of Combinatorial Designs, 2nd edition (Eds. C. J. Colbourn, J. H. Dinitz), CRC Press, Boca Raton (2007), 543–547. 3. M. Hall, A Combinatorial Problem on Abelian Groups, Proc. Amer. Math. Soc., 3 (1952), 584–587. Received: December 12, 2007 Final version received: October 14, 2008

Graphs and Combinatorics (2009) 25:27–34 Digital Object Identifier (DOI) 10.1007/s00373-008-0818-4

Graphs and Combinatorics © Springer-Verlag 2009

The Interchange Graphs of Tournaments with Minimum Score Vectors Are Exactly Hypercubes An–Hang Chen1 , Jou–Ming Chang1 , Yue–Li Wang2, 1 Department of Information Management, National Taipei College of Business, Taipei,

Taiwan, ROC. e-mail: [email protected]; [email protected]

2 Department of Information Management, National Taiwan University of Science and

Technology, Taipei, Taiwan, ROC. e-mail: [email protected]

Abstract. A Δ-interchange is a transformation which reverses the orientations of the arcs in a 3-cycle of a digraph. Let T (S) be the collection of tournaments that realize a given score vector S. An interchange graph of S, denoted by G(S), is an undirected graph whose vertices are the tournaments in T (S) and an edge joining tournaments T, T  ∈ T (S) provided T  can be obtained from T by a Δ-interchange. In this paper, we find a set of score vectors of tournaments for which the corresponding interchange graphs are hypercubes. Key words. Tournaments, Score vectors, Interchange graphs, Hypercubes.

1. Introduction An n-tournament, denoted by Tn , is an oriented complete graph with vertices p1 , p2 , . . . , pn such that every two distinct vertices pi and p j are joined by exactly one of the arcs ( pi , p j ) or ( p j , pi ). If an arc ( pi , p j ) is in Tn , we say that pi beats p j (or p j is beaten by pi ) and write pi → p j . The score of a vertex pi in a tournament, denoted si , is the number of vertices beaten by pi . The score vector S = (s1 , s2 , . . . , sn ) of Tn is the list of scores of vertices of Tn in nondecreasing order. Let T (S) be the collection of tournaments that realize a given score vector S, and denote |T (S)| as its cardinality. Landau [6] proved that T (S) = ∅ if and only if   i for i = 1, 2, . . . , n − 1 (1) s1 + s2 + · · · + si  2 with equality for i = n. A tournament is called strong if, for every vertex, there exist directed paths to all other vertices. A score vector S is said to be strong if there is a strong tournament in T (S), or equivalently, there is no integer i with i < n such that the equality holds in (1) (cf. [4]). In fact, an easy consequence directly  All correspondence should be addressed to Professor Yue–Li Wang, Department

of Information Management, National Taiwan University of Science and Technology, 43, Section 4, Kee-Lung Road, Taipei, Taiwan, ROC (Phone: 886–2–27376768, e-mail: [email protected]).

28

A.-H. Chen et al.

obtained from Landau’s condition is that a score vector S is strong if and only if every tournament T ∈ T (S) is strong. Let Sn denote the set consisting of all strong score vectors of length n. For S = (s1 , s2 , . . . , sn ) and S  = (s1 , s2 , . . . , sn ) in Sn , define S  S  if and only if k k  i=1 si  i=1 si for all 1  k  n. It is easy to verify that (Sn , ) is a partially ordered set. Brualdi and Li [1] proved that (Sn , ) has a minimum element denoted by Sˆn = (1, 1, 2, 3, . . . , n − 3, n − 2, n − 2) (in particular, Sˆ3 = (1, 1, 1) and Sˆ4 = (1, 1, 2, 2)) and |T ( Sˆn )| = 2n−2 . They further conjectured that |T (S)| > 2n−2 for every other score vector S ∈ Sn \{ Sˆn }. Later on, the conjecture had been confirmed by Gibson in [3]. A cycle of length k in a digraph is called a k-cycle. A Δ-interchange (also called Δ-reversal [8]) is a transformation which reverses the orientations of the arcs in a 3-cycle of a digraph. For a given score vector S, an undirected graph G(S) is called the interchange graph of S if its vertices are the tournaments in T (S) and an edge joining tournaments T, T  ∈ T (S) provided T  can be obtained from T by a Δ-interchange. Moon [9] pointed that a tournament Tn with score vector S = (s1 , s2 , . . . , sn )  out n si  3-cycles, and it follows that G(S) is a regular graph has exactly k = n3 − i=1 2 of degree k. Brualdi and Li [1] also showed that such a graph G(S) is connected (i.e., every pair of tournaments in T (S) can be transformed to each other by a finite sequence of Δ-interchanges.) In particular, if S is a strong score vector, then G(S) has the following interesting properties. Theorem 1 [1]. Suppose S ∈ Sn with n  3. Then (a) G(S) is 2-connected; (b) G(S) is bipartite; (c) G(S) has diameter greater than or equal to n − 2. Till now, no more discussions about the properties of interchange graphs of tournaments appeared in the literature except a study on the topic related to Markov chains in [5] and [8]. In this paper, we shall show that {G( Sˆn ) : n  3} forms a class of graphs called hypercubes. Hypercubes are one of the most popular, versatile and efficient topological structures of interconnection networks [10]. The m-dimensional hypercube Q m is a graph with 2m vertices in which each vertex corresponds to an m-tuple (b1 , b2 , . . . , bm ) on the set {0, 1}m and two vertices are linked by an edge if and only if they differ in exactly one coordinate. It is well-known that Q m is an m-regular m-connected bipartite graph with diameter m. Intuitively, all properties stated in Theorem 1 are conformable to those of Q m . Accordingly, we shall prove that G( Sˆn ) is isomorphic to Q n−2 . 2. Preliminaries From now on, for any m-tuple b ∈ {0, 1}m , we simply write b = b1 b2 · · · bm instead of b = (b1 , b2 , . . . , bm ), and for convenience we say that b is a binary string. Let Tn be any tournament with n  3. If three vertices pi , p j and pk with i < j < k

The Interchange Graphs of Tournaments & Hypercubes

29

are contained in a 3-cycle of Tn , then there are two possible distinct cycles. One is pi → p j → pk → pi and the other is its reversal. The former, denoted by i, j, k, is called the type-0 triplet and the latter, denoted by i, k, j, is called the type-1 triplet since they can be viewed as even permutation and odd permutation, respectively. Let C(T ) be the set consisting of all 3-cycles in a tournament T . We now define the following lexicographical order on C(T ). For any two 3-cycles with representation a, b, c and d, e, f  in C(T ), define by a, b, c  d, e, f  if (i) a < d, or (ii) a = d and b < e, or (iii) a = d, b = e and c ≤ f . Suppose |C(T )| = r . To represent the list of all 3-cycles of T , we use an r -tuple c1 c2 · · · cr where each ci ∈ {0, 1} indicates that the i-th cycle in the order is either type-0 or type-1 triplet. For example, consider three tournaments of size 5 with the same score vector (1, 1, 2, 3, 3) as shown in Fig. 1. The tournament T51 contains three 3-cycles 1, 2, 3, 2, 3, 5 and 3, 5, 4, where 1, 2, 3 and 2, 3, 5 are type-0 triplets, and 3, 5, 4 is a type-1 triplet. Obviously, 1, 2, 3  2, 3, 5  3, 5, 4. Thus, the list of cycles for T51 is 001. The reader can easily verify the lists of cycles for T52 and T53 . The following characterization implies that the interchange graph G( Sˆn ) is (n − 2)-regular. Lemma 1. If T ∈ T ( Sˆn ) for n  3, then |C(T )| = n − 2. that every tournament with score vector (s1 , s2 , . . . , sn ) has exactly Proof. Recall n si  n ˆ − i=1 2 3-cycles and Sn = (1, 1, 2, 3,· · · , n − 3, n − 2, n − 2). Thus, 3        n n−2 n−2 |C(T )| = − 0 + 0 + 1 + 3 + 6 + ··· + + = n − 2. 3 2 2  We will show that for the other strong score vectors, there are more than n − 2 3-cycles. To show the result, we rely on the following lemma proved in [2, p. 174] (see also Lov´asz [7, Problem 6.13] for a stronger form). Lemma 2 [2]. If T is a strong tournament of order n  4, then there exists a vertex v of T such that T \{v} is a strong tournament.

p3

p1

p4

p3 p2

p5 〈1,2,3〉 〈2,3,5〉 〈3,5,4〉 =001

p3

p1

p4

p2 p5

〈1,2,3〉 〈2,3,4〉 〈3,4,5〉 =000

p1

p4

p2 p5

〈1,2,5〉 〈2,5,3〉 〈2,5,4〉 =011

Fig. 1. The lists of all 3-cycles in distinct tournaments with the same score vector (1, 1, 2, 3, 3)

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A.-H. Chen et al.

It is well-known that S3 = {(1, 1, 1)} and S4 = {(1, 1, 2, 2)}. Tournaments realizing these two particular score vectors have one 3-cycle and two 3-cycles, respectively. For S5 , the strong score vectors are (2, 2, 2, 2, 2), (1, 2, 2, 2, 3) and (1, 1, 2, 3, 3), and we can easily verify that the numbers of 3-cycles in each tournament for these score vectors are 5, 4 and 3, respectively. In general, we can prove the following. Lemma 3. Suppose Sn ∈ Sn \{ Sˆn } for n  5 and let Tn ∈ T (Sn ) and Tˆn ∈ T ( Sˆn ). Then |C(Tn )| > |C(Tˆn )|. Proof. The proof is by induction on n. For the base case n = 5, the statement is clear. We now suppose that n  6 and the lemma holds for Sn−1 . Let Tn be an arbitrary tournament realizing Sn = (s1 , s2 , . . . , sn ). By Lemma 2, we may let Tn−1 = Tn − v for some vertex v such that Tn−1 remains strong and has score vector Sn−1 = (s1 , s2 , . . . , sn−1 ). Denote Sˆn = (ˆs1 , sˆ2 , . . . , sˆn ) and Sˆn−1 = (ˆs1 , sˆ2 , . . . , sˆn−1 ). Let / T ( Sˆn−1 ) then Tˆn−1 ∈ T ( Sˆn−1 ). By induction hypothesis, if Tn−1 ∈ 0 < |C(Tn−1 )| − |C(Tˆn−1 )|       n−1   n−1   si sˆi n−1 n−1 − − = − 2 3 2 3 =

n−1    sˆi i=1

2

i=1

 1 1  2 1 sˆi − si2 + si − sˆi 2 2 2 n−1

=

i=1

=

i=1

  si − 2

1 2

n−1



n−1

n−1

i=1

i=1



sˆi2 − si2 .

i=1

Let Tˆn ∈ T ( Sˆn ). We claim 1  2 1  2 |C(Tn )| − |C(Tˆn )| = (ˆsi ) − (si ) > 0. 2 2 n

n

i=1

i=1

The inequality holds even if the second term is maximum among all possible valn ues of i=1 (si )2 . For convenience, two vertices pi , p j ∈ Tn \{v} with i < j are called an inversion pair of Tn with respect to v if pi → v and v → p j . Since (si )2 + (s j )2  (si − 1)2 + (s j + 1)2 when i < j, reversing the two arcs in an invern sion pair does not obtain a smaller value of i=1 (si )2 . We consider the following two cases. / T ( Sˆn−1 ). Since Tn is strong, we assume that there are k vertices, Case 1: Tn−1 ∈ that the k vertices are 1  k  n − 2, in Tn−1 that beat v. We may further nassume (si )2 is as large as possible. pn−k , pn−k+1 , . . . , pn−1 since in this case the sum i=1

The Interchange Graphs of Tournaments & Hypercubes

31

Thus, n 

(si )2 

n−1−k 

i=1

n−1 

n−1 

si2 + 2

i=1

(si + 1)2 + (n − 1 − k)2

i=n−k

i=1

=

n−1 

si2 +

si + k + (n − 1 − k)2 .

(2)

i=n−k

Also, since Sˆn = (1, 1, 2, 3,· · · , n − 3, n − 2, n − 2), we have n n−1 n−1    (ˆsi )2 = sˆi2 − (n − 3)2 + 2(n − 2)2 = sˆi2 + n 2 − 2n − 1. i=1

i=1

(3)

i=1

Subtracting (2) from (3) gives n n   (ˆsi )2 − (si )2 i=1



i=1

n−1 

sˆi2

2

+ n − 2n − 1 −

i=1

=

n−1



si2

n−1 

+2

2

si + k + (n − 1 − k)

i=n−k

i=1

n−1 n−k−1 

   sˆi2 − si2 − 2 si + 2 si − k 2 − (2n − 3)k + 2

i=1



n−1 

i=1

i=1

    

n−1

   n−1 n−k−1 sˆi2 − si2 − 2 +2 + 1 − k 2 − (2n − 3)k + 2 2 2 i=1

=

n−1



 sˆi2 − si2 > 0.

i=1

Case 2: Tn−1 ∈ T ( Sˆn−1 ). In this case, if there exists an integer k with 1  k  n−2 such that v → pi for all 1  i  k and p j → v for all k + 1  j  n − 1, then / T ( Sˆn ), Tn contains at least one inversion pair.  Moreover, Tn ∈ T ( Sˆn ). Since Tn ∈ n (si )2 , since reversing the arcs of an inversion pair does not decrease the value of i=1 we assume that Tn contains only one inversion pair occurring at pk and pk+1 . Thus, n k−1 n−1    2 (si )2  si2 + (sk + 1)2 + sk+1 + (si + 1)2 + k 2 i=1

i=1

=

n−1  i=1

i=k+2

si2 + 2sk + 1 + 2

n−1  i=k+2

si + (n − k − 2) + k 2

32

A.-H. Chen et al.

=

n−1 

sˆi2

+ 2(k − 1) + 1 + 2

i=1

=

n−1 

n−2 

(i − 1) + 2(n − 3) + (n − k − 2) + k 2

i=k+2

sˆi2 + (n 2 − 2n − 1) − 2.

i=1

Therefore, the result

n

si )2 i=1 (ˆ

−

n

 2 i=1 (si )

 2 directly follows from (3).



Combining Lemma 1 and Lemma 3 gives the following theorem. Theorem 2. For n  3, T ∈ T ( Sˆn ) if and only if |C(T )| = n − 2. 3. Main Result We are now in a position to state our main result and show its validity. We claim that there is a one-to-one correspondence between T ( Sˆn ) and {0, 1}n−2 . Define f : T ( Sˆn ) → {0, 1}n−2 by f (T ) = b1 b2 · · · bn−2 , where T is a tournament realizing Sˆn and b1 b2 · · · bn−2 is the list of types of triplets corresponding to the cycles in (C(T ), ). Based on |T ( Sˆn )| = 2n−2 = |{0, 1}n−2 |, we show that f is onto to establish the bijection. Actually, we provide the inverse function of f as follows. Let g : {0, 1}n−2 → T (Sn ) be a function, where Sn is a score vector of length n. Given a binary string b ∈ {0, 1}n−2 , we add a “0” at the beginning and at the end of b, and call the resulting string the extended binary string of b, denoted by  b = b1 b2 · · · bn . Regarding the n symbols of  b as vertices (i.e., each vertex pi is associated with a symbol bi ), we define a tournament on them by the following rule: the arcs going to the right are between vertices with symbol “0” such that there is no vertex with symbol “0” between them, and all other arcs go to the left. From the above construction, it is easy to see directly that g(b) forms a tournament of n vertices. Moreover, we observe that if the number of arcs going to the right in the construction is k, then the number of 3-cycles with type-0 triplet in g(b) is k − 1. Also, every vertex with symbol “1” together with the nearest vertices with symbol “0” in the front and rear form a 3-cycles with type-1 triplet in g(b). For example, consider the case n = 5 and b = 011. Since  b = 00110, we can easily check that p1 → p2 and p2 → p5 are arcs going to the right, and all remaining arcs go to the left. As a result, the tournament g(011) is T53 as shown in Fig. 1. Lemma 4. Let b ∈ {0, 1}n−2 be a binary string for n  3 and suppose that T is the tournament constructed from g(b). Then |C(T )| = n − 2. Proof. The proof is by induction on n. For n = 3, g(0) and g(1) are tournaments representing by the 3-cycles p1 → p2 → p3 → p1 and p1 → p3 → p2 → p1 , respectively. We now consider n  4 and suppose that the lemma holds for any binary string of length no more than n − 3. Let b be any binary string of length n − 2

The Interchange Graphs of Tournaments & Hypercubes

33

and  b = b1 b2 · · · bn be the extended binary string of b. Recall that b1 = bn = 0 in  b. Let b be the substring of b consisting of n − 3 symbols in the front and denote b as its extended binary string. Let T = g(b) and T  = g(b ). By induction hypothesis, |C(T  )| = n − 3. Define j = max{i | bi = 0 and 1  i  n − 2}. Clearly, p j → pn−1 in T  . The number of 3-cycles in T can be calculated as follows: b. In this case, the last two symbols of  b are set to 0. MoreCase 1: bn−1 = 0 in  over, the preceding n − 1 symbols of  b are the same as those of b . Thus, all 3-cycles in T  are preserved in T . We now consider arcs incident with pn in T . According to our constructing rule, pn−1 → pn and no more arcs can go to the right. Thus, the unique 3-cycle with type-0 triplet  j, n − 1, n is newly created in the induction step. By the same argument, we easily see that no more 3-cycles with type-1 triplet in T can be produced. Therefore, |C(T )| = |C(T  )| + 1. b. In this case, the preceding n − 2 symbols of  b are the same Case 2: bn−1 = 1 in    b and in b are the same, as those of b . Since the numbers of arcs going to the right in  no more 3-cycles with type-0 triplet in T can be produced in the induction step. On the other hand, since  b can be viewed as a string obtained from b by inserting a symbol “1” between b j and bn−1 which are set to 0 in b , the unique 3-cycle with type-1 triplet  j, n, n − 1 is newly created in T . Since no other 3-cycles of T  are  disturbed in T , we have |C(T )| = |C(T  )| + 1. Theorem 3. The function f : T ( Sˆn ) → {0, 1}n−2 defined above is a bijection. Proof. As mentioned before, we only need to prove that f is a surjection because |T ( Sˆn )| = 2n−2 = |{0, 1}n−2 |. For any binary string b of length n − 2, let T be the tournament constructed from g(b). By Lemma 4, T has exactly n − 2 3-cycles. Thus,  T realizing Sˆn directly follows from Theorem 2 and g = f −1 . Moreover, from the above transformation, we can see that for any two tournaments T, T  ∈ G( Sˆn ), T  can be obtained from T by a Δ-interchange if and only if the two corresponding binary strings f (T ) and f (T  ) differ in exactly one bit. As a result, we have the following corollary. Corollary 1. The graph G( Sˆn ) is isomorphic to Q n−2 . Acknowledgements. This research was partially supported by National Science Council under the Grants NSC95–2221–E–260–025 and NSC95–2115–M–141–001–MY2. The authors would like to thank anonymous referees for their careful reading of the paper and their constructive comments, which have significantly improved the presentation of this article.

References 1. Brualdi, R.A., Li, Q.: The interchange graph of tournaments with the same score vector. In: Bondy, J.A., Murty, U.S.R., (eds.), Progress in Graph Theory. (Academic Press, Toronto, 1984) pp. 128–151 2. Chartrand, G., Zhang, P.: Introduction to Graph Theory. McGraw-Hill, Boston (2005) 3. Gibson, P.M.: A bound for the number of tournaments with specified scores. J. Comb. Theory, Ser. B 36, 240–243 (1984)

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4. Harary, F., Moser, L.: The theory of round robin tournaments. Amer. Math. Monthly 73, 231–246 (1966) 5. Kannan, R., Tetali, P., Vempala, S.: Simple Markov-chain algorithms for generating bipartite graphs and tournaments. Random Struct. Algorithms 14, 293–308 (1999) 6. Landau, H.G.: On dominance relations and the structure of animal societies, III: the condition for a score structure. Bull. Math. Biol. 15, 143–148 (1953) 7. Lov´asz, L.: Combinatorial Problems and Exercises, 2nd edition. North-Holland, Amsterdam, (1993) 8. McShine, L.: Random sampling of labeled tournaments. Electron. J. Comb. 7(1), #R8 (2000) 9. Moon, J. W.: Topics on Tournaments. Holt, Rinehart and Winston, New York (1968) 10. Saad, Y., Schultz, M.H.: Topological properties of hypercubes. IEEE Trans. Comput. 37, 867–872 (1988) Received: January 26, 2007 Final version received: October 19, 2008

Graphs and Combinatorics (2009) 25:35–40 Digital Object Identifier (DOI) 10.1007/s00373-008-0817-5

Graphs and Combinatorics © Springer-Verlag 2009

Multigraphs with Δ

3 are Totally-(2Δ1)-Choosable

Daniel W. Cranston University of Illinois, Urbana-Champaign.

Abstract. The total graph T (G) of a multigraph G has as its vertices the set of edges and vertices of G and has an edge between two vertices if their corresponding elements are either adjacent or incident in G. We show that if G has maximum degree Δ(G), then T (G) is (2Δ(G) − 1)-choosable. We give a linear-time algorithm that produces such a coloring. The best previous general upper bound for Δ(G) > 3 was  23 Δ(G) + 2, by Borodin et al. When Δ(G) = 4, our algorithm gives a better upper bound. When Δ(G) ∈ {3, 5, 6}, our algorithm matches the best known bound. However, because our algorithm is significantly simpler, it runs in linear time (unlike the algorithm of Borodin et al.). Key words. Choosability, List coloring, Total coloring, Multigraph

Throughout this paper, G is a loopless multigraph. For convenience, we refer to edges and vertices as elements of the graph. The total graph T (G) of a graph G has as its vertices the set of edges and vertices of G and has an edge between two vertices if their corresponding elements are either adjacent or incident in G. Let L be an assignment of lists to the vertices of a graph G. If G has a proper coloring such that each vertex v gets a color from its list L(v), then we say that G has an L-coloring. If G always has an L-coloring when each vertex has a list of size k, then we say that G is k-list-colorable (or k-choosable). In this paper, we study the problem of list-coloring a total graph. If a total graph T (G) is k-list-colorable, we say that G is totally-k-list-colorable (or totally-k-choosable). Often, our algorithm will greedily color all but a few edges and vertices of G; we generally call this uncolored subgraph H . This motivates the following definition. For a graph G and a subgraph H , we use G − H to mean G − (V (H ) ∪ E(H )) (thus, edge uv may be present in G − H even if one or both of vertices u and v are missing). We say a graph algorithm runs in linear time if for fixed maximum degree the algorithm runs in time linear in the number of vertices of the graph. Let Δ(G) denote the maximum vertex degree of a graph G. Juvan et al. [4] showed that if Δ(G) = 3, then G is totally-5-choosable. Skulrattanakulchai and Gabow [5] used these ideas to show that if Δ(G) = 3, then we can construct a total-5-listcoloring in linear time. In this paper, we extend these ideas further to show that if Δ(G) ≥ 3, then we can construct a total-(2Δ(G)−1)-list-coloring in linear time. The  Present address: Center for Discrete Mathematics and Theoretical Computer Science

(DIMACS), Rutgers. e-mail: [email protected]

36

D. W. Cranston

best previous upper bound for Δ(G) > 3 was  23 Δ(G) + 2, by Borodin et al. [1]. When Δ(G) = 4, our algorithm gives a new upper bound. When Δ(G) ∈ {3, 5, 6}, our algorithm matches the best known bound. However, because our algorithm is significantly simpler, it runs in linear time (unlike the algorithm of Borodin et al.). In Lemma 1, we show how to greedily construct a total-(2Δ(G)−1)-coloring for almost all of G. The rest of this paper shows that we can extend the coloring to all of G. Lemma 1. Let G be a connected multigraph. If G contains a vertex v with d(v) < Δ(G), or G contains an edge with multiplicity at least 3, then G is totally-(2Δ(G)−1)choosable. Proof. Let k = Δ(G). Suppose that d(v) < k or that vertex v is incident to edge e with multiplicity at least 3. The subdivision graph S(G), is formed by replacing each edge of G with a path of length 2. (The subdivision graph S(G) has the same vertex set as the total graph T (G), but it has fewer edges. If we begin with S(G) and add an edge between each pair of vertices at distance 2, we form T (G).) For each element x ∈ V (G) ∪ E(G), let f (x) be the distance from x to v in S(G). We will color all the elements of G sequentially. Let N (x) be the set of vertices in S(G) that are distance at most 2 from x; these are the elements that can restrict the color of x in T (G). Note that we always have |N (x)| ≤ 2k. Greedily color each element x in decreasing order of f (x). If f (x) ≥ 2, let u and w be the first and second vertices after x on a shortest path from x to w in S(G). Since f (w) < f (u) < f (x), vertices u and w will be uncolored at the time that we color x. Thus at most 2k − 2 colors are restricted from being used on x; hence we have a color available for x. The elements x with f (x) = 1 are edges incident to v. In the case of a multiple edge, the last elements with distance 1 that we color are the copies of this edge, and finally we color v. When at least one edge remains uncolored, at most 2k − 2 restrictions are imposed. For the last edge, the multiplicity or the degree restriction on v implies that at most 2k − 3 restrictions are imposed. For v also, in either case at most 2k − 2 restrictions are imposed, so we can complete the coloring.  For any cycle C, by Lemma 1 we can totally-(2Δ(G) − 1)-color G − E(C). Our plan is to greedily total-color all of G except for a few edges and vertices; we call these uncolored elements H. Lemmas 2 and 3 (from Juvan et al. [4]) show how to extend the coloring to H for various choices of H. For convenience, Juvan et al. define halfedges to be edges with only one endpoint. We use halfedge to describe an edge of H which has only one endpoint in H . A halfedge is colored similarly to an edge; the only difference is that a halfedge has only one endpoint in H , so it has at most Δ(G) − 1 incident edges in H . Lemma 2 ([4]). Let H be a cycle with a halfedge attached to each vertex. If L is a list assignment for H such that ⎧ ⎨ 5, if t is a full edge, |L(t)| ≥ 4, if t is a vertex, ⎩ 2, if t is a halfedge, then H admits an L-total-coloring.

Multigraphs with Δ ≥ 3 are Totally-(2Δ−1)-Choosable

(a)

(b)

37

(c)

Fig. 1. The figures for Lemma 3. (a) A double edge with each endpoint incident to a thick halfedge. (b) A 3-cycle with two vertices incident to thick halfedges and the third vertex incident to a thin halfedge. (c) A 4-cycle with two nonadjacent vertices incident to thick halfedges and the other two vertices incident to thin halfedges.

When G has girth at least 5, we will show (in Lemma 5) that G contains an induced cycle C with a matching incident to the vertices of C. In this case, we will greedily color all the elements of G except for C and the incident matching. By treating the edges of the matching as halfedges, we use Lemma 2 to finish the coloring (we give the details in Theorem 1). The next two lemmas consider the cases when G has girth at most 4. In each case we find a small subgraph H and greedily total-color G − H ; in Lemmas 3 and 4 we show that we are able to extend the coloring to H . In Lemma 3 we refer to thick halfedges and thin halfedges. Both are halfedges as described above; the only difference is that thick halfedges will receive lists of size 3, whereas thin halfedges will receive lists of size 2. Thick halfedges always appear in pairs; they designate nonincident halfedges in H that correspond to incident edges in the larger graph. Lemma 3 ([4]). Let H be isomorphic to one of the multigraphs in Figure 1. If L is a list assignment for H such that ⎧ 5, ⎪ ⎪ ⎪ ⎨ 4, |L(t)| ≥ ⎪ 3, ⎪ ⎪ ⎩ 2,

if t if t if t if t

is a proper edge, is a vertex, is a thick halfedge is a thin halfedge,

then H admits an L-total-coloring such that the two thick halfedges receive distinct colors. Juvan et al. were the first to show that a graph G with Δ(G) = 3 is totally5-choosable. Most of the subgraphs H that we encounter in our proof also arose in their proof (as seen in Lemmas 2 and 3). However, there are a few additional subgraphs that we must consider as we prove our generalization of their result. We handle these subgraphs in the following lemma. Lemma 4. Let H be isomorphic to one of the multigraphs in Figure 2. If L is a list assignment for H such that

38

D. W. Cranston

⎧ ⎨ 5, |L(t)| ≥ 4, ⎩ 2,

if t is a proper edge, if t is a vertex, if t is a halfedge,

(1)

Then H admits an L-total-coloring. Proof. If H is congruent to the multigraph in Figure 2a, then let e1 and e2 be the parallel edges. Let the halfedges be e3 (incident to vertex v1 ) and e4 (incident to vertex v2 ). We may assume that there are equalities in (1); otherwise we discard colors from the lists. Since |L(v1 )| + |L(e4 )| > |L(e1 )|, either v1 and e4 have a common color c available, or there exists a color c ∈ (L(v1 ) ∪ L(e4 )) \ L(e1 ). If color c is available on both v1 and e4 , we use it on both elements; otherwise use color c on the element where it is available (either v1 or e4 ) and color the other element arbitrarily. Note that L(e1 ) has 4 elements other than the colors used on v1 and e4 . Hence, after greedily coloring e3 , v2 , and e2 in order, a color remains available for e1 . If H ∼ = K 4 (as shown in Figure 2b), then greedily color the vertices of H in any order. Note that each edge now has at least 3 colors available (since each edge lost at most one color to each endpoint); let L (e) denote the list of remaining available colors on each edge e after coloring the vertices of H . Suppose that we cannot give distinct colors to the edges. By Hall’s Theorem there exists a set E 1 of edges such that | ∪e∈E 1 L (e)| < |E 1 |. Set E 1 must contain at least 4 edges, since each edge has at least 3 colors available. Among any 4 edges of K 4 there are two nonincident edges, call them e1 and e2 . Since |L(e1 ) ∪ L(e2 )| < |E 1 | ≤ 6, edges e1 and e2 must have a common available color, c. Use color c on edges e1 and e2 . The four remaining uncolored edges form a 4-cycle. Note that each uncolored edge has at least 2 ¨ Rubin, and Taylor [2] proved that cycles of even length are colors available. Erdos, edge-2-choosable, so we can finish the coloring. If H ∼ = K 3,3 (as shown in Figure 3b), then greedily color the vertices of H . Note that each edge now has at least 3 colors available. Galvin [3] proved that a bipartite multigraph H is edge-Δ(H )-choosable. Hence, we can finish the coloring. 

(a)

(b)

(c)

Fig. 2. The figures for Lemma 4. (a) A double edge with each endpoint incident to a thin halfedge. (b) A complete graph on 4 vertices. (c) A complete bipartite graph with each part of size 3.

Our final lemma is structural. We use it to show that every multigraph G contains a subgraph H for which we can extend a total-(2Δ(G) − 1)-list-coloring of G − H to such a coloring of G. For convenience we consider a double-edge to be a cycle of length 2.

Multigraphs with Δ ≥ 3 are Totally-(2Δ−1)-Choosable

39

Lemma 5. If G is a regular multigraph, then G contains an induced cycle C with the following property. If the cycle C contains any pair of vertices with a common neighbor not on C, then |V (C)| ≤ 4. Furthermore, we can find such a cycle in linear time. Proof. If G has a multiple edge, then G has a cycle of size 2. We can find it in linear time. So we assume G is simple. Choose an arbitrary vertex v. Using breadth-first-search, find a shortest cycle D through v. If there exist vertices w and x on D with a common neighbor y not on D, then either w and x are adjacent or w and x have a common neighbor z on D (otherwise we could find a shorter cycle through v). In either case let C be the 3-cycle or 4-cycle containing w, x, and y. If D does not contain such a pair w, x, then let C = D.  By combining Lemmas 1 through 5, we prove our main result. Theorem 1. If G is a multigraph with maximum degree Δ(G), then G is totally(2Δ(G) − 1)-choosable. Furthermore, given lists of size 2Δ(G) − 1, we can construct a coloring in linear time. Proof. If G is disconnected, then we color each component separately. If G is not Δ(G)-regular or if G contains an edge with multiplicity at least 3, then by Lemma 1 we can color G. If G contains an edge uv with multiplicity 2, then let e1 and e2 be additional edges incident to u and v, respectively. We view e1 and e2 as halfedges (thick if they have a common endpoint, and thin otherwise). This subgraph (call it H ) is isomorphic to either Figure 1a or Figure 2a. By Lemma 1, we can greedily color G − E(H ). We will use either Lemma 3 or Lemma 4 to complete the coloring. To apply Lemma 3 or Lemma 4, we must verify that each uncolored vertex, edge, and halfedge has a sufficient number of available colors. We do this as follows. Let k = Δ(G). An uncolored vertex (u or v) is incident to at most k − 3 colored edges and at most k−2 colored vertices. So it has at least (2k−1)−(k−3)−(k−2) = 4 available colors. An uncolored edge (one of the uv edges) is not incident to any colored vertices; each endpoint is incident to at most k −3 colored edges. So an edge has at least (2k − 1) − 2(k − 3) = 5 available colors. A thin halfedge (e1 or e2 ) is incident to a single colored vertex; one endpoint is incident to at most k − 1 colored edges and the other endpoint is incident to at most k − 3 colored edges. So a thin halfedge has at least (2k − 1) − 1 − (k − 1) − (k − 3) = 2 available colors. A thick halfedge has one additional available color, since it is incident to the other thick halfedge, which is uncolored. Hence, we can assume that G is a regular simple graph. Find a cycle C as described in Lemma 5. By Lemma 1, greedily color G − E(C). If |V (C)| > 4, then uncolor the vertices of C and uncolor one edge incident to each vertex of C. We treat each uncolored incident edge as a halfedge, and finish the coloring by Lemma 2. To apply Lemma 2, we must verify that the lists of colors are big enough. However, the analysis is similar to that given above, so we omit it.

40

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Thus, if |V (C)| > 4, then Lemma 2 applies and we can finish the coloring. So we may assume that |V (C)| ≤ 4. Suppose that |V (C)| = 3. Uncolor an edge incident to each vertex of C. If the three uncolored incident edges have a common endpoint, then also uncolor this common endpoint. The uncolored subgraph is isomorphic to K 4 . By Lemma 4, we can finish the coloring. If the uncolored subgraph is not isomorphic to K 4 , then it is isomorphic to a subgraph in Lemma 2 or Lemma 3 (Figure 1b). In each case, we can finish the coloring. To apply Lemma 2, 3, or 4 we must verify that the lists of available colors are big enough. Again, we omit the case analysis. Finally, suppose |V (C)| = 4. If two adjecent vertices of C have a common neighbor not on C, then instead we let C be this 3-cycle and we are in the above case. Otherwise, uncolor an edge incident to each vertex of C. If at most one pair of these incident edges has a common endpoint, then we can finish the coloring by Lemma 2 or Lemma 3 (Figure 1c). If two pairs of these incident edges have common endpoints, call these endpoints u and v. If u and v are adjacent, then we have a subgraph isomorphic to K 3,3 , so by Lemma 4 (Figure 2c) we can finish the coloring. If u and v are not adjacent, then instead replace C with the cycle induced by (V (C) ∪ u)\{w}, where w is a vertex of C but w is not adjacent to u. Because edge uv is not present in G, we can choose one edge incident to each vertex of this new 4-cycle so that at most one pair of incident edges has a common endpoint. Now we are in the previous case, and we can finish by Lemma 2 or Lemma 3 (Figure 1c).  Acknowledgements Thanks to Douglas Woodall for bringing this problem to my attention and for his insights that greatly improved the exposition of this paper. Thanks to Doug West and two referees, whose comments each improved the exposition. Thanks to Kevin Milans for helpful conversation.

