A STRONGLY REGULAR N-Full GRAPH OF SMALL ORDER Vu Ha Van Dept of Math, Yale Univ, New Haven, CT 05620, U.S.A. Abstract For every n positive integer we show the construction of a strongly regular graph of order at most 2n+2 which contains every graph of ordered n as a subgraph. The estimation concerning the construction is best possible.
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INTRODUCTION
A graph is called n-full if it contains every graph of order n as an induced subgraph. It is easy to show that a n-full graph should contains at least 2(n−1)/2 vertices, and it is proved by probabilistic method that there are n-full graphs of order (roughly) n2 2n/2 , but no construction of this order is known [2]. Suppose that we want to find a n-full graph satisfying a given property. If the set of graphs having this property is small, then the probabilistic method may not be effective any longer, and hence the existance of such a n-full graph is questionable. In this paper we deal with the following property. A graph is strongly regular with parameters (v, k, λ, µ) if it is a regular graph with valency k on v vertices, moreover any two adjacent vertices have precisely λ common neighbours and any two non-adjacent vertices have precisely µ common neighbours. This notion was introduced by Bose and plays an important role in algebraic combinatorics (see for example [4]). Our purpose is to present a n-full graph H which is strongly regular and contains at most 2n+2 vertices (Theorem 1). It shows that the strong regularity of H does not influence the structure of its small subgraphs. Another interesting point is that the order of H is relatively close to the lower bound, although strongly regular graphs are very rare. To prove the n-fullness of this graph, we shall give a quick algorithm which embedds an arbitrarily given graph G of order n into H, using only c.n2 operations. We should mention here the result of Bollob´as and Thomason solving the same problem in [2]. They proved that every sufficiently large Paley graph (well-known strongly regular graphs) is n-full. The main tool of their proof is the famous Weil’s theorem, known as Riemann’s hypothesis on curves over finite fields. However this proof is rather an “existance” proof which does not give any explicit embedding. Moreover the Paley graphs require much more vertices (roughly n2 4n ) to be n-full.
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In addition we shall shows that the estimation on the fullness of our graph is best possible (Theorem 2). So it requires new method to construct a strongly regular n-full graph with order closer to the lower bound. Nevertherless, even the existance of such graph is not clear.
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PRELIMINARIES
Denote by VPn the n dimensional vector space over GF (2) with the inner product P . Let P0 be the following product: P0 (a, b) = a1 bn + a2 bn−1 + . . . + an b1 where a = (a1 , a2 , . . . , an ) and b = (b1 , b2 , . . . , bn ) are two vectors of V n , where V n is the set of vectors of the space. If not specially mentioned ab will be understood as P0 (a, b) in the rest of the paper for the sake of brevity. Two vectors are called orthogonal if their product is zero. Let S be a subset of vertors of V n , denote by S ⊥ the set of vectors orthogonal to every vector of S and Span S the subspace spanned by S. A vector is called isotropic if it is orthogonal to itself and a subset S is called totaly isotropic if S ⊆ S ⊥ . For a subspace T of VPn denote by dim T and i(T ) the dimension of T and the maximun dimension of a totally isotropic subspace of T , respectively. It is a well known fact in linear algebra that dim VPn is n/2 for every non-singular product P [1]. Consider the space VPn0 where n is an even positive number. If n is even, it is easy to see that every vector of VPn0 is isotropic (any product with this property are called self-orthogonal). Let G be a graph on the set {1, 2, . . . , k}. Let δij = 1 if (i, j) is an edge and 0 otherwise. We call the graph G (n,P)-representable if there are k non-zero, different vectors a1 , . . . , ak in VPn satisfying P (ai , aj ) = δij for every pair 1 ≤ i, j ≤ k. In this case we say that a1 , a2 , . . . , ak represent G. Denote by G(VPn ) the graph of non-zero vectors of VPn , where (a, b) is an edge iff P (a, b) = 1, that is, b ∈ / a⊥ . Consider the graph G(VPn0 ), the abovementioned note implies that deg a = |{b; b ∈ / a⊥ }| = 2n−1 . On the other hand, if a and b are two vertices (vectors) then the number of their common neighbours is the number of vectors in VPn0 not contained in a⊥ ∪ b⊥ , which is 2n−2 and thus G(VPn0 ) is a (2n − 1, 2n−1 , 2n−2 , 2n−2 ) strongly regular graph. It is also very easy to show that G(VPn ) are quasi-random by using one of the characterizations of quasi-random graphs in [3].
