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We exhibit explicit, combinatorially defined graphs satisfying the kth extension axiom: Given any set of k distinct vertices and any partition of it into two pieces, there exists another vertex adjacent to all of the vertices in the first piece and to none in the second. Quisani:1 I’ve been reading about zero-one laws, and many of the results involve extension axioms. In the simple case of graphs, by which I mean undirected graphs without loops or multiple edges, the kth extension axiom2 says that, for any k distinct vertices x1, . . . , xk and any subset S ⊆ {1, . . . , k}, there is another vertex adjacent to xα for all α ∈ S and for no other α. I know that each of these axioms is true in almost all sufficiently large finite graphs. (Of course, “sufficiently large” ∗Mathematics Department, University of Michigan, Ann Arbor, MI 48109–1109, U.S.A., [email protected] †Microsoft Research, One Microsoft Way, Redmond, WA 98052, U.S.A., [email protected] 1We thank Yuri Gurevich for lending us his graduate student, Quisani. 2Extension axioms are also called adjacency axioms, and graphs that satisfy the kth extension axiom are also called k-existentially closed. depends on k.) So there are lots of these graphs, but I’d like to see some actual examples. Authors: Well, just take a big set of vertices, flip coins to decide which pairs to join by edges, and chances are you’ll get what you want. Q: Yes, but I’d like a reasonably regular-looking graph, not something totally random. A: There are strongly regular graphs that satisfy extension axioms. In fact, Cameron and Stark [3] show that there are lots of them. Q: What does “strongly regular” mean? A: Every vertex has the same number of neighbors (i.e., the graph is regular), every pair of adjacent vertices has the same number of common neighbors, and similarly for pairs of non-adjacent vertices. Q: That sounds good; so they got rid of the randomness. A: No, part of their construction involves randomization. Q: So they don’t get a really explicit example? That’s what I’d want — an example that I can get my hands on and really see why the k-extension axiom holds. Have people given explicit, non-randomizing constructions of graphs that satisfy extension axioms? More precisely, are there explicitly defined sequences of finite graphs such that, as you go further in the sequence, more and more of the extension axioms are true? A: Yes. It was shown in both [1] and [2] that Paley graphs have the property you want.3 Furthermore, they are strongly regular. Q: What are Paley graphs? A: Take a prime p (or a prime power) that is congruent to 1 modulo 4 and form a graph whose vertices are the elements of the field Fp of size p. Join two distinct vertices by an edge if their difference is a square in Fp. The theorem is that this graph will satisfy the kth extension axiom provided p is sufficiently large compared to k. Q: Three questions: How large is sufficiently large? Can you explain why these Paley graphs satisfy extension axioms? And what’s the purpose of having p ≡ 1 (mod 4)? A: The easiest question to answer is the last. The congruence is needed to ensure that −1 is a square in Fp, which in turn is needed to make sure the Paley graph is an undirected graph; x is joined to y if and only if y is joined to x. If p were ≡ 3 (mod 4) then we’d have a Paley tournament instead of a graph. Sufficiently large is exponentially larger than k, roughly k24k. (For comparison, a random graph has a good chance of satisfying the kth extension axiom when 3The result isn’t explicitly stated in [2], but it’s proved in the course of proving Theorem 3, which says that every finite graph occurs as an induced subgraph in all sufficiently large Paley graphs. the number of vertices is somewhat larger than k22k; see [4].) Unfortunately, it’s not easy to explain why the Paley graphs satisfy the extension axioms. Both [1] and [2] invoke non-trivial results from number theory in the proofs of the extension axioms. Q: So are there no explicit examples where one can directly see why the extension axioms hold? A: Actually, we have such examples.4 They’re not as pretty as the Paley graphs, and they’re larger (for a given k), but we can