References 1. Borodin, O.V., Kostochka, A.V., Woodall, D.R.: List Edge and List Total Colourings of Multigraphs. J. Combin. Theory Ser. B 71, 184–204 (1997) ¨ P., Rubin, A.L., Taylor, H.: Choosability in graphs. Congr. Numer. 26, 125–157 2. Erdos, (1979) 3. Galvin, F.: The list chromatic index of a bipartite multigraph. J. Combin. Theory Ser. B 63, 153–158 (1995) 4. Juvan, M., Mohar, B., Skrekovski, R.: List Total Colourings of Graphs. Combinatorics, Probability and Computing 7, 181–188 (1998) 5. Skulrattanakulchai, S., Gabow, H.N.: Coloring Algorithms on Subcubic Graphs. Internat. J. Found. Comput. Sci. 15 no. 1, 21–40 (2004) Received: November 17, 2006 Final version received: October 26, 2008

Graphs and Combinatorics (2009) 25:41–47 Digital Object Identifier (DOI) 10.1007/s00373-008-0824-6

Graphs and Combinatorics © Springer-Verlag 2009

A New Bound on the Total Domination Subdivision Number O. Favaron1 , H. Karami2 , R. Khoeilar2,∗ , S. M. Sheikholeslami2,† 1 Univ Paris-Sud, LRI, UMR 8623, Orsay, F-91405, France. e-mail: [email protected] 2 Department of Mathematics, Azarbaijan University of Tarbiat Moallem, Tabriz, I.R. Iran.

e-mail: [email protected]

Abstract. A set S of vertices of a graph G = (V, E) without isolated vertex is a total dominating set if every vertex of V (G) is adjacent to some vertex in S. The total domination number γt (G) is the minimum cardinality of a total dominating set of G. The total domination subdivision number sdγt (G) is the minimum number of edges that must be subdivided (each edge in G can be subdivided at most once) in order to increase the total domination number. In this paper we prove that for every simple connected graph G of order n ≥ 3, sdγt (G) ≤ 3 + min{d2 (v); v ∈ V and d(v) ≥ 2} where d2 (v) is the number of vertices of G at distance 2 from v. Key words. Total domination number, Total domination subdivision number.

1. Introduction Let G = (V (G), E(G)) be a simple graph of order n with no isolated vertices. The neighborhood of a vertex u is denoted by N G (u) and its degree |N G (u)| by degG (u). The minimum and maximum degrees of G are respectively denoted by δ(G) and Δ(G) (briefly V, E, N (u), deg(u), δ, Δ when no ambiguity on the graph is possible). The distance between two vertices u and v is the length of a shortest path joining them. We denote by N2 (v) the set of vertices at distance 2 from the vertex v and put d2 (v) = |N2 (v)| and δ2 (G) = min{d2 (v); v ∈ V (G)}. A set S of vertices of G is a total dominating set (TDS for short) if it is a dominating set of G such that the subgraph G[S] has no isolated vertex. The minimum cardinality of a total dominating set, denoted by γt (G), is called the total domination number of G. A γt (G)-set is a total dominating set of G of cardinality γt (G). When an edge uv of G is subdivided by inserting a new vertex x between u and v, the total domination number does not decrease (Observation 1). The total domination subdivision number sdγt (G) is the minimum number of edges of G that must be subdivided in order to increase the total domination number. This concept was first introduced in [10] for domination, and then extended to total domination in [7]. Since the total ∗ Research supported by the Research Office of Azarbaijan University of Tarbiat Moallem. † Corresponding author.

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domination number of the graph K 2 does not change when its only edge is subdivided, in the study of the total domination subdivision number we must assume that one of the components of the sgraph has order at least 3. If G 1 , . . . , G s are the γt (G i ) and if G 1 , . . . , G r are the components components of G, then γt (G) = i=1 of G of order at least 3, then sdγt (G) = min {sdγt (G i ) ; 1 ≤ i ≤ r }. Hence, it is sufficient to study sdγt (G) for connected graphs. The parameter sdγt can take large values, although it cannot exceed 3 for trees. The reader can find in [8] a construction showing that there exist connected graphs of arbitrarily large order n satisfying sdγt (G) > log2 n/3. Therefore an interesting problem is to find good upper bounds on sdγt (G) in terms of the order and possibly of other parameters of G. Some bounds are already known. For instance it has been proved that for any graph G of order n, sdγt (G) ≤ n − γt (G) + 1 [5] and sdγt (G) ≤ 2n/3 [6]. Recently, Karami et al. [9] proved that for any simple connected graph G of order n ≥ 3, sdγt (G) ≤ n − δ + 2

(1)

with equality if and only if G is isomorphic to K n (n ≥ 4). Our purpose in this paper is to improve this bound by considering δ2 and Δ instead of δ. As a corollary of the main result we show that sdγt (G) ≤ n − Δ + 2. We will use the following observations and properties. Observation 1. Let G be simple connected graph G of order n ≥ 3 and S ⊆ E(G). If G  is obtained from G by subdividing the edges of S, then γt (G  ) ≥ γt (G). Observation 2. Let G  be constructed from G by subdividing a set of edges with a set T of subdivising vertices and let D be a γt (G  )-set such that D ∩ T = ∅. Since every vertex of D ∩ T is adjacent in G  with a vertex of D ∩ V (G), the set D \ T is dominating in G. Hence when we claim that D \ T is a TDS of G, we just have to check that in G this set has no isolated vertex. Theorem 1 [8]. For any connected graph G with adjacent vertices u and v, each of degree at least two, sdγt (G) ≤ d(u) + d(v) − |N (u) ∩ N (v)| − 1 = |N (u) ∪ N (v)| − 1. To simplify the writing, we say that a set A is smaller than a set B when |A| < |B|. We use [11] for terminology and notation which are not defined here. 2. The New Upper Bound We make use of the following lemmas in the proof of Theorem 3. A support vertex is a vertex adjacent with a leaf. Lemma 1. Let G be a simple connected graph. If v ∈ V (G) is a support vertex contained in a triangle, then sdγt (G) ≤ 2.

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Proof. Assume N (v) = {v1 , . . . , vdeg(v) }, deg(v1 ) = 1 and v2 v3 ∈ E(G). Let G  be obtained from G by subdividing the edges vv1 and v2 v3 , with vertices x, y, respectively. Let D be a γt (G  )-set. Without loss of generality we may assume x, v ∈ D. In order to totally dominate y, we must have D ∩ {v2 , v3 } = ∅. Then obviously  D \ {x, y} is a TDS for G smaller than D. Hence, sdγt (G) ≤ 2. Lemma 2. Let G be a simple connected graph of order n ≥ 3. If v ∈ V (G) is a support vertex and has a neighbor u with N (u) \ N [v] = ∅, then sdγt (G) ≤ 2 + |N (u) \ N [v]|. Proof. If N (u) ∩ N (v) = ∅, then the result follows by Lemma 1. Assume N (v) ∩ N (u) = ∅. Let N (v) = {u = v1 , v2 , . . . , vdeg(v) } where deg(v2 ) = 1, and N (u) \ N [v] = {y1 , y2 , . . . , yk }. Let G  be obtained from G by subdividing the edge vvi with a vertex u i for i = 1, 2, and the edge uy j with a vertex z j for 1 ≤ i ≤ k. Let Z be the set of the k + 2 subdividing vertices and let D be a γt (G  )-set. Without loss of generality we may assume v, u 2 ∈ D. If u ∈ D, then obviously D \ Z is a γt (G)-set. Let u ∈ D. In order to dominate u, we must have D ∩ (Z \ {u 2 }) = ∅. Then (D \ Z ) ∪ {u} is a TDS of G smaller than D. Thus sdγt (G) ≤ 2 + |N (u) \ N [v]| and the proof is complete.  Lemma 3. Let G be a simple connected graph of order n ≥ 3. If G has a vertex v ∈ V (G) which is contained in a triangle vuw such that N (u) ∪ N (w) ⊆ N [v], then sdγt (G) ≤ 3. Proof. Let G  be obtained from G by subdividing the edges vu, vw, uw with vertices x, y, z, respectively. Let D be a γt (G  )-set. In order to totally dominate x, y and z, the set D must contain at least one endvertex of each edge uv, vw and wu. Hence |D ∩ {u, v, w}| ≥ 2. By Observation 2, if D ∩ {x, y, z} = ∅ then D \ {x, y, z} is a TDS of G smaller than D. Suppose now D ∩ {x, y, z} = ∅. To totally dominate v, D contains a vertex in N G (v) \ {u, w}. If v ∈ D then, since D ∩ {u, w} = ∅, the set D \ {u, w} is a TDS of G smaller than D. If v ∈ / D, then {u, w} ⊆ D and (D \ {u, w}) ∪ {v} is a TDS of G  smaller than D. Thus in all cases |D| > γt (G) and the proof is complete. Lemma 4. Let G be a simple connected graph of order n ≥ 3. If G has a vertex v ∈ V (G) which is contained in a triangle vuw such that N (u) ⊆ N [v] and N (w) \ N [v] = ∅, then sdγt (G) ≤ 3 + |N (w) \ N [v]|. Proof. Let N (w) \ N [v] = {w1 , . . . , wk } and let G  be obtained from G by subdividing the edges vu, vw, uw with vertices x, y, z, respectively, and for each i between 1 and k, the edge wwi with the vertex z i . We put T = {z 1 , z 2 , . . . , z k } and also consider the graph G  obtained from G by only inserting the vertices of T . By Observation 1, γt (G  ) ≥ γt (G  ) ≥ γt (G). Let D be a γt (G  )-set. As in Lemma 3, |D ∩ {u, v, w}| ≥ 2 and if D ∩ {x, y, z} = ∅ then D \ {x, y, z} is a TDS of G  smaller than D. We now suppose that D ∩ {x, y, z} = ∅.

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If {v, u} ⊆ D then, since v is not isolated in G[D \ {u}] and N G (u) ⊆ N G [v], D \ {u} is a TDS of G  smaller than D. If {v, u}  D, then {v, w} ⊆ D or {u, w} ⊆ D. If D ∩ T = ∅, then D \ T is a TDS of G smaller than D. Suppose now D ∩ T = ∅. Then the set D \ {w} if {v, w} ⊆ D, or (D \ {u, w}) ∪ {v} if {u, w} ⊆ D, is a TDS of G smaller than D.  In all cases we showed that γt (G  ) > γt (G), which completes the proof. Lemma 5. Let G be a simple connected graph of order n ≥ 3 and v a vertex of degree at least 2 of G such that (i) N (y) \ N [v] = ∅ for each y ∈ N (v), (ii) there exists a pair α, β of vertices in N (v) such that (N (α) ∩ N (β)) \ N [v] = ∅. Then sdγt (G) ≤ 3 + |N2 (v)|. Proof. Let v1 , v2 be any pair of adjacent vertices of N (v) satisfying (ii) if such a pair exists, otherwise any pair of vertices of N (v) satisfying (ii). Let S = {v1 , v2 , . . . , vk } be a largest subset of N (v) containing v1 , v2 and such that every pair α, β of verk N (v )) \ N [v]. We put N (v ) \ N [v] = tices of S satisfies (ii), and let K = (∪i=1 i i {vi1 , vi2 , . . . , vii } for 1 ≤ i ≤ k. Let G  be obtained from G by subdividing the edges vv1 and vv2 with respectively z 1 and z 2 , and for each i between 1 and k, the edges vi vi j , 1 ≤ j ≤ i , with a set Ti of i vertices. We put T = ∪1≤i≤k Ti . When v1 and v2 are adjacent, we also subdivide the edge v1 v2 with a vertex u. Let D be a γt (G  )-set. In what follows, the expression D \ {u} means D when u does not exists, / E(G). that is when v1 v2 ∈ Case 1. v ∈ / D. Then, to totally dominate z 1 and z 2 , D contains v1 and v2 . Let G 1 be obtained from G by only inserting the vertices of T3 ∪ · · · ∪ Tk (when k = 2 then G 1 = G). If D ∩ (T1 ∪ T2 ) ≥ 2, then (D \ (T1 ∪ T2 ∪ {z 1 , z 2 , u})) ∪ {v} is a TDS of G 1 smaller than D. If |D ∩ (T1 ∪ T2 )| ≤ 1, let without loss of generality D ∩ T1 = ∅. If (T2 ∪ {u, z 1 , z 2 }) ∩ D = ∅, then (D \ (T2 ∪ {z 1 , z 2 , u, v1 })) ∪ {v} is a TDS of G 1 smaller than D. If (T2 ∪ {u, z 1 , z 2 }) ∩ D = ∅, then v has a neighbor in D \ {v1 , v2 } and (D \ {v1 , v2 }) ∪ {v} is a TDS of G 1 smaller than D. Case 2. v ∈ D. We consider two subcases. Subcase 2.1. D ∩ (N (v) \ S) = ∅. Let y ∈ D ∩ (N (v) \ S). If D ∩ S = ∅, let z ∈ D ∩ S (the vertex z may be one of v1 , v2 or not). Let Z be the set of vertices subdividing the edges of G incident with z and let G 2 be the graph obtained from G by inserting all the subdivision vertices used to construct G  except those of Z . If D ∩ Z = ∅, then D \ {z} is a TDS of G 2 smaller than D. If D ∩ Z = ∅, then D \ Z is a TDS of G 2 smaller than D.

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If D ∩ S = ∅ (in this case, v1 v2 ∈ / E(G) and u does not exists), D must contain the set K to totally dominate all the vertices of T in G  . Moreover, by the definition of S, every vertex of N (v) \ S has a neighbor in K . Let X = D ∩ T and X  = N G  (X ) ∩ S. Then |X  | ≤ |X | and (D \ (X ∪ {z 1 , z 2 , v})) ∪ X  is a TDS of G smaller than D. Subcase 2.2. D ∩ (N (v) \ S) = ∅. If {z 1 , z 2 } ⊆ D, then (D \ {z 1 , z 2 , u}) ∪ {v1 } is a TDS of G 3 smaller than D, where the graph G 3 is obtained from G by only inserting the set T of subdivision vertices. If |D ∩ {z 1 , z 2 }| = 1, let z 1 ∈ D and z 2 ∈ D. If D ∩ S = ∅, then D \ {z 1 , u} is a TDS of the graph G 3 . If D ∩ S = ∅ (in which case u does not exist), then K ⊆ D and every vertex of N (v) \ S has a neighbor in K . As in Subcase 2.1, let X = D ∩ T and X  = N G  (X ) ∩ S. Then |X  | ≤ |X | and (D \ (X ∪ {z 1 , v})) ∪ X  ∪ {v1 } is a TDS of G smaller than D. If D ∩ {z 1 , z 2 } = ∅, then v has a neighbor in D ∩ (S \ {v1 , v2 }), say v3 ∈ D. If D ∩ T3 = ∅ then D \ T3 is a TDS of G 4 smaller than D, where G 4 is obtained from G by inserting the vertices of (T \ T3 ) ∪ {z 1 , z 2 , u}. We now assume D ∩ T3 = ∅. If D ∩ {v1 , v2 } = ∅, let v1 ∈ D. If D ∩ (T1 ∪ {u}) = ∅, then D \ (T1 ∪ {u}) is a TDS of G 5 smaller than D, where the graph G 5 is obtained from G by inserting the vertices of T \ T1 . If D ∩ (T1 ∪ {u}) = ∅, then D \ {v1 } is a TDS of G 5 . Finally let / E(G), for otherwise u could not be dominated. By D ∩ {v1 , v2 } = ∅. Then v1 v2 ∈ the choice of the pair v1 , v2 , the set S is independent. In order to totally dominate v1 , D ∩ T1 = ∅. Let G 6 be obtained from G by inserting the subdividing vertices of the set T \ (T1 ∪ T3 ). Since D ∩ T3 = ∅, the set (D \ (T1 ∪ {v3 })) ∪ {v1 } is a TDS of G 6 smaller than D. In each case, we found a graph G i constructed from G by inserting a subset of the subdivision vertices used to construct G  and such that γt (G i ) < γt (G  ). By Observation 1, γt (G  ) > γt (G i ) ≥ γt (G). Since G  was obtained by inserting at most 3 + |T | ≤ 3 + |N2 (v)| vertices, sdγt (G) ≤ 3 + |N2 (v)|.  Lemma 6. Let G be a simple connected graph of order n ≥ 3 and v a vertex of degree at least 2 of G such that (i) N (y) \ N [v] = ∅ for each y ∈ N (v), (ii) (N (α) ∩ N (β)) \ N [v] = ∅ for every pair α, β of vertices in N (v). Then sdγt (G) ≤ 3 + |N2 (v)|. Proof. Let N (v) = {v1 , . . . , vk } and M = N (v1 ) \ N [v] = {w1 , . . . , w p }. It follows from the hypothesis that each y ∈ N (v)\{v1 } has a neighbor in M. Let T be a largest subset of N (v)\{v1 } such that for each subset T1 ⊆ T , |N (T1 )\(N [v]∪ M)| ≥ |T1 |. By the definition of T , |N2 (v)| ≥ |M|+|T | and for every vertex u in U = N (v)\(T ∪{v1 }), N (u) \ N [v] ⊆ M ∪ N (T ). Moreover M dominates N (v) by (ii).

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Case 1. |U | ≤ 1. By Theorem 1, sdγt (G) ≤ |N (v) ∪ N (v1 )| − 1 = |T | + |U | + 1 + |M| + 1 − 1 ≤ |U | + 1 + |N2 (v)| ≤ |N2 (v)| + 2. Case 2. |U | ≥ 2. Let T = ∅ or, without loss of generality, T = {v2 , . . . , vs }. Let G  be obtained from G by subdividing the |M| + |T | + 3 edges v1 w j for 1 ≤ j ≤ k and vvi for 1 ≤ i ≤ s + 2 (1 ≤ i ≤ 3 if T = ∅). Let Z 1 be the set of the vertices subdividing the edges v1 w j for 1 ≤ j ≤ k, Z 2 the set of the vertices subdividing the edges vvi for 1 ≤ i ≤ s + 2 and Z = Z 1 ∪ Z 2 . Let D be a γt (G  )-set. First assume that v ∈ D. Then {v1 , . . . , vs+2 } ⊆ D to totally dominate the vertices of Z 2 and since N ({vs+1 , vs+2 })\ N [v] ⊆ M ∪ N (T ), (D\(Z ∪{vs+1 , vs+2 }))∪ {v} is a TDS of G smaller than D. / D. Then M ⊆ D. If D ∩ Z 2 = ∅, then Now assume that v ∈ D and v1 ∈ (D \ (Z 2 ∪ Z 1 ∪ {v})) ∪ {v1 } is a TDS of G smaller than D. If D ∩ Z 2 = ∅, then D contains a vertex v  ∈ U \ {vs+1 , vs+2 } (note that v  has a neighbor in M). If D ∩ Z 1 = ∅, then (D \ (Z 1 ∪{v})) ∪{v1 } is a TDS of G smaller than D. If D ∩ Z 1 = ∅, then D \ {v} is a TDS of G smaller than D. Finally assume that v ∈ D and v1 ∈ D. If D ∩ Z = ∅, then D \ Z is a TDS of G smaller than D. If D ∩ Z = ∅, then D contains a neighbor of v belonging to U \ {vs+1 , vs+2 } and D \ {v1 } is a TDS of G smaller than D. Therefore γt (G) < γt (G  ) and sdγt (G) ≤ |M| + |T | + 3 ≤ |N2 (v)| + 3.  We are now ready to prove our main result. Theorem 3. Let G be a simple connected graph of order n ≥ 3. Then sdγt (G) ≤ 3 + min{d2 (v) ; v ∈ V and d(v) ≥ 2}. Proof. If G is a star K 1,n−1 then sdγt (G) = 2. Otherwise, let v be a vertex of degree at least 2 of G such that d2 (v) is minimum. The result is a consequence of Lemmas 1 and 2 if v is a support vertex, of Lemmas 3 and 4 if some neighbor u of v different from a leaf satisfies N (u) ⊆ N [v], and of Lemmas 5 and 6 if N (y) ∩ N [v] = ∅ for every y ∈ N (v).  Corollary 1. Let G be a connected graph of minimum degree at least 2. Then sd γt (G) ≤ δ2 (G) + 3. For a vertex v of degree Δ, |N2 (v)| ≤ n − Δ − 1. Therefore the following improvement of the bound (1) is an immediate corollary of Theorem 3. Corollary 2. Let G be a connected graph of order n ≥ 3. Then sdγt (G) ≤ n − Δ + 2.

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The complete graphs K n (n ≥ 4), which form the extremal class for (1), are obviously also extremal for Theorem 3 and Corollary 2. Another example of graphs which are extremal for Theorem 3, Corollary 2 and Corollary 1, although not for (1), is given by the complete split graphs K m ∨ K n−m for 1 ≤ m ≤ n − 3. As shown in [9], these graphs form the class of graphs such that γt (G) = 2 and sdγt (G) = 3. One can wonder whether the new bound is also an improvement for the two other bounds cited in the introduction. When G is a cycle Cn with n = 12k, then 2n/3 = 8k, n − γt (G) = 6k and δ2 (G) + 3 = 5. Therefore for cycles, the bound of ˝ Theorem 3 is much better than the other ones. This is not always the case. Erdos and Rényi [3] constructed with the the aid of projective planes an infinite family √ of graphs with diameter 2 and Δ asymptotically equal to n. For these graphs, δ2 + 3 = n − Δ + 2 > 2n/3. To conclude the paper, let us mention the following conjecture proposed in [4] and established in some classes of graphs. Conjecture 1. For any connected graph of order n ≥ 3, sdγt (G) ≤ γt (G) + 1. Since by Theorem 2 in [2], γt (G) ≤ n − Δ + 1 if G is connected of order at least 3, this conjecture, if true, would imply Corollary 2. References 1. Bhattacharya, A., Vijayakumar, G.R.: Effect of edge-subdivision on vertex-domination in a graph. Discuss. Math. Graph Theory 22, 335–347 (2002) 2. Cockayne, E.J., Dawes, R.M., Hedetniemi, S.T.: Total domination in graphs. Networks 10, 211–219 (1980) ˝ P., Rényi, A.: On a problem in the theory of graphs. Magyar Tud. Akad. Mat. 3. Erdos, ˝ 7, 623–641 (1962) Kutató Int. Kozl. 4. Favaron, O., Karami, H., Khoeilar, R., Sheikholeslami, S.M.: On the total subdivision number in some classes of graphs, J. Comb. Optim. (to appear) 5. Favaron, O., Karami, H., Sheikholeslami, S.M.: Total domination and total domination subdivision numbers. Australas. J. Combin. 38, 229–235, (2007) 6. Favaron, O., Karami, H., Sheikholeslami, S.M.: Bounding the total domination subdivision number of a graph in terms of its order (submitted) 7. Haynes, T.W., Hedetniemi, S.T., van der Merwe, L.C.: Total domination subdivision numbers. J. Combin. Math. Combin. Comput. 44, 115–128 (2003) 8. Haynes, T.W., Henning, M.A., Hopkins, L.S.: Total domination subdivision numbers of graphs, Discuss. Math. Graph Theory 24, 457–467, (2004) 9. Karami, H., Khodkar, A., Sheikholeslami, S.M.: An upper bound for total domination subdivision numbers of graphs. Ars Combin. (to appear) 10. Velammal, S.: Studies in Graph Theory: Covering, Independence, Dominationand Related Topics, Ph.D. Thesis (Manonmaniam Sundaranar University, Tirunelveli, 1997) 11. West, D.B.: Introduction to Graph Theory (Prentice-Hall, Inc, 2000) Received: January 28, 2008 Final version received: October 16, 2008

Graphs and Combinatorics (2009) 25:49–64 Digital Object Identifier (DOI) 10.1007/s00373-008-0819-3

Graphs and Combinatorics © Springer-Verlag 2009

Further 6-sparse Steiner Triple Systems A. D. Forbes, M. J. Grannell, T. S. Griggs Department of Mathematics, The Open University, Walton Hall, Milton Keynes, MK7 6AA, UK. e-mail: [email protected], [email protected], [email protected]

Abstract. We give a construction that produces 6-sparse Steiner triple systems of order v for all sufficiently large v of the form 3 p, p prime and p ≡ 3 (mod 4). We also give a complete list of all 429 6-sparse systems with v < 10000 produced by this construction. Key words. Steiner triple system, 6-sparse, Pasch configuration, Mitre configuration, Crown configuration.

1. Introduction A Steiner triple system of order v, STS(v), is a pair (V, B) where V is a set of cardinality v of elements, or points, and B is a collection of triples, also called blocks, which has the property that every pair of distinct elements of V occurs in precisely one triple. It is well known that an STS(v) exists if and only if v ≡ 1 or 3 (mod 6). Such values are called admissible. For any two points a and b in an STS(v), (V, B), we define the cycle graph G a,b as follows. The vertex set of G a,b is V \{a, b, a ∗ b}, where we denote by x ∗ y the third point in a block containing the pair {x, y}. The edge set of G a,b is the set of pairs {x, y} such that either {x, y, a} is a block or {x, y, b} is a block. Clearly, G a,b is a set of disjoint cycles {Cn 1 , Cn 2 , . . . , Cnr }, where n 1 + n 2 + · · · + nr = v − 3 and for i = 1, 2, . . . , r , n i is even and n i ≥ 4. A configuration in the context of a Steiner triple system is a set of triples, also called blocks, which has the property that every pair of distinct elements occurs in at most one triple. If C is a configuration, we denote by P(C) its set of points. Two configurations C and D are said to be isomorphic, in symbols C ∼ = D, if there exists a bijection φ : P(C) → P(D) such that for each triple T ∈ C, φ(T ) is a triple in D. For a Steiner triple system (V, B), the set B itself may be regarded as a configuration with P(B) = V . The degree of a point in a configuration is the number of blocks of the configuration which contain that point. We sometimes write blocks with set brackets and commas omitted, so that for example {0, 1, 3} might be written as 013. In this paper we will be concerned with configurations having n blocks and n + 2 points. Such configurations are of particular interest because of the following result proved in [4].

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A. D. Forbes et al. Table 1. Configurations having n blocks and n + 2 points, 4 ≤ n ≤ 6.

n 4 5 5 6 6 6 6 6

Name Pasch mitre 6-cycle crown

Blocks 012, 034, 135, 245 012, 034, 135, 236, 456 012, 034, 135, 245, 056 012, 034, 135, 246, 257, 367 012, 034, 135, 236, 147, 567 012, 034, 135, 236, 146, 057 012, 034, 135, 236, 146, 247 012, 034, 135, 236, 147, 257

Comment

contains Pasch contains Pasch contains Pasch contains mitre

Theorem 1. For every integer d ≥ 3 and for every integer n satisfying n ≥ d2 there exists v0 (n, d) such that for all admissible v ≥ v0 (n, d), every STS(v) contains a configuration having n blocks and n + d points. Here, the value of d is sharp. For d = 2, the theorem does not hold. Indeed, the ¨ [3]: For every integer k ≥ 4, there case d = 2 is the subject of a conjecture of Erdos exists v0 (k) such that if v > v0 (k) and if v is admissible, then there exists an STS(v) with the property that it contains no configurations having n blocks and n + 2 points for any n satisfying 4 ≤ n ≤ k. Such an STS(v) is said to be k-sparse. Clearly, a k-sparse system is also k  -sparse for every k  satisfying 4 ≤ k  ≤ k. Up to isomorphism, there is only one configuration having four blocks and six points, namely the Pasch configuration, also known as a quadrilateral; this is shown in Table 1. The existence of 4-sparse (better known as anti-Pasch) STS(v)s for all admissible v, except v = 7 and 13, was established in [1, 8, 10, 7]. There is, up to isomorphism, only one Pasch-free configuration having five blocks and seven points, namely the mitre. This is also shown in Table 1. In [2, 9, 5], culminating in recent work by Fujiwara and Wolfe [6, 12], it is established that anti-mitre systems exist for all admissible orders except v = 9. Systems which are 5-sparse, that is, both anti-Pasch and anti-mitre, are known for v ≡ 1, 19 (mod 54), except possibly v = 109, and for many other sporadic values [9, 6]. Also we are aware that there exists a 5-sparse STS(109) [13]. Substantial further progress has recently been made by Wolfe in [14], where it is shown that 5-sparse STS(v)s exist for almost all admissible v (meaning arithmetic set density 1 in the set of all admissible orders), and in [15], where existence for all v ≡ 3(mod 6) with v ≥ 21 is established. There are, up to isomorphism, two Pasch-free and mitre-free configurations having six blocks and eight points, of which one is the 6-cycle. The other configuration is called the crown, a word that is suggested by the following diagram. sXX s XX  B   XXX  s  X  B  X  XX XXs s  s B B  B  Crown configuration s Bs

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Thus a system is 6-sparse if and only if it contains no Pasch configurations, no mitres, no 6-cycles and no crowns. In [4] we presented the first known non-trivial examples of 6-sparse Steiner triple systems. Our results depended on two basic theorems. The first of these is the following. Theorem 2. Suppose that v is a prime congruent to 7 modulo 12 and that χ is a multiplicative character of GF(v) of order 6. Suppose also that α ∈ G F(v) has the property that χ (α) = −1, 0, 1 and that χ (1−α)χ (α) = ±1. Let G denote the group comprising all mappings on G F(v) having the form x → ax + b for a, b ∈ G F(v) with χ (a) = 1. Then the orbit generated by the block {0, 1, α} under the action of G forms a block transitive STS(v). Using Theorem 2, we obtained 29 6-sparse systems with 27 different prime orders v ≡ 7(mod 12). Furthermore, by employing Weil’s theorem on bounding character sums [11, page 43] we were able to show that our list of such systems is complete. The other theorem from [4] asserts that the standard product construction preserves 6-sparseness under certain conditions. Theorem 3. Suppose that S = (V, B) is a block transitive Steiner triple system of order v, with α and χ as in Theorem 2 and V = G F(v). Suppose also that S ∗ = (W, B ∗ ) is a Steiner triple system of order w. For each block of B ∗ , arbitrarily fix the order of the points, so that B ∗ may be regarded as a set of ordered triples (i, j, k). Put V  = V × W and let B  = {{ai , bi , ci } : {a, b, c} ∈ B, i ∈ W } ∪ {{xi , y j , (xβ + yγ )k } : x, y ∈ G F(v), (i, j, k) ∈ B ∗ }, where β, γ = 0 are fixed parameters in G F(v). Then S  = (V  , B  ) is a Steiner triple system of order vw. Furthermore, if both S and S ∗ are 6-sparse, if α 2 ∈ {α − 1, 1 − α, α + 1, 3α − 1},

(1)

and if χ (β), χ (γ ), χ (β/γ ) = ±1, then S  is also 6-sparse. Having shown that (1) holds for each of our original 29 block transitive systems, we can repeatedly apply Theorem 3, choosing, for example, β = α and γ = 1/α, to establish that there are infinitely many 6-sparse Steiner triple systems. In this paper we prove a theorem analogous to Theorem 2 for the case v = 3 p, where p is prime and p ≡ 3(mod 4). Using this theorem we are able to construct 6-sparse Steiner triple systems of order 3 p for all sufficiently large primes p ≡ 3(mod 4). 2. Steiner Triple Systems with v

9 (mod 12)

For the remainder of the paper, p will always denote a prime congruent to 3 modulo 4, and θ will denote the quadratic character modulo p. Thus if x ≡ 0(mod p), θ (x) = (x/p), the Legendre symbol, and if x ≡ 0(mod p), θ (x) = 0.