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RESULTS AND PROOFS
Theorem 1. For every k positive integer G(VP2k ) is (2k − 1)-full 0 The proof of the theorem is based on the following lemma. Lemma 1. Every graph with 2k − 1 vertices is (2k, P0 )-representable for every k positive integer. 2
Proof. We use induction. The case k = 1 is trivial. Assume that the statement holds for k − 1, we will show that it also holds for k, k > 1. Let G be an arbitrary graph on the vertex set {1, 2, . . . , 2k − 1}; two cases are to be considered. 1. G is a coclique. Choose 2k − 1 different, non-zero vectors from V 2k such that their last k coordinates are 0; clearly it is possible for k > 1. These vectors are pairwise orthogonal and thus represent G. 2. G is not a coclique. Without lack of generality we can suppose that (2k − 2, 2k − 1) is an edge of G. For every i, j < 2k − 2 let: 0 δi,j = δi,j − δ2k−2,i δ2k−1,j − δ2k−2,j δ2k−1,i (∗)
On the vertex set {1, 2, . . . , 2k − 3} define an auxiliary graph G0 as follows: (i, j) is an 0 = 1. edge iff δi,j Note that G0 has 2k − 3 vertices, thus by the induction hypothesis G0 is (2k − 2)representable. So in VP2k−2 we can find 2k − 3 different, non-zero vectors a01 , a02 , . . . , a02k−3 0 0 . such that a0i a0j = δi,j Let ai = (δ2k−2,i , a0i , δ2k−1,i ) for all i < 2k −2; a2k−2 = (0, 0, ..0, 1) and a2k−1 = (1, 0, ...0, 0) (with 2k − 1 zeros). Obviously, a1 , a2 , . . . , a2k−2 , a2k−1 are different, non-zero vectors of V 2k . To complete the proof it suffices to show that these vectors represent G. Let’s check all the products: a2k−2 a2k−1 = 1 = δ2k−2,2k−1 since (2k − 2, 2k − 1) is an edge in G a2k−2 ai = δ2k−2,i for every i < 2k − 2 a2k−1 ai = δ2k−1,i for every i < 2k − 2 ai aj = δ2k−2,i δ2k−1,j + a0i a0j + δ2k−1,i δ2k−2,j = δi,j for every 1 ≤ i, j ≤ 2k − 3 by (∗) The proof of the lemma is complete. Note that a graph G is an induced subgraph of G(VPn ) iff G is (n,P)-representable, thus the lemma implies Theorem 1. Since every graph with even vertices can be seen as an induced subgraph of a graph with one more vertex, the following corollary is immediate. Corollary 1. For every r positive integer there is a strongly regular r-full graph of order at most 2r+2 . Remark. Let f (n) be the maximum time one needs to find a representation of a graph of order n, the algorithm provided in Lemma 1 implies that f (n + 1) ≤ f (n) + cn, hence f (n) ≤ 12 cn2 , where c is a constant. This means that we can find very quickly where a copy of a given small graph is placed in G(VPn0 ). In the case of Paley graphs this question seems to be hopeless because of the difficulties concerning large primes. Finding a representation of a graph of order n − 1 in G(VPn0 is equivalent to finding a solution of the system of 21 n(n − 1) equations ai aj = δij . Since we have n(n − 1) variables which are the coordinates of ai , the gap of the number of equations and that of variables 3
is rather lagre. So one may ask if the estimation in Theorem 1 could be improved, i.e., if G(VPn0 ) could be m-full for some m > n − 1. Although the possibility seems promising the following theorem surprisingly says that no improvement can be carried out, or in other words Theorem 1 is best possible. Theorem 2. For every k positive integer and P non-singular self-orthogonal product G(VP2k ) is not 2k-full. Proof. Let H2k be a graph on {1, 2, . . . 2k} with k − 1 disjoint edges (3, 4), . . . , (2k − 1, 2k) (H2 has no edges). The following lemma is needed. Lemma 2. Suppose that H2k , k > 1 is (n,P)-representable for some n under the above conditions on P . Let a1 , a2 , . . . a2k represent H2k . Then in VPn the following inequality holds: 2k−2 i(Span({ai }2k i=1 )) > i(Span ({ai }i=1 )) Proof. Note that the edges of H2k are disjoint and P (a2k−1 , a2k ) = 1, thus Span (a2k−1 , a2k ) ⊥ Span {ai }2k−2 i=1 and their intersection contains the origin only. Let S be an isotropic subspace of Span {ai }2k−2 i=1 then Span (S, a2k−1 ) is an isotropic subspace of 2k Span {ai }i=1 and dim S < dim Span (S, a2k−1 ) since P is self-orthogonal, which complete the proof. Proof of theorem 2. Indirectly assume that H2k is 2k-representable and k > 1 (the case k = 1 is trivial). First note that if a1 , a2 , . . . a2k represent H2k then a1 , a2 , . . . , a2k−2 represent H2k−2 . On the other hand, i(VP2k ) = k as mentioned in section 2. Using this fact throw out the edges of H2k step by step, by lemma 2 one can easily see that at the last step H2 should be represented by two vectors a1 and a2 where i(Span {a1 , a2 }) ≤ 1, therefore P (a1 , a2 ) = 1, which is a contradiction because H2 has no edges. Thus our proof is complete. Remark. For an arbitrary inner product P one can show that the graph G(VPn ) is not (n + 2)-full. The counter example is the path of lenght n + 2 (containing n + 2 vertices). It is not difficult to prove that if this path is represented in a vector space then the n + 1 vectors representing the inner vertices of the path and one of the two endpoints must be linearly independent, and so the space must have dimension at least n + 1.
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REFERENCES
[1] Babai-Frankl: Linear algebraic methods in combinatorics (Preliminary version). [2] Bollob´as-Thomason: Graph which contains all small graphs, Europ. J. Combinatorics (1981), 2, 13-15. [3] Alon-Spencer -Erd˝os: The probabilistic method, Wiley Interscience Publication, 1992. [4] A.E. Brouwer, A.M. Cohen, and A. Neumaier: Distance-regular graphs, Springer Verlag, Berlin, 1989. e-mail:
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