Triply Existentially Complete Triangle-Free Graphs
aa r X i v : . [ m a t h . C O ] M a y Triply Existentially CompleteTriangle-Free Graphs
Chaim Even-Zohar ∗ Nati Linial † August 26, 2018
Abstract
A triangle-free graph G is called k -existentially complete if for every induced k -vertexsubgraph H of G , every extension of H to a ( k + 1)-vertex triangle-free graph can be realizedby adding another vertex of G to H . Cherlin [11, 12] asked whether k -existentially completetriangle-free graphs exist for every k . Here, we present known and new constructions of 3-existentially complete triangle-free graphs. It is well known that the Rado graph R [15, 22] is characterized by being existentially complete.Namely, for every two finite disjoint subsets of vertices A, B ⊂ V ( R ) there is an additional vertex x that is adjacent to all vertices in A and to none in B . Many variations on this theme suggestthemselves, specifically concerning finite and H -free graphs. First, the aforementioned extensionproperty of R suggests a search of small finite graphs that satisfy this condition whenever | A | + | B | ≤ k . It is known that Paley graphs and finite random graphs of order exp( O ( k )) have thisproperty [3, 5, 10, 15, 7].The Rado graph is also homogeneous , that is to say, every isomorphism between two of itsfinite subgraphs can be extended to an automorphism of R . Henson [18] has discovered the genericinfinitely countable triangle-free graph R that is homogeneous as well. It is uniquely defined bythe following property. Given two finite disjoint subsets A, B ⊂ V ( R ) with A independent , thereis an additional vertex x that is adjacent to all vertices in A and to none in B .The graph R suggests an extremely interesting question that was raised and studied by Cher-lin [11, 12]. Namely, do there exist finite graphs with a similar property? We say that a triangle-freegraph is k - existentially complete if it satisfies this condition whenever | A | + | B | ≤ k . Are there k -existentially complete triangle-free graphs for every k ? A weaker variant of this question hadbeen raised by Erd˝os and Pach [21, 14]. Similar questions for graphs and other combinatorialstructures appear in [1, 8, 6]. In the literature one occasionally finds the shorthand k -e.c. as wellas the alternative term k -existentially closed .Cherlin’s question can be viewed as an instance of a much wider subject in graph theory, namelyunderstanding the extent to which the local behavior of infinite graphs can be emulated by finiteones. Here are some other instances of this general problem.The infinite d -regular tree T d is the ultimate d -regular expander in at least two senses. It hasthe largest possible number of edges emanating from every finite set of vertices. It also has thelargest spectral gap that a d -regular graph can have. This observation suggests the search for finitearbitrarily large Ramanujan graphs, and for the limits on expansion in finite d -regular graphs. T d is also, of course, acyclic that leads to the question how large the girth can be in a finite d -regulargraph.There are several examples of local conditions that can be satisfied in an infinite graph but notin a finite one. A nice example comes from an article of Blass, Harari and Miller [4]. They define ∗ Department of Mathematics, Hebrew University, Jerusalem 91904, Israel.e-mail: [email protected] . † Department of Computer Science, Hebrew University, Jerusalem 91904, Israel.e-mail: [email protected] . Supported by ISF and BSF grants. H . As they observe, there is an infinite graph with alllinks isomorphic to H = , but this is impossible for finite graphs.For all the currently known k -existentially complete triangle-free graphs, the parameter k isbounded by 3. The question is wide open for k ≥
4. The known finite graphs with the R -likeextension property do not seem adjustable to the triangle-free case. In fact we tend to believe thatthere is some absolute constant k such that no triangle-free graph is k -existentially complete.For general p ≥