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Theorem 4. Let p = 2s + 1 ≥ 7 be a prime such that p ≡ 3(mod 4) and let v = 3 p. Let τ be an integer modulo v such that τ ≡ 0(mod 3) and τ is a primitive root modulo p. Let ω = τ 2 mod v. Choose α modulo v such that either (i) α ≡ 0(mod 3) and θ (α − 1) = 1, or (ii) α ≡ 1(mod 3) and θ (−α) = 1. Then, with all arithmetic modulo v, {{m, m + ωi , m + αωi } : i = 0, 1, . . . , s − 1, m = 0, 1, . . . , v − 1} ∪ {{n, n + 13 v, n + 23 v} : n = 0, 1, . . . , 13 v − 1} is the set of blocks of an STS(v), defined on {0, 1, . . . , v − 1}, which is generated by {0, 1, α} and {0, v/3, 2v/3} under the action of the group of mappings G = {x → ωi x + m mod v, i = 0, 1, . . . , s − 1, m = 0, 1, . . . , v − 1}. Proof. In this proof and the remarks which follow we shall tacitly assume that unless otherwise specified all arithmetic is performed modulo v. Clearly, the orbit of the starter block {0, p, 2 p} under the action of G is {{n, n + p, n + 2 p} : n = 0, 1, . . . , p − 1}. Let

(x) = {x ωi mod v : i = 0, 1, . . . , s − 1} and observe that for any x modulo v, we have θ (xω) = θ (x), xω ≡ x(mod 3) and

(x) = {y mod v : θ (y) = θ (x) and y ≡ x(mod 3)}. Therefore we can prove the theorem by showing that the six differences ±1, ±α and ±(1 − α) generated by the triple {0, 1, α} have distinct combinations of quadratic character modulo p and residue class modulo 3. Since θ (−1) = −1, this is possible if and only if α satisfies (i) or (ii) in the statement of the theorem.  The choice of τ is immaterial, subject to τ ≡ 0(mod 3) and τ being a primitive root modulo p. To see this, suppose τ  ≡ 0(mod 3) is also a primitive root modulo p and let ω = (τ  )2 . Then τ  ≡ τ t (mod p) for some t with (t, p − 1) = 1 and it is plain that for any x,

(x) = {x (ω )i mod v : i = 0, 1, . . . , s − 1}. If α ≡ 0(mod 3), the four STS(v)s generated by the blocks {0, 1, δ} and {0, v/3, 2v/3} for δ ∈ {α, 1 − α, 1/(1 − α), 1 − 1/(1 − α)} are isomorphic under the mappings x → 1 − x, x → (x − 1)/(α − 1) and x → (α − x)/(α − 1). If α ≡ 1(mod 3), the four STS(v)s generated by the blocks {0, 1, δ} and {0, v/3, 2v/3} for δ ∈ {α, 1 − α, 1/α, 1 − 1/α} are isomorphic under the mappings x → 1 − x, x → x/α and x → 1 − x/α. The above observations may be used to reduce the size of a search for 6-sparse systems obtained from Theorem 4. A complete list, up to isomorphism, of such 6-sparse Steiner triple systems for v < 10000 is given in Table 2. Systems with the

Further 6-sparse Steiner Triple Systems

53

Table 2. 6-sparse systems with v ≡ 9(mod 12). v 489 501 1077 1101 1149 1329 1437 1461 1461 1509 1569 1641 1689 1857 1857 1929 1929 1941 1941 1977 2157 2157 2181 2217 2229 2361 2433 2589 2649 2649 2721 2733 2733 2733 2733 2841 2949 2949 2949 2973 2973 3057 3093 3093 3117 3117 3189 3261 3261

α

v

α

v

α

v

α

v

α

v

α

v

α

135 160 75 379 328 309 12 13 42 490 232 223 276 141 328 502 508 3 736 519 36 186 9 193 880 979 594 684 421 609 534 24 240 585 682 447 711 906 919 288 309 954 445 610 345 579 318 9 409

3837 3849 3909 3909 3981 4101 4101 4101 4281 4317 4317 4317 4317 4317 4317 4353 4353 4377 4377 4377 4449 4497 4569 4569 4569 4593 4593 4629 4629 4677 4677 4677 4677 4677 4677 4701 4701 4701 4701 4749 4749 4749 4821 4821 4821 4821 4821 4857 4857

880 1263 544 1063 1627 265 427 561 204 201 432 658 693 744 993 660 1057 58 184 409 94 430 370 1402 1837 117 1210 366 1699 12 78 99 126 583 1240 76 337 430 499 418 1239 1294 43 565 826 1240 1587 163 1057

5277 5277 5277 5349 5361 5361 5361 5469 5469 5469 5469 5493 5493 5541 5541 5541 5601 5613 5613 5613 5613 5637 5721 5853 5853 5853 5853 5937 5937 5961 5961 5997 5997 6009 6009 6009 6033 6033 6033 6081 6081 6117 6117 6117 6189 6189 6189 6249 6249

1377 1486 2074 15 835 1075 1377 84 415 1114 1516 430 1576 1104 1707 2344 1065 1470 1900 2218 2343 880 1594 376 435 1677 2064 1365 1606 358 1540 643 1372 360 900 1167 126 792 2251 457 1360 604 2373 2490 63 1429 2224 69 561

6429 6429 6429 6537 6537 6609 6609 6717 6753 6753 6861 6861 6933 6933 7017 7017 7017 7041 7041 7041 7041 7053 7053 7053 7053 7113 7149 7197 7197 7233 7269 7341 7341 7377 7377 7401 7401 7509 7509 7509 7593 7593 7593 7617 7617 7617 7617 7617 7629

129 1462 2097 915 1068 31 810 954 1551 2184 385 604 933 3030 81 240 1117 351 1009 1305 1392 520 985 1650 2227 2404 714 966 1138 1794 85 1390 1597 891 2287 87 3546 907 1293 1762 103 219 1108 85 223 231 816 864 141

7977 7977 7977 7989 7989 7989 7989 8013 8013 8013 8013 8013 8013 8049 8049 8061 8061 8061 8061 8061 8061 8061 8061 8097 8121 8121 8121 8133 8133 8133 8133 8157 8193 8193 8301 8301 8301 8301 8301 8301 8373 8373 8373 8373 8409 8409 8409 8457 8529

1960 2404 2944 402 657 2298 3429 348 496 549 793 1009 2353 570 1173 3 18 57 439 576 1270 1333 1531 666 307 1231 1347 292 1386 2764 3225 2062 160 1153 700 835 871 994 1011 2398 537 1657 2697 2913 630 1927 2554 685 57

8637 8637 8637 8637 8637 8661 8661 8661 8661 8661 8709 8709 8709 8709 8709 8709 8709 8709 8709 8781 8781 8781 8781 8781 8781 8817 8817 8817 8817 8913 8913 8913 8997 8997 8997 8997 8997 8997 8997 8997 8997 8997 9033 9033 9033 9057 9057 9057 9057

919 1393 2046 2077 4141 490 1011 1254 1918 2901 42 99 250 705 1296 1395 1695 2010 3925 366 498 685 979 2251 3706 571 1552 1969 2991 694 1725 3289 150 351 367 753 955 1227 2253 2857 3295 3606 273 1582 3421 397 720 1308 2643

9357 9357 9357 9357 9357 9357 9357 9357 9489 9489 9489 9489 9501 9501 9501 9501 9501 9501 9561 9561 9561 9573 9573 9573 9573 9573 9573 9573 9573 9609 9609 9609 9609 9609 9609 9609 9609 9753 9753 9753 9777 9777 9813 9813 9897 9921 9921 9921 9921

18 390 403 1033 1516 2152 2403 2643 1048 1191 1809 4314 168 471 486 1605 2514 3609 148 4164 4273 54 162 391 687 1093 1350 2085 2202 306 721 1191 1260 1731 1783 2994 3166 2193 3313 3454 364 903 1743 3049 1206 96 910 3514 3865

54

A. D. Forbes et al. Table 2. continued

v 3261 3309 3309 3453 3513 3513 3561 3669 3669 3669 3693 3693 3693

α

v

α

v

α

v

α

v

α

v

α

v

α

735 390 940 802 223 598 313 87 231 520 102 544 838

4881 4881 4989 5001 5001 5001 5097 5097 5097 5097 5169 5241 5241

942 1761 336 919 1530 1608 70 633 1227 1747 915 538 2160

6249 6261 6261 6261 6261 6297 6297 6297 6333 6333 6333 6333 6429

2653 907 1200 1219 1422 286 1278 1983 135 648 810 2242 72

7629 7653 7653 7653 7653 7653 7737 7773 7773 7773 7773 7941 7977

876 162 366 498 1440 1612 1192 1327 2185 2239 3270 1864 1107

8529 8529 8529 8553 8553 8553 8553 8553 8637 8637 8637 8637 8637

471 507 3192 444 568 1189 1738 2931 52 232 432 523 744

9069 9201 9201 9201 9201 9201 9201 9237 9237 9237 9249 9249 9249

2761 486 595 946 1327 2146 2365 648 693 2287 556 3069 3339

9957 9957 9957 9957 9969 9969 9969 9993

99 144 2194 4138 619 2410 2565 2443

same value of v are pairwise non-isomorphic, as can be seen by examining the structure of the cycle graphs G 0,1 , G 0,α , G 1,α and G 0,v/3 . We refer to a system created by Theorem 4 as a two-generator system with parameters v and α. The special mitres and Pasch configurations that are shown in [4] to be unavoidable in all systems with sufficiently large order obtained from Theorem 2 do not form in the two-generator systems of Theorem 4. We now prove that there is no such blocking mechanism to prevent the formation of 6-sparse two-generator systems of arbitrarily large orders. Theorem 5. For all sufficiently large v with v = 3 p, p prime and p ≡ 3(mod 4), there exists α such that the two-generator system of Theorem 4 with parameters v and α is 6-sparse. The proof of this theorem makes use of the following lemmas, the last of which relies on extensive computations. Lemma 1. Let n be a positive integer, let p be a prime, let a1,1 x1 + a1,2 x2 + · · · + a1,n xn ≡ c1 (mod p) a2,1 x1 + a2,2 x2 + · · · + a2,n xn ≡ c2 (mod p) ... an,1 x1 + an,2 x2 + · · · + an,n xn ≡ cn (mod p)

(2)

be a set of linear congruences modulo p and let A = [ai, j ] be the corresponding matrix of coefficients. Suppose |A| ≡ 0 (mod p). Then there exists a unique solution of (2) in GF( p). Furthermore, the solution is formally the same as that obtained by solving (2) over the rationals. Proof. This is well known.



Further 6-sparse Steiner Triple Systems

55

Lemma 2. Let S = (V, B) be a two-generator Steiner triple system with parameters v and α containing one of the configurations Pasch, mitre, 6-cycle, crown. Let V = {0, 1, . . . , v − 1} and let Γ = {G1 , G2 , . . . , G13 }, where G1 = {{0, 1, α}, {0, x1 , x2 }, {1, x2 , x3 }, {α, x1 , x3 }}, G2 = {{0, 1, α}, {0, x1 , x2 }, {0, x3 , x4 }, {1, x1 , x3 }, {α, x2 , x4 }}, G3 = {{0, 1, α}, {1, x1 , x2 }, {1, x3 , x4 }, {0, x1 , x3 }, {α, x2 , x4 }}, G4 = {{0, 1, α}, {α, x1 , x2 }, {α, x3 , x4 }, {0, x1 , x3 }, {1, x2 , x4 }}, G5 = {{0, 1, α}, {0, x1 , x2 }, {0, x3 , x4 }, {x5 , 1, x1 }, {x5 , α, x3 }, {x5 , x2 , x4 }}, G6 = {{0, 1, α}, {1, x1 , x2 }, {1, x3 , x4 }, {x5 , 0, x1 }, {x5 , α, x3 }, {x5 , x2 , x4 }}, G7 = {{0, 1, α}, {α, x1 , x2 }, {α, x3 , x4 }, {x5 , 0, x1 }, {x5 , 1, x3 }, {x5 , x2 , x4 }}, G8 = {{0, 1, α}, {0, x1 , x2 }, {0, x3 , x5 }, {1, x1 , x4 }, {α, x1 , x5 }, {x2 , x3 , x4 }}, G9 = {{0, 1, α}, {0, x1 , x2 }, {0, x3 , x5 }, {α, x1 , x4 }, {1, x1 , x5 }, {x2 , x3 , x4 }}, G10 = {{0, 1, α}, {1, x1 , x2 }, {1, x3 , x5 }, {0, x1 , x4 }, {α, x1 , x5 }, {x2 , x3 , x4 }}, G11 = {{0, 1, α}, {1, x1 , x2 }, {1, x3 , x5 }, {α, x1 , x4 }, {0, x1 , x5 }, {x2 , x3 , x4 }}, G12 = {{0, 1, α}, {α, x1 , x2 }, {α, x3 , x5 }, {0, x1 , x4 }, {1, x1 , x5 }, {x2 , x3 , x4 }}, G13 = {{0, 1, α}, {α, x1 , x2 }, {α, x3 , x5 }, {1, x1 , x4 }, {0, x1 , x5 }, {x2 , x3 , x4 }}. Then there is a G ∈ Γ such that G ⊂ B for some x1 , x2 , . . . , xn ∈ V , where n = |G|−1. Proof. By Theorem 4, v = 3 p, p prime, p ≡ 3(mod 4), and S is generated by blocks {0, 1, α} and {0, p, 2 p}. Let X be one of the configurations Pasch, mitre, 6-cycle, crown. Suppose X ⊂ B. Observe that G1 is a Pasch configuration, G2 , G3 and G4 are mitres, G5 , G6 and G7 are 6-cycles, G8 , G9 , . . . , G13 are crowns and the block of G ∈ Γ labelled {0, 1, α} is one of two intersecting blocks which map to each other under an automorphism of G. Since X cannot contain two intersecting blocks belonging to the orbit of {0, p, 2 p}, it is straightforward to verify (perhaps by drawing diagrams) that there exists an automorphism of S which maps X to some G ∈ Γ for some x1 , x2 , . . . , x|G|−1 ∈ V .  Lemma 3. Let p be prime and suppose that the polynomial f (x) is not a constant multiple of a square over G F( p). Then        √  θ ( f (x)) = O( p).   x∈G F( p) Proof. This is a special case of the theorem on page 43 of [11].



In the next lemma we introduce a set of polynomials, Λ. In Lemma 5 we investigate certain sets of linear congruences. The coefficients of these congruences involve a parameter, α. We wish to show that there exists an α such that either the congruences have no solution, or the solution satisfies certain conditions that can be

56

A. D. Forbes et al.

expressed in the form θ (ρ(α)) = 1 for certain rational functions ρ(x). We then find that there is a set of polynomials Λ with the property that if θ (λ(α)) = 1 for all λ(x) ∈ Λ, then θ (ρ(α)) = −1 for at least one of the functions ρ(x). Actually, to deal with questions of existence and uniqueness of solutions, slightly more than this is required, and the key property of Λ is that given in Lemma 4. The set Λ given in this lemma was obtained by considering the numerators and denominators of the functions ρ(x). It is not feasible to explain why each individual polynomial is included in Λ. However, we give below, following the proof of Lemma 5, several examples to illustrate the method. In particular, Example 1 explains why −x 3 +5x 2 −6x +3 ∈ Λ. Lemma 4. Let Λ = {x, x − 1, x + 1, −2x + 1, 2x − 3, −x + 3, x 2 + 1, −x 2 − 2, −x 2 − x + 1, x 2 − x + 1, −x 2 + x + 1, −x 2 + 2x − 2, −x 2 + 3x − 3, −2x 2 + 3x − 2, 3x 2 − 4x + 2, 2x 2 − 4x + 3, −2x 2 + 3x − 3, 3x 2 − 5x + 3, x 2 − 2x + 3, x 2 − 3x + 1, −x 3 + x 2 − 1, −x 3 + 2x 2 − x − 1, −x 3 + 3x 2 − 2x + 1, x 3 − 2x 2 + 3x − 3, x 3 − 3x 2 + 6x − 3, x 3 − 3x + 3, −x 3 + 5x 2 − 6x + 3, −x 3 + 3x 2 − 4x + 1}. Given any positive number N , for all sufficiently large prime p, there exist at least N numbers α, distinct modulo p, such that θ (λ(α)) = 1 for all λ(x) ∈ Λ. Proof. Let 

π(x) =

(1 + θ (λ(x)))

λ(x)∈Λ

and =



π(x).

x∈GF( p)

Then = p+





θ ( f (x)),

f (x) x∈GF( p)

where f (x) in the outer sum runs through all 2|Λ| − 1 non-empty products of polynomials λ(x) ∈ Λ. It is easily checked that over the rationals the polynomial  λ(x) has non-zero discriminant. Hence, assuming that p is sufficiently large, λ(x)∈Λ f (x) is never a constant multiple of a square over GF( p). So by Lemma 3 we have √  = p − O( p). Since both π(x) and the number of factors of π(x) which are equal to 1 are bounded as p → ∞, it follows that for each fixed N and for p sufficiently large, there exist N distinct values of α modulo p such that θ (λ(α)) = 1 for all λ(x) ∈ Λ. 

Further 6-sparse Steiner Triple Systems

57

Lemma 5. Let v = 3 p, p prime, p ≡ 3(mod 4). Let Λ be the set of polynomials in Lemma 4. Then there exists a polynomial Q(x) such that if α ≡ 0 (mod 3), if θ (λ(α)) = 1 for all λ(x) ∈ Λ and if Q(α) ≡ 0 (mod p), then there exists a 6-sparse two-generator Steiner triple system with parameters v and α. Proof. Let v = 3 p, p prime, p ≡ 3(mod 4) and suppose α satisfies the conditions of the lemma with Q(x) to be chosen later. Observe that x − 1 ∈ Λ; therefore θ (α − 1) = 1, as required by Theorem 4, and hence there exists a two-generator Steiner triple system S = (V, B) with parameters v and α. We show that with a suitable choice of Q(x) S is 6-sparse. Let Γ = {G1 , G2 , . . . , G13 } be the set of configurations in Lemma 2. Let G ∈ Γ and let G have n + 1 blocks. For d = 1, 2, . . . , n, let (ad , bd , cd ) be the dth block of G \ {{0, 1, α}} in some ordering. Then if G ⊂ B, we have the following set of 3n linear congruences modulo 3 p in variables x1 , x2 , . . . , xn , m 1 , m 2 , . . . , m n and the variables ωd for those d where the corresponding congruences have the first alternative on the right:  (a1 , b1 , c1 ) ≡  (a2 , b2 , c2 ) ≡ ..., (an , bn , cn ) ≡



(m 1 , m 1 + ω1 , m 1 + αω1 ) or (m 1 , m 1 + p, m 1 + 2 p), (m 2 , m 2 + ω2 , m 2 + αω2 ) or (m 2 , m 2 + p, m 2 + 2 p), (m n , m n + ωn , m n + αωn ) or (m n , m n + p, m n + 2 p).

On eliminating the m d we have 2n congruences modulo 3 p: (bi − ai , ci − ai ) ≡ (ωi , αωi ) or ( p, 2 p),

i = 1, 2, . . . , n.

(3)

Since there are six permutations of (ai , bi , ci ) and two possible congruences for each, there are 12n possible sets of congruences represented by (3). (Although this number can be reduced somewhat, we prefer, for simplicity, to present the results of our original computations, which do not exploit additional symmetries in (3).) Thus by Lemma 2, if S contains a Pasch, mitre, 6-cycle or crown configuration, there exists a G ∈ Γ and a corresponding set of congruences (3) which has, for some orderings of the blocks of G and some choice of the alternatives on the right of (3), a solution modulo 3 in which all the ωd present satisfy ωd ≡ 1(mod 3) and a solution modulo p in which all the ωd present satisfy θ (ωd ) = 1. To show that this cannot happen, we examine each of the 12n possible sets of congruences (3) for each configuration G ∈ Γ . Denote this collection of congruence sets by Φ0 . Thus |Φ0 | = 123 + 3 · 124 + 9 · 125 = 2303424. As an immediate first step, we eliminate from Φ0 all sets where there are two intersecting blocks in the orbit of {0, p, 2 p}, for such configurations cannot occur in S. This leaves a collection Φ1 of 584064 sets: 864 for G1 , 7776 each for G2 , G3 and G4 , 62208 each for G5 , G6 , . . . , G13 . For example, take the crown configuration

58

A. D. Forbes et al.

G8 . Denote the blocks other than {0, 1, α}, by A, B, C, D, and E, where {A, B} and {C, D} are pairs of parallel blocks. Then we have the following possibilities for blocks in the orbit of {0, p, 2 p}: none, 65 ; block E, 65 ; one or both of {A, B}, 2 · 65 + 65 ; one or both of {C, D}, 2 · 65 + 65 . So the total number of legitimate congruence sets that arise from G8 is 8 · 65 = 62208. Next, we eliminate from Φ1 all cases where (3) has no solution modulo 3. We assume that α = 0 and that ωd = 1 for all multipliers ωd present. We also assume that p = 1. For if a set of congruences (3) has a solution modulo 3 with p = 2 and includes the pairs {b j − a j ≡ p, c j − a j ≡ 2 p} for those j ∈ {1, 2, . . . , n} where the block {a j , b j , c j } is in the orbit of {0, p, 2 p}, then the set of congruences obtained by interchanging b j and c j has the same solution with p = 1, and, of course, both sets of congruences are identical modulo p. After performing the computations we are left with the collection Φ2 of 3320 congruence sets, partitioned as follows: G1 , 32; G2 , G3 , G4 , 168 each; G5 , G6 , G7 , 384 each; G8 , G13 , 344 each; G9 , G12 , 224 each; G10 , G11 , 248 each. In all cases the solution modulo 3 is unique. We deal with Φ2 by examining each congruence set modulo p. For a given congruence set, let t be the number of blocks in the orbit of {0, p, 2 p} and note that 0 ≤ t ≤ 2. Recall that the configuration has n + 1 blocks. So there are 2n congruences, n point variables, x1 , x2 , . . . , xn , and n − t multiplier variables, ωd . We select 2n − t congruences by excluding t (possibly none) of the 2t congruences that involve p. Suppose t = 0. With the congruences (3) written in matrix form Dx = e, we find that in every case |D| is a polynomial in α, d(α), say, which is not identically zero. Assuming that p is sufficiently large and α is chosen such that d(α) ≡ 0(mod p), we can obtain the unique solution x = D−1 e (modulo p), where x = (x1 , x2 , . . . , xn , ω1 , ω2 , . . . , ωn ) and the elements of D−1 e are rational functions of α. Next we attempt to compute the quadratic characters of the multipliers ω j and the ratios ω j/ωk on the assumption that θ (λ(α)) = 1 for each λ(x) ∈ Λ. In all except four cases we find that at least one multiplier or ratio of multipliers is not a quadratic residue modulo p, and hence G cannot occur in S. Example 1 illustrates this point. The remaining four cases correspond to a Pasch configuration and three 6-cycles, where in the solution of the congruences (3) the xi are such that every block is of the form {a, b, c} with (a, b, c) ≡ (0, 1, α)(mod p) and the ωi are all ≡ 1(mod p). Since each ωi is also congruent to 1 modulo 3, it follows that each ωi is equal to 1. This in turn implies that the configuration contains repeated blocks. See Example 2 below. Now suppose t > 0. We find that it is always possible to choose 2n − t congruences from (3) such that when they are written in matrix form Dx = e, |D| = d(α) is not identically zero. So if α is chosen such that d(α) ≡ 0(mod p), then we get a unique solution for the 2n − t variables, x = D−1 e. The excluded t = 1 or 2 congruences have the form bi −ai ≡ 0(mod p) or ci −ai ≡ 0(mod p) for some i. So suppose for these i that the solution x = D−1 e gives ai = ai (α), bi = bi (α) and ci = ci (α) for rational functions ai (α), bi (α) and ci (α). We either have: (i) for all t excluded congruences, bi (α) − ai (α) or ci (α) − ai (α) is identically zero for all α; or (ii) for one of the excluded congruences there exists α such that bi (α) − ai (α) ≡ 0(mod p) or ci (α) − ai (α) ≡ 0(mod p).

Further 6-sparse Steiner Triple Systems

59

In case (i), the excluded congruences may be ignored and we proceed as for t = 0, where it turns out always that, assuming θ (λ(α)) = 1 for all λ(x) ∈ Λ, either θ (ω j ) = −1 for some multiplier ω j or θ (ω j /ωk ) = −1 for some ratio ω j /ωk of multipliers. Hence G does not occur in S. Example 3 illustrates this case. In case (ii), by clearing the denominator we obtain an additional constraint, which takes the form q(α) ≡ 0(mod p) for some polynomial q(x). Then if α is chosen such that q(α) ≡ 0(mod p), the congruences (3) will be inconsistent and hence G will not occur in S. See Example 4. To complete the proof we set Q(x) equal to the least common multiple of all the determinant polynomials d(x) and constraint polynomials q(x) encountered in the preceding analysis. Observe that if p is sufficiently large, none of the functions d(x) and q(x) depend on p, and hence Q(x) is independent of p.  Whilst it is not feasible within the space limitations of this paper to give details of all the cases that occur in the proof of Lemma 5, the main features of the method can be illustrated by a few examples. Example 1. Let G be the 6-cycle configuration G5 with the blocks ordered as written: {{0, 1, α}, {0, x1 , x2 }, {x5 , α, x3 }, {x3 , 0, x4 }, {x1 , x5 , 1}, {x4 , x2 , x5 }}. Suppose all these blocks belong to the orbit of {0, 1, α}. The congruences to be solved modulo 3 p are (0, x1 , x2 ) ≡ (m 1 , m 1 + ω1 , m 1 + αω1 ), (x5 , α, x3 ) ≡ (m 2 , m 2 + ω2 , m 2 + αω2 ), (x3 , 0, x4 ) ≡ (m 3 , m 3 + ω3 , m 3 + αω3 ), (x1 , x5 , 1) ≡ (m 4 , m 4 + ω4 , m 4 + αω4 ), (x4 , x2 , x5 ) ≡ (m 5 , m 5 + ω5 , m 5 + αω5 ), or, after eliminating m 1 , m 2 , m 3 , m 4 , m 5 , (x1 , x2 ) ≡ (ω1 , αω1 ), (α − x5 , x3 − x5 ) ≡ (ω2 , αω2 ), (−x3 , x4 − x3 ) ≡ (ω3 , αω3 ),

(4)

(x5 − x1 , 1 − x1 ) ≡ (ω4 , αω4 ), (x2 − x4 , x5 − x4 ) ≡ (ω5 , αω5 ). Setting α = 0 and ω1 = ω2 = ω3 = ω4 = ω5 = 1, we solve this set of congruences modulo 3 to obtain the unique solution: x1 = 1,

x2 = 0,

x3 = x4 = x5 = 2.

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Therefore we consider the congruences (4) modulo p, and for this purpose we put them into matrix form: ⎤⎡ ⎤ ⎡ ⎤ ⎡ x1 0 −1 0 0 0 0 1 0 0 0 0 ⎢ 0⎥ ⎢ 0 −1 0 0 0 α 0 0 0 0 ⎥ ⎢ x2 ⎥ ⎢α⎥ ⎢ 0 0 0 0 1 0 1 0 0 0⎥⎢ x ⎥ ⎥⎢ 3⎥ ⎢ ⎥ ⎢ ⎢ 0⎥ ⎢ 0 0 −1 0 1 0 α 0 0 0 ⎥ ⎢ x ⎥ ⎥⎢ 4⎥ ⎢ ⎥ ⎢ ⎢ ⎥ ⎢ 0 0 1 0 0 0 0 1 0 0 ⎥ ⎢ x5 ⎥ ⎥ ⎢ ⎥ ≡ ⎢ 0 ⎥ (mod p). ⎢ ⎢ 0⎥ ⎢ 0 0 1 −1 0 0 0 α 0 0 ⎥ ⎢ ω1 ⎥ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎢ 0⎥ ⎢ 1 0 0 0 −1 0 0 0 1 0 ⎥ ⎢ ω2 ⎥ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎢ 1⎥ ⎢ 1 0 0 0 0 0 0 0 α 0 ⎥ ⎢ ω3 ⎥ ⎦⎣ ⎦ ⎣ ⎦ ⎣ 0 0 −1 0 1 0 0 0 0 0 1 ω4 0 0 0 0 1 −1 0 0 0 0 α ω5 There are ten congruences and ten variables, the determinant of the system is −α(α 3 − 5α 2 + 6α − 3), and we assume that x(x 3 − 5x 2 + 6x − 3) is a factor of Q(x). Hence there is a unique solution modulo p: α 4 − 2α 3 + 3α − 3 α 5 − 2α 4 + 3α 2 − 3α −α 4 + 2α 2 − 2α , x2 = , x3 = 3 , 3 2 3 2 α − 5α + 6α − 3 α − 5α + 6α − 3 α − 5α 2 + 6α − 3 α 5 − α 4 − 2α 3 + 4α 2 − 2α α 4 − 3α 3 + 3α 2 − 2α x4 = , x5 = , 3 2 α − 5α + 6α − 3 α 3 − 5α 2 + 6α − 3 α 4 − 2α 3 + 3α − 3 −2α 3 + 3α 2 − α α 4 − 2α 2 + 2α ω1 = 3 , ω2 = 3 , ω3 = 3 , 2 2 α − 5α + 6α − 3 α − 5α + 6α − 3 α − 5α 2 + 6α − 3 −α 3 + 3α 2 − 5α + 3 −α 4 + 2α 3 − α 2 − α ω4 = 3 , ω5 = 3 . 2 α − 5α + 6α − 3 α − 5α 2 + 6α − 3 x1 =

Since x, x − 1, 1 − 2x and −x 3 + 5x 2 − 6x + 3 are in Λ, we can assume that θ (α) = θ (α − 1) = θ (1 − 2α) = θ (−α 3 + 5α 2 − 6α + 3) = 1. Hence we can compute the quadratic character of ω2 ,

   −2α 3 + 3α 2 − α α(α − 1)(1 − 2α) = −1, θ (ω2 ) = θ =θ α 3 − 5α 2 + 6α − 3 α 3 − 5α 2 + 6α − 3 and deduce that the configuration does not occur in S. Example 2. Let G be the Pasch configuration G1 with the blocks ordered as written: {{0, 1, α}, {0, x1 , x2 }, {x3 , 1, x2 }, {x3 , x1 , α}}. Suppose all these blocks belong to the orbit of {0, 1, α}. The congruences to be solved modulo 3 p are (0, x1 , x2 ) ≡ (m 1 , m 1 + ω1 , m 1 + αω1 ), (x3 , 1, x2 ) ≡ (m 2 , m 2 + ω2 , m 2 + αω2 ), (x3 , x1 , α) ≡ (m 3 , m 3 + ω3 , m 3 + αω3 ),

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or, after eliminating m 1 , m 2 , m 3 , (x1 , x2 ) ≡ (ω1 , αω1 ), (1 − x3 , x2 − x3 ) ≡ (ω2 , αω2 ), (x1 − x3 , α − x3 ) ≡ (ω3 , αω3 ).

(5)

Setting α = 0 and ω1 = ω2 = ω3 = 1, we solve this set of congruences modulo 3 to obtain the unique solution x1 = 1, x2 = x3 = 0. Therefore we consider the congruences (5) modulo p, and for this purpose we put them into matrix form: ⎤⎡ ⎤ ⎡ ⎤ ⎡ x1 0 −1 0 0 1 0 0 ⎢ 0 −1 0 α 0 0 ⎥ ⎢ x2 ⎥ ⎢ 0 ⎥ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎢ 0 0 1 0 1 0 ⎥ ⎢ x3 ⎥ ⎢ 1 ⎥ ⎢ 0 −1 1 0 α 0 ⎥ ⎢ ω ⎥ ≡ ⎢ 0 ⎥ (mod p). ⎥⎢ 1⎥ ⎢ ⎥ ⎢ ⎣ −1 0 1 0 0 1 ⎦ ⎣ ω ⎦ ⎣ 0 ⎦ 2 α 0 01 0 0α ω3 The determinant of the system is 2α(α − 1), and we assume that x(x − 1) is a factor of Q(x). Hence there is a unique solution modulo p : x1 = 1, x2 = α, x3 = 0, ω1 = ω2 = ω3 = 1. In fact, this is one of the four configurations where the system of congruences (3) has a legitimate solution and, as previously explained, it does not exist in S. The other three configurations where the congruences have legitimate solutions modulo 3 p are the 6-cycles G5 , G6 and G7 , with the blocks, all in the orbit of {0, 1, α}, ordered as written: G5 : {{0, 1, α}, {0, x3 , x4 }, {x5 , 1, x1 }, {x5 , x2 , x4 }, {0, x2 , x1 }, {x5 , x3 , α}}, G6 : {{0, 1, α}, {0, x5 , x1 }, {x2 , x5 , x4 }, {x3 , 1, x4 }, {x2 , 1, x1 }, {x3 , x5 , α}}, G7 : {{0, 1, α}, {0, x1 , x5 }, {x2 , x4 , x5 }, {x3 , x4 , α}, {x2 , x1 , α}, {x3 , 1, x5 }}. Example 3. Let G be the mitre configuration G2 with the blocks ordered as written: {{0, 1, α}, {0, x1 , x2 }, {x1 , x3 , 1}, {x2 , x4 , α}, {0, x4 , x3 }}. Suppose the second, third and fourth blocks belong to the orbit of {0, 1, α} and the fifth belongs to the orbit of {0, p, 2 p}. The congruences to be solved modulo 3 p are (0, x1 , x2 ) ≡ (m 1 , m 1 + ω1 , m 1 + αω1 ), (x1 , x3 , 1) ≡ (m 2 , m 2 + ω2 , m 2 + αω2 ), (x2 , x4 , α) ≡ (m 3 , m 3 + ω3 , m 3 + αω3 ), (0, x4 , x3 ) ≡ (m 4 , m 4 + p, m 4 + 2 p), or, after eliminating m 1 , m 2 , m 3 , m 4 , (x1 , x2 ) ≡ (ω1 , αω1 ), (x3 − x1 , 1 − x1 ) ≡ (ω2 , αω2 ), (x4 − x2 , α − x2 ) ≡ (ω3 , αω3 ), (x4 , x3 ) ≡ ( p, 2 p).

(6)

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Setting α = 0 and ω1 = ω2 = ω3 = p = 1, we solve this set of congruences modulo 3 to obtain this unique solution: x1 = 1, x2 = 0, x3 = 2, x4 = 1. Therefore we consider the congruences (6) modulo p, and for this purpose we put them into matrix form: ⎡ ⎤ ⎤ ⎡ 0 −1 0 0 0 1 0 0 ⎡ ⎤ ⎢ 0⎥ ⎢ 0 −1 0 0 α 0 0 ⎥ x1 ⎢ ⎥ ⎥⎢ x ⎥ ⎢ ⎢ 0⎥ ⎢ 1 0 −1 0 0 1 0 ⎥ ⎢ 2 ⎥ ⎢ ⎥ ⎥ ⎢ x3 ⎥ ⎢ ⎢ 1⎥ ⎢ 1 0 0 0 0 α 0⎥⎢ ⎥ ⎢ 0 1 0 −1 0 0 1 ⎥ ⎢ x4 ⎥ ≡ ⎢ 0 ⎥ (mod p). ⎢ ⎥ ⎥⎢ω ⎥ ⎢ ⎢α⎥ ⎢ 0 1 0 0 0 0 α⎥⎢ 1⎥ ⎢ ⎥ ⎥⎣ω ⎦ ⎢ 2 ⎣ 0⎦ ⎣ 0 0 0 −1 0 0 0 ⎦ ω3 0 0 0 −1 0 0 0 0 There are eight congruences but only seven variables. So we exclude the last congruence and work with just the first seven. The determinant of the reduced system is α 2 (α − 1), and we assume that x(x − 1) is a factor of Q(x). Hence there is a unique solution modulo p: 1 α 1 1 α , x2 = , x3 = x4 = 0, ω1 = , ω2 = , ω3 = 1−α 1−α 1−α α−1 α−1 and, furthermore, this solution is consistent with the excluded congruence, x3 ≡ 0(mod p). However, we can compute the quadratic character of ω1 : x1 =

θ (ω1 ) = θ (1 − α) = −1, since x − 1 ∈ Λ, and therefore deduce that this configuration does not occur in S. Example 4. Let G be the crown configuration G8 with the blocks ordered as written: {{0, 1, α}, {1, x1 , x4 }, {x2 , x3 , x4 }, {x1 , α, x5 }, {x3 , 0, x5 }, {0, x2 , x1 }}, and suppose only the last block belongs to the orbit of {0, p, 2 p}. The congruences to be solved modulo 3 p are (1, x1 , x4 ) ≡ (m 1 , m 1 + ω1 , m 1 + αω1 ), (x2 , x3 , x4 ) ≡ (m 2 , m 2 + ω2 , m 2 + αω2 ), (x1 , α, x5 ) ≡ (m 3 , m 3 + ω3 , m 3 + αω3 ), (x3 , 0, x5 ) ≡ (m 4 , m 4 + ω4 , m 4 + αω4 ), (0, x2 , x1 ) ≡ (m 5 , m 5 + p, m 5 + 2 p), or, after eliminating m 1 , m 2 , m 3 , m 4 , m 5 , (x1 − 1, x4 − 1) ≡ (ω1 , αω1 ), (x3 − x2 , x4 − x2 ) ≡ (ω2 , αω2 ), (α − x1 , x5 − x1 ) ≡ (ω3 , αω3 ), (−x3 , x5 − x3 ) ≡ (ω4 , αω4 ), (x2 , x1 ) ≡ ( p, 2 p).