explain what they are and how they work, without appealing to any deep theorems. Q: Great! Show me. A: OK. Given k, we’ll construct a graph whose vertices are certain matrices of 0’s and 1’s. These matrices will have r = 2k(k−1) + 1 rows and c columns, where c is chosen large enough so that 2 ≥ 2k2 ( rc k − 1 ) Q: Wait a minute; let me check that such a c exists. Yes. The left side is exponential in c while the right side, despite the exponential dependence on k, is only a polynomial in c when k is fixed, so any sufficiently large c will do. A: Right. Having fixed suitable r and c, let the vertices of our graph be r by c matrices of 0’s and 1’s in which a majority of the r rows are identical. That is, in each of our matrices, at least k(k − 1) + 1 of the rows are identical. To define the edges of our graph requires some preliminary terminology. We consider constraints, which a vertex may or may not satisfy. A constraint5 is given by a pair (A, F) where A is a set of k − 1 locations in our matrix (i.e., k − 1 pairs (i, j) with 1 ≤ i ≤ r and 1 ≤ j ≤ c) and F is a family of at most k functions from A to {0, 1}. We say that a vertex V satisfies a constraint (A, F) if the entries in V at the locations in A form an element of F, i.e., if (∃ f ∈ F)(∀(i, j) ∈ A) Vi j = f (i, j). We need to estimate the number of constraints (A, F). There are ( rc k−1 ) possibilities for A. For each fixed A, there are 2k − 1 functions from A to {0, 1}, so there are 2(k−1)k sequences of k such functions. Every possible second component F of a constraint (A, F), except for F = ∅, is the range of such a sequence. So the number of F’s for a fixed A is certainly at most 2k 2 , and the total number of 4The first version of these examples was derived by the second author from a construction, introduced for quite different purposes, in [8]. 5Readers familiar with combinatorial set theory will notice a similarity between the notion of constraint and Hausdorff’s construction [7] of large independent families of sets. constraints is no more than ( rc k − 1 ) 2 2 ≤ 2. Therefore, we can fix a function C from the set of c-component vectors of 0’s and 1’s onto the set of constraints. Since the notion “c-component vector of 0’s and 1’s” will be needed repeatedly, we abbreviated it as “row vector,” which makes sense since these are the vectors that occur as rows in our matrices. Q: You’re not choosing C at random, are you? A: No. We promised an explicit construction, with no randomization. To get a definite C, list all the row vectors in lexicographic order, and, after choosing some reasonable notation for constraints, list the constraints lexicographically also. Then let C map the nth element of the first list to the nth element of the second list, cycling back to the beginning of the second list if the first list is longer (which in fact it will be). Q: OK. It’s an unpleasantly arbitrary C, but I agree it’s not random. Why do you cycle back to the beginning rather than, say, just repeating the last element? A: The cycling is irrelevant in this argument, but we’ll want it for another purpose later. You’re quite right about the arbitrariness of C. Any surjection C will work for this proof, so, if you can think of a nicer explicit C, feel free to use it. But remember, we warned you that these graphs won’t be as pretty as Paley graphs. Using C, every vertex V of our graph determines a constraint V∗ as follows. A majority of the rows of V are the same row vector, which we call the majority row of V; apply C to that vector to get a constraint V∗. Now define a directed graph by putting an arrow from V to W whenever W satisfies the constraint V∗. Q: I thought you were going to produce an undirected graph. A: We will; the directed graph is only an auxiliary construction. The undirected graph has an edge joining V and W if, of the two possible directed edges, V to W and W to V , either both are present or neither is present. That is, V is adjacent to W just in case (V satisfies W∗) ⇐⇒ (W satisfies V∗). We’ll show that the graph so defined satisfies the kth extension axiom, but in order to do so we’ll need the following preliminary information. Claim. Let V1, . . . ,Vk be k distinct vertices of our graph, and let S be any subset of {1, . . . , k}. There is a constraint that is satisfied by Vα for all α ∈ S and for none of the other α’s. Proof. It suffices to find a set A of k − 1 locations (i, j) that separate the Vα’s, in the sense that, whenever α and β are distinct indices in {1, . . . , k}, then Vα and Vβ differ at some location in A. Once we have such an A, we have k distinct functions fα : A→ {0, 1} defined by fα(i, j) = (Vα)i j. Then let F = { fα : α ∈ S } and observe that (A, F) can serve as the desired constraint. So it remains only to produce an appropriate A. Q: That would be trivial if you allowed ( k 2 ) locations in A, rather than only k − 1. You could just choose, for each Vα and Vβ, one location where they differ. A: Right, and in fact, if you don’t want to worry about getting |A| down to k − 1, you could rewrite this whole story with ( k 2 ) in place of k − 1, starting with the definition of c (where the exponent k2 would also have to be adjusted). But in fact, it’s not hard to achieve |A| = k − 1. Proceed by induction on k, the case k = 1 being vacuous. For k > 1, start by choosing a location (i, j) where some Vα and Vβ differ. So our set of k vertices is partitioned into two nonempty subsets, according to the (i, j) entries. Let these subsets consist of a and b elements, so a + b = k. By induction hypothesis, we can find a − 1 locations sufficient to separate any two vertices from the first class and b − 1 locations sufficient to separate any two vertices from the second class. Together with (i, j), that gives us (a − 1) + (b − 1) + 1 = k − 1 locations that separate all the vertices. Q: The claim you just proved gives a sort of extension axiom for the auxiliary, directed graph. The constraint from the claim is C(w) for some row vector w. An r× c matrix W having all its rows equal to w would be a vertex of your graph, and there would be a directed edge from W to Vα if and only if α ∈ S . So if you could arrange for this W to satisfy all the constraints V∗ α, then W would be adjacent, in the undirected graph, to Vα if and only if α ∈ S . Unfortunately, I don’t see how you can arrange that. The constraints V∗ α might contradict each other. A: That’s right, so we have to be a little sneakier. Suppose we’re given distinct vertices V1, . . . ,Vk and a set S ⊆ {1, . . . , k} as above, and we want a vertex W adjacent to Vα if and only if α ∈ S . First, fix an arbitrary (not random!) vertex W ′; for definiteness, let it be the matrix of all zeros. Let T = {α ∈ {1, . . . , k} : W ′ satisfies V∗ α}. Apply the claim with the given vertices Vα but with S replaced by the complement of the symmetric difference of S and T , i.e., by {α : (α ∈ S ) ⇐⇒ (α ∈ T )}. The constraint given by the claim is, as you noted, C(w) for some row vector w. Let W ′′ be the vertex that has all its rows equal to w. So we have, thanks to the choice of W ′′ and the definition of T , Vα satisfies (W ′′)∗ ⇐⇒ ((α ∈ S ) ⇐⇒ (W ′ satisfies V∗ α) ) . Since ⇐⇒ is an associative and commutative operation on truth values, this can be rewritten as α ∈ S ⇐⇒ ((Vα satisfies (W ′′)∗) ⇐⇒ (W ′ satisfies V∗ α) ) . (1) Q: You’d be done if W ′ and W ′′ were equal, but that would require a miracle. It’s true that W ′ was arbitrary, but W ′′ depends on T which depends on the choice of W ′, so I see no chance to use the arbitrariness of W ′ to make it match W ′′. A: Absolutely right; there’s no reason to think W ′ and W ′′ are equal. But we can combine them into a single W that inherits the desirable features of both. For each α, whether W ′ satisfies the constraint V∗ α (and thus whether α ∈ T ) depends only on the entries of the matrix W ′ in k−1 locations, namely the locations in the first component A of the constraint V∗ α = (A, F). So at most k(k − 1) entries of W ′ are involved in the satisfaction or non-satisfaction of the k constraints V∗ α. Define W to agree with W ′ in those entries and with W ′′ at all other locations. We’ve kept enough entries of W ′ in W to ensure that (W satisfies V∗ α) ⇐⇒ (W ′ satisfies V∗ α). (2) On the other hand, W agrees with W ′′ at all but at most k(k − 1) entries. Since there are r = 2k(k − 1) + 1 rows, the majority of the rows of W are identical to the rows of W ′′, namely the w that we chose when constructing W ′′. So