3, Henson’s universal countable K p -free graph R p is defined by the analogousextension property, where the subset A is only required to be K ( p − -free. There is a simpleconnection between the existential completeness properties of K p -free graphs for different valuesof p . Thus, if, as we suspect, no finite k -existentially complete triangle-free graphs exist, then nofinite K -free graph can be ( k + 1)-existentially complete, etc. This follows since the link of everyvertex in a ( k + 1)-existentially complete K p -free graph is a k -existentially complete K ( p − -freegraph. Finite random ( p − K p -freeand ( p − µ it ispossible that every 2 nonadjacent vertices have at least µ common neighbors? Similarly, canevery independent set of 3 vertices have µ > n evenwe show that there are at least 2 n / (16+ o (1)) n vertices. We show (Corollary 6)that in these graphs the average degree can be as small as O (log n ) and as high as n/
4. For all µ ,we find 2 Ω( n ) distinct 3ECTF graphs of order ≤ n in which every two nonadjacent vertices haveat least µ common neighbors (Corollary 13). However, some independent triplets in these graphsmay have only one common neighbor.The constructions are presented in three main phases. In each of Sections 3, 4 and 5 wedescribe a basic construction, which we then extend in various ways. Table 1 exhibits three mainparameters of the graph families under discussion: the number of vertices, the vertex degrees, and µ , the minimum number of common neighbors of nonadjacent pairs. Graph Vertices Degrees µ Albert A ( n ), n ≥ n n + 1 2 A M , M ∈ { , } m × n * 2 m + 2 n m + 1 , n + 1 2 Hypercube C k +1 (Erd˝os [13], Pach [21]) 2 k +1 (cid:0) kk (cid:1) ** (cid:0) kk (cid:1) C k − ( m ), k ≥ m ≥ m k − m (cid:0) kk (cid:1) ** (cid:0) kk (cid:1) C k,j , 1 ≤ j ≤ k k + j (cid:0) k + jk + j (cid:1) ** 2 (cid:0) k − k − j (cid:1) Twisted G ( m , ..., m ), m i ≥ P i m i P i m i + 1 2 G T ( m, k ), T tournament* | T | m k − | T | m (cid:0) kk (cid:1) ** (cid:0) kk (cid:1) * For asymptotically almost every such M or T . ** Upto a multipilcative constant, for k → ∞ . Table 1: Constructions of 3ECTF graphsAlthough this work is meant to be self-contained, the reader is encouraged to consult partIII of Cherlin’s article [12] for more background of the subject and many additional details. To2implify matters we maintain a graph-theoretic terminology, and refrain from using Cherlin’s viewof maximal triangle-free graphs as combinatorial geometries. We believe that this presentationmakes the structure and the symmetries of the graphs more transparent.Notation: We denote the fact that vertices x, y are adjacent in the graph under discussion by x ∼ y . The neighborhood of x is N ( x ) = { y ∈ V | x ∼ y } . Here are some definitions and useful reductions that are due to Cherlin.
Definition (Extension Properties, Section 11.1 of [12]) . Let k be a positive integer. The followingproperties of a triangle-free graph G = ( V, E ) are defined thus:1. ( E k ), [also known as k -existentially complete]: For every B ⊆ A ⊆ V where | A | ≤ k and B is independent, there exists a vertex v that is adjacent to each vertex of B and to no vertexof A \ B .2. ( E ′ k ): For every B ⊆ A ⊆ V where A is independent of cardinality exactly k there exists avertex v , adjacent to each vertex of B and to no vertex of A \ B .Also, every independent set with fewer than k vertices is contained in an independent set ofcardinality k .3. ( Adj k ): Every independent set of cardinality ≤ k has a common neighbor. An (
Adj ) graph is also called maximal triangle-free , since the addition of any edge would createa triangle. In other words, the graph has diameter ≤
2. In a twin-free graph no two vertices havethe same neighborhood.We have the following implications for triangle-free graphs (Cherlin [12], Lemmas 11.2-11.4).
Lemma 1.
For k ≥ , the properties ( E k ) and ( E ′ k ) are equivalent. Lemma 2.
For k ≥ , property ( E k ) is equivalent to the conjunction of ( Adj k ) and ( E ) . Lemma 3.