(7)

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Setting α = 0 and ω1 = ω2 = ω3 = ω4 = p = 1, this set of congruences has a unique solution modulo 3: x1 = x3 = x5 = 2,

x2 = x4 = 1.

For solving modulo p, we put (7) in matrix form: ⎤ ⎤ ⎡ −1 −1 0 0 0 0 1 0 0 0 ⎡ ⎤ x1 0 0 −1 0 α 0 0 0⎥⎢ ⎥ ⎢ −1 ⎥ ⎢ 0 x2 ⎥ ⎢ 0⎥ ⎢ 0 1 −1 0 0 0 1 0 0⎥ ⎥ ⎥⎢ ⎢ ⎢ x3 ⎥ ⎥⎢ ⎢ 0⎥ ⎢ 0 ⎢ ⎥ 1 0 −1 0 0 α 0 0 ⎥ ⎥ ⎢ x4 ⎥ ⎢ ⎢ ⎢ α⎥ ⎢ 1 0 0 0 0 0 0 1 0⎥ ⎥ ⎥⎢ ⎢ ⎢ ⎢ x5 ⎥ ≡ ⎢ 0 ⎥ ⎥ (mod p). ⎢ 1 0 0 0 −1 0 0 α 0 ⎥ ⎥ ⎥⎢ ⎢ ⎢ ω1 ⎥ ⎢ ⎥ ⎥ ⎢ ⎢ 0 0⎥ 0 1 0 0 0 0 0 1⎥⎢ ⎥ ⎥ ⎢ ⎢ ω 2⎥ ⎢ 0⎥ ⎢ 0 0 1 0 −1 0 0 0 α ⎥ ⎢ ⎦ ⎦ ⎣ ω3 ⎦ ⎣ ⎣ 0 0 −1 0 0 0 0 0 0 0 ω4 0 −1 0 0 0 0 0 0 0 0 ⎡

There are ten congruences but only nine variables. We temporarily remove the last congruence and consider only the first nine. However, the determinant of this reduced system is identically zero. So we take instead the first eight and the tenth congruences, omitting the ninth. This system has determinant d(α) = (α − 1)2 , which is not zero modulo p provided x − 1 is a factor of Q(x). The unique solution modulo p of this system is then x1 = 0, ω1 = −1,

1 − 2α + α 2 − α 3 −α 2 , x4 = 1 − α, , x3 = 2 α−1 (α − 1) 2α − 1 α2 , ω2 = , ω = α, ω = 3 4 α−1 (α − 1)2

x2 =

x5 = α 2 ,

which is inconsistent with the omitted congruence, x2 ≡ 0(mod p), unless q(α) ≡ 0(mod p), where q(x) = 1 − 2x + x 2 − x 3 . Since we can assume that x − 1 and q(x) are factors of Q(x), this configuration does not occur in S. Proof of Theorem 5. The result follows from Lemma 4 and Lemma 5. Choose N greater than the degree of Q(x). Take p so large that it does not divide any of the coefficients of Q(x) and is sufficiently large for Lemma 4 to apply. Then by Lemma 4 we select an α which is not a root of Q(x) modulo p and is such that θ (λ(α)) = 1 for all λ(x) ∈ Λ. If necessary we add a multiple of p to α to obtain a value that is congruent to 0 modulo 3. Now apply Lemma 5.  Finally, we briefly address a question which naturally arises. Can Theorem 4 be used to create 7-sparse Steiner triple systems? We suspect not. In our research we have been unable to find any 6-sparse system which avoids the 7-block, 9-point configuration {012, 034, 135, 246, 257, 168, 078}, obtained by adding a diagonal to the ‘window frame’.

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References 1. Brouwer, A.E.: Steiner triple systems without forbidden subconfigurations. Mathematisch Centrum Amsterdam, ZW 104/77, (1977) ˇ an, ˇ J.: Anti-mitre Steiner triple systems. 2. Colbourn, C.J., Mendelsohn, E., Rosa, A., Sir´ Graphs Combin. 10, 215–224 (1994) ¨ P.: Problems and results in combinatorial analysis. Colloquio Internazionale sulle 3. Erdos, Teorie Combinatorie (Rome 1973), Tomo II, pp. 3–17. Atti dei Convegni Lincei, No. 17. Accad. Naz. Lincei, Rome (1976) 4. Forbes, A.D., Grannell, M.J., Griggs, T.S.: On 6-sparse Steiner triple systems. J. Combin. Theory Ser. A 114, 235–252 (2007) 5. Fujiwara, Y.: Constructions for anti-mitre Steiner triple systems. J. Combin. Des. 13, 286–291 (2005) 6. Fujiwara, Y.: Infinite classes of anti-mitre and 5-sparse Steiner triple systems. J. Combin. Des. 14, 237–250 (2006) 7. Grannell, M.J., Griggs, T.S., Whitehead, C.A.: The resolution of the anti-Pasch conjecture. J. Combin. Des. 8, 300–309 (2000) 8. Griggs, T.S., Murphy, J.P., Phelan, J.S.: Anti-Pasch Steiner triple systems. J. Combin. Inform. System Sci. 15, 79–84 (1990) 9. Ling, A.C.H.: A direct product construction for 5-Sparse Steiner triple systems. J. Combin. Des. 5, 443–447 (1997) 10. Ling, A.C.H., Colbourn, C.J., Grannell, M.J., Griggs, T.S.: Construction techniques for anti-Pasch Steiner triple systems. J. London Math. Soc. (2) 61, 641–657 (2000) 11. Schmidt, W.M.: Equations over Finite Fields. Lecture Notes in Mathematics 536. Berlin Heidelberg New York: Springer (1976) 12. Wolfe, A.: The resolution of the anti-mitre Steiner triple system conjecture. J. Combin. Des. 14, 229–236 (2006) 13. Wolfe, A.: Private communication (2006) 14. Wolfe, A.: 5-sparse Steiner triple systems of order n exist for almost all admissible n. Electron. J. Combin. 12, #R68, 42 pp. (electronic) (2005) 15. Wolfe, A.: The existence of 5-sparse Steiner triple systems of order n ≡ 3(mod 6), n∈ / {9, 15}. J. Combin. Theory Ser. A 115, 1487–1503 (2008) Received: March 17, 2007 Final Version received: October 10, 2008

Graphs and Combinatorics (2009) 25:65–79 Digital Object Identifier (DOI) 10.1007/s00373-008-0823-7

Graphs and Combinatorics © Springer-Verlag 2009

Distance-Regular Graph with c2 > 1 and a1

0 < a2

Akira Hiraki Division of Mathematical Sciences, Osaka Kyoiku University, Osaka 582-8582, Japan

Abstract. Let Γ be a distance-regular graph of diameter d ≥ 3 with c2 > 1. Let m be an integer with 1 ≤ m ≤ d − 1. We consider the following conditions: (SC)m : For any pair of vertices at distance m there exists a strongly closed subgraph of diameter m containing them. (B B)m : Let (x, y, z) be a triple of vertices with ∂Γ (x, y) = 1 and ∂Γ (x, z) = ∂Γ (y, z) = m. Then B(x, z) = B(y, z). (C A)m : Let (x, y, z) be a triple of vertices with ∂Γ (x, y) = 2, ∂Γ (x, z) = ∂Γ (y, z) = m and |C(z, x) ∩ C(z, y)| ≥ 2. Then C(x, z) ∪ A(x, z) = C(y, z) ∪ A(y, z). In [12] we have shown that the condition (SC)m holds if and only if both of the conditions (B B)i and (C A)i hold for i = 1, . . . , m. In this paper we show that if a1 = 0 < a2 and the condition (B B)i holds for i = 1, . . . , m, then the condition (C A)i holds for i = 1, . . . , m. In particular, the condition (SC)m holds. Applying this result we prove that a distance-regular graph with classical parameters (d, b, α, β) such that c2 > 1 and a1 = 0 < a2 satisfies the condition (SC)i for i = 1, . . . , d −1. In particular, either (b, α, β) = (−2, −3, −1 − (−2)d ) or (b, α, β) = −3, −2, − 1+(−3) 2

d

holds.

Key words. Distance-regular graph, Classical parameters, Strongly closed subgraph.

1. Introduction Our notation and terminologies are standard. The reader is referred to the next section or [1, 2] for the definitions. The known distance-regular graphs have many subgraphs of high regularity. For example the Odd graphs, the doubled Odd graphs, the doubled Grassmann graphs, the Hamming graphs, the dual polar graphs and the Hermitian forms graphs satisfy the following condition: “For any pair of vertices there exists a strongly closed subgraph containing them whose diameter is the distance between them.” Here we do not assume that a strongly closed subgraph is distance-regular. Our problem is what kinds of conditions are sufficient for the existence of strongly closed subgraphs. Several results have been obtained (for example see [4–6, 11, 12, 17] and [19–21]). We remark that a strongly closed subgraph in this paper is called a weak-geodetically closed subgraph in [19–21]. Let Γ be a distance-regular graph of diameter d ≥ 3 and valency k ≥ 3. Let m be an integer with 1 ≤ m ≤ d − 1. We consider the following three conditions:

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(SC)m : For any pair of vertices at distance m there exists a strongly closed subgraph of diameter m containing them. (B B)m : Let (x, y, z) be a triple of vertices with ∂Γ (x, y) = 1 and ∂Γ (x, z) = ∂Γ (y, z) = m. Then B(x, z) = B(y, z). (C A)m : Let (x, y, z) be a triple of vertices with ∂Γ (x, y) = 2, ∂Γ (x, z) = ∂Γ (y, z) = m and |C(z, x) ∩ C(z, y)| ≥ 2. Then C(x, z) ∪ A(x, z) = C(y, z) ∪ A(y, z). It is clear that B(x, z) = B(y, z) if and only if C(x, z) ∪ A(x, z) = C(y, z) ∪ A(y, z). We showed that if the condition (SC)m holds, then the condition (B B)m holds (see [5, Proposition 2.1] or Lemma 10). We remark that the condition (SS)m in [5] is the same as the condition (B B)m . In [8, Theorem 1] we proved that if the condition (SC)m holds, then the condition (SC)i holds for all i = 1, . . . , m. It follows that if the condition (SC)m holds, then the condition (B B)i holds for all i = 1, . . . , m. Suppose the condition (B B)m does not hold. Then there exists a triple (x, y, z) of vertices with ∂Γ (x, y) = 1, ∂Γ (x, z) = ∂Γ (y, z) = m such that B(x, z) = B(y, z). Take w ∈ B(x, z)\B(y, z). Then ∂Γ (x, y) = ∂Γ (z, w) = 1, ∂Γ (x, w) = m + 1 and ∂Γ (x, z) = ∂Γ (y, z) = ∂Γ (y, w) = m. Such a quadruple (x, y, z, w) of vertices is called a parallelogram of length m + 1 in [20]. Conversely if there exists a parallelogram (x, y, z, w) of length m + 1, then (x, y, z) is a triple of vertices with ∂Γ (x, y) = 1, ∂Γ (x, z) = ∂Γ (y, z) = m such that B(x, z) = B(y, z). Therefore the condition (B B)m holds if and only if there is no parallelogram of length m + 1. For the case c2 > 1 and a1 = 0 C.-W. Weng proved that if there is no parallelogram of length i with i ≤ m + 1, (i.e., the condition (B B)i holds for all i = 1, . . . , m) then the condition (SC)i holds for all i = 1, . . . , m in [20, Theorem 1]. We consider the general case and give a sufficient condition for that the condition (SC)m holds. Assume c2 > 1. In [12, Theorem 1] we have proved that the condition (SC)m holds if and only if both of the conditions (B B)i and (C A)i hold for i = 1, . . . , m. In this paper we consider the case a1 = 0 < a2 and prove the following result. Proposition 1. Let Γ be a distance-regular graph of diameter d ≥ 3 such that c2 > 1 and a1 = 0 < a2 . Let m be an integer with 1 ≤ m ≤ d − 1. Suppose the condition (B B)i holds for i = 1, . . . , m. Then the condition (C A)i holds for i = 1, . . . , m. As a direct consequence of these results we obtain the following theorem. Theorem 2. Let Γ be a distance-regular graph of diameter d ≥ 3 such that c2 > 1 and a1 = 0 < a2 . Let m be an integer with 1 ≤ m ≤ d − 1. Then the following conditions are equivalent. (i) The condition (SC)m holds. (ii) The condition (SC)i holds for i = 1, . . . , m. (iii) The condition (B B)i holds for i = 1, . . . , m. Let Γ be a distance-regular graph with classical parameters (d, b, α, β) such that b < −1 which is called negative type (see Section 5 for the definitions). In [21] C.-W. Weng studied such distance-regular graphs and classified them if d ≥ 4, c2 > 1 and

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a1 = 0. In [16, Theorem 1.1] Y.-J. Pan, M.-H. Lu and C.-W. Weng showed that if c2 > 1 and a1 = 0 < a2 , then such a graph has no parallelograms of length i + 1 for i = 2, . . . , d − 1 (i.e., the condition (B B)i holds for i = 2, . . . , d − 1). It follows, by Theorem 2, that the condition (SC)i holds for i = 2, . . . , d − 1. Using this fact we can prove the following result. Theorem 3. Let Γ be a distance-regular graph of diameter d ≥ 3 such that c2 > 1 and a1 = 0 < a2 . Suppose Γ has classical parameters (d, b, α, β). Then Γ satisfies the either (b, α, β) = (−2, −3, −1 − condition (SC)i for i =   1, . . . , d − 1. In dparticular, 1+(−3) d holds. (−2) ) or (b, α, β) = −3, −2, − 2 This paper is organized as follows. In Section 2, we recall some definitions and basic terminologies for distance-regular graphs and strongly closed subgraphs. We collect several known results for strongly closed subgraphs and give some consequences. In Section 3 we prove Proposition 1 and Theorem 2. In Section 4 we recall several results obtained by counting strongly closed subgraphs. In Section 5 we study distance-regular graphs with classical parameters (d, b, α, β). Finally we prove Theorem 3. 2. Preliminaries First we recall our notation and terminologies. Let Γ = (V Γ, EΓ ) be a connected graph with usual distance ∂Γ and diameter d = d(Γ ). For a vertex u in Γ we denote by Γ j (u) the set of vertices which are at distance j from u, where Γ−1 (u) = Γd+1 (Γ ) = ∅. For two vertices x and y in Γ with ∂Γ (x, y) = i, let C(x, y) := Γi−1 (x) ∩ Γ1 (y), A(x, y) := Γi (x) ∩ Γ1 (y), B(x, y) := Γi+1 (x) ∩ Γ1 (y). Definition 4. Let i be an integer with 0 ≤ i ≤ d. (i) We say ci (Γ )-exists if ci (Γ ) = |C(x, y)| is a constant whenever ∂Γ (x, y) = i. (ii) We say ai (Γ )-exists if ai (Γ ) = |A(x, y)| is a constant whenever ∂Γ (x, y) = i. (iii) We say bi (Γ )-exists if bi (Γ ) = |B(x, y)| is a constant whenever ∂Γ (x, y) = i. A connected graph Γ of diameter d is said to be distance-regular if ci (Γ )-exists and bi (Γ )-exists for all i = 0, . . . , d. Then Γ is a regular graph of valency k = k(Γ ) = b0 (Γ ) and ai (Γ )-exists with ai (Γ ) = k(Γ ) − ci (Γ ) − bi (Γ ) for all i = 0, . . . , d. We remark that c0 (Γ ) = a0 (Γ ) = bd (Γ ) = 0 and c1 (Γ ) = 1. The constants ci (Γ ), ai (Γ ) and bi (Γ ) (i = 0, . . . , d) are called the intersection numbers of Γ. Let u be a vertex of Γ. Then it is well known that |Γi (u)| =

b0 (Γ ) · · · bi−1 (Γ ) c1 (Γ ) · · · ci (Γ )

(1)

for i = 1, 2, . . . , d which depends only on i rather than the choice of u (see [2, § 4.1A]).

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We refer the reader to [1, 2] for more background information about distanceregular graphs The rest of this paper let Γ be a distance-regular graph of diameter d = d(Γ ) ≥ 2 and valency k = k(Γ ) ≥ 3. We write ci , ai and bi the intersection numbers ci (Γ ), ai (Γ ) and bi (Γ ) of Γ. Next we recall the definition of strongly closed subgraphs and collect several known facts for strongly closed subgraphs. Definition 5. Let Δ be a non-empty subset of vertices in Γ. We identify Δ with the induced subgraph on it. Let x ∈ Δ. A subgraph Δ is called strongly closed with respect to x if C(x, y) ∪ A(x, y) ⊆ Δ for any y ∈ Δ. A subgraph Δ is called strongly closed if it is strongly closed with respect to z for any z ∈ Δ. It is clear that a strongly closed subgraph is geodetically closed. The following are well known basic facts. Lemma 6. Let Δ ⊆ V Γ, x, y ∈ Δ and z ∈ V Γ. (i) If Δ is strongly closed with respect to x and ∂Γ (x, z) + ∂Γ (z, y) ≤ ∂Γ (x, y) + 1, then z ∈ Δ. (ii) If Δ is strongly closed, then ci (Δ)-exists and ai (Δ)-exists with ci (Δ) = ci and ai (Δ) = ai for i = 0, 1, . . . , d(Δ). (iii) Suppose Δ is a strongly closed subgraph of diameter m := d(Δ). If Δ is regular, then it is distance-regular with ci (Δ) = ci and bi (Δ) = bi − bm for i = 0, 1, . . . , m.  We remark that there exist several examples of non-regular strongly closed subgraphs in a distance-regular graph (see [17, Theorem 1.4–1.5], [9, §2] and [10]). Let G be a connected graph. We define the n-subdivision graph of G, denote by G (n) , the graph obtained from G by replacing each edge by a path of length n. The following result is proved by H. Suzuki in [17, Theorem 1.1]. Theorem 7. Let Γ be a distance-regular graph with r = max{i | (ci , ai , bi ) = (c1 , a1 , b1 )}. Let Δ be a strongly closed subgraph of diameter m. Then one of the following holds. (i) Δ is distance-regular. (ii) 2 ≤ m ≤ r. (iii) Δ is distance-biregular. Moreover r ≡ m ≡ 0 (mod 2) and c2i−1 = c2i for all i with 2i ≤ m. (3) (iv) Δ is the 3-subdivision graph K k+1 of a complete graph K k+1 or the 3-subdivision (3)

graph Mk of a Moore graph Mk . Moreover m = r + 2 ∈ {5, 8}, a1 = 0 and (cr +1 , ar +1 , br +1 ) = (cr +2 , ar +2 , br +2 ) = (1, 1, k − 2). In particular, c2 = 1 and (cm−1 , am−1 , bm−1 ) = (cm , am , bm ) except the case (i).



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Since the intersection of strongly closed subgraphs is also strongly closed unless it is the empty set, we obtain the following lemma as a direct consequence of this theorem. Lemma 8. Let m be an integer with 2 ≤ m ≤ d − 1. Let (x, y) be a pair of vertices in Γ with ∂Γ (x, y) = m. Suppose c2 > 1 and the condition (SC)m holds. Then the following hold. (i) There exists a unique strongly closed subgraph Δ = Δ(x, y) of diameter m containing x and y. In particular, Δ is distance-regular with ci (Δ) = ci , ai (Δ) = ai and bi (Δ) = bi − bm for i = 0, . . . , m. (ii) If a strongly closed subgraph Ψ contains x and y, then Δ(x, y) ⊆ Ψ. In particular, Ψ satisfies the condition (SC)m . Proof. Since c2 > 1, any strongly closed subgraph of Γ is distance-regular by Theorem 7. (i) Since the condition (SC)m holds, there exists a strongly closed subgraph Δ = Δ(x, y) of diameter m containing x and y. Then Lemma 6 (iii) shows that Δ is a distance-regular graph with ci (Δ) = ci , ai (Δ) = ai and bi (Δ) = bi − bm for i = 1, . . . , m. Let Δ be any strongly closed subgraph of diameter m containing x and y. Set Λ = Δ ∩ Δ . Since x, y ∈ Λ ⊆ Δ, Λ is a strongly closed subgraph of diameter m, and hence it is a distance-regular graph with the same intersection numbers as Δ. It follows that Λ = Δ and thus Δ ⊆ Δ . By symmetry Δ ⊇ Δ and the uniqueness of Δ = Δ(x, y) is proved. (ii) Set Λ = Δ ∩ Ψ. Then Λ is a strongly closed subgraph of diameter m con taining x and y. It follows, by (i), that Λ = Δ and thus Δ ⊆ Ψ. Since a strongly closed subgraph of diameter 1 is a clique of size a1 + 2, it is straightforward to see the following fact. Lemma 9. The following conditions are equivalent. (i) (ii) (iii) (iv) (v)

The condition (B B)1 holds. Γ contains no induced subgraph K 2,1,1 . 0 Γ1 (u) is a disjoint union of a1b+1 cliques of size a1 + 1 for each vertex u in Γ. Each edge lies on a clique of size a1 + 2.  The condition (SC)1 holds.

We have claimed in Section 1 that the condition (B B)i holds if and only if there is no parallelogram of length i + 1. Similarly it is straightforward to see that the condition (C A)i holds if and only if there is no 6-tuple (u, v, w, x, y, z) of vertices such that ∂Γ (x, z) = ∂Γ (y, z) = i, ∂Γ (x, u) = ∂Γ (y, u) = ∂Γ (x, v) = ∂Γ (y, v) = ∂Γ (z, w) = 1, ∂Γ (x, y) = 2, ∂Γ (z, u) = ∂Γ (z, v) = i − 1, ∂Γ (x, w) ≤ i and ∂Γ (y, w) = i + 1. The next lemma suggests us that the definitions of the conditions (B B)i and (C A)i are natural to observe strongly closed subgraphs (see [12, Lemma 5]).

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Lemma 10. Let m be an integer with 2 ≤ m ≤ d − 1. (i) Let Λ be a strongly closed subgraph of diameter m. Let x, y and z be vertices in Λ with ∂Γ (x, z) = m. Then C(x, z)∪ A(x, z) ⊇ C(y, z)∪ A(y, z) and B(x, z) ⊆ B(y, z). If ∂Γ (y, z) = m, then C(x, z) ∪ A(x, z) = C(y, z) ∪ A(y, z) and B(x, z) = B(y, z). (ii) Suppose the condition (SC)m holds. Then the conditions (B B)m and (C A)m hold. (i) Since d(Λ) = m, we have B(x, z) ∩ Λ = ∅ and Γ1 (z) ∩ Λ = C(x, z) ∪ A(x, z). Hence we obtain C(y, z) ∪ A(y, z) ⊆ Γ1 (z) ∩ Λ = C(x, z) ∪ A(x, z) and B(x, z) ⊆ B(y, z). The second assertion follows by comparing the sizes of both sides. (ii) Let (x, y, z) be a triple of vertices with ∂Γ (x, z) = ∂Γ (y, z) = m. Let Δ be a strongly closed subgraph of diameter m containing x and z. If ∂Γ (x, y) = 1, then y ∈ A(z, x) ⊆ Δ. Thus B(x, z) = B(y, z) by (i). The condition (B B)m holds. If ∂Γ (x, y) = 2 and there exist two distinct vertices w and w in C(z, x) ∩C(z, y), then w, w  ∈ C(z, x) ⊆ Δ and thus y ∈ Γ1 (w) ∩Γ1 (w  ) ⊆ Δ. Hence C(x, z) ∪ A(x, z) = C(y, z) ∪ A(y, z) by (i). The conditions (C A)m holds. 

Proof.

Let (x, z) be a pair of vertices at distance m in Γ. Let Ψ0 = {x, z}. Define Ψi = {y | ∂Γ (w, y) + ∂Γ (y, w  ) ≤ ∂Γ (w, w  ) + 1 for some w = w  ∈ Ψi−1 } for any integer i by inductive procedure. Then {x, z} = Ψ0 ⊆ Ψ1 ⊆ · · · ⊆ Ψi ⊆ · · · holds. Suppose there exists a strongly closed subgraph Δ of diameter m containing x and z. It follows, by Lemma 6 (i) and induction on i, that Ψi ⊆ Δ for any i. Let y ∈ Ψi for some i such that ∂Γ (z, y) = m. Then y ∈ Δ and C(x, z) ∪ A(x, z) = Γ1 (z) ∩ Δ = C(y, z) ∪ A(y, z),

B(x, z) = B(y, z).

This gives us a necessary condition for that the condition (SC)m holds. The conditions (B B)m and (C A)m are some specific cases. Another specific case of this fact has been studied in [4, 5]. Their main results show that a necessary and sufficient condition for the condition (SC)m holds is that C(x, z)∪ A(x, z) = C(y, z)∪ A(y, z) holds for some specific triple (x, y, z) of vertices with ∂Γ (x, z) = ∂Γ (y, z) = m (see [4, 5] for more detailed). In [12] we proved that for distance-regular graphs with c2 > 1 the condition (SC)m holds if and only if both of the conditions (B B)i and (C A)i hold for i = 1, . . . , m (see [12, Theorem 1]). This is a specialization of the results in [4, 5] for the case c2 > 1. In [8, Theorem 1] we have proved that if the condition (SC)m holds, then the condition (SC)i holds for i = 1, . . . , m. Hence we have the following result. Proposition 11. Let Γ be a distance-regular graph of diameter d ≥ 3 and c2 > 1. Let m be an integer with 1 ≤ m ≤ d − 1. Then the following conditions are equivalent.

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(i) The condition (SC)m holds. (ii) The condition (SC)i holds for all i with 1 ≤ i ≤ m. (iii) The conditions (B B)i and (C A)i hold for all i with 1 ≤ i ≤ m. Proof. This is a direct consequence of [8, Theorem 1] and [12, Theorem 1].



To close this section we prove the following result. Lemma 12. Let m be an integer with 1 ≤ m ≤ d − 1. Suppose the condition (B B)s holds for s = 1, . . . , m. Then for any integers i, j, h with 1 ≤ i ≤ j ≤ h ≤ m the following hold. (i) Let x, y and z be vertices such that ∂Γ (x, z) = j and ∂Γ (x, y)+∂Γ (y, z) ≤ j +1. Then B(x, z) ⊆ B(y, z). (ii) There is no quadruple (x, y, z, w) of vertices such that ∂Γ (x, w) = h + 1, ∂Γ (z, w) = 1, ∂Γ (x, z) = h, ∂Γ (y, w) = ∂Γ (y, z) = j and ∂Γ (x, y) = h+1− j. (iii) There is no quadruple (x, y, z, w) of vertices such that ∂Γ (x, w) = h + 1, ∂Γ (z, w) = i, ∂Γ (x, z) = h + 1 − i, ∂Γ (y, w) = j, ∂Γ (x, y) = h + 1 − j and ∂Γ (y, z) = j + 1 − i. (iv) Let u and v be vertices at distance h + 1. Then there are no edges in C(u, v). Proof.

(i) If ∂Γ (x, y) + ∂Γ (y, z) = j, then the assertion follows by the triangle inequalities. Suppose ∂Γ (x, y) + ∂Γ (y, z) = j + 1 and we prove the assertion by the induction on i := ∂Γ (x, y). If i = 1, then the assertion follows by the condition (B B) j . Assume i ≥ 2. Let w ∈ C(x, y). Then ∂Γ (x, w) + ∂Γ (w, z) ≤ (∂Γ (x, y) − 1) + (∂Γ (y, z) + 1) = j + 1 and ∂Γ (w, y) + ∂Γ (y, z) = 1 + (∂Γ (x, z) − ∂Γ (x, w)) ≤ 1 + ∂Γ (w, z).

Hence B(x, z) ⊆ B(w, z) ⊆ B(y, z) by the inductive hypothesis. The assertion is proved. (ii) If such a quadruple (x, y, z, w) exists, then w ∈ B(x, z)\B(y, z) which contradicts (i). (iii) We prove the assertion by induction on i. The case i = 1 is proved in (ii). Suppose i ≥ 2 and there exists a quadruple (x, y, z, w) of vertices as in the statement. Let z  ∈ C(w, z). Then z  ∈ B(x, z) ⊆ B(y, z) by (i). The quadruple (x, y, z  , w) of vertices contradicts the inductive hypothesis. The desired result is proved. (iv) Suppose there exists an edge ( p, p  ) in C(u, v). Then the quadruple (u, p, p  , v) of vertices contradicts (ii). The lemma is proved.  The similar result of Lemma 12 is proved in [11, Lemma 7]. Lemma 12 (iv) had been already proved in [7, Lemma 3.3].

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3. Proof of Theorem 2 In this section we prove Proposition 1 and Theorem 2. Throughout this section Γ denotes a distance-regular graph of diameter d ≥ 3, valency k ≥ 3, c2 > 1 and a1 = 0 < a2 . A 5-tuple (x0 , x1 , x2 , x3 , x4 ) of distinct vertices is called a pentagon if ∂Γ (xi , xi+1 ) = 1 for i = 0, 1, 2, 3, 4, where indices are given by modulo 5. Then we have ∂Γ (xi , xi+2 ) = 2 for i = 0, 1, 2, 3, 4 since a1 = 0. Let p0 and p2 be vertices at distance 2. Take p1 ∈ C( p0 , p2 ), p3 ∈ A( p0 , p2 ) and p4 ∈ C( p0 , p3 ). Then ( p0 , p1 , p2 , p3 , p4 ) forms a pentagon. Let Δ be a strongly closed subgraph in Γ such that p0 , p2 ∈ Δ. Then p1 ∈ C( p0 , p2 ) ⊆ Δ, p3 ∈ A( p0 , p2 ) ⊆ Δ, p4 ∈ C( p0 , p3 ) ⊆ Δ and hence this pentagon is contained in Δ. This simple fact is a key of our proof. Lemma 13. Let u be a vertex and (x0 , x1 , x2 , x3 , x4 ) be a pentagon in Γ. Let m and t be integers with 1 ≤ t ≤ m. Suppose the condition (B B)i holds for i = 1, . . . , m. Then the following hold. (i) If x0 ∈ Γt−1 (u), x1 ∈ Γt (u) and x2 ∈ Γt+1 (u), then x3 ∈ Γt+1 (u) and x4 ∈ Γt (u). (ii) If x0 , x1 , x2 , x3 ∈ Γt (u), then x4 ∈ Γt+1 (u). (iii) If x1 , x3 ∈ Γt−1 (u) and x0 , x2 , x4 ∈ Γt (u), then B(x0 , u) = B(x2 , u) = B(x4 , u). (iv) If x1 , x3 , x4 ∈ Γt−1 (u) and x0 , x2 ∈ Γt (u), then B(x0 , u) = B(x2 , u). (i) We prove the assertion by induction on t. The case t = 1 is clear since a1 = 0. Suppose t ≥ 2. Let v ∈ C(x0 , u) ⊆ C(x1 , u) ⊆ C(x2 , u). Then we have x3 ∈ Γt (v) by the inductive hypothesis. Thus u ∈ B(x2 , v) = B(x3 , v) by the condition (B B)t . It follows that x3 ∈ Γt+1 (u) and x4 ∈ Γt (u). The assertion follows. (ii) We prove the assertion by induction on t. Suppose x4 ∈ Γt+1 (u) and derive a contradiction. If t = 1, then no such pentagon exists since a1 = 0. Assume t ≥ 2. The condition (B B)t implies that B(x0 , u) = B(x1 , u) = B(x2 , u) = B(x3 , u). We prove C(x0 , u) = C(x1 , u) = C(x2 , u) = C(x3 , u). If there exists u  ∈ C(x1 , u)\C(x0 , u), then u  ∈ A(x0 , u) as B(x0 , u) = B(x1 , u). Thus we have x4 ∈ B(u, x0 ) = B(u  , x0 ) by the condition (B B)t . It follows, by (i), that x3 ∈ Γt+1 (u  ). Then u  ∈ B(x3 , u)\B(x1 , u) which is a contradiction. Hence we obtain C(x1 , u) = C(x0 , u), and C(x2 , u) = C(x3 , u) by symmetry. If there exists u  ∈ C(x0 , u)\C(x3 , u). Then u  ∈ A(x3 , u) as B(x3 , u) = B(x0 , u). Since ∂Γ (u  , x4 ) ≤ ∂Γ (u  , x0 ) + ∂Γ (x0 , x4 ) = t, x4 ∈ B(u, x3 )\B(u  , x3 ) which contradicts the condition (B B)t . Then C(x0 , u) = C(x3 , u). The claim is proved. Let v ∈ C(x0 , u) = C(x1 , u) = C(x2 , u) = C(x3 , u). Then the vertex v and the pentagon (x0 , x1 , x2 , x3 , x4 ) contradict the inductive hypothesis. The desired result is proved. (iii) We have B(x0 , u) = B(x4 , u) by the condition (B B)t . We prove B(x0 , u) = B(x2 , u). Let w ∈ B(x2 , u). Then x1 , x3 ∈ C(u, x2 ) ⊆ C(w, x2 ) and thus

Proof.