A graph has property ( E ) iff it is maximal triangle-free, twin-free, and contains ananti-triangle. In order to bridge between Lemmas 2 and 3, Cherlin investigates triangle-free graphs thatare (
Adj ) and ( E ) but not ( E ). In Sections 11.4-11.5 of [12] these exceptions are described interms of linear combinatorial geometries, which can be interpreted as graphs of a certain circularstructure. In general, circular graphs can be defined as a set of arcs in a cyclically ordered set, whereadjacency means disjointness of the corresponding arcs. Such graphs, defined by disjointedness ina family of subsets, are sometimes called (general) Kneser graphs (e.g. [20]).Here and below we only consider finite graphs. The circular graph O n − is formed by all arcsof n consecutive elements in Z n − . It was independently introduced several times, e.g., Erd˝osand Andr´asfai [2], Woodall [23], Pach [21], and van den Heuvel [19]. In terms of rational completegraphs [17], O n − is equivalent to K (3 n − /n , being ”almost” a triangle. As shown in [21, 9],these are the only finite triangle-free twin-free graphs where every independent set has a commonneighbor.By Lemma 11.15 of [12], the circular graphs { O n − } n ≥ are the only finite triangle-free graphsthat are ( Adj ) and ( E ) but not ( E ). Moreover, the 5-cycle O and the edge O are the onlymaximal triangle-free graphs that are twin-free and do not contain an anti-triangle. Taking intoaccount these exceptions, Lemmas 2 and 3 easily imply the following corollary. Corollary 4 (Cherlin, Sections 11.1–11.5 of [12]) . For k ≥ , a finite triangle-free graph G is k -existentially complete if and only if the following conditions hold.1. Every independent set of cardinality ≤ k has a common neighbor.2. There do not exist two vertices x, y with N ( x ) = N ( y ) .3. G is not isomorphic to O n − for n ≥ .
3n Section 12 of [12], the notion of (
Adj k ) is refined to the k -th multiplicity of G . This is done byconsidering the smallest possible number of common neighbors of an independent set of k vertices: µ k ( G ) = min A ⊆ V ( G ) independent | A | = k { b ∈ V ( G ) | N ( b ) ⊇ A } Cherlin proves the following chain of implications for 3ECTF graphs: µ ( G ) ≥ ⇒ µ ( G ) ≥ ⇒ µ ( G ) ≥ ⇒ µ ( G ) ≥ ⇒ µ ( G ) ≥ µ ( HS ) = 2 and µ ( HS ) = 6. However, by a beautiful spectral calculation([12], Section 12.3) no strongly regular graph has property 4ECTF. Perhaps the Higman–Simsconstruction should be viewed as a sporadic example. On the other hand, here we introduce alarge collection of 3ECTF graphs with arbitrarily large µ ( G ) and µ ( G ) = 1. Thus the two lowestlevels in this hierarchy are not very restrictive. We turn to describe an infinite sequence of 3ECTF graphs which Cherlin attributes to MichaelAlbert. The Clebsch graph is a triangle-free strongly regular graph on 16 vertices. It can berepresented as the union of four 4-cycles, where each vertex is adjacent as well to its antipodesin the other cycles. One can check directly that this graph is 3-existentially complete, e.g. byCorollary 4. Albert’s construction is the extension of the Clebsch graph to any number of 4-cycles.Formally, Albert’s 3ECTF graphs sequence A ( n ) is defined by V ( A ( n )) = { ( i, x ) | i ∈ { , , ..., n } , x ∈ Z } ( i, x ) ∼ ( i, x + 1) for all i, x ( i, x ) ∼ ( i ′ , x + 2) for all x, i = i ′ , where addition is in Z .This construction was thoroughly generalized by Cherlin, to Albert geometries. Here, we offera different viewpoint of these graphs. Let m, n ≥ m × n zero-one matrix M issaid to be shattered if the submatrix corresponding to any three rows or three columns containsall four possible patterns aaa, aab, aba, baa . Namely, it must contain at least one of the strings 000and 111, and one of 001 and 110, and so on. The Albert graph A M of a shattered matrix M isobtained from an m -matching and an n -matching. The corresponding entries of M tell us how toconnect these 2 m + 2 n vertices. V ( A M ) = { a , ..., a m } ∪ { b , ..., b m } ∪ { c , ..., c n } ∪ { d , ..., d n } a i ∼ b i for all ic j ∼ d j for all ja i ∼ c j , b i ∼ d j if M ij = 1 a i ∼ d j , b i ∼ c j if M ij = 0 . For example, when M is the 4 × A M is the Clebsch graph. Albert’s construc-tion corresponds to larger identity matrices. Proposition 5.
If the matrix M is shattered, then the Albert graph A M is 3ECTF.Proof. We first observe that A M is triangle-free. Of any three vertices at least two must eithercome from U = { a , ..., a m } ∪ { b , ..., b m } , or from W = { c , ..., c n } ∪ { d , ..., d n } . But an edge in U must be of the form ( a i , b i ), and a i and b i have no common neighbors.We proceed to verify the conditions in Corollary 4. It is easy to see that A M is twin-free andnot an O n − . Property ( Adj ) follows since M is shattered. An independent triplet in U mustconsist of either a i or b i , either a j or b j and either a k or b k for some distinct i, j, k . A commonneighbor exists thanks to the appropriate pattern in the restriction of M to the rows i , j , and k .For an independent set with two vertices in U and one in W , the neighbor of the latter inside W is adjacent also to the first two. 4he constructions of Albert Geometries in Examples 13.1 of [12], come from explicit shatteredmatrices. Random matrices can be used as well. Thus, simple counting arguments yield Corollary 6.
1. Almost every zero-one m × n matrix with m ≥ C log n and n ≥ C log m is shattered. Here C > is some absolute constant.2. For even n , the number of n -vertex 3ECTF graphs is at least n / (16+ o (1)) . This gives some insight on the possible behavior of 3ECTF graphs. On the one hand, taking m = n we get a regular graph of degree | V | / m = Θ(log n ) the graphhas Θ(log | V | ) vertices of degree close to | V | / | V | ) neighbors.As for multiplicities, in every Albert graph µ ( G ) = 2 and µ ( G ) = 1. Recall that the Clebsch graph is a Cayley graph of Z , with generators the unit vectors and theall-ones vector. Equivalently, x ∼ y iff their Hamming distance d ( x, y ) is 1 or 4. Following Franekand R¨odl [16], we denote this graph by (cid:10) Z , { , } (cid:11) . Note that it also equals (cid:10) Z , { , } (cid:11) . Here, weconsider the graphs C k +1 = (cid:10) Z k +12 , { k + 1 , k + 2 , ..., k + 1 } (cid:11) which Erd˝os [13] used in the study of Ramsey numbers. As mentioned, C is the Clebsch graph.The extension properties of these graphs were studied by Pach [21]. For future use, we record avariant of his argument in the following lemma. Lemma 7. If x, y, z ∈ Z n satisfy d ( x, y ) ≤ a + b, d ( x, z ) ≤ a + c, d ( y, z ) ≤ b + c. for some integers a, b, c ≥ , then there is some v ∈ Z n for which d ( v, x ) ≤ a, d ( v, y ) ≤ b, d ( v, z ) ≤ c. Proof.