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∂Γ (w, x1 ) = ∂Γ (w, x3 ) = t. Suppose ∂Γ (w, x0 ) = t − 1. Then x0 ∈ Γt−1 (w), x1 , x3 ∈ Γt (w) and x2 ∈ Γt+1 (w) which contradicts (i). Suppose ∂Γ (w, x4 ) = t − 1. Then x4 ∈ Γt−1 (w), x1 , x3 ∈ Γt (w) and x2 ∈ Γt+1 (w) which contradicts (i). Suppose ∂Γ (w, x0 ) = ∂Γ (w, x4 ) = t. Then x3 , x4 , x0 , x1 ∈ Γt (w) and x2 ∈ Γt+1 (w) which contradicts (ii). Hence B(x2 , u) ⊆ B(x0 , u) = B(x4 , u). The desired result follows by |B(x2 , u)| = |B(x0 , u)| = |B(x4 , u)|. (iv) Let w ∈ B(x2 , u). Then x1 , x3 ∈ C(u, x2 ) ⊆ C(w, x2 ) and ∂Γ (w, x1 ) = ∂Γ (w, x3 ) = t. By the condition (B B)t−1 , we have w ∈ B(x3 , u) = B(x4 , u) and thus ∂Γ (w, x4 ) = t. Suppose ∂Γ (w, x0 ) = t − 1. Then x0 ∈ Γt−1 (w), x1 , x3 ∈ Γt (w) and x2 ∈ Γt+1 (w) which contradicts (i). Suppose ∂Γ (w, x0 ) = t. Then x3 , x4 , x0 , x1 ∈ Γt (w) and x2 ∈ Γt+1 (w) which contradicts (ii). Hence ∂Γ (w, x0 ) = t + 1 and B(x2 , u) ⊆ B(x0 , u). The desired result follows by  |B(x2 , u)| = |B(x0 , u)|. The lemma is proved. Here we recall the proof of Lemma 10. Let Δ be a strongly closed subgraph of diameter t and x, y, z ∈ Δ with ∂Γ (x, z) = ∂Γ (y, z) = t. Then C(x, z) ∪ A(x, z) = Γ1 (z) ∩ Δ = C(y, z) ∪ A(y, z) and hence B(x, z) = (Γ1 (z) \ Δ) = B(y, z). The conditions (B B)t and (C A)t are special cases of this fact. Lemma 13 also treats some special cases of this fact. Proof of Proposition 1. Let j be a positive integer with j ≤ m. Let (x, y, z) be a triple of vertices with ∂Γ (x, y) = 2, ∂Γ (x, z) = ∂Γ (y, z) = j such that |C(z, x) ∩ C(z, y)| ≥ 2. Then j ≥ 2. Let p1 = p3 ∈ C(z, x) ∩ C(z, y). Then ∂Γ ( p1 , p3 ) = 2 by Lemma 12 (iv). Let p4 ∈ A( p1 , p3 ) and p0 ∈ C( p1 , p4 ). Then ( p0 , p1 , x, p3 , p4 ) and ( p0 , p1 , y, p3 , p4 ) are pentagons. If p0 ∈ Γ j−2 (z), then p0 ∈ Γ j−2 (z), p1 , p3 ∈ Γ j−1 (z) and x ∈ Γ j (z) which contradicts Lemma 13 (i). Hence we obtain p0 ∈ Γ j−2 (z) and p4 ∈ Γ j−2 (z) by symmetry. If both of p0 and p4 are contained in Γ j−1 (z), then z and ( p3 , p4 , p0 , p1 , x) contradicts Lemma 13 (ii). By symmetry we may assume that p0 ∈ Γ j (z) and p4 ∈ Γ j−1 (z) ∪ Γ j (z). Then we obtain B(x, z) = B( p0 , z) = B(y, z) by Lemma 13 (iii),(iv). Therefore the condition (C A) j holds. The proposition is proved.  Proof of Theorem 2. Suppose the condition (B B)i holds for i = 1, . . . , m. Then the condition (C A)i holds for i = 1, . . . , m by Proposition 1. Hence Proposition 11 shows that the three conditions are equivalent. The theorem is proved.  4. Strongly Closed Subgraphs In this section we recall several results obtained by counting strongly closed subgraphs which satisfy some conditions. Lots of such kind of results had been obtained (see, for example, [2, Corollary 4.3.12], [3, §4], [5, Proposition 4.1] and [21, §9]). We will apply these results to prove Theorem 3 in the next section.

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Lemma 14. Let m be an integer with m ≥ 3. Let Γ be a distance-regular graph of diameter m such that c2 > 1. Let ψ = 1 + (c2 + a2 ) +

(c2 + a2 )(c2 + a2 − 1 − a1 ) . c2

(2)

Suppose the condition (SC)2 holds. Then the following holds. (i) Let Δ be a strongly closed subgraph of diameter 2. Then Δ is a distanceregular graph with b0 (Δ) = c2 + a2 , ci (Δ) = ci and ai (Δ) = ai for i = 1, 2. In particular, Δ has ψ vertices. (ii) The number of strongly closed subgraphs of diameter 2 containing a fixed vertex is (cm + am )(cm + am − 1 − a1 ) . (c2 + a2 )(c2 + a2 − 1 − a1 )

(3)

(iii) The number of strongly closed subgraphs of diameter 2 containing a fixed edge is cm + am − 1 − a1 . c2 + a2 − 1 − a1

(4)

(iv) The number of strongly closed subgraphs of diameter 2 in Γ is |V Γ |(cm + am )(cm + am − 1 − a1 ) . ψ(c2 + a2 )(c2 + a2 − 1 − a1 ) Proof.

(5)

(i) Since c2 > 1, the first assertion follows by Theorem 7 and Lemma 6 (iii). Take u ∈ Δ. Then |Δ| = |{u}| + |Γ1 (u) ∩ Δ| + |Γ2 (u) ∩ Δ| = ψ

by (1). The second assertion is proved. (ii) Let Lx be the set of strongly closed subgraphs of diameter 2 containing a fixed vertex x. We count the size of the set Px = {(y, Λ) | Λ ∈ Lx , y ∈ Γ2 (x) ∩ Λ} in two ways. For any y ∈ Γ2 (x) there exists a unique strongly closed subgraph of diameter 2 containing x and y by Lemma 8 (i). For any Λ ∈ Lx there are |Γ2 (x) ∩ Λ| =

(c2 + a2 )(c2 + a2 − 1 − a1 ) c2

choices for y. Hence we obtain (c2 + a2 )(c2 + a2 − 1 − a1 ) b0 b1 · 1 = |Px | = |Lx | · . c2 c2 We remark that b0 = cm + am and b1 = b0 − 1 − a1 . The desired result is proved.

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(iii) Let Lx,x  be the set of strongly closed subgraphs of diameter 2 containing a fixed edge (x, x  ). We count the size of the set Px,x  = {(y, Λ) | Λ ∈ Lx,x  , y ∈ B(x, x  ) ∩ Λ} in two ways. The desired result follows similar to the proof of (ii). (iv) Let L be the set of strongly closed subgraphs of diameter 2 in Γ. We count the size of the set P = {(x, y, Λ) | Λ ∈ L, x, y ∈ Λ, ∂Γ (x, y) = 2} in two ways. Let x ∈ V Γ and y ∈ Γ2 (x). Then there exists a unique strongly closed subgraph Δ(x, y) of diameter 2 containing x and y by Lemma 8 (i). Conversely take any Λ ∈ L. Then there are ψ vertices in Λ by (i) and there 2 −1−a1 ) choices for y. Therefore are (c2 +a2 )(c2c+a 2 |V Γ | ·

b0 b1 (c2 + a2 )(c2 + a2 − 1 − a1 ) · 1 = |P| = |L| · ψ · . c2 c2

Since b0 = cm + am and b1 = b0 − 1 − a1 , the desired result is proved.



Lemma 15. Let m be an integer with m ≥ 3. Let Γ be a distance-regular graph of diameter m such that c2 > 1. Let u and w be vertices at distance m − 1 in Γ. If the c +a condition (SC)m−1 holds, then C(u, w) ∪ A(u, w) is a disjoint union of n = m−11+a1m−1 cliques of size 1 + a1 . In particular, (cm−1 + am−1 )(c2 + a2 − 1 − a1 ) ≤ (1 + a1 )(cm + am − 1 − a1 ).

(6)

Moreover the following conditions are equivalent. (i) The equality holds in (6). (ii) For any triple (u, w, z) of vertices with ∂Γ (u, w) = m − 1 and z ∈ B(u, w), each strongly closed subgraph of diameter 2 containing w and z contains exactly one of n cliques in C(u, w) ∪ A(u, w). (iii) There exists a triple (u, w, z) of vertices with ∂Γ (u, w) = m −1 and z ∈ B(u, w) such that each strongly closed subgraph of diameter 2 containing w and z contains exactly one of n cliques in C(u, w) ∪ A(u, w). Proof. Let Ψ be a strongly closed subgraph of diameter m − 1 containing u and w. Then k(Ψ ) = cm−1 + am−1 and C(u, w) ∪ A(u, w) = Γ1 (w) ∩ Ψ is a disjoint union of n cliques of size 1 + a1 by Proposition 11, Lemma 8 (ii) and Lemma 9. Let {Y1 , . . . , Yn } be the set of n cliques in C(u, w)∪ A(u, w). Take yi ∈ Yi for i = 1, . . . , n. Let z ∈ B(u, w). Then ∂Γ (yi , z) = 2 by Lemma 12 (iv). Set Λi be a strongly closed subgraph of diameter 2 containing yi and z for i = 1, . . . , n. Suppose Λi = Λ j for some i = j. Then yi , y j ∈ Ψ ∩ Λi with ∂Γ (yi , y j ) = 2. It follows, by Lemma 8, that z ∈ Λi = Δ(yi , y j ) ⊆ Ψ which contradicts ∂Γ (u, z) > d(Ψ ). Hence Λ1 , . . . , Λn

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are distinct. Lemma 14 (iii) gives us the number of strongly closed subgraphs of diameter 2 containing the edge (w, z). Therefore cm−1 + am−1 cm + am − 1 − a1 = n ≤ 1 + a1 c2 + a2 − 1 − a1 and (6) holds. Since Yi = ({yi } ∪ A(w, yi )) ⊆ Λi , the second assertion follows.  5. Distance-Regular Graphs with Classical Parameters In this section we prove Theorem 3. First we recall some definitions and terminologies. Let Γ be a distance-regular graph of diameter d ≥ 3. We say Γ has classical parameters (d, b, α, β) if its intersection numbers are     i i −1 1+α , (7) ci = 1 1        d i i − β −α (8) bi = 1 1 1 for i = 0, . . . , d, where     j j = := b j−1 + · · · + b + 1 1 1 b denotes the Gaussian binomial coefficient with basis b. Then we have         i −1 i d i − − β −1+α ai = 1 1 1 1        i i −1 i a1 − α = −1 + 1 1 1

(9)

for i = 1, . . . , d. For more information on distance-regular graphs with classical parameters (d, b, α, β) we refer the reader to [2, §6.1]. It has been known that b is an integer with b ∈ {−1, 0} for a distance-regular graph with classical parameters (d, b, α, β) (see [2, Proposition 6.2.1]). By (9) we have β = 1 − α(bd−1 + · · · + b) if a1 = 0. Let Γ be a distance-regular graph of diameter d ≥ 3 with classical parameters (d, b, α, β) such that c2 > 1. In [20, Theorem 1] C.-W. Weng proved that if a1 = 0 and b < −1, then for any pair (x, y) of vertices there exists a regular strongly closed subgraph of diameter ∂Γ (x, y) (i.e., the condition (SC)i holds for all i = 1, . . . , d − 1). In [16, Lemma 3.3, Theorem 1.1] Y.-J. Pan, M.-H. Lu and C.-W. Weng showed that if a1 = 0 < a2 , then b < −1 and that Γ has no parallelograms of length i + 1 for i = 2, . . . , d − 1 (i.e., the condition (B B)i holds for i = 2, . . . , d − 1). Hence by using this result and applying Theorem 2 we obtain the following result. Corollary 16. Let Γ be a distance-regular graph of diameter d ≥ 3 with classical parameters (d, b, α, β) such that c2 > 1 and a1 = 0 < a2 . Then b < −1 and the condition (SC)i holds for i = 1, . . . , d − 1.

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Proof. The conditions (B B)1 and (SC)1 hold as a1 = 0. By [16, Lemma 3.3, Theorem 1.1] we have that b < −1 and the condition (B B)i holds for i = 2, . . . , d − 1. Hence Theorem 2 implies that the condition (SC)i holds for i = 2, . . . , d − 1.  Lemma 17. Let Γ be a distance-regular graph of diameter d = 3 with classical parameters (d, b, α, β) such that c2 > 1. Suppose the condition (SC)2 holds. Then (c2 − b − 1)(c2 − 2 − (1 − b)a1 ) ≤ 0.

(10)

In particular, if b < −1, then c2 ≤ 2 + (1 − b)a1 . Proof. It follows, by (7) and (9), that c2 = (b + 1)(α + 1), c3 = (b2 + b + 1)(α(b + 1) + 1), 2

(11) 2

a2 = (b + 1)(a1 − α(b + 1)), a3 = (b + b + 1)(a1 − α(b + 1) ).

(12)

Put m = 3 in Lemma 15. Then 0 ≥ (c2 + a2 )(c2 + a2 − 1 − a1 ) − (1 + a1 )(c3 + a3 − 1 − a1 ) = b2 (b + 1)α ( (b − 1)(1 + a1 ) + (b + 1)α ) = b2 (c2 − b − 1)(c2 − 2 − (1 − b)a1 ) by (11) and (12). The desired result is proved.



Lemma 18. Let Γ be a distance-regular graph of diameter d = 3 with classical parameters (d, b, α, β) such that c2 > 1 and a1 = 0 < a2 . Then c2 = 2, a2 = b2 − 1, c3 = (b2 + b + 1)(2 − b), a3 = (b2 + b + 1)(b2 − 1), (13) where α =

1−b b+1

and β = b2 − b + 1. In particular, b ∈ {−2, −3}.

Proof. Corollary 16 and Lemma 17 imply that b < −1 and c2 = 2. Then α = 1−b b+1 and (13) follow by (11) and (12). Hence b0 = c3 + a3 = b4 + b2 + 1. Let x be a vertex in Γ. Then   960 b0 b1 b2 1 f (b) + , (14) |Γ3 (x)| = = c1 c2 c3 2 2−b where f (b) = −b9 − b8 − 4b7 − 7b6 − 15b5 − 30b4 − 60b3 − 120b2 − 240b − 480. Hence 2 − b is a positive divisor of 960. Thus there are only finitely many possible values for b. For each possible value for b we can obtain all intersection numbers of Γ . Then we have the values in (2) and (5) which have to be integers. Hence we obtain b ∈ {−2, −3, −4} since the other parameters are ruled out by Lemma 14 (iv). If b = −4, then a strongly closed subgraph Λ of diameter 2 must be a distanceregular graph with k(Λ) = 17, a1 (Λ) = 0, and c2 (Λ) = 2. But no such graph exists by integrality of the multiplicity of eigenvalues. The lemma is proved. 

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Proof of Theorem 3. The first assertion follows by Corollary 16. If d = 3, then the desired result is proved by Lemma 18. Suppose d ≥ 4. Let Λ be a strongly closed subgraph of diameter 3. Then Λ is a distance-regular graph with ci (Λ) = ci , ai (Λ) = ai and bi (Λ) = bi − b3 for i = 1, 2, 3. It follows that Λ has classical parameters (d  , b , α  , β  ) with (d  , b , α  , β  ) = (3, b, α, b2 − b + 1). Therefore the desired result  is proved by Lemma 18 since β = 1 − α(bd−1 + · · · + b). Let d be a positive integer. Let q be a prime power and Fq 2 be the finite field with elements. Let H be the set of all d × d Hermitian matrix over Fq 2 . The Hermitian forms graph H er (d, q) is the graph whose vertex set is H and two vertices x and y are adjacent whenever rank(x − y) = 1. Then the Hermitian forms graph H er (d, q) is a distance-regular graph of diameter d with

q2

 q i−1 q i − (−1)i ci = q +1

and

bi =

q 2d − q 2i q +1

for

i = 0, . . . , d.

(15)

The Hermitian forms graph H er (d, q) has classical parameters (d, b, α, β), where b = −q, α = −1 − q and β = −1 − (−q)d (see [2, §6.2 §9.5.C]). Ivanov-Shpectorov proved the following result which characterized the Hermitian forms graphs H er (d, 2) by their intersection numbers (see [14], [2, Theorem 9.5.8]). We remark that the Hermitian forms graphs H er (d, q) can be characterized by their intersection numbers for any q (see [15], [18]). Theorem 19. Let Γ be a distance-regular graph of diameter d ≥ 3 whose intersection numbers are as in (15) with q = 2. Then Γ is the Hermitian forms graph H er (d, 2). This result shows that a distance-regular graph with classical parameters (d, b, α, β) such that (b, α, β) = (−2, −3, −1 − (−2)d ) is the Hermitian forms graph H er (d, 2). We will studya distance-regular graph with classical parameters (d, b, α, β) such d in the next paper [13]. that (b, α, β) = −3, −2, − 1+(−3) 2 Acknowledgement. This work was supported by the Grant-in-Aid for Scientific Research, the Ministry of Education, Science and Culture, JAPAN

References 1. Bannai, E., Ito, T.: Algebraic Combinatorics I. California: Benjamin-Cummings, 1984 2. Brouwer, A.E., Cohen, A.M., Neumaier, A.: Distance-Regular Graphs. Heidelberg: Springer Verlag, Berlin, 1989 3. Hiraki, A.: Distance-regular subgraphs in a distance-regular graph, III. Europ. J. Combin. 17, 629–636 (1996) 4. Hiraki, A.: Distance-regular subgraphs in a distance-regular graph V. Europ. J. Combin. 19, 141–150 (1998)

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5. Hiraki, A.: Distance-regular subgraphs in a distance-regular graph, VI. Europ. J. Combin. 19, 953–965 (1998) 6. Hiraki, A.: An application of a construction theory of strongly closed subgraphs in a distance-regular graph. Europ. J. Combin. 20, 271–278 (1999) 7. Hiraki, A.: Strongly closed subgraphs in a regular thick near polygon. Europ. J. Combin. 20, 789–796 (1999) 8. Hiraki, A.: A distance-regular graph with strongly closed subgraphs. J. Alg. Combin. 14, 127–131 (2001) 9. Hiraki, A.: A characterization of the doubled Grassmann graphs, the doubled Odd graphs, and the Odd graphs by strongly closed subgraphs. Europ. J. Combin. 24, 161–171 (2003) 10. Hiraki, A.: A characterization of the Hamming graph by strongly closed subgraphs. European J. Combin. 29, 1603–1616 (2008) 11. Hiraki, A.: A characterization of some distance-regular graphs by strongly closed subgraphs. European J. Combin (to appear). 12. Hiraki, A.: Strongly closed subgraphs in a distance-regular graph with c2 > 1. Graphs and Combin. 24, 537–550 (2008) 13. Hiraki, A.: Distance-regular graphs with classical parameters, (in preparation) 14. Ivanov, A.A., Shpectorov, S.V.: A characterization of the Hermitian forms. Geom. Dedicata 12, 75–85 (1982) 15. Ivanov, A.A., Shpectorov, S.V.: A characterization of the association schemes of Hermitian forms. J. Math. Soc. Japan 43(1), 25–48 (1991) 16. Pan, Y.-J., Lu, M.-H., Weng, C.-W.: Triangle-free distance-regular graphs. J. Alg. Combin. 27, 23–34 (2008) 17. Suzuki, H.: On strongly closed subgraphs of highly regular graphs. Europ. J. Combin. 16, 197–220 (1995) 18. Terwilliger, P.: Kite-free distance-regular graphs. Europ. J. Combin. 16, 405–414 (1995) 19. Weng, C.-W.: D-bounded distance-regular graphs. Europ. J. Combin. 18, 211–229 (1997) 20. Weng, C.-W.: Weak-geodetically closed subgraphs in distance-regular graphs. Graphs and Combin. 14, 275–304 (1998) 21. Weng, C.-W.: Classical distance-regular graphs of Negative type. J. Combin. Theory Ser. B 76, 93–116 (1999) Received: December 22, 2007 Final version received: September 22, 2008

Graphs and Combinatorics (2009) 25:81–90 Digital Object Identifier (DOI) 10.1007/s00373-008-0828-2

Graphs and Combinatorics © Springer-Verlag 2009

Transitive, Locally Finite Median Graphs with Finite Blocks Wilfried Imrich1 , Sandi Klavˇzar2, 1 Montanuniversit¨at Leoben, A-8700 Leoben, Austria.

e-mail: [email protected]

2 Faculty of Mathematics and Physics, University of Ljubljana, Jadranska 19,

1000 Ljubljana, Slovenia. e-mail: [email protected]

Abstract. The subject of this paper are infinite, locally finite, vertex-transitive median graphs. It is shown that the finiteness of the -classes of such graphs does not guarantee finite blocks. Blocks become finite if, in addition, no finite sequence of -contractions produces new cutvertices. It is proved that there are only finitely many vertex-transitive median graphs of given finite degree with finite blocks. An infinite family of vertex-transitive median graphs with finite intransitive blocks is also constructed and the list of vertex-transitive median graphs of degree four is presented. Key words. Median graphs, Infinite graphs, Vertex-transitive graphs. AMS Subject Classification (2000). 05C75, 05C12.

1. Introduction Let u and v be vertices of a connected graph G. The set I (u, v) of all vertices in G that lie on shortest u, v-paths is called the interval between u and v. G is a median graph if for every triple of vertices u, v, w of G there exists a unique vertex in I (u, v) ∩ I (u, w) ∩ I (v, w). This vertex is called the median of u, v, w. The structure of median graphs is well understood, see the survey [10], recent papers [4–6, 13], and references therein. Most of the numerous studies of these graphs focus on finite graphs, nonetheless, [1, 2, 14] are important references for the infinite case. There is at least one aspect of median graphs that is uninteresting on finite graphs but becomes intriguing for infinite graphs—regularity. The variety of k-regular finite median graphs is very modest: the k-cube Q k is the only such graph [12]. The situation significantly changes in the case of infinite graphs. For instance, any finite or infinite median graph G of largest degree d gives rise to a d-regular median graph as follows. Let u be an arbitrary vertex of G of degree smaller than d. Then, at u, attach to G an infinite rooted tree in which the root is of degree d − d(u) and any  Supported by the Ministry of Science of Slovenia under the grant P1-0297. The author is

also with the Faculty of Mathematics and Natural Sciences, University of Maribor, Slovenia and the Institute of Mathematics, Physics and Mechanics, Ljubljana.

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other vertex is of degree d. In fact, Bandelt and Mulder [2] observed that there are 2ℵ0 cubic median graphs. Hence one has to impose additional structure to get more insight into infinite median graphs. Natural candidates for a condition that is stronger than regularity are vertextransitivity and distance-transitivity. By another result of [2], the only distancetransitive median graphs are hypercubes and regular trees. Thus we focus on vertex-transitivity. Suppose that G is an infinite median graph. Following Tardif [14] let us call G compact if G does not contain any isometric ray, that is, a one-sided infinite path that is isometric. Tardif proved that a compact median graph contains a finite subcube that is invariant by any automorphism of G. Therefore a compact median graph is vertex-transitive if and only it is a cube. Since finite graphs are compact, we can formulate this remark as a lemma. Lemma 1. A vertex-transitive median graph is compact if and only if it is a hypercube. Consequently, median graphs of interest to us must be infinite, but not compact. This is assured, for example, if they are locally finite. We thus focus on infinite, locally finite, vertex-transitive median graphs in this paper. The simplest such graphs are homogeneous trees, integer lattices, and Cartesian products of such graphs by hypercubes. Homogeneous trees are special cases of graphs with finite blocks. In our case all blocks have to be median graphs. We will first show that even if all the -classes of a vertex-transitive median graph are finite, the graph need not have finite blocks. On the other hand, if we assume in addition that no finite sequence of -contractions produces new cut-vertices, then blocks are finite. As our main theorem we then prove that there only finitely many vertextransitive median graphs of given finite degree with finite blocks. We also construct an infinite family of vertex-transitive median graphs with finite intransitive blocks. The paper concludes with the list of vertex-transitive median graphs of degree four. There are 13 such graphs. 2. Preliminaries All graphs in this paper are finite or infinite simple graphs, that is, graphs without loops or multiple edges. The two way infinite path will be denoted P∞ and the k-regular tree with Tk . G  H denotes the Cartesian product of the graphs G and H ; Q n is the n-th power of K 2 with respect to the Cartesian product. By a ladder of a graph G we mean an induced subgraph of G isomorphic to Pn  K 2 for some n ≥ 1. Median graphs have already been defined in the introduction. It follows easily from the definition that median graphs are bipartite and that K 2,3 is a forbidden subgraph for median graphs. We also note that the Cartesian product of median graphs is a median graph again, see e.g. [9]. We say two edges e = x y and f = uv of G are in the relation , in symbols ef , if d(x, u) + d(y, v) = d(x, v) + d(y, u) .

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In median graphs  is an equivalence relation. If ab ∈ E(G) we set Fab = {uv | ab  uv}. In other words, Fab is the -class of ab. We further define the sets Wab = {w | w ∈ G, d(w, a) < d(w, b)}, Wba = {w | w ∈ G, d(w, b) < d(w, a)}, Uab = {u | u ∈ Wab , u is adjacent to a vertex in Wba }, Uba = {u | u ∈ Wba , u is adjacent to a vertex in Wab }. As such these sets are sets of vertices, but by abuse of language we shall use the same notation for the subgraphs they induce. We further recall that a subgraph H of a graph G is convex, if all shortest uv-paths in G between any two vertices in u, v ∈ H are also in H . Now everything is ready for a characterization of median graphs by Mulder [11]. The formulation given here is that of [9]. It is a variant of Mulder’s results that is also true for infinite graphs. Theorem 2. Let ab be an edge of a connected, bipartite graph G. Then G is a median graph if and only if the following three conditions are satisfied: (i) Fab is a matching defining an isomorphism between Uab and Uba . (ii) Uab is convex in Wab and Uba in Wba . (iii) Wab and Wba are median graphs. We wish to note that the removal of the edges in Fab separates G into the two connected components Wab and Wba , that no two edges on a shortest path are in the relation , and that any two edges e, f of a median graph that are in the relation  are connected by a ladder. 3. Regular Median Graphs with Finite Blocks It is a folklore fact that G is a median graph if and only if every block of G is median. Hence it is natural to wonder how (median) blocks can be combined to obtain infinite, locally finite vertex-transitive median graphs. Consider first the median graph G of Fig. 1. G has finite -classes but contains an infinite block. (In fact, G is 2-connected.) Clearly G is not vertex-transitive. However, it has only two orbits, namely the set of vertices of degree two and those of degree four. We form a new graph G 1 by identifying each vertex of degree two in G with a vertex of degree four of a copy of G and every vertex of degree four with a vertex of degree two of a copy of G. The construction is illustrated in Fig. 2 where an identification of the first type and another of the second type are performed in black vertices. In addition to the vertices of degree six the new graph G 1 will again have vertices of degrees two and four. By the same construction as before we identify these

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Fig. 1. An infinite 2-connected median graph with finite -classes.

Fig. 2. How to construct G 1 .

vertices with vertices of copies of G to obtain a graph G 2 . Continuing ad infinitum the limit graph G ∞ is clearly vertex-transitive (of degree 6). All of its blocks are isomorphic to G and every cut-vertex identifies vertices of two blocks, one vertex of degree four with one of degree two. Since the -classes of a graph are those of its blocks it is clear that G ∞ has only finite -classes. This construction shows that vertex-transitivity alone does not ensure finite blocks, even if all -classes are finite. A different condition though assures finiteness of blocks: Proposition 3. Let G be a median graph with finite -classes. If no finite sequence of -contractions produces new cut-vertices, then every block of G is finite. Proof. Suppose that B is an infinite block of G and let e = ab be an edge of B. Then e is not a bridge, for otherwise e would contract to a cut-vertex. Hence there exists an edge e0 ∈ Uab . Let e1 ∈ Uba be the edge opposite to e0 and let F be the -class containing e0 , e1 . Let G 1 be the graph obtained from G by contracting Fab and denote with f 1 the edge of G 1 obtained from e0 and e1 . Let B1 be the subgraph of G 1 induced by the contracted block B. By the theorem assumption, B1 is a block of G 1 . Since B1 is median and 2-connected, f 1 is contained in a 4-cycle of G 1 , say C. Let e2 be the edge opposite to f 1 in C. Now contract the -class containing the other two edges of C. Let f 2 be the contracted edge corresponding to f 1 and e2 . Continuing this procedure we obtain a sequence of edges e0 , e1 , e2 , e3 , . . . of G that are all in the same -class, in contradiction to its finiteness. 

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We have observed above that vertex-transitivity does not ensure finite blocks. On the other hand, we are going to prove that there are only finitely many vertextransitive median graphs of a given degree that have finite blocks. We begin with the observation that k-regular median graphs have arbitrarily many orbits if they are sufficiently large. This result is a consequence of the tree-like structure of median graphs, and in particular of the fact that every finite median graph G has a subcube that is invariant under all automorphisms; see [9, Theorem 2.42]. Proposition 4. Every finite median graph of largest degree k on at least (2k)k vertices has at least k orbits. Proof. Let G be a finite k-regular median graph and Q c a subcube that is invariant under all automorphisms of G. Clearly 0 ≤ c ≤ k. Therefore, no vertex of Q c can be mapped by an automorphism into any vertex in V (G)\V (Q c ). Thus, V (Q c ) consists of one or more orbits under the action of the automorphism group of G on V (G). This implies that the set L 1 of vertices at distance 1 from Q c consists of one or more orbits too. By induction this holds for the set L r of vertices of any distance r from Qc. If G has fewer than k orbits, then we infer from the above that G has at most |Q c | + |L 1 | + · · · + |L k−2 | vertices. Since |Q c | ≤ 2k and |L r | ≤ |Q c | · (k − 1)r we obtain the estimate |G| ≤ 2k [1 + (k − 1) + (k − 1)2 + · · · + (k − 1)k−2 ] < 2k · k k−1 < (2k)k .  Note that this result is not true for general isometric subgraphs of hypercubes, as the cycle C2r shows. It has degree 2, can be arbitrarily large, but has only one orbit. (For the definition of partial cubes, their relation to median graphs, and other properties see [3, 7–9] and references therein). Theorem 5. For given k there are only finitely many k-regular vertex-transitive median graphs with finite blocks. Proof. We distinguish finite median graphs, infinite median graphs with vertextransitive blocks and infinite median graphs with at least one intransitive block. (a) As already mentioned, every regular finite median graph is a hypercube, thus Q k is the only finite k-regular vertex-transitive median graph. (b) Let G be a k-regular infinite vertex-transitive median graph with finite, vertex-transitive blocks. Then every block is a hypercube. By transitivity every vertex is a cut-vertex. Let v be an arbitrary vertex of G and n i (v) the number of blocks incident with v that are isomorphic to Q i . Clearly i
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By transitivity n i (v) = n i (w) for any w ∈ G. Hence v determines G and there are only finitely many graphs G for given k. (c) Let G be a k-regular infinite vertex-transitive median graph with finite blocks and B a block with intransitive automorphism group. Since B is finite it has only finitely many orbits, say Oi , 1 ≤ i ≤ k B . Let di denote the degree of the vertices in Oi . For every copy B of B we denote the corresponding orbits by Oi . By transitivity every vertex v of B must be incident, for given i, 1 ≤ i ≤ k B , with i i a vertex  in the orbit Oi of a block B isomorphic to B; compare Fig. 2. Also, set d B = 1≤i≤k B di . Clearly, d B ≤ k. If v is incident with a block C different from the B i , 1 ≤ i ≤ k B , then a similar construction yields blocks Ci isomorphic to C that are incident to v and the degree sum dC with d B + dC ≤ k. Clearly v can be incident to no more than k blocks and each block can have no more than k orbits. To complete the proof we recall that every connected median graph of more than (2k)k vertices has more than k orbits by Proposition 4, because there are only finitely many graphs (and thus median graphs) on any given number of vertices. 

In the last case of the above proof regular infinite vertex-transitive median graphs with finite intransitive blocks were treated. For further illustration construct an infinite family Mm , m ≥ 2, of such graphs as follows: Let L m be the ladder Pm K 2 , then every block B of Mm will be isomorphic to L m . Note that L m has k = m/2 orbits. At every vertex of Mm we identify k blocks B such that every orbit is represented. Then Mm is a vertex-transitive median graph of degree 2 + 3(k − 1). For instance, M2 is shown in Fig. 3.

Fig. 3. 5-regular vertex-transitive median graph with intransitive blocks.

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4. 4-Regular Vertex-Transitive Median Graphs In this section we provide a complete list of 4-regular vertex-transitive median graphs. We first consider the case when the blocks are finite. Proposition 6. The 4-regular vertex-transitive median graphs with finite blocks are Q 4 , T4 , and the graphs G 1 , G 2 , G 3 of Fig. 4. Proof. Let G be a 4-regular vertex-transitive median graph with finite blocks. In G is finite then it is Q 4 . Let G be infinite and u an arbitrary vertex of G. Then u cannot lie in a Q 4 , for otherwise its degree would be greater than 4. If u lies in no 4-cycle, then G is the 4-regular infinite tree T4 . Suppose next that G contains 4-cycles but no Q 3 . If u lies in exactly one 4-cycle, then G is the graph G 1 from Fig. 4. Suppose u lies in at least two 4-cycles. If u is the intersection of two 4-cycles, G must be the graph G 2 from the same figure. Otherwise, two 4-cycles must intersect in an edge uv. Let uvv u and uvv1 u 1 be the 4-cycles containing u. Note that if u 1 v ∈ E(G) or u v1 ∈ E(G) then we get a K 2,3 . This means that the subgraph consisting of these two 4-cycles is induced. Therefore, considering the edge u 1 v1 we find two new vertices u 2 , v2 such that u 1 v1 v2 u 2 is a 4-cycle. The edge u 2 u would lead to a Q 3 . Continuing by induction we see that u i u G2

G1

G3

Fig. 4. 4-regular vertex-transitive median graphs with finite blocks.

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cannot be an edge, for otherwise G would contain an odd cycle or an isometric even cycle. The convex closure of the latter cycle must be a hypercube (cf. [10]) which is not possible. Thus we obtain an infinite ladder, which is 2-connected. Assume finally that G contains 3-cubes and let Q = Q 3 be a cube containing u. Let v be the neighbor of u not in Q. Then the edge uv is in no 4-cycle containing an edge of Q, for otherwise we end up with an infinite block. So G must be the graph  G 3 from Fig. 4. Now the list of all 4-regular vertex-transitive median graphs. Theorem 7. Let G be a vertex-transitive, 4-regular median graph. Then G is one of the following graphs: the Q 4 , the 4-regular tree, P∞ P∞ , or one of the graphs G 1 , . . . , G 10 from Figs. 4 and 5.

Fig. 5. Additional 4-regular vertex-transitive median graphs.

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Proof. From Proposition 6 we already know all vertex-transitive, 4-regular median graphs with finite blocks. So suppose that u is an arbitrary vertex of G and that G contains infinite blocks. Suppose that G contains 4-cycles but no Q 3 . Then from the proof of Proposition 6 we know that u lies in an infinite ladder L. Let v be the fourth neighbor of u, that is, the neighbor of u not in L. If u lies in exactly two 4-cycles we get the graph G 4 from Fig. 5. Suppose u lies in exactly three 4-cycles and that no three 4-cycles share a common edge, see Fig. 6. Then there are two possibilities how the situation can be extended, see Fig. 6 again. This gives us the graphs G 5 and G 6 from Fig. 5. It the three 4-cycles that contain u share an edge, the graph G 7 appears. Note that G 7 = T3  K 2 . The last subcase is when u lies in four 4-cycles. Then the two possibilities as shown in Fig. 7 lead to a K 2,3 and the 3-cube minus a vertex, respectively. The latter case is not possible because the convex closure of it would be a Q 3 . Hence in this subcase we obtain P∞ P∞ . Assume now that u lies in a Q = Q 3 . Let v be the neighbor of u that is not in Q. The case when uv is a bridge was treated in Proposition 6. Hence assume there is a path between v and Q different from the edge vu. A shortest such path must be of length 2, for otherwise the degree of G would be larger than 4. (Having in mind that the convex closure of an isometric cycle in a median graph is a hypercube.) Hence we have a 4-cycle C attached to the 3-cube Q. By the transitivity, every vertex of Q

Fig. 6. The situation when a vertex is in three 4-cycles.

Fig. 7. Two forbidden configurations.

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must have such an attachment. Suppose that C is not in a 3-cube. Then there are two ways how this can be done, either the attachments are done on the edges of the same -class of Q or in two edges from one class and in two edges from another. This gives the graphs G 8 and G 9 . Finally, if C lies in a 3-cube Q , then |Q ∩ Q| = 4, for otherwise the degree of G would be more than four. But then G 10 is the graph.  (Note that G 10 is the Cartesian product of C4 by P∞ .) 5. Concluding Remarks The (unique) extendibility in a tree-like fashion used in this paper is a consequence of the fact that a median graph has no convex cycles of length more than four. In fact, as we already mentioned, the convex closure of such a cycle (having in mind that convex subgraphs are isometric) is a hypercube. We have no example of a strip that is a median graph and not of the form Q n P∞ . However, we cannot show that no other such strips exist, except for strips that are Cayley graphs or strips that have an infinite -class. This will be treated in a forthcoming paper. References 1. Bandelt, H.-J.: Retracts of hypercubes. J. Graph Theory 8, 501–510 (1984) 2. Bandelt, H.-J., Mulder, H.M.: Infinite median graphs, (0, 2)-graphs, and hypercubes. J. Graph Theory 7, 487–497 (1983) 3. Breˇsar, B., Klavˇzar, S.: Crossing graphs as joins of graphs and Cartesian products of median graphs. SIAM J. Discrete Math. 21, 26–32 (2007) ˇ 4. Breˇsar, B., Klavˇzar, S., Skrekovski, R.: On cube-free median graphs. Discrete Math. 307, 345–351 (2007) 5. Breˇsar, B., Tepeh Horvat, A.: On the geodetic number of median graphs. Discrete Math. 308, 4044–4051 (2008) 6. Choe, Y.B., Huber, K.T., Koolen, J.H., Kwon, Y.S., Moulton, V.: Counting vertices and cubes in median graphs of circular split systems. European J. Combin. 29, 443–456 (2008) 7. Deza, M., Laurent, M.: Geometry of Cuts and Metrics. Springer-Verlag, Berlin (1997) 8. Eppstein, D.: The lattice dimension of a graph. European J. Combin. 26, 585–592 (2005) 9. Imrich, W., Klavˇzar, S.: Product Graphs: Structure and Recognition. Wiley-Interscience, New York (2000) 10. Klavˇzar, S., Mulder, H.M.: Median graphs: characterizations, location theory and related structures. J. Combin. Math. Combin. Comp. 30, 103–127 (1999) 11. Mulder, H.M.: The structure of median graphs. Discrete Math. 24, 197–204 (1978) 12. Mulder, H.M.: n-Cubes and median graphs. J. Graph Theory 4, 107–110 (1980) 13. Peterin, I.: A characterization of planar median graphs. Discuss. Math. Graph Theory 26, 41–48 (2006) 14. Tardif, C.: On compact median graphs. J. Graph Theory 23, 325–336 (1996) Received: March 30, 2008 Final version received: November 18, 2008

Graphs and Combinatorics (2009) 25:91–99 Digital Object Identifier (DOI) 10.1007/s00373-008-0838-0

Graphs and Combinatorics © Springer-Verlag 2009

Bounding the Size of Equimatchable Graphs of Fixed Genus Ken-ichi Kawarabayashi1 , Michael D. Plummer2 1 National Institute of Informatics 2 - 1 - 2 Hitotsubashi, Chiyoda-ku Tokyo 101-8430,

Japan. e-mail: k [email protected]

2 Department of Mathematics Vanderbilt University Nashville, TN 37240, USA.

e-mail: [email protected]

Abstract. A graph G is said to be equimatchable if every matching in G extends to (i.e., is a subset of) a maximum matching in G. In an earlier paper with Saito, the authors showed that there are only a finite number of 3-connected equimatchable planar graphs. In the present paper, this result is extended by showing that in a surface of any fixed genus (orientable or non-orientable), there are only a finite number of 3-connected equimatchable graphs having a minimal embedding of representativity at least three. (In fact, if the graphs considered are non-bipartite, the representativity three hypothesis may be dropped.) The proof makes use of the Gallai-Edmonds decomposition theorem for matchings. Key words. Embedded graph, Genus, Matching, Equimatchable, Surface. MR Subject Classification: 05C10, 05C70.

1. Introduction All graphs herein are assumed to be simple; i.e., there are no loops or multiple edges. A set M of edges in a graph G is a matching if no two members of M share a vertex. A matching M in G is maximum if, among all matchings in G, it is one of largest cardinality. A matching M in G is perfect if every vertex of G is covered by an edge of M. Matching M is a k-matching if |M| = k. A graph in which every matching extends to a perfect matching is said to be randomly matchable. More generally, a graph in which every matching extends to a maximum matching is called equimatchable. Randomly matchable graphs were first characterized by Sumner [20] (see also [Theorem 1; 8] for a different proof) who showed that the only such graphs are either even and complete (K 2n ) or balanced complete bipartite (K n,n ). The more interesting family of equimatchable graphs were first characterized by Meng [11] and Lewin [9], but unfortunately neither of these characterizations gives much insight into the structure of such graphs nor do they provide a good characterization of equimatchable graphs from the point of view of complexity. That is to say, neither of these characterizations provides a polynomial algorithm for verifying membership in the class and for verifying non-membership in the class.

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In [8] this difficulty was overcome with the presentation of a good algorithm for membership in the class of equimatchable graphs. But this result, rather technical in nature, provided little real insight into the structure of equimatchable graphs in general. In [6], the present authors, together with A. Saito, determined that there are exactly twenty-three 3-connected planar equimatchable graphs. In the present paper we will obtain an extension of this result by showing that any 3-connected equimatchable graph minimally embedded in a surface  of Euler characteristic χ () has bounded order, provided it has an embedding of representativity at least three there, and hence that there are only a finite number of 3-connected equimatchable graphs minimally embeddable with representativity three in a surface of fixed genus. In the special case when a graph has a perfect matching, one can study the property of n-extendability. (See [15].) Let n ≥ 1 be an integer. A graph G having at least 2n + 2 vertices is said to be n-extendable if every matching M ⊆ E(G) with |M| = n, extends to (i.e., is a subset of) a perfect matching in G. Results somewhat similar in flavor to those of the present paper, but dealing rather with the concept of n-extendability of embedded graphs, have also recently been obtained by Aldred and the present authors [1]. A surface is a connected compact Hausdorff space which is locally homeomorphic to an open disc in the plane. If a surface  is obtained from the sphere by adding some number g ≥ 0 of handles (respectively, some number g > 0 of crosscaps),  is said to be orientable of genus g = g() (resp. non-orientable of genus g = g()). An embedding of a graph G on the orientable surface (resp. non-orientable surface)  is minimal if G cannot be embedded on any orientable (resp. non-orientable) surface   where g(  ) < g() (resp. g(  ) < g()). A graph G is said to have orientable genus g (respectively, non-orientable genus g) if G minimally embeds on a surface of orientable genus g (respectively, non-orientable genus g). In this situation we write γ (G) = g (respectively, γ (G) = g). An embedding of a graph G on surface  is said to be 2-cell if every face of the embedding is homeomorphic to a disc. For 2-cell embeddings, we have the important classical result of Euler: Theorem 1.1. If a connected graph G is 2-cell embedded on surface  having genus g (resp. non-orientable genus g) and if the embedded G has |V (G)| = p vertices, |E(G)| = q edges and |F(G)| = f faces, then p − q + f = 2 − 2g (resp. p − q + f = 2 − g). It is known that every minimal embedding of a graph in an orientable surface must be 2-cell ([22]) and that there exists a minimal embedding of any graph in a non-orientable surface which is 2-cell ([14]). The representativity (or face-width) of a graph embedded on a surface  is the smallest number k such that  contains a noncontractible closed curve that intersects the graph in at least k points. We shall also make use of the concept of “Euler Contribution”. Let v be any vertex of a graph G minimally embedded on an orientable surface of genus g (resp. embedded on a non-orientable surface of genus g).

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Define the Euler contribution of vertex v to be φ(v) = 1 −

deg v 1 deg v , + 2 fi i=1

where the sum runs over the face angles at vertex v and f i denotes the size of the ith face at v. (One should keep in mind here that a face may contribute more than one face angle at a vertex v. There is an embedding of K 5 on the torus with this property, for example.) The next result is essentially due to Lebesgue [7]. Lemma 1.2. If a connected graph G is minimally  embedded on a surface of orientable genus g (resp. non-orientable genus g), then v φ(v) = 2 − 2g (resp. 2 − g). In the present paper, we show that if a 3-connected equimatchable graph of genus g (respectively, non-orientable genus g) is non-bipartite, or if it is bipartite and has representativity at least three in a minimal embedding, then the order of G is bounded above by a certain function of g (respectively, g), and hence there are only a finite number of such graphs with this genus. 2. Existence of Vertex-isolating Matchings in Embedded Graphs Our first result deals with arbitrary 3-connected graphs minimally embedded on a surface (orientable or non-orientable) of fixed genus. Given a vertex v ∈ V (G), a matching M ⊆ E(G) is said to isolate v if M covers N (v), but not v. As is customary, we shall denote the minimum degree in graph G by δ(G). Theorem 2.1. Suppose G is 3-connected of orientable genus g with |V (G)| > max{8, 24g − 24} or of non-orientable genus g with |V (G)| > max{8, 12g − 24}. Then (i) 3 ≤ δ(G) ≤ 6; (ii) if δ(G) = 3, for every vertex v ∈ V (G) with deg v = 3 there is a matching Mv ⊆ E(G) with |Mv | ≤ 3 which isolates v; (iii) while if 4 ≤ δ(G) ≤ 6, then for every vertex v ∈ V (G) such that deg v = δ(G), there is either a matching Mv ⊆ E(G) with |Mv | ≤ 4 which isolates v or else such an Mv which isolates a neighbor of v. Proof. We shall prove the result for orientable embeddings. The proof in the non-orientable case is completely analogous to that in the orientable case. Let G be a connected graph minimally embedded on a surface of genus g ≥ 0 and suppose that |V (G)| > max{8, 24g − 24}. It is then a simple consequence of Euler’s formula that G has a vertex v with deg v ≤ 6 and hence since G is 3-connected, that 3 ≤ deg v ≤ 6. (See Lemma 2.2(a) of [21]). Our strategy will now be to show that either there is a vertex v in G and a matching Mv with |Mv | ≤ 4 isolating v or a neighbor of v or else we get a contradiction of Lemma 1.2.

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1. Suppose first that δ(G) = 3 and that v is a vertex of degree three. Let the three neighbors of v be x1 , x2 and x3 . 1.1. First, assume that there is at least one triangle at v. Without loss of gen/ {v, x1 , x2 }, then erality, denote this triangle by vx1 x2 v. If x3 has a neighbor y ∈ Mv = {x1 x2 , x3 y} is a matching of size 2 which isolates vertex v. So suppose there is no such y. Then G ∼ = K 4 and hence |V (G)| = 4, a contradiction. 1.2. Suppose then that there is no triangle at v. Since δ(G) ≥ 3, there is a / {v, x1 , x2 , x3 } which is adjacent to x1 . Since x2 is adjacent to neither vertex x4 ∈ / x1 nor x3 , choose a vertex x5 ∈ N (x2 ) − {v, x1 , x3 , x4 }. If x3 has a neighbor y ∈ {v, x1 , x2 , x4 , x5 }, then Mv = {x1 x4 , x2 x5 , x3 y} is a 3-matching isolating vertex v. So suppose that N (x3 ) = {v, x4 , x5 }. 1.2.1. Suppose first that N (x1 ) = {v, x4 , x5 }. If N (x2 ) = {v, x4 , x5 }, then either {x4 , x5 } is a 2-cut in G or else |V (G)| = 6. But either of these situations results in / {v, x1 , x3 , x4 , x5 }. a contradiction. So we may suppose that x2 has a neighbor y ∈ But then Mv = {x1 x5 , x2 y, x3 x4 } is a 3-matching isolating v. / {v, x2 , x3 , x4 , x5 }. But then the 31.2.2. So suppose x1 has a neighbor y ∈ matching Mv = {x1 y, x2 x5 , x3 x4 } isolates vertex v. 2. Next suppose δ(G) = 4 and v is a vertex of degree four. Let the neighbors of v be x1 , x2 , x3 and x4 . 2.1. Suppose there are at least three triangles at v. First suppose that some three of these triangles share a common neighbor of v, say without loss of generality, they share vertex x1 . So without loss of generality, suppose that {x1 x2 , x1 x3 , x1 x4 } ⊆ E(G). Then if x2 is adjacent to x3 , x3 is adjacent to x4 or x2 is adjacent to x4 , we have a matching Mv of size two which isolates v. So suppose that x2 is not adjacent to x3 , x3 is not adjacent to x4 and x2 is not adjacent to x4 . Then, since deg x2 ≥ 4, there exists a vertex y1 ∈ N (x2 ) such that y1 ∈ / {v, x1 , x2 , x3 , x4 } and since deg x4 ≥ 4, / {v, x1 , x2 , x3 , x4 , y1 }. But then there exists a vertex y2 ∈ N (x4 ) such that y2 ∈ Mv = {x1 x3 , x2 y1 , x4 y2 } isolates v and |Mv | = 3. So we may assume that there are two (but not three) triangles at v which share one of the xi ’s, say without loss of generality, x1 . Renumbering the neighbors of v if necessary, we may then suppose that vx1 x2 v and vx1 x4 v are two such triangles and that x1 is not adjacent to x3 . Now if either x2 is adjacent to x3 or x3 is adjacent to x4 , we obtain a matching Mv isolating v with |Mv | = 2. So let us suppose that x2 is not adjacent to x3 and x3 is not adjacent to x4 . But then, since there are at least three triangles at v, x2 is adjacent to x4 . Then, arguing as in the preceding paragraph, we can find a matching Mv which isolates v such that |Mv | = 3. 2.2. Suppose next that there are exactly two triangles at v. If these two triangles do not share an edge, then clearly there exists a 2-matching isolating v. So suppose they do share an edge. Without loss of generality, suppose these triangles are vx1 x2 v / {v, x1 , x2 , x3 }. and vx2 x3 v. Since deg x4 ≥ 4, there exists a y1 ∈ N (x4 ) such that y1 ∈ / {v, x2 , x3 , x4 , y1 }, then Mv = {x1 y2 , x2 x3 , x4 y1 } isoNow if x1 has a neighbor y2 ∈ lates v with |Mv | = 3. So we may suppose that N (x1 ) ⊆ {v, x2 , y1 }. But then deg x1 ≤ 3, a contradiction. 2.3. Now suppose there is exactly one triangle at v; say without loss of generality / {v, x1 , x2 , x4 }. But vx1 x2 v. Since deg x3 ≥ 4, there is a y1 ∈ N (x3 ) such that y1 ∈

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then since deg x4 ≥ 4, there is a y2 ∈ N (x4 ) such that y2 ∈ / {v, x1 , x2 , x3 , y1 } and then Mv = {x1 x2 , y1 x3 , y2 x4 } isolates v and has size three. 2.4. So finally we may suppose that there are no triangles at v. Then since / {v, x2 , x3 , x4 } and a y2 ∈ N (x2 ) such δ(G) = 4, there is a y1 ∈ N (x1 ) such that y1 ∈ / {v, x2 , x3 , x4 , y1 }. Again, since there are no triangles at v and deg x3 ≥ 4, that y2 ∈ / {v, x1 , x2 , x4 , y1 , y2 }. Similarly, there is there is a vertex y3 ∈ N (x3 ) such that y3 ∈ / {v, x1 , x2 , x3 , y1 , y2 }. If y3 = y4 , then there is a a vertex y4 ∈ N (x4 ) such that y4 ∈ matching Mv of size four isolating v. So suppose that y3 = y4 . Moreover, we are done unless N (x3 ) = N (x4 ) = / {v, x2 , x3 , x4 , {v, y1 , y2 , y3 }. Consider N (x1 ). If there is a y5 ∈ N (x1 ) such that y5 ∈ y1 , y2 , y3 }, then Mv = {x1 y5 , y1 x4 , x2 y2 , x3 y3 } isolates v and has size four. So we may suppose that N (x1 ) = {v, y1 , y2 , y3 } and by symmetry N (x2 ) = {v, y1 , y2 , y3 }. If y1 has a neighbor z 1 , z 1 ∈ / {v, x1 , x2 , x3 , x4 , y2 , y3 }, then Mv = {y1 z 1 , x2 y2 , vx3 , y3 x4 } is a matching of size four which isolates x1 (a neighbor of v). Hence we may assume that N (y1 ) ⊆ {v, x1 , x2 , x3 , x4 , y2 , y3 } and by symmetry N (y2 ) ⊆ {v, x1 , x2 , x3 , x4 , y1 , y3 }. But since |V (G)| > 8, vertex y3 must be then a cutvertex of G, contradicting the assumption that G is 3-connected. 3. Suppose next that δ(G) = 5 and that v is a vertex of degree five. Let the neighbors of v be x1 , . . . , x5 . Suppose there is a triangle at v, say without loss of generality, vx1 x2 v. 3.1. Suppose first that x3 is adjacent to x4 . Then if x5 has a neighbor y ∈ / {v, x1 , x2 , x3 , x4 }, Mv = {x1 x2 , x3 x4 , x5 y} isolates v with |Mv | = 3. So suppose that N (x5 ) ⊆ {v, x1 , x2 , x3 , x4 } and hence that N (x5 ) = {v, x1 , x2 , x3 , x4 }. Now if / {v, x2 , x3 , x4 , x5 }, then Mv = {x1 y, x3 x4 , x2 x5 } there is a y ∈ N (x1 ) such that y ∈ isolates v and has size three. So suppose N (x1 ) ⊆ {v, x2 , x3 , x4 , x5 } and hence that N (x1 ) = {v, x2 , x3 , x4 , x5 }. Furthermore, by symmetry, we may also suppose that N (x4 ) = {v, x1 , x2 , x3 , x5 }. Similarly we may suppose that N (x2 ) = {v, x1 , x3 , x4 , x5 } and N (x3 ) = {v, x1 , x2 , x4 , x5 }. But then G is isomorphic to K 6 , contradicting the assumption that |V (G)| > 8. 3.2. So suppose that x3 is not adjacent to x4 and by symmetry that x4 is not adjacent to x5 and x3 is not adjacent to x5 . Since deg x3 ≥ 5, there is a y1 ∈ N (x3 ), / {v, x1 , x2 , x4 , x5 }. Similarly, x4 has a neighbor y2 , y2 ∈ / {v, x1 , x2 , x3 , x5 }. Morey1 ∈ / over, since deg x4 ≥ 5, we may choose y2 = y1 . Now if x5 has a neighbor y3 ∈ {v, x1 , x2 , x3 , x4 , y1 , y2 }, then Mv = {x1 x2 , y3 x5 , x3 y1, x4 y2 } isolates v and has size four. So suppose that N (x5 ) ⊆ {v, x1 , x2 , y1 , y2 } and hence N (x5 ) = {v, x1 , x2 , y1 , y2 }. / {v, x1 , x2 , x3 , x4 , x5 , y1 , y2 }, then Mv = {x1 y3 , x2 x5 , x3 y1 , If x1 has a neighbor y3 ∈ x4 y2 } isolates v and has size four. So suppose N (x1 ) ⊆ {v, x2 , x3 , x4 , x5 , y1 , y2 }. If x1 is adjacent to x3 , then Mv = {x1 x3 , x2 x5 , x4 y2 } isolates v and has size three. If x1 is adjacent to x4 , then Mv = {x1 x4 , x2 x5 , x3 y1 } isolates v and has size three. So let us suppose that x1 is not adjacent to x3 and x1 is not adjacent to x4 . Hence N (x1 ) = {v, x2 , x5 , y1 , y2 }. By symmetry, we may also assume that x2 is not adjacent to x3 or x4 and hence that N (x2 ) = {v, x1 , x5 , y1 , y2 }. Since deg x3 ≥ 5, there exists a vertex y3 ∈ N (x3 ) / {v, x1 , x2 , x3 , x4 , x5 , y1 , y2 }. But then Mv = {x3 y3 , x5 y1 , x4 y2 , x1 x2 } such that y3 ∈ isolates v and |Mv | = 4.

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So we may suppose there are no triangles at v. We will return to this case in our summary below. 4. Suppose now that δ(G) = 6 and that v is a vertex of degree six. If all the faces at v are triangular, then we label the neighbors of v as x1 , . . . , x6 such that Mv = {x1 x2 , x3 x4 , x5 x6 } isolates v. So we may assume that no vertex of degree six is incident with six triangular faces. So let us now summarize. Suppose that there is no vertex in G with an isolating matching of size at most four. Then, by Cases 1 and 2, we may suppose that δ(G) ≥ 5. First suppose δ(G) = 5 and suppose v is any vertex of degree five. Then by Case 3, we may suppose that there is no triangle at v and hence no triangular face at v. But then φ(v) ≤ 1 − 5/2 + 5/4 = −1/4. Now consider the rest of the vertices in G. Suppose w is such a vertex. First, suppose w is such that deg w = 6. Then as in Case 4 we may suppose there is at least one face at w which has size at least four. But then φ(w) ≤ 1−6/2 +5/3+1/4 = −1/12. On the other hand, if w has  deg w = k ≥ 7, then φ(w) ≤ 1 − k/2 + k/3 = 1 − k/6 ≤ −1/6. So it follows that v φ(v) ≤ (−1/12)|V (G)|.  If δ(G) = 6, clearly the argument in the preceding paragraph also shows that v φ(v) ≤ (−1/12)|V (G)|. Thus in all cases in which G does not contain a vertex v and an isolating matching Mv with |Mv | ≤ 4, we have 2 − 2g =



φ(v) ≤ (−1/12)|V (G)|,

v

and so |V (G)| ≤ 24g − 24, a contradiction and the theorem is proved.



3. Main Result In order to present our main result we need to introduce one additional matching concept. A graph G is said to be factor-critical if G − v contains a perfect matching for every v ∈ V (G). Theorem 3.1. Let G be a 3-connected equimatchable graph of genus g (respectively, non-orientable genus g). ¯ Then if G is non-bipartite or if G is bipartite and the represen¯ f 4 (g)}, ¯ tativity of the embedding is at least three, |V (G)| ≤ max{ f 1 (g), f 2 (g), f 3 (g), where √    8 7 + 1 + 48g (4g + 3) + 9, f 1 (g) = 2 3    √  8 f 2 (g) = 4 1 + g (4g + 3) + 9, 3 √    8 7 + 1 + 24g¯ (2g¯ + 3) + 9, ¯ = f 3 (g) 2 3

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and

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 8 f 4 (g) ¯ = 4 + 2 2g¯ (2g¯ + 3) + 9. 3

Proof. We treat only the orientable case. The non-orientable case follows quite similarly. Case 1. Suppose G has a perfect matching. Then G is randomly matchable and hence by [20] either G ∼ = K 2n . But then in the former case, by [16]√|V (G)| ≤ = K n,n or G ∼ √ 4(1 + g) ( < f 2 (g)), and in the latter case, by [18] |V (G)| ≤ (1/2)(7 + ( 1 + 48g) ( < f 1 (g)). Case 2. So suppose G does not contain a perfect matching. Then let the GallaiEdmonds decomposition of V (G) be denoted by V (G) = (D, A, C) where, as usual, D is the set of all vertices v in V (G) such that v is left unmatched by some maximum matching in G, A is the set of all neighbors of vertices in D which are not themselves in D, and C = V (G) − D − A. Then by [8, Lemma 1], C = ∅ and A is independent. Case 2.1. Suppose first that A = ∅. Then G = D and G is factor-critical by the Gallai-Edmonds Structure Theorem (cf: [3–5] and Theorem 3.2.1 of [10]). If |V (G)| ≤ {max{8, 24g − 24}, then |V (G)| < f 1 (g) and |V (G)| < f 2 (g) and we are done. So suppose |V (G)| > max{8, 24g − 24}. Then by Theorem 2.1 there exists a vertex v and a matching Mv with |Mv | ≤ 4 which isolates v. Let G  = G − (V (Mv ) ∪ {v}) and let M  be a maximal matching in G  . Then M = Mv ∪ M  is a maximal matching in G by [6, Lemma 1.1]. But G is factor-critical, so M  must be a perfect matching in G  . It then follows that each component of G  must be randomly matchable and hence each such component is either a K 2n or a K n,n . Thus the size of each component of G  is bounded above by the respective bounds given in Case 1. So it remains only to obtain a bound on the number of components of G  . Suppose |Mv | = 4 and hence |V (M  v)| = 8. (The proof for 2 ≤ |Mv | ≤ 3 is similar.) Suppose that G  has more than 83 (4g+3) components. Since G is 3-connected, each of these components contains at least three vertices of attachment in V (Mv ). Then by the Pigeonhole Principle, there must be a triple of vertices {v1 , v2 , v3 } ⊆ V (Mv ) such that at least 4g+3 of the components of G  each have {v1 , v2 , v3 } as vertices of attachment. But now contract these 4g +3 components each to a single vertex and we have a minor isomorphic to K 3,4g+3 embedded in a surface of genus g. However, by [16], γ (K 3,4g+3 ) = g+1 and we have a contradiction. So the number of components must  be bounded above by 83 (4g + 3) and it follows that |V (G)| ≤ max{ f 1 (g), f 2 (g)}. Case 2.2. So suppose A = ∅. Suppose there is a component D1 of the subgraph induced by D which has at least three vertices. (It follows from the Gallai-Edmonds theorem that each component of the subgraph induced by D is of odd order.) Then since G is 3-connected, there are two independent edges e and f joining D1 to A. One

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can then extend {e, f } to a maximum matching M of G, since G is equimatchable. But this contradicts the fact that by the Gallai-Edmonds theorem, every maximum matching of G contains a near-perfect matching of D1 . Thus all the components of D are singletons. On the other hand, the Gallai-Edmonds theorem also says that every maximum matching of G matches each vertex of A to a vertex of D. Hence A must be an independent set and thus G is bipartite. Choose a ∈ A. Since the representativity of the embedding of G is at least three and G is 3-connected, by Proposition 5.5.12 of [13] (see also Proposition 3.3 of [19]) there is a cycle Ca in G covering N (a), but not a. Moreover, cycle Ca is even since G is bipartite. Choose every second edge of cycle Ca to form a matching Ma isolating a. Extend Ma to a maximal matching M in G. But then M contradicts the fact that a ∈ A. Concluding Remarks In Theorem 3.1 above we assume that the embedding has representativity three when G is bipartite. Such embeddings are called polyhedral embeddings. (See [2].) A graph in which for every vertex v there exists a cycle covering N (v), but not v, is said to have the wheel neighborhood property. An embedding of a graph G has the wheel neighborhood property if and only if G is simple, 3-connected and the embedding has representativity at least three. (See Proposition 3.3 of [19].) It is known that the decision problem as to whether a graph has an embedding of representativity at least three is NP-complete and, in fact, this problem remains NP-complete even when the graph G is assumed to be 6-connected. (See [12].) Mohar believes (personal communication) that this problem remains NP-complete when the graphs are restricted to be bipartite. References 1. Aldred, R.E.L., Kawarabayashi, K., Plummer, M.D.: On the matching extendability of graphs in surfaces. J. Combin. Theory Ser. B 98, 105–115 (2008) 2. Barnette, D.: “Polyhedral maps on 2-manifolds,” Convexity and Related Combinatorial Geometry (Norman, Oklahoma, 1980), New York (1982), pp. 7–19 3. Edmonds, J.: Paths, trees and flowers. Canad. J. Math. 17, 449–467 (1965) 4. Gallai, T.: Kritische Graphen II. Magyar Tud. Akad. Mat. Kutató Int. Közl. 8, 373–395 (1963) 5. Gallai, T.: Maximale Systeme unabhängiger Kanten. Magyar Tud. Akad. Mat. Kutató Int. Közl. 9, 401–413 (1964) 6. Kawarabayashi, K., Plummer, M.D., Saito, A.: On two equimatchable graph classes. Discrete Math. 266, 263–274 (2003) 7. Lebesgue, H.: Quelques conséquences simples de la formule d’Euler. J. Math. 9, 27–43 (1940) 8. Lesk, M., Pulleyblank, W.R., Plummer, M.D.: “Equimatchable graphs”, Graph Theory ˝ B. and Combinatorics (Proceedings of Cambridge Conference in Honor of Paul Erdos) Bollobás (Editor), Academic Press, London (1984), pp. 239–254 9. Lewin, M.: Matching-perfect and cover-perfect graphs. Israel J. Math. 18, 345–347 (1974) 10. Lovász, L., Plummer, M.D.: Matching Theory, Akadémiai Kiadó Budapest and Annals of Discrete Mathematics, Vol. 29, North-Holland, Amsterdam, 1986

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11. Meng, D.H.-C.: Matchings and coverings for graphs, PhD thesis, Michigan State University, East Lansing, 1974 12. Mohar, B.: Existence of polyhedral embeddings of graphs. Combinatorica 21, 395–401 (2001) 13. Mohar, B., Thomassen, C.: Graphs on Surfaces. Johns Hopkins Univ. Press, Baltimore and London, (2001) 14. Parsons, T., Pica, G., Pisanski, T., Ventre, A.: Orientably simple graphs. Math. Slovaca 37, 391–394 (1987) 15. Plummer, M.: On n-extendable graphs. Discrete Math. 31, 201–210 (1980) 16. Ringel, G.: Das Geschlecht des vollständigen paaren Graphen. Math. Sem. Univ. Hamburg 28, 139–150 (1965) 17. Ringel, G.: Der vollständige paare Graph auf nichtorientierbaren Flächen. J. Reine Angew. Math. 220, 89–93 (1965) 18. Ringel, G., Youngs, J.W.T.: Solution of the Heawood map-coloring problem. Proc. Nat. Acad. Sci. U.S.A. 60, 438–445 (1968) 19. Robertson, N., Vitray, R.: “Representativity of surface embeddings”, Paths, Flows and VLSI-Layout B. Korte et al. (Editors), Springer-Verlag, Berlin (1990), 293–328 20. Sumner, D.P.: Randomly matchable graphs. J. Graph Theory 3, 183–186 (1979) 21. Thomassen, C.: Tilings of the torus and the Klein bottle and vertex-transitive graphs on a fixed surface. Trans. Amer. Math. Soc. 323, 605–635 (1991) 22. Youngs, J.W.T.: Minimal imbeddings and the genus of a graph. J. Math. Mech. 12, 303–315 (1963) Received: December 12, 2007 Final version received: December 27, 2008

Graphs and Combinatorics (2009) 25:101–109 Digital Object Identifier (DOI) 10.1007/s00373-008-0821-9

Graphs and Combinatorics © Springer-Verlag 2009

A Comment on Ryser’s Conjecture for Intersecting Hypergraphs Toufik Mansour1 , Chunwei Song2 , Raphael Yuster1 1 Department of Mathematics, University of Haifa, Haifa 31905, Israel.

e-mail: [email protected], e-mail: [email protected]

2 School of Mathematical Sciences, LMAM, Peking University, 1000871 Beijing, P.R. China.

e-mail: [email protected]

Abstract. Let τ (H) be the cover number and ν(H) be the matching number of a hypergraph H. Ryser conjectured that every r -partite hypergraph H satisfies the inequality τ (H) ≤ (r − 1)ν(H). This conjecture is open for all r ≥ 4. For intersecting hypergraphs, namely those with ν(H) = 1, Ryser’s conjecture reduces to τ (H) ≤ r − 1. Even this conjecture is extremely difficult and is open for all r ≥ 6. For infinitely many r there are examples of intersecting r -partite hypergraphs with τ (H) = r − 1, demonstrating the tightness of the conjecture for such r . However, all previously known constructions are not optimal as they use far too many edges. How sparse can an intersecting r -partite hypergraph be, given that its cover number is as large as possible, namely τ (H) ≥ r − 1? In this paper we solve this question for r ≤ 5, give an almost optimal construction for r = 6, prove that any r -partite intersecting hypergraph with τ (H ) ≥ r − 1 must have at least (3 − √1 )r (1 − o(1)) ≈ 2.764r (1 − o(1)) edges, and 18

conjecture that there exist constructions with (r ) edges. Key words. Hypergraph, Ryser’s conjecture, Covering, Matching, r -partite, Intersecting. Mathematics Subject Classification. 05C65, 05D05, 05C75.

1. Introduction For a hypergraph H = (V, E), the (vertex) cover number, denoted by τ (H), is the minimum size of a vertex set that intersects every edge. The matching number, denoted by ν(H), is the maximum size of a subset of edges whose elements are pairwise-disjoint. Clearly, ν(H) ≤ τ (H) for any hypergraph: given t pairwise-disjoint edges, we need at least t vertices to cover them. In the graph-theoretic case, König’s Theorem [2] asserts that the converse non-trivial inequality also holds for bipartite graphs. Thus, if H is a bipartite graph then τ (H) = ν(H). Ryser conjectured the following hypergraph generalization of König’s Theorem for hypergraphs. A hypergraph is called r -partite if its vertex set can be partitioned into r parts, and every edge contains precisely one vertex from each part. In particular, r -partite hypergraphs are r -uniform. Ryser conjectured that every r -partite hypergraph H satisfies τ (H) ≤ (r − 1)ν(H).

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This conjecture turns out to be notoriously difficult. Indeed, only the case r = 3 has been proved by Aharoni [1] using topological methods. A hypergraph is called intersecting if any two edges have nonempty intersection. Clearly, H is intersecting if and only if ν(H) = 1. For intersecting hypergraphs, Ryser’s conjecture amounts to: Conjecture 1 If H is an r -partite intersecting hypergraph then τ (H) ≤ r − 1. Conjecture 1 is still wide open. It has been proved for r = 4, 5 by Tuza [3, 4]. We note that the case r = 3 of Conjecture 1 was first proved by Tuza in [4], before Aharoni’s general proof for the case r = 3. Conjecture 1 (if true) is tight in the sense that for infinitely many r there are constructions of intersecting r -partite hypergraphs with τ (H) = r − 1. Indeed, whenever r = q + 1 and q is a prime power, consider the finite projective plane of order q as a hypergraph. This hypergraph is r -uniform and intersecting. To make it r -partite one just needs to delete one point from the projective plane. This truncated projective plane gives an intersecting r -partite hypergraph with cover number r − 1, with q 2 + q = r (r − 1) vertices, and with q 2 = (r − 1)2 edges. However, the projective plane construction is not the “correct” extremal construction, not only because it does not apply to all r , but also because it is not the smallest possible. Although the projective plane construction only contains r (r − 1) vertices (and this is clearly optimal since otherwise some vertex class would have size less than r − 1 resulting in a cover number less than r − 1), should an extremal example contain so many (namely, (r − 1)2 ) edges? In order to understand the extremal behavior of intersecting r -partite hypergraphs, it is desirable to construct the sparsest possible intersecting r -partite hypergraph with cover number as large as possible, namely at least r − 1. More formally, let f (r ) be the minimum integer so that there exists an r -partite intersecting hypergraph H with τ (H) ≥ r − 1 and with f (r ) edges. (we write τ (H) ≥ r − 1 instead of τ (H) = r − 1 to allow for the possibility that Conjecture 1 is false; also note that trivially τ (H) ≤ r since the set of vertices of any edge forms a cover). A trivial lower bound for f (r ) is 2r − 3. Indeed, the edges of an intersecting hypergraph with at most 2r − 4 edges can greedily be covered with r − 2 vertices. We prove, however, the following non-trivial lower bound. Theorem 2 f (r ) ≥ (3 −

√1 )r (1 − o(1)) 18

≈ 2.764r (1 − o(1)).

Although we do not have a matching upper bound, we conjecture that a linear (in r ) number of edges indeed suffice. Conjecture 3 f (r ) = (r ). Computing precise values of f (r ) seems to be a difficult problem. Trivially, f (2) = 1. It is also easy to see that f (3) = 3. Indeed, a 3-partite intersecting hypergraph with only two edges has cover number 1. The hypergraph whose edges are (a1 , b1 , c1 ), (a1 , b2 , c2 ), (a2 , b1 , c2 ) is a 3-partite intersecting hypergraph with cover number 2. The next theorem establishes the first non-trivial values of f (r ), namely r = 4, 5, in addition to upper and lower bounds in the case r = 6. More specifically, we prove:

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Theorem 4 f (4) = 6, f (5) = 9, and 12 ≤ f (6) ≤ 15. Comparing our constructions with the projective plane construction, we see that in the case r = 4, 5, 6 the latter has 9, 16 and 25 edges respectively. Thus, the projective plane construction is far from being optimal. Our constructions also have the property that the number of vertices they contain is r (r − 1), which, as mentioned earlier, is optimal. In the rest of this paper we prove Theorems 2 and 4. 2. Proof of Theorem 2 Throughout this section we assume that H is an r -partite intersecting hypergraph with τ (H) ≥ r − 1. Recall that the degree of a vertex v in a hypergraph is the number of edges containing v. Consider the following greedy procedure, starting with the original hypergraph H. As long as there is a vertex x of degree at least 4 in the current hypergraph, we delete x and all of the edges containing x from the current hypergraph, thereby obtaining a smaller hypergraph. Vertices that become isolated are also deleted. Denote by H3 the hypergraph obtained at the end of the greedy procedure and denote by X 4 the set of vertices deleted by the greedy procedure. Notice that H3 is either the empty hypergraph or else it is an r -partite intersecting hypergraph, every vertex of which has degree at most 3. We then continue in the same manner, where as long as there is a vertex x of degree 3 in the current hypergraph, we delete x and all of the edges containing x from the current hypergraph. Again, vertices that become isolated are also deleted. Denote by H2 the hypergraph obtained at the end of this second greedy procedure and denote by X 3 the set of vertices deleted in the second greedy procedure. Notice that H2 is either the empty hypergraph or else it is an r -partite intersecting hypergraph, every vertex of which has degree at most 2. We first claim that H3 contains at most 2r + 1 edges. Indeed, if H is any edge, then every vertex of H appears in at most two other edges. Thus, there are at most 2r other edges in addition to H . Similarly, H2 contains at most r + 1 edges. Let, therefore, the number of edges of H3 be denoted by γ r and hence 0 ≤ γ ≤ 2 + 1/r . Consider first the case γ ≤ 1. In this case we can cover the edges of H3 greedily with a set U of at most γ r/2 vertices. Now, since U ∪ X 4 is a cover of H and since τ (H) ≥ r − 1, we have that |X 4 | ≥ r − 1 − γ r/2. As every vertex of X 4 was greedily selected to appear in four distinct edges of H − H3 we have that the number of edges of H is at least 4|X 4 | + γ r ≥ 4(r − 1 − γ r/2) + γ r ≥ (4 − γ )r − 6 ≥ 3r − 6 which is even better than the bound in the statement of the theorem. We may now assume that 1 < γ ≤ 2 + 1/r . Since H2 has at most r + 1 edges, we have that |X 3 | ≥ (γ r − r − 1)/3. The number of edges of H2 is γ r − 3|X 3 |. It follows that there is a cover of H3 whose size is at most     1 γ γ r − 3|X 3 | |X 3 | + ≤ + r (1 + o(1)). 2 6 3

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As such a cover, together with X 4 , is a cover of H, and since τ (H) ≥ r − 1, we have that   5 γ r (1 − o(1)). |X 4 | ≥ − 6 3 We therefore have that the number of edges of H, which is at least 4|X 4 | + γ r , is at least 10 − γ r (1 − o(1)). 3

(1)

There is, however, another way to bound from below the number of edges of H. For i = 1, 2, 3 let αi r 2 denote the number of vertices of H3 having degree i. Since the sum of the degrees in H3 is γ r 2 we have: α1 + 2α2 + 3α3 = γ . Consider a specific edge H of H3 and let rβiH be the number of vertices in H with degree i for i = 1, 2, 3. Clearly, β1H + β2H + β3H = 1. As H intersects every edge we must have rβ2H + 2rβ3H ≥ γ r − 1. It follows that: 2β1H + β2H = 2 − β2H − 2β3H ≤ 2 − γ + 1/r. In particular, 

(2β1H + β2H ) ≤ γ r (2 − γ + 1/r ).

H ∈H3

On the other hand, by definition we have that   rβ1H = α1r 2 , rβ2H = 2α2 r 2 . H ∈H3

H ∈H3

It follows that 2α1 + 2α2 ≤ γ (2 − γ + 1/r ). Hence, 2 2 γ 1 γ 2 + α1 + α2 ≤ + α1 + α2 ≤ 3 3 3 3 3 3 1 1 2 γ γ + γ (2 − γ + 1/r ) = γ − γ + . 3 3 3 3r α1 + α2 + α3 =

Since r 2 (α1 + α2 + α3 ) is the number of vertices of H3 we have that H3 has at most   1 2 2 γ − γ r (1 + o(1)) 3

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vertices. In particular, there is a vertex class consisting of at most   1 γ − γ 2 r (1 + o(1)) 3 vertices. Since any vertex class of H3 , together with X 4 , form a cover of H, we have that   1 2 |X 4 | ≥ 1 − γ + γ r (1 − o(1)). 3 We therefore have that the number of edges of H, which is at least 4|X 4 | + γ r , is at least   4 (2) 4 − 3γ + γ 2 r (1 − o(1)). 3 Comparing (1) and (2) we see √ that the minimum of the maximum of both of them is attained when γ = 1 + 1/ 2, and in this case the number of edges of H is at least  (3 − √1 )r (1 − o(1)), as required. 18

3. Proof of Theorem 4 3.1. The Case r = 4 We need to show first that f (4) > 5. Assume the contrary and let H be a 4-partite intersecting hypergraph with only 5 edges and with τ (H) ≥ 3. No vertex can appear in three or more edges, since such a vertex v, and a vertex u intersecting the (at most two) edges in which v does not appear form a cover of size  2, a contradiction. Thus, every vertex has degree at most 2. Now, since there are 25 nonempty intersections of pairs of edges of  H, we have, by the inclusion-exclusion principle that H contains at most 5 · 4 − 25 = 10 vertices. But this means that some vertex class contains at most two vertices, again resulting in τ (H) ≤ 2, a contradiction. We construct a 4-partite intersecting hypergraph with 6 edges and with τ (H) = 3. Consider the four vertex classes V1 = {a1 , a2 , a3 }, V2 = {b1 , b2 , b3 }, V3 = {c1 , c2 , c3 }, and V4 = {d1 , d2 , d3 }. The 6 edges are (a1 , b1 , c1 , d1 ), (a1 , b2 , c2 , d2 ), (a2 , b1 , c2 , d3 ), (a2 , b2 , c3 , d1 ), (a3 , b3 , c2 , d1 ), and (a3 , b1 , c3 , d2 ). It is easy to check that any two edges intersect and that two vertices cannot cover all 6 edges. 3.2. The Case r = 5 We need to show first that f (5) > 8. Assume the contrary and let H be a 5-partite intersecting hypergraph with only 8 edges and with τ (H) ≥ 4. Notice first that there is no vertex with degree 4 or greater, since if v is such a vertex, then the (at most) four remaining edges not containing v can always be greedily covered with two additional vertices, resulting in cover number at most 3, a contradiction. Thus, we may assume that the degree of each vertex is at most 3. Clearly we can assume that H has at least 20 vertices, as otherwise there is a vertex class with at most three vertices, again contradicting the assumption that τ (H) ≥ 4.

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Let xi denote the number of vertices with degree i, for i = 1, 2, 3. Since the sum of the degrees of all vertices is 8 · 5 = 40 we have that x1 + 2x2 + 3x3 = 40. We claim also that each vertex class has at most one vertex with degree 3. Indeed, if there were two such vertices in the same vertex class, then they both cover 6 edges, and the remaining two edges can be covered by an additional vertex, contradicting the assumption that τ (H)   ≥ 4. Thus, we have that x3 ≤ 5. As there are 28 = 28 intersections, we have that x2 + 3x3 ≥ 28. Finally, since there are at least 20 vertices, we have x1 + x2 + x3 ≥ 20. Now, it follows that x1 + 2x2 + 4x3 ≥ 48 which implies that x3 ≥ 8 which contradicts x3 ≤ 5. Next we construct a 5-partite intersecting hypergraph with 9 edges and with τ (H) = 4. Consider the five vertex classes V1 = {1, 2, 3, 4}, V2 = {5, 6, 7, 8}, V3 = {9, 10, 11, 12}, V4 = {13, 14, 15, 16}, and V5 = {17, 18, 19, 20}. The 9 edges are divided into two parts: A = {(1, 5, 9, 13, 17), (2, 6, 10, 14, 17), (3, 7, 10, 13, 18), (1, 6, 11, 15, 18), (2, 7, 9, 15, 19)}, B = {(4, 5, 10, 15, 20), (4, 7, 11, 16, 17), (4, 8, 9, 14, 18), (4, 6, 12, 13, 19)}. First, notice that the constructed hypergraph is, indeed, 5-partite, and intersecting. Also note that τ (H) ≤ 4 by considering, for example, the cover {4, 17, 18, 19}. It remains to show that there is no cover of size 3. There is only one vertex with degree 4, and it is vertex 4. Vertex 4 precisely covers the set B. Notice, however, that no vertex covers three edges of A. This means that any cover containing 4 must have size at least 4. We now only need to rule out the possibility of a cover of size 3, each vertex of which has degree 3. The vertices of degree 3 are {1, 2, 6, 7, 9, 10, 13, 15, 17, 18}. However, each of them appears at most one time in B, hence any three of them cannot cover all the vertices of B. 3.3. The Case r = 6 We first show that f (6) > 11. Like before, assume the contrary and let H be a 6-partite intersecting hypergraph with only 11 edges and with τ (H) ≥ 5. Notice first that there is no vertex with degree 5 or greater, since if v is such a vertex, then the (at most) 6 remaining edges not containing v can always be greedily covered with 3 additional vertices, resulting in cover number at most 4, a contradiction. Thus, we may assume that the degree of each vertex is at most 4. Clearly we can assume that H has at least 30 vertices, as otherwise there is a vertex class with at most 4 vertices, again contradicting the assumption that τ (H) ≥ 5. Let xi denote the number of vertices with degree i, for i = 1, 2, 3, 4. Since the sum of the degrees of all vertices is 11·6 = 66 we have that x1 +2x2 +3x3 +4x4 = 66. Again, notice that the assumption that τ (H) ≥ 5 implies that each vertex class has at most one vertex with degree 4, at most two vertices of degree 3, and if a vertex class contains a vertex of degree 4, it does not contain a vertex of degree 3. Thus, x4 ≤ 6 and x3 ≤ 12 −2x4 . Hence x3 + 3x4 ≤ 18. As there are 55 = 11 2 intersections, we have that x 2 + 3x 3 + 6x 4 ≥ 55. Combining this with the fact x1 + x2 + x3 + x4 ≥ 30 we have that x1 + 2x2 + 4x3 + 7x4 ≥ 85. This implies that x3 + 3x4 ≥ 19, contradicting the fact that x3 + 3x4 ≤ 18.

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Next we create a 6-partite intersecting hypergraph H = (V, E) with 30 vertices and 15 edges as follows. The six vertex classes of V are: V1 = {a1 , a3 , a4 , a6 , a8 }, V2 = {b1 , b2 , b4 , b8 , b12 }, V3 = {c1 , c2 , c4 , c7 , c11 }, V4 = {d1 , d2 , d3 , d9 , d11 }, V5 = {e1 , e2 , e3 , e5 , e7 }, V6 = { f 1 , f 2 , f 3 , f 5 , f 7 }. (The fact that the selection of indices in each set is not consecutive simplifies the description that follows.) We construct E in several steps so as to guarantee that 1. |H ∩ Vi | = 1, ∀H ∈ E, 1 ≤ i ≤ 6, so that H is a 6-partite. 2. H ∩ H = ∅, ∀H, H ∈ E. 3. τ (H) ≥ 5. Step 1: A “cyclic" construction. Throughout the whole procedure we try to let the edges, albeit intersecting, repeat as little as possible. H1 = {a1 , b1 , c1 , d1 , e1 , f 1 },

H2 = {a1 , b2 , c2 , d2 , e2 , f 2 },

H3 = {a3 , b1 , c2 , d3 , e3 , f 3 },

H4 = {a4 , b4 , c4 , d1 , e2 , f 3 },

H5 = {a3 , b4 , c1 , d2 , e5 , f 5 },

H6 = {a6 , b2 , c4 , d3 , e5 , f 1 }.

Note that by construction, |Hi ∩ H j | = 1,

1 ≤ i < j ≤ 6,

Hi ∩ H j ∩ Hk = ∅,

1 ≤ i < j < k ≤ 6.

Therefore, a minimum cover of H1 , . . . , H6 has size 3. Now we consider the pairwise intersections and take the union of every three mutually disjoint pairs (the union 6 Hi ), for instance {H1 ∩ H2 , H3 ∩ H4 , H5 ∩ of the three pairs forms exactly i=1 H6 } = {a1 , f 3 , e5 }. The following list L1 thus contains all the minimum covers of H1 , . . . , H6 : L1 = {{a1 , f 3 , e5 }, {a1 , a3 , c4 }, {a1 , d3 , b4 }, {b1 , e2 , e5 }, {b1 , d2 , c4 }, {b1 , b2 , b4 }, {d1 , c2 , e5 }, {d1 , d2 , d3 }, {d1 , b2 , a3 }, {c1 , c2 , c4 }, {c1 , e2 , d3 }, {c1 , b2 , f 3 }, { f 1 , c2 , b4 }, { f 1 , e2 , a3 }, { f 1 , d2 , f 3 }} . Step 2: Any additional edge must contain a cover of H1 , . . . , H6 . Selecting carefully an element from L1 each time, we construct the edges H7 through H10 . H7 = {a1 , b4 , c7 , d3 , e7 , f 7 }, H9 = {a8 , b2 , c1 , d9 , e7 , f 3 },

H8 = {a8 , b8 , c2 , d1 , e5 , f 7 }, H10 = {a3 , b8 , c7 , d9 , e2 , f 1 }.

Up to now, |Hi ∩ H j | = 1, f or 1 ≤ i < j ≤ 10. Moreover, since the (unique) intersecting element of any pair of edges constructed in Step 2 has a subscript index in {7, 8, 9}, Hi ∩ H j ∩ Hk = ∅, 1 ≤ i < j < k ≤ 10, if either |{i, j, k} ∩ [6]| = 3 or |{i, j, k} ∩ {7, 8, 9, 10}| ≥ 2. Notice also that any minimum cover that covers H1 − H10 consists of at least four vertices. Step 3: Five additional edges that force an increase in the size of the minimum cover. H11 = {a1 , b8 , c11 , d11 , e5 , f 3 },

H12 = {a8 , b12 , c1 , d3 , e2 , f 3 },

H14 = {a4 , b8 , c1 , d3 , e2 , f 1 },

H15 = {a8 , b8 , c1 , d3 , e2 , f 5 }.

H13 = {a8 , b4 , c2 , d11 , e1 , f 1 },

Notice that the 15 constructed edges indeed form an intersecting 6-partite hypergraph. It remains to show that:

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Proposition 5 τ (H) = 5. Proof. Suppose the proposition is false, and let C = {x, y, z, w} be a cover of size 4. We use a sequence of arguments to deduce that this is impossible. For convenience, for 7 ≤ i ≤ 15, let Hi ∗ be the collection of those vertices of Hi with index subscript no bigger than 6. That is, H7∗ = {a1 , b4 , d3 }, H8∗ = {c2 , d1 , e5 }, H9∗ = {b2 , c1 , f 3 }, ∗ = {a , e , f }, ∗ = {a , e , f }, ∗ = {c , d , e , f }, H10 H H12 3 2 1 1 5 3 1 3 2 3 11 ∗ ∗ ∗ = {c , d , e , f }. H13 = {b4 , c2 , e1 , f 1 }, H14 = {a4 , c1 , d3 , e2 , f 1 }, H15 1 3 2 5 First, notice that if C ∩ {b12 , c11 , d11 } = ∅, then |C\{b12 , c11 , d11 }| ≤ 3. So C\{b12 , c11 , d11 } must be a triple, and furthermore it must be one of the triples ∗ are in L1 . But no element of L1 may cover H7 , H8 , H9 and H10 since H7∗ − H10 pairwise disjoint. Hence, C ∩ {b12 , c11 , d11 } = ∅. Second, assume C ∩ {a8 , b8 } = ∅. Without loss of generality, let x = C ∩ {a8 , b8 }. Then {y, z, w} ∈ L1 . • Case 1: x = a8 . The fact that {y, z, w} must cover H7∗ implies that {y, z, w} ∩ ∗ implies that {y, z, w}∩{a , e , f } = ∅; {a1 , b4 , d3 } = ∅; {y, z, w} must cover H10 3 2 1 ∗ implies that {y, z, w} ∩ {a , e , f } = ∅. The only tri{y, z, w} must cover H11 1 5 3 ∗ = ple in L1 satisfying these three requirements is {a1 , a3 , c4 }. But then H14 {a4 , c1 , d3 , e2 , f 1 } is left uncovered. Impossible. • Case 2: x = b8 . The fact that {y, z, w} must cover H7∗ implies that {y, z, w} ∩ {a1 , b4 , d3 } = ∅; {y, z, w} must cover H9∗ implies that {y, z, w} ∩ {b2 , c1 , f 3 } = ∅; ∗ implies that {y, z, w} ∩ {b , c , e , f } = ∅. The only {y, z, w} must cover H13 4 2 1 1 ∗ = triple in L1 satisfying these three requirements is {b1 , b2 , b4 }. But then H12 {c1 , d3 , e2 , f 3 } is left uncovered. Impossible. Hence, C ∩ {a8 , b8 } = ∅. Third, assume C ∩ {c7 , e7 , f 7 , d9 } = ∅. Let C ∩ {c7 , e7 , f 7 , d9 } = x. Then {y, z, w} ∗ − H ∗ . The fact that {y, z, w} must be a triple in L1 , and it must also cover H11 15 ∗ ∗ must cover H11 implies that {y, z, w} ∩ {a1 , e5 , f 3 } = ∅; {y, z, w} must cover H13 ∗ implies that {y, z, w} ∩ {b4 , c2 , e1 , f 1 } = ∅; {y, z, w} must cover H15 implies that {y, z, w} ∩ {c1 , d3 , e2 , f 5 } = ∅. The only triple in L1 satisfying these three requirements is {a1 , b4 , d3 }. But then H8 , H9 , H10 are not yet covered. This can not be fixed by any additional one vertex as H8 ∩ H9 ∩ H10 = ∅. Hence we must have C ∩ {c7 , e7 , f 7 , d9 } = ∅. Last, we have by now established that the index of any vertex in C is in {1, 2, 3, 4, 5, 6}. Furthermore, |C ∩ Hi∗ | = 1, for 7 ≤ i ≤ 10. In particular, let x = C ∩ H7∗ , we discuss each of the three possibilities. (i)

∗ , H ∗ − H ∗ . Let y ∈ H ∗ . x = a1 . Then {y, z, w} must cover H3 − H6 , H8∗ − H10 12 15 8 • Case 1: y = c2 . The fact {z, w} needs to cover H4 − H6 ⇒ {z, w} ∩ {b4 , e5 , c4 } = ∅; {z, w} needs to cover H9∗ ⇒ {z, w} ∩ {b2 , c1 , f 3 } = ∅; ∗ ⇒ {z, w} ∩ {a , e , f } = ∅. But this is clearly {z, w} needs to cover H10 3 2 1 impossible. • Case 2: y = d1 . The fact {z, w} needs to cover H3 , H5 and H6 ⇒ {z, w} ∩ {a3 , e5 , d3 } = ∅; {z, w} needs to cover H9∗ ⇒ {z, w} ∩ {b2 , c1 , f 3 } = ∅; ∗ ⇒ {z, w} ∩ {b , c , e , f } = ∅. Impossible. {z, w} needs to cover H13 4 2 1 1

A Comment on Ryser’s Conjecture for Intersecting Hypergraphs

•

(ii)

x y •

• •

(iii)

x • •

•

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Case 3: y = e5 . The fact {z, w} needs to cover H9∗ ⇒ {z, w}∩{b2 , c1 , f 3 } = ∗ ⇒ {z, w} ∩ {a , e , f } = ∅; {z, w} needs ∅; {z, w} needs to cover H10 3 2 1 ∗ ∗ to cover H13 ⇒ {z, w} ∩ {b4 , c2 , e1 , f 1 } = ∅; {z, w} needs to cover H15 ⇒ {z, w} ∩ {c1 , d3 , e2 , f 5 } = ∅. Putting the four requirements all together, {z, w} is forced to equal {c1 , f 1 }. But then H4 is left uncovered. Impossible. ∗ , H ∗ , H ∗ . Let = b4 . Then {y, z, w} must cover H1 − H3 , H6 , H8∗ − H12 14 15 ∗ ∈ H8 . Case 1: y = c2 . The fact {z, w} needs to cover H9∗ ⇒ {z, w}∩{b2 , c1 , f 3 } = ∗ ⇒ {z, w} ∩ {a , e , f } = ∅; {z, w} needs ∅; {z, w} needs to cover H10 3 2 1 ∗ ∗ ⇒ to cover H11 ⇒ {z, w} ∩ {a1 , e5 , f 3 } = ∅; {z, w} needs to cover H15 {z, w} ∩ {c1 , d3 , e2 , f 5 } = ∅. The four requirements force {z, w} to equal { f 3 , e2 }. But then H1 is left uncovered. Impossible. Case 2: y = d1 . The fact {z, w} needs to cover H2 , H3 and H6 ⇒ {z, w} ∩ ∗ ⇒ {z, w} ∩ {a , e , f } = ∅; {c2 , b2 , d3 } = ∅; {z, w} needs to cover H10 3 2 1 ∗ {z, w} needs to cover H11 ⇒ {z, w} ∩ {a1 , e5 , f 3 } = ∅. Impossible. Case 3: y = e5 . The fact {z, w} needs to cover H1 − H3 ⇒ {z, w} ∩ {a1 , b1 , c2 } = ∅; {z, w} needs to cover H9∗ ⇒ {z, w} ∩ {b2 , c1 , f 3 } = ∅; ∗ ⇒ {z, w} ∩ {a , e , f } = ∅. Impossible. {z, w} needs to cover H10 3 2 1 ∗ , H ∗ . Let y ∈ H ∗ . = d3 . Then {y, z, w} must cover H1 , H2 , H4 , H5 , H8∗ − H11 13 8 Case 1: y = c2 . The fact {z, w} needs to cover H1 , H4 and H5 ⇒ {z, w} ∩ {d1 , b4 , c1 } = ∅; {z, w} needs to cover H9∗ ⇒ {z, w} ∩ {b2 , c1 , f 3 } = ∅; ∗ ⇒ {z, w} ∩ {a , e , f } = ∅. Impossible. {z, w} needs to cover H10 3 2 1 Case 2: y = d1 . The fact {z, w} needs to cover H9∗ ⇒ {z, w}∩{b2 , c1 , f 3 } = ∗ ⇒ {z, w} ∩ {a , e , f } = ∅; {z, w} needs ∅; {z, w} needs to cover H10 3 2 1 ∗ ⇒ {z, w} ∩ {a , e , f } = ∅; {z, w} needs to cover H ∗ ⇒ to cover H11 1 5 3 13 {z, w} ∩ {b4 , c2 , e1 , f 1 } = ∅. The four requirements force {z, w} to be { f 3 , e2 }. But then H2 is left uncovered. Impossible. Case 3: y = e5 . The fact {z, w} needs to cover H1 , H2 and H4 ⇒ {z, w} ∩ {a1 , d1 , e2 } = ∅; {z, w} needs to cover H9∗ ⇒ {z, w} ∩ {b2 , c1 , f 3 } = ∅; ∗ ⇒ {z, w} ∩ {b , c , e , f } = ∅. Impossible. {z, w} needs to cover H13 4 2 1 1

In conclusion, our assumption is contradicted. Hence, τ (H) = 5.



References 1. Aharoni, R.: Ryser’s conjecture for tripartite 3-graphs. Combinatorica 21, 1–4 (2001) ˝ 2. Konig, D.: Theorie der endlichen und unendlichen Graphen. Leipzig, 1936; reprinted Chelsea and New York, (1950) 3. Tuza, Zs.: Some special cases of Ryser’s conjecture, manuscript (1979) 4. Tuza, Zs.: Ryser’s conjecture on transversals of r -partite hypergraphs. Ars Combinatoria 16, 201–209 (1983) Received: September 21, 2007 Final version received: October 24, 2008

Graphs and Combinatorics (2009) 25:111–114 Digital Object Identifier (DOI) 10.1007/s00373-008-0825-5

Graphs and Combinatorics © Springer-Verlag 2009

On the Nullity of Bipartite Graphs G. R. Omidi1,2,∗ 1 Department of Mathematical Sciences, Isfahan University of Technology, Isfahan,

84156-83111, Iran.

2 School of Mathematics, Institute for Research in Fundamental Sciences (IPM),

P. O. Box: 19395-5746, Tehran, Iran. e-mail: [email protected]

Abstract. The nullity of a graph is the multiplicity of the eigenvalue zero in its spectrum. We obtain some lower bounds for the nullity of graphs and we then find the nullity of bipartite graphs with no cycle of length a multiple of 4 as a subgraph. Among bipartite graphs on n vertices, the star has the greatest nullity (equal to n − 2). We generalize this by showing that among bipartite graphs with n vertices, e edges and maximum degree  which do not have any cycle of length a multiple of 4 as a subgraph, the greatest nullity is n − 2e/. Key words. Nullity of a graph, Maximum matchings, Bipartite graphs. AMS Subject Classification: 05C50.

1. Introduction Let G be a simple graph with n vertices and the adjacency matrix A(G). The characteristic polynomial of the adjacency matrix A(G) is said to be the characteristic polynomial of the graph G; we denote it by χG (λ). Since A(G) is a real symmetric matrix, its eigenvalues are real numbers. The multiset of eigenvalues of A(G) is called the adjacency spectrum of G. The number of zero eigenvalues in the spectrum of the graph G is called its nullity and is denoted by η(G). A matching of G is a collection of independent edges of G. A maximum matching is a matching with the maximum possible number of edges. The size of a maximum matching of G, i.e., the maximum number of independent edges in G, is denoted by m = m(G). It is clear that m ≤ n/2. The maximum matching with m = n/2 is called the complete matching. The problem of characterizing all graphs with nonzero nullity was first posed by Collatz and Sinogowitz almost 50 years ago (see [2]). Some recent results on the nullity of graphs in specific situations can be found in [4–6]. In [4], it was shown that among trees on n-vertices with maximum degree , the greatest nullity is n − 2(n − 1)/ where x denotes the smallest integer a ≥ x. In this paper first we obtain some lower bounds for the nullity of graphs and we give some families of graphs which meet these bounds. We focus on the nullity of bipartite graphs and ∗ This research was in part supported by a grant from IPM (No.87050016).

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we obtain the nullity of bipartite graphs with no cycle of length a multiple of 4 as a subgraph. Moreover, we show that among bipartite graphs with n vertices, e edges and maximum degree  which do not have any cycle of length a multiple of 4 as a subgraph, the greatest nullity is n − 2e/. 2. The Nullity of Bipartite Graphs In this section first we find some lower bounds for the nullity of graphs. We then obtain the nullity of bipartite graphs with no cycle of length a multiple of 4 and we will see that these graphs have the minimum nullity among bipartite graphs on n vertices with maximum matching m. n ci x n−i . Then the coefficient of x n−i is Let χG (x) = i=0  (−1)k(H ) 2c(H ) , (1) ci = H

where the sum is over all subgraphs H of G consisting of disjoint edges and cycles and having i vertices. If H is such a subgraph then k(H ) is the number of components in it and c(H ) is the number of its cycles (see [3]). Lemma 1. Let G be a graph on n ≥ 1 vertices and let m be the size of its maximum matchings. i) If G is a bipartite graph, then η(G) ≥ n − 2m. Moreover, if the number of maximum matchings of G is an odd number, then η(G) = n − 2m. ii) If G is a non-bipartite graph and g is the size of its minimum odd cycle, then η(G) ≥ n − 2mg/(g − 1). Moreover, η(G) = n − 2mg/(g − 1) if and only if G has 2m/(g − 1) disjoint cycles Cg . Proof.

(i) The graph G does not have any odd cycle and each even cycle has two complete matchings. Let H be a subgraph of G consisting of disjoint edges and cycles and having i vertices. Replacing each cycle of H by one of its complete matching we get a matching of G on i vertices. Therefore for i > 2m, G does not have a subgraph consisting of disjoint edges and cycles on i vertices and so by (1), ci = 0 for i > 2m. Hence we have η(G) ≥ n − 2m. Now let the number of maximum matchings of G be an odd number and let H have i = 2m vertices. It is clear that (−1)k(H ) 2c(H ) = (−1)m if H is a maximum matching and (−1)k(H ) 2c(H ) is an even number, otherwise. Hence  by (1), c2m = H (−1)k(H ) 2c(H ) is an odd number and so η(G) = n − 2m. (ii) Let H be a subgraph of G consisting of disjoint edges and cycles and having i vertices. Replacing each cycle of H by one of its maximum matching we get a matching of G on i − l vertices where l is the number of odd cycles of H . It is clear that l ≤ i/g. On the other hand i − l ≤ 2m and so i ≤ 2mg/(g − 1). Therefore for i > 2mg/(g − 1), G does not have any subgraph consisting of disjoint edges and cycles on i vertices and so ci = 0 for i > 2mg/(g − 1). Hence by (1) we have η(G) ≥ n − 2mg/(g − 1). If H has i = 2mg/(g − 1)

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vertices, then l = i/g and so H is a disjoint union of i/g cycles Cg . Therefore by (1), c2mg/(g−1) = r (−2)i/g where r is the number of subgraph H of G consisting 2m/(g − 1) disjoint cycles Cg . Since η(G) ≥ n − 2mg/(g − 1), we have η(G) = n − 2mg/(g − 1) if and only if c2mg/(g−1) = r (−2)i/g > 0. Therefore η(G) = n − 2mg/(g − 1) if and only if G has 2m/(g − 1) disjoint cycles Cg .  Theorem 1. If G is a bipartite graph on n ≥ 1 vertices with no cycle of length a multiple of 4 as a subgraph and if m is the size of its maximum matchings, then its nullity is equal to η(G) = n − 2m. Proof. Let H be a subgraph of G consisting of disjoint edges and cycles and having 2m vertices. Since G does not have any cycle of length a multiple of 4, m − k(H ) is an even number. So c2m > 0, for even m and c2m < 0, otherwise. Hence c2m = 0 and so using Lemma 1, the proof is complete.  Corollary 1 [3]. Let T be a tree on n ≥ 1 vertices and let m be the size of its maximum matchings. Then its nullity is equal to η(T ) = n − 2m. Proof. This is an immediate consequence of Theorem 1.



Corollary 2. Let G be a bipartite graph on n ≥ 1 vertices with no cycle of length a multiple of 4 as a subgraph. The nullity η(G) of G is zero if and only if it contains a perfect matching. Proof. This is a direct consequence of Theorem 1.



Lemma 2 [1]. Every nonempty regular bipartite graph has a perfect matching. Corollary 3. Let G be a nonempty regular bipartite graph on n ≥ 1 vertices with no cycle of length a multiple of 4 as a subgraph. Then we have η(G) = 0. Proof. This is an immediate consequence of Lemma 2 and Corollary 2.



3. Upper bound of η(G) In this section we identify the greatest nullity of bipartite graphs with n vertices, e edges and maximum degree  which do not have any cycle of length a multiple of 4 as a subgraph. A k-edge coloring of G is a partition {E 1 , E 2 , . . . , E k } of edges, where E i denotes the (possibly empty) set of edges assigned color i. An edge coloring is proper if adjacent edges receive distinct colors. Thus a proper k-edge-coloring is a k-edgecoloring {M1 , M2 , . . . , Mk } in which each subset Mi is a matching. A graph is k-edge-colorable if it has a proper k-edge-coloring. The edge chromatic number, χ  (G), of a graph G is the minimum k for which G is k-edge colorable.

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Theorem 2 [1]. Let G be a bipartite graph with maximum degree . Then χ  (G) = . Theorem 3. Let G be a bipartite graph with n ≥ 1 vertices, e edges and maximum degree  and let G not have any cycle of length a multiple of 4 as a subgraph. Then η(G) ≤ n − 2e/. Proof. By Theorem 2, we have χ  (G) = . Let {M1 , M2 , . . . , M } be the proper -edge-coloring of G and let m be the size of its maximum matchings. Then e =   i=1 |Mi | ≤ m and so by Theorem 1, we have η(G) = n − 2m ≤ n − 2e/. Since η(G) is an integer, we have η(G) ≤ n − 2e/.



Corollary 4 [4]. Let T be a tree on n ≥ 1 vertices with maximum degree . Then η(G) ≤ n − 2(n − 1)/. Proof. This is an immediate consequence of Theorem 3.



There are infinite families of bipartite graphs with the greatest nullity η(G) = n − 2e/. In [4], it has been shown that for all n ≥ 1 and  ≥ 3, there exist trees T on n ≥ 1 vertices with maximum degree , such that η(G) = n − 2(n − 1)/. By Corollary 3, each regular bipartite graph with no cycle of length a multiple of 4 as a subgraph has η(G) = n − 2e/ = 0. For example η(C2l ) = n − 2e/ = 0. The characterization of bipartite graphs with the greatest nullity η(G) = n − 2e/ seems to be an interesting problem. Acknowledgements. This work was partially supported by IUT (CEAMA).

References 1. Bondy, J.A., Murty, U.S.R.: Graph Theory with Applications. American Elsevier Publishing, New York, 1976 2. Collatz, L., Sinogowitz, U.: Spectren endlicherGrafen, Abh. Math. Sera. Univ. Hamburg 21, 63–77 (1957) 3. Cvetkovic, D.M., Doob, M., Sachs, H.: Spectra of Graphs. Third edition, Johann Abrosius Barth Verlag, 1995 4. Fiorini, S., Gutman, I., Sciriha, I.: Trees with maximum nullity, Linear Algebra Appl. 397, 245–251 (2005) 5. Sciriha I., Gutman, I.: On the nullity of line graphs of trees, Discrete Math. 232, 35–45 (2001) 6. Xuezhong, T., Liu, B.: On the nullity of unicyclic graphs, Linear Algebra Appl. 408, 212–220 (2005) Received: April 28, 2008 Final version received: October 17, 2008

Graphs and Combinatorics (2009) 25:115–121 Digital Object Identifier (DOI) 10.1007/s00373-008-0820-x

Graphs and Combinatorics © Springer-Verlag 2009

Frankl’s Conjecture and the Dual Covering Property R. S. Shewale, Vinayak Joshi, V. S. Kharat Department of Mathematics, University of Pune, Pune- 411 007, India. e-mail: [email protected], [email protected], [email protected]

Abstract. We have proved that the Frankl’s Conjecture is true for the class of finite posets satisfying the dual covering property. Key words. Dual covering property, Frankl’s Conjecture. AMS Subject Classification (2000) 06A06, 06B05, 06C10.

Peter Frankl conjectured the following in 1979, now known as union-closed sets conjecture or Frankl’s Conjecture: Conjecture A. Let F be a collection of subsets of a finite set X such that F ∪ G ∈ F holds for each F, G ∈ F, that is, F is a union-closed family. If |F| ≥ 2 then there is an element x in X such that at least |F| /2 members F ∈ F satisfy x ∈ F. Poonen [10] formulated Frankl’s Conjecture in the language of lattice theory as follows, in which, an element j of a lattice L is said to be join-irreducible if j = a ∨ b implies j = a or j = b for a, b ∈ L. The set of join-irreducible elements of L is denoted by J (L). Conjecture B. (see Poonen [10]) In every finite lattice L with |L| ≥ 2, there is a join-irreducible element j such that |{x ∈ L : j ≤ x}| ≤ |L| /2. Poonen [10] essentially proved the following theorem. Theorem 1. Suppose L is a lattice with at least two elements. If, for all x ∈ L, the interval [0, x], where 0 is the minimum element in L, is complemented, then |χ (L)| ≤ 1/2 1 min{|Vx | : x ∈ J (L)}, where Vx = {y ∈ L : x ≤ y}. holds, where χ (L) = |L|  This research was supported by the Board of College and University Development,

University of Pune, via the projects BCUD/494 and SC-66.

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Using this theorem, it has also been proved that Frankl’s conjecture is true for finite geometric lattices and relatively complemented lattices. In fact, till date, the conjecture is shown to be true for the following classes of finite lattices: 1.

Distributive lattices, Geometric lattices, Relatively Complemented lattices (see Poonen [10]) 2. Modular lattices (see Abe and Nakano [2]) 3. Lower Semimodular lattices (see Reinhold [11]) 4. Strong semimodular lattices (see Abe [1]) 5. Lower quasi-semimodular lattices (see Abe and Nakano [3]). First, we have a few definitions mentioned below. We say that x is covered by y and write x ≺ y, if x ≤ z ≤ y implies x = z or z = y. An element which covers the least element, denoted by 0, if exists, is called an atom. Dually, a lower cover of 1 is called a dual atom. In a poset P, the set {c ∈ P : c ≤ a} denoted by (a] is called as the principal ideal and dually we have principal filter denoted by [a). For other known concepts in lattice theory one is referred to Gr¨atzer [5] and Maeda and Maeda [9]. Knil [8] posed the conjecture for meet semilattices as follows: Conjecture C. Let L be a meet semilattice with |L| ≥ 2. Does there always exists a join-irreducible element j ∈ L such that |[ j)| ≤ |L|/2? In this paper, we have made an attempt to answer this conjecture for a finite poset satisfying the dual covering property. Now, we have a poset-theoretic version of Frankl’s Conjecture as follows: Conjecture D. Let P be a finite poset with |P| ≥ 2. Does there always exists a join-irreducible element j ∈ P such that |[ j)| ≤ |P|/2? We have following two definitions of a join-irreducible element in a finite poset P. Definition 1. Let P be a poset. An element j ∈ P is join-irreducible if j covers a unique element. Definition 2. Let P be a poset. An element j ∈ P is join-irreducible if it is not obtained as join of elements different from j. The set of all non-zero join-irreducible elements of P is denoted by J (P). Note that, if a finite poset happens to be a lattice then these two definitions coincide with each other. Remark 1. It is readily observed that if one considers the Definition 1, then the conjecture is not true in the case of posets. In fact, in the poset depicted in Figure 1, the

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Fig. 1.

only join-irreducible elements are j1 , j2 according to Definition 1 and the conjecture fails. Note that, in a finite lattice every element is the join of join-irreducible elements. However, this is not true with Definition 1 of join-irreducible elements (see Figure 1). Hence, in this paper we consider the Definition 2 of join-irreducible elements in a poset which enables us to mention the following lemma. Lemma 1. Let P be a finite poset. Let a, b ∈ P be the elements such that a ≤ b then there exists a join-irreducible element j ∈ P such that j ≤ a and j ≤ b. From the above lemma, it is clear that, in a finite poset every element is a join of join-irreducible elements. Now, we proceed with some definitions and terminologies in a poset P. Consider a subset A of P. The set Au = {x ∈ P | x ≥ a for every a ∈ A} is called the upper cone of A. Dually, we have the concept of the lower cone A of A. The set Au shall mean {Au } and Au shall mean {A }u . The lower cone {a} is simply denoted by a  and {a, b} is denoted by (a, b) . Similar notations are used for upper cones. Further, for A, B ⊆ P, {A ∪ B}u is denoted by {A, B}u and for x ∈ P, the set {A ∪ {x}}u is denoted by {A, x}u . Similar notations are used for lower cones. We note that Au = A , Auu = Au and {a u } = {a} = a  . Moreover, A ⊆ Au and A ⊆ Au . If A ⊆ B then B  ⊆ A and B u ⊆ Au . Let L be a lattice. A pair of elements a, b ∈ L is called dual modular, denoted by (a, b)M L∗ , if the following condition holds: “If b ≤ c ≤ a ∨ b, then there exists an element d ∈ L such that d ≤ a and d ∨ b = c”. The following definition of a dual modular pair in a poset is due to Thakare, Maeda and Waphare [13]. For more details on modular pairs in posets; see Waphare and Joshi [16]. Let P be a poset. A pair of elements a, b ∈ P is called dual modular, denoted by (a, b)M ∗ , if the following condition holds:

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“If b ≤ c, for c ∈ (a, b)u then there exists an element d ∈ P such that d ≤ a and (d, b)u = c ”. We say that a poset P is modular if (a, b)M ∗ holds for all a, b ∈ P. Note that if a poset P happens to be a lattice then (a, b)M ∗ is equivalent to (a, b)M L∗ . A lattice L has the covering property if for any atom p ∈ L and p ≤ a, for a ∈ L, then a ≺ a ∨ p. Dually, we have the concept of the dual covering property in lattices. Before we prove our result, we note the following result due to Maeda and Maeda [9]. Lemma 2. A lattice L with 1 has the dual covering property if and only if (d, x)M L∗ holds for all dual atoms d ∈ L and for all x ∈ L. Thakare, Maeda and Waphare [13] introduced the definition of dual covering property in a poset as follows: “ A poset P is said to satisfy the dual covering property, if for any dual atom d ∈ P and a ≤ d, for a ∈ P, (d, a)M ∗ holds.” Now we are ready to prove the Frankl’s Conjecture for most weaker class, that is, the class of finite posets satisfying dual covering property. Theorem 2. If a finite bounded poset P satisfies the dual covering property then Frankl’s Conjecture D holds for it. Proof. If 1 is join-irreducible in P, then we are through. If not, then for any dual atom d there exists j ∈ J (P) such that j < 1 and j ≤ d. Then ( j, d)u = P. Take x ∈ P such that j ≤ x, then clearly x ∈ (d, j)u . By the dual atomic covering property, as j ≤ x ∈ (d, j)u , there exists y ∈ P such that y ≤ d and x  = (y, j)u . It is clear that j ≤ y; otherwise x = y and so j ≤ x = y ≤ d, a contradiction to the fact that j ≤ d. Thus, for a given x with j ≤ x, there exists y ≤ d such that x = j ∨ y. |P| .  In fact, we have a one-to-one map from [ j) into (d]. Therefore |[ j)| ≤ 2 Remark 2. In fact, the proof shows that even if there is a pair of an atom and a dual atom which is dual modular, the poset satisfies the Frankl’s Conjecture. |P| 2 for every join-irreducible element j of P, then P (Dedekind MacNeille Completion of P) is Boolean. For this purpose, we need following definitions and results and we also note that if a poset happens to be a lattice then these concepts coincide with the corresponding known concepts in lattices. A poset P with 0 is called (1) section semicomplemented (in brief SSC), if for a, b ∈ P, a ≤ b implies the existence of c ∈ P such that 0 < c ≤ a and (c, b) = {0} and (2) atomistic, if for a, b ∈ P, a ≤ b, there exists an atom p such that p ≤ a Now, we prove that if a poset P satisfies dual covering property and |[ j)| =

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and p ≤ b (see Thakare, Pawar and Waphare [14]). It is clear that, a finite bounded poset is atomistic iff it is SSC. An element x ∗ of a poset P with 0 is called the pseudocomplement of x ∈ P, if (x, x ∗ ) = {0} and if (x, y) = {0} for y ∈ P then y ≤ x ∗ . A poset is called pseudocomplemented if each element of it has the pseudocomplement (see Venkatanarasimhan [15]). An element s  of a poset P is called a complement of s ∈ P, if (s, s  )u = (s, s  )u = P. A poset P is called complemented if every element in P has a complement (see Chajda [4]). Let P be a poset. A pair of elements a, b ∈ P is called distributive, denoted by (a, b)D, if the following condition holds: “ {(a, b)u , x} = {(a, x) , (b, x) }u for every x ∈ P.” A poset P is said to be distributive if (a, b)D holds for all a, b ∈ P (see Waphare and Joshi [16]). A distributive complemented poset is called Boolean; see Halas [6]. Theorem 3. (Waphare and Joshi [16]) Let a, b be elements of an atomistic poset P. Then the following conditions are equivalent. (α) (a, b)D. (β) If p is an atom and p ∈ (a, b)u then p ≤ a or p ≤ b. Theorem 4. (Joshi [7]) Let P be a finite poset with 0. Then the completion by cuts of P is a Boolean lattice if and only if P is a distributive SSC poset. |P| 2 for every join-irreducible element j ∈ P, then P (Dedekind MacNeille Completion of P) is Boolean.

Theorem 5. Let P be a poset satisfying the dual covering property and if |[ j)| =

Proof. In view of Theorem 4, it is enough to show that P is distributive and atomistic. Claim 1. P is atomistic. It is sufficient to show that every join-irreducible element is an atom. Suppose that j ∈ P is a join-irreducible element which is not an atom. Then there exists an |P| = |[ p)|, a contradiction. Thus atom p ∈ P such that p < j. But then |[ j)| = 2 P is an atomistic poset. Claim 2. For an atom p ∈ P there is a unique dual atom d p ∈ P which is a pseudocomplement of P and vice versa. Since P is an atomistic poset and 1 ≤ d p for a dual atom d p ∈ P, there exists an atom p ∈ P such that p ≤ d p . Hence ( p, d p )u = P. For any x ∈ P with p ≤ x we

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have p ≤ x ∈ ( p, d p )u . By the dual covering property, there exists y ∈ P such that y ≤ d p and x l = (y, p)u . Clearly, this defines a one-to-one and onto map from [ p) |P| = |P\[ p)| . Thus for every atom p, there is unique dual into P\[ p), as |[ p)| = 2 atom d p such that p ≤ d p and conversely. Now, we prove d p is a pseudocomplement of an atom p. Let ( p, a) = {0}, for a ∈ P. We have to show that a ≤ d p . Suppose a ≤ d p . Since P is an atomistic poset, there exists an atom q ≤ a and q ≤ d p . But there is only one atom p with the property that p ≤ d p . Hence p = q. But then p ≤ a, a contradiction to p ≤ a. Thus, a ≤ d p . Claim 3. P is distributive. For an atom p ∈ P, suppose p ∈ (a, b)u , where a, b ∈ P. In view of Theorem 3, to show P is distributive, it is sufficient to show that either p ≤ a or p ≤ b. Suppose p ≤ a and p ≤ b, and so { p, a} = {0} and { p, b} = {0}. By Claim 2, we have a dual atom d p which is also pseudocomplement of p. Therefore, we must have a ≤ d p , b ≤ d p , which means that p ∈ (a, b)u ⊆ d p , a contradiction to the fact that p ≤ d p .  Thus, P is distributive and atomistic, hence by Theorem 4, P is Boolean. Remark 3. We note that a finite distributive poset need not be a Boolean poset though it is atomistic, a situation contrary to the case of lattices (see Joshi [7]). In the context of Frankl’s Conjecture, it is worth mentioning here that many of the known results for lattices follow from Theorem 2. If we consider the dual concepts (usually denoted by ∗ ) of the respective concepts that appeared in Figure 3.12 of Stern [12] (page 133), we have the following Figure 2 and an easy observation verifies our claim and also it can be observed that the Frankl’s Conjecture is true for all these classes.

Fig. 2.

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Acknowledgements. The authors are grateful to the referees for their valuable comments and suggestions.

References 1. Abe, T.: Strong semimodular lattices and Frankl’s Conjecture. Algebra Universalis 44, 379–382 (2000) 2. Abe, T., Nakano, B.: Frankl’s Conjecture is True for Modular Lattices. Graph. Comb. 14, 305–311 (1998) 3. Abe, T., Nakano, B.: Lower semimodular types of lattices: Frankl’s Conjecture holds for lower quasi-modular Lattices. Graph. Comb. 16, 1–16 (2000) 4. Chajda, I.: Complemented ordered sets. Arch. Math. (Brno). 28, 25–34 (1992) 5. Gr¨atzer, G.: General Lattice Theory, second edition. Birkhauser Verlag, Berlin, (1998) 6. Halas, R.: Some properties of Boolean ordered sets. Czech. Math. Journal (Praha), 46(121), 93–98 (1996) 7. Joshi, V.: On completion of section semicomplemented posets. Southeast Asian Bull. Math. 31, 881–892 (2007) 8. Knil, E.: Graph Generated Union-closed Families of Sets, Ph. D. Thesis, University of Colorado at Bouldre, (1991) 9. Maeda, F., Maeda, S.: Theory of Symmetric Lattices. Springer-Verlag, Berlin, Heidelberg, New York (1970) 10. Poonen, B.: Union-closed families. J. Comb. Theory Ser. A 59, 253-268 (1992) 11. Reinhold, J.: Frankl’s conjecture is true for lower semimodular lattices. Graph. Comb. 16, 115–116 (2000) 12. Stern, M.: Semimodular Lattices, Theory and Applications. Cambridge University Press, (1999) 13. Thakare, N.K., Maeda, S., Waphare, B.N.: Modular pairs, covering property and related results in posets. J. Indian Math. Soc. (N.S.) 70, 50–87 (2003) 14. Thakare, N.K., Pawar, M.M., Waphare, B.N.: Modular pairs, standard elements, neutral elements and related results in posets. J. Indian Math. Soc. (N.S.) 71, 13–53 (2004) 15. Venkatanarasimhan, P.V.: Pseudo-complements in posets. Proc. Amer. Math. Soc. 28, 9–17 (1971) 16. Waphare, B.N., Joshi, V.: On Distributive Pairs in Posets. Southeast Asian Bull. Math. 31, 1205–1233 (2007) Received: August 25, 2007 Final version received: October 4, 2008

Graphs and Combinatorics (2009) 25:123–128 Digital Object Identifier (DOI) 10.1007/s00373-008-0827-3

Graphs and Combinatorics © Springer-Verlag 2009

List Point Arboricity of Dense Graphs Lingyan Zhen, Baoyindureng Wu College of Mathematics and System Sciences, Xinjiang University, Urumqi, Xinjiang 830046, P.R. China. e-mail: [email protected]

Abstract. Let G be a simple graph. The point arboricity ρ(G) of G is defined as the minimum number of subsets in a partition of the point set of G so that each subset induces an acyclic subgraph. The list point arboricity ρl (G) is the minimum k so that there is an acyclic L-coloring for any list assignment L of G which |L(v)| ≥ k. So ρ(G) ≤ ρl (G) for any graph G. Xue and Wu proved that the list point arboricity of bipartite graphs can be arbitrarily large. As an analogue to the well-known theorem of Ohba for list chromatic number, we obtain ρl (G + K n ) = ρ(G + K n ) for any fixed graph G when n is sufficiently large. As a consequence, if ρ(G) is close enough to half of the number of vertices in G, then ρl (G) = ρ(G). Particularly, we determine that ρl (K 2(n) ) =  2n 3 , where K 2(n) is the complete n-partite graph with each partite set containing exactly two vertices. We also conjecture that for a graph G with n vertices, if ρ(G) ≥ n3 then ρl (G) = ρ(G). Key words. List coloring, Point arboricity, List point arboricity.

1. Introduction All graphs considered here are finite, undirected and simple. We refer to [8] for unexplained terminology and notations. As usual, K n and Pn denote the complete graph and the path of order n, respectively. We say two graphs G and H are vertexdisjoint if they have no vertex in common, and denote their join by G + H , which is obtained from their union by joining each vertex of G to each vertex of H . The point arboricity of a graph G is defined as the minimum number of subsets in a partition of the point set of G so that each subset induces an acyclic subgraph. The notion of list coloring of graphs was introduced by Vizing [7] and independently ¨ Rubin and Taylor [3]. Borodin et. al [1] defined a similar concept for the by Erdos, point arboricity of a graph. A list assignment of a graph G is a function L defined on V (G) such that L(v) ⊆ N is the list of colors available for the vertex v ∈ V (G), where N is the set of natural numbers. For a given positive integer k, if |L(v)| = k for every vertex v ∈ V (G), we say L is a k-list assignment of G. For a list assignment L of G, we say c is an L-coloring if c(v) ∈ L(v) for every vertex v ∈ V (G). Set c(L) = {c(v)|v ∈ V (G)}, that is the set of colors chosen for the vertices of G under c. An L-coloring c is called acyclic if for each color i ∈ c(L), G[Vi ] is acyclic, or is ∗ Research supported by NSFC (No.10601044) and XJEDU2006S05.

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a forest, where Vi is the set of vertices v of G with c(v) = i. In this case, we say G is acyclic L-colorable. If G is acyclic L-colorable for any possible k-list assignment L, G is called acyclic k-list colorable. The list point arboricity of a graph G, denoted by ρl (G), is the minimum number k for which G is acyclic k-list colorable. So, it is trivial to see that ρ(G) ≤ ρl (G). Xue and Wu [9] proved that the list point arboricity of biparticle graphs can be arbitrarily large. Ohba [4] proved that if the chromatic number of a graph G is close to the number of vertices in G, then the chromatic number χ (G) of G coincides with its list chromatic number χl (G). It is surprising that the similar results hold for point arboricity and list point arboricity of graphs. We shall use Ohba’s idea to prove the following two theorems. Theorem 1.1. For any graph G, there exists a non-negative integer n 0 = n 0 (G) such that ρ(G + K n ) = ρl (G + K n ), for any integer n with n ≥ n 0 . Theorem 1.2. If |V (G)| ≤ 2ρ(G) +

√

2ρ(G) − 1, then ρ(G) = ρl (G).

We also get some interesting by-products. 2. Main Results Before proving the theorems, we shall introduce some lemmas. Let G be a graph with a list-assignment L. Let X = {x1 , x2 , . . . , x p } ⊆ V (G). Set L(X ) = L(x1 ) ∪ L(x2 ) ∪ · · · ∪ L(x p ). G[X ] denotes the subgraph of G induced by X . Lemma 2.1. If a graph G is not acyclic L-colorable, then there exists a set X ⊆ V (G) such that |X | > 2|L(X )|. Proof. By contradiction, suppose |X | ≤ 2|L(X )| for all subset X of V (G). We consider the bipartite graph H with bipartition (V (G), C), where C is the two copies of L(V (G)) and for any c ∈ C there is an edge between c and v precisely if c ∈ L(v). Obviously, N H (S) = 2|L(S)| for all S ⊂ V (G). Consider the subset X of V (G). By the assumption, N H (X ) = 2|L(X )| ≥ |X |, so there is a matching M that saturates V (G) by Hall’s theorem. We color v by the color matched with it in M. Since C is two copies of L(V (G)), each element c of L(V (G)) appears at most two times as an end vertices of some edges of the matching M. That is, each color of L(G) is assigned to at most two vertices of V (G). Thus we get an acyclic L-coloring of V (G). A contradiction.  Lemma 2.2. Let G be a graph and F p be a forest with p vertices for a positive integer p. If G is acyclic k-list colorable and ( p − 1)(|V (G)| + p) ≤ 2 p(k + 1), then G + F p is acyclic (k + 1)-list colorable. Proof. We prove that G + F p is acyclic L-colorable by induction on p. Let L be a (k + 1)-list assignment of G + F p .

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For the induction basis, first let p = 1 and V (F1 ) = {u}. We color u by one of the elements in L(u), say c. For any v ∈ V (G), let L (v) = L(v) − {c}. Since |L (v)| ≥ |L(v)| − 1 ≥ k for any v ∈ V (G), G is acyclic L -colorable by the assumption of the lemma. Thus, G + F1 is acyclic L-colorable. Now, assume p ≥ 2 and let U = V (F p ).  Case 1. L(u) = ∅. u∈U  Take a color c ∈ u∈U L(u). Then, assign c to each vertex of F p . For any v ∈ V (G), let L (v) = L(v)\{c}. Since |L (v)| ≥ |L(v)| − 1 ≥ k for any v ∈ V (G), G has an acyclic L -coloring . Thus, G + F p is acyclic L-colorable.  Case 2. L(u) = ∅. u∈U

By contradiction, suppose that G + F p has no acyclic L-coloring. By Lemma 2.1, there must exist X ⊆ V (G + F p ) with |X | > 2|L(X )|. Here, we take X to be maximal. Subcase 2.1. U ⊆ X . By the assumption of Case 2, each color in L(U ) appears in the lists of at most p p − 1 vertices of U , and hence we have |L(X )| ≥ |L(U )| ≥ p−1 (k + 1). On the other p hand, by the assumption of the lemma, we have |X | ≤ |V (G + F p )| ≤ 2 p−1 (k + 1). Thus we have |X | ≤ 2|L(X )|, a contradiction. Subcase 2.2. U \ X = ∅. Let Y = V (G + F p )\X and L (y) = L(y)\ L(X ) for any y ∈ Y . Next, we shall see that G + F p has a acyclic L-coloring by showing that there is an acyclic L-coloring of G[X ] and an acyclic L -coloring of G[Y ]. (This is a contradiction.) First, we show that G[X ] has an acyclic L-coloring. Note that the function f ( p) = (1− 1p )(|V (G)|+ p) is an increasing function, and since ( p−1)(|V (G)|+ p) ≤ 2 p(k +1), we have ( p−2)(|V (G)|+ p−1) ≤ 2( p−1)(k +1). By induction hypothesis, G + F p−1 is acyclic (k + 1)-list colorable. Since G[X ] ⊆ G + F p−1 , G[X ] has an acyclic L-coloring. Now, we show that G[Y ] is acyclic L -colorable. Otherwise, by Lemma 2.1, there is a set S ⊆ Y such that |S| > 2|L (S)|. But, we have |L(X ∪ S)| = |L(X )|+|L (S)| < 1 1 1 2 |X | + 2 |S| = 2 |X ∪ S|. This contradicts the maximality of X . Thus, we have  |S| ≤ 2|L (S)| for any S ⊆ Y , and hence G[Y ] is acyclic L -colorable. Let K 2(n) denote the complete n-partite graph with all partite sets of size 2, that is, ¨ Rubin and Taylor[3] proved that χl (K 2(n) ) = χ (K 2(n) ) = K 2,2,...,2 with n 2’s. Erdos, n. Using Lemma 2.2, we obtain Corollary 2.3. ρ(K 2(n) ) = ρl (K 2(n) ) =  2n 3 . 2n Proof. Since  2n 3  = ρ(K 2(n) ) ≤ ρl (K 2(n) ), it suffices to show that ρl (K 2(n) ) ≤  3 . 2n We prove it by induction on n. First, it is trivial that ρl (K 2(n) ) =  3  when n = 1, 2. For n = 3, let V (K 2(3) ) = {u 1 , v1 , u 2 , v2 , u 3 , v3 } and E(K 2(3) ) = {u i v j | 1 ≤ i, j ≤ 3} \ {u 1 v1 , u 2 v2 , u 3 v3 }. Since G[{u 1 , u 2 , v2 }] ∼ = P3 ∼ = G[{v1 , u 3 , v3 }], G[{u 1 , u 2 , v2 }] is acyclic 1-list colorable and G[{v1 , u 3 , v3 }] is a forest with 3 vertices.

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Since (3 − 1)(|V (G[{u 1 , u 2 , v2 }])| + 3) = 12 = 2 · 3(1 + 1), by Lemma 2.2, P3 + P3 is acyclic 2-list colorable. Thus K 2(3) is acyclic 2-list colorable since K 2(3) is a subgraph of P3 + P3 . Now, we assume n > 3. To show ρl (K 2(n) ) ≤  2n 3 , it is enough to show that ρl (K 2(n−3) + P3 + P3 ) ≤  2n , since K is a spanning subgraph of K 2(n−3) + P3 + P3 . 2(n) 3 2(n−3) By induction hypothesis ρl (K 2(n−3) ) ≤  3  and since (3−1)(2(n −3)+3) ≤ 2× 2(n−3) 3( 2(n−3) 3  + 1), we conclude that K 2(n−3) + P3 is acyclic ( 3  + 1)-list colorable by Lemma 2.2. Moreover, since |V (K 2(n−3) + P3 )| = 2n − 3 and (3-1)(2n − 3 + 3) ≤ 2n 2×3( 2n−3 3 +1), again by Lemma 2.2, K 2(n−3) + P3 + P3 is acyclic  3 -list colorable. 2n Thus ρl (K 2(n) ) ≤  3 .  A graph G is called k-degenerate if every induced subgraph of G contains a vertex of degree at most k. The degeneracy of G, denoted by deg(G), is the minimum number k for which G is k-degenerate. Namely, deg(G) = max{δ(H ), H is taken over every subgraph of G }. Chartrand and Kronk [2] obtained that deg(G)+1  for a graph G. Xue and Wu [9] proved that it is ρ(G) ≤ 1 +  deg(G) 2 = 2 . Thus for any graph also true for list point arboricity, namely, ρl (G) ≤  deg(G)+1 2 G with n vertices, ρl (G) ≤  n2  since deg(G) ≤ (G) ≤ n − 1. But for convenience of readers, we provide an independent proof. Lemma 2.4. For any graph G with n vertices, ρl (G) ≤  n2 . Proof. We prove this lemma by induction on |V (G)|. First, it is obvious that ρl (G) ≤  n2  when n = 1, 2. Now, we assume n ≥ 3 and let L be a  n2 -list assignment of G. If there exist v1 and v2 in G such that L(v1 ) ∩ L(v2 ) = ∅, we take a color c ∈ L(v1 ) ∩ L(v2 ) and assign c to v1 and v2 and let L (v) = L(v) \ {c} for any v ∈ V (G)\{v1 , v2 }. Since |L (v)| ≥ |L(v)| − 1 ≥  n−2 2 , by induction hypothesis, G − {v1 , v2 } has an acyclic L -coloring. Thus G is acyclic L-colorable. Otherwise, for any two vertices u and v, L(u) ∩ L(v) = ∅. Then it is trivial that G is acyclic L-colorable.  Let G be a graph and let c : V (G) → {1, 2, . . . , ρ(G)} be a ρ(G)-coloring of G such that G[Vi ] is a forest for each i, where Vi = {v ∈ V (G) : c(v) = i}. Let |Vi | = ai for i = 1, 2, · · · , ρ(G). We assume a1 ≥ a2 ≥ · · · ≥ aρ(G) . Note that ai ≥ 2 for each i, 1 ≤ i ≤ ρ(G) − 1. Lemma 2.5. If (a1 − 1)|V (G)| ≤ 2a1 ρ(G),

(1)

then ρ(G) = ρl (G). Proof. We prove this lemma by induction on ρ(G). First, if ρ(G) = 1 then G is a forest, and thus ρl (G) = 1 = ρ(G). Now assume that ρ(G) ≥ 2 and thus a1 ≥ 2. By (1) we have |V (G)| ≤

2a1 ρ(G). a1 − 1

(2)

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Since ρ(G) ≤ ρl (G), it suffices to show that G is acyclic ρ(G)-list colorable. If a1 = 2, then ρ(G) =  |V (G)| 2 , by Lemma 2.4 G is acyclic ρ(G)-list colorable. Now, we assume a1 ≥ 3, we have |V (G − V1 )| = |V (G)| − a1 2a1 ρ(G) − a1 a1 − 1 a1 (2ρ(G) − a1 + 1) = a1 − 1 a1 ≤ (2ρ(G) − 2) a1 − 1 2a2 (ρ(G) − 1) ≤ a2 − 1 ≤

By induction hypothesis, this inequality implies that (G − V1 ) is acyclic (ρ(G) − 1)list colorable since V2 is the largest color class of G − V1 . As (G − V1 ) is acyclic (ρ(G) − 1)-list colorable and the relation (2) holds, Lemma 2.2 implies G is acyclic ρ(G)-list colorable.  Suppose graphs G and H are vertex-disjoint. We know that χ (G + H ) = χ (G)+ χ (H ). But for point arboricity, ρ(G + H ) = ρ(G) + ρ(H ) does not hold in general. For instance, if we take G ∼ = = H ∼ = K n for an odd number n ≥ 3, then G + H ∼ . It is trivial to see that ρ(G + H ) ≤ K 2n , but ρ(K 2n ) = n and ρ(K n ) = n+1 2 ρ(G) + ρ(H ). Next we present a lower bound for ρ(G + H ) when H is a complete graph. Lemma 2.6. For any graph G and a positive integer n, ρ(G + K n ) ≥

ρ(G)+n . 2

Proof. Let k = ρ(G + K n ) and c be an acyclic k-coloring of G + K n with Vi = {v ∈ V (G + K n ) : c(v) = i} for i = 1, 2, · · · , ρ(G). Since G[Vi ] is a forest, it contains at most two vertices of K n . So if k j = |{i : 1 ≤ i ≤ k and |Vi ∩ V (K n )}| = j}| and k j = 0 then j ≤ 2. Furthermore, we have k = k0 + k1 + k2 , k1 + 2k2 = n and .  k0 + k1 ≥ ρ(G). Therefore, k ≥ 21 k0 + k1 + k2 ≥ ρ(G)+n 2 Proof of Theorem 1.1. Let t = |V (G)|. Let f : V (G) → {1, . . . , ρ(G + K n ) = ρ} be an optimal acyclic coloring of G + K n , with color class of sizes a1 ≥ · · · ≥ aρ(G) . . Clearly, 2 ≤ a1 ≤ t + 2 for large enough n. By Lemma 2.6, ρ(G + K n ) ≥ ρ(G)+n 2 2a1 2(t+2) ρ(G)+n Observe that n + t ≤ t+1 ≤ a1 −1 ρ for large enough n. Thus, Lemma 2.5 2 implies that ρ(G + K n ) = ρl (G + K n ).  Proof of Theorem 1.2. Since ai ≥ 2 for i = 2, 3, . . . , ρ(G) − 1 and aρ (G) ≥ 1, we have a1 ≤ |V (G)| − (2ρ(G) − 3). Hence,   a1 ≤ |V (G)| − (2ρ(G) − 3) ≤ 2ρ(G) + 2ρ(G) − 1 − (2ρ(G) − 3) = 2ρ(G) + 2.

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√ 2ρ(G) + 2 a1 × 2ρ(G) 2ρ(G) ≥ √ a1 − 1 2ρ(G) + 1 2ρ(G) = 2ρ(G) + √ 2ρ(G) + 1 2ρ(G) − 1 > 2ρ(G) + √ 2ρ(G) + 1  = 2ρ(G) + 2ρ(G) − 1 ≥ |V (G)|.

Therefore, by Lemma 2.5, ρ(G) = ρl (G).



Ohba [4] conjectured that if G has at most 2χ (G) + 1 vertices then χl (G) = χ (G). Reed and Sudakov [5] proved the weaker version of the conjecture by replacing 2χ (G)+1 with 53 χ (G)− 43 . They [6] also proved that this conjecture is asymptotically correct. Motivated from the above conjecture and the result of Corollary 2.3, we raise the following conjecture. Conjecture. For a graph G with n vertices, if ρ(G) ≥

n 3

then ρl (G) = ρ(G).

Acknowledgements. We would like to thank the referee for her/his many helpful comments. In particular, the authors are greatly indebted to the referee, who pointed out a flaw in our original proof of Theorem 1.1, and provided us the current elegant proof.

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