Define the vector m by the coordinate-wise majority vote of x , y , and z . Note that thethree vectors x ′ = x + m , y ′ = y + m , and z ′ = z + m have disjoint supports. We find a v ′ satisfyingthe claim for these three vectors and let v = v ′ + m .If the Hamming weights satisfy w ( x ′ ) ≤ a , w ( y ′ ) ≤ b and w ( z ′ ) ≤ c , then we are done bytaking v ′ = 0. Otherwise, by assumption, at most one of these inequalities can be violated, say w ( x ′ ) > a . We take v ′ to have weight w ( v ′ ) = w ( x ′ ) − a and satisfy x ′ ≥ v ′ coordinate-wise.Obviously d ( v ′ , x ′ ) ≤ a . But also, d ( v ′ , y ′ ) ≤ w ( v ′ ) + w ( y ′ ) = w ( x ′ ) − a + w ( y ′ ) = d ( x ′ , y ′ ) − a ≤ b, and similarly d ( v ′ , z ′ ) ≤ c . Proposition 8.
The graphs C k +1 are 3ECTF.Proof. To see that C k +1 is triangle-free, suppose x ∼ y ∼ z , then d ( x, z ) ≤ d ( x, ¯ y ) + d (¯ y, z ) ≤ k + k = 2 k , where ¯ y is y ’s antipode, namely ¯ y = y + (1 , ..., x z .Now we check the conditions of Corollary 4. Clearly N ( x ) is not equal to N ( y ) for x = y ,since both are distinct Hamming balls of the same radius. Property ( Adj ) follows using Lemma 7.Indeed, if the set { x, y, z } is independent, apply the lemma with a = b = c = k . For v as in thelemma, ¯ v is at least 2 k + 1 away from each of the three vectors. Proposition 9. µ ( C k +1 ) = (cid:0) kk (cid:1) , µ ( C k +1 ) = 1 . roof. We check that every two nonadjacent vertices in C k +1 have at least (cid:0) kk (cid:1) common neighbors.Let x and y be two vectors of even Hamming distance 2 t where 1 ≤ t ≤ k . We count some of thecommon neighbors of x and y . Of the 3 k + 1 − t coordinates in which they agree, we flip 2 k + 1 − t coordinates of our choice, and keep the other k − t unchanged. The remaining 2 t coordinates aredivided equally, t as in x and t as in y . Each resulting vector is at Hamming distance 2 k + 1 fromboth x and y , which hence have at least (cid:18) k + 1 − tk − t (cid:19)(cid:18) tt (cid:19) distinct joint neighbors. This expression is decreasing in t , and hence always ≥ (cid:0) kk (cid:1) , with equalityfor t = k .For d ( x, y ) = 2 t − k + 1 − ( t −
1) of the 3 k + 1 − (2 t − t − t and t −
1. This yields2 (cid:18) k + 1 − (2 t − k − t (cid:19)(cid:18) t − t (cid:19) common neighbors. The minimum, (cid:0) kk (cid:1) , is again attained at t = k .Finally, it is easy to demonstrate three vectors in C k +1 with a single joint neighbor. Takethree vectors, x , y , and z , of Hamming weight k and disjoint supports. For a common neighbor v we have 3(2 k + 1) ≤ d ( x, v ) + d ( y, v ) + d ( z, v ) ≤ · k + 3 = 3(2 k + 1)since in 3 k coordinates at most two of the distances can contribute. But equality is reached onlyby the all-ones vector v = (1 , ..., C k +1 constitutes an example for 3ECTF graphs with µ ( C k +1 ) → ∞ . Here µ = n / − o (1) , where n = 2 k +1 is the number of vertices. Also, thesegraphs are n λ − o (1) -regular, where λ = log − ≈ . C k +1 . For x ∈ Z k denote parity ( x ) = P ki =1 x i mod 2. We first partition C k +1 as follows. V = { x ∈ Z k +12 | parity ( x ) = x = x } V = { x ∈ Z k +12 | parity ( x ) = x = x } V = { x ∈ Z k +12 | parity ( x ) = x = x } V = { x ∈ Z k +12 | parity ( x ) = x = x } Of course, we may forget the first two coordinates, and regard the elements of each V i as Z k − .The adjacencies within each V i correspond to Hamming distances 2 k − k + 1 , ..., k − V i ’s are adjacent iff their Hamming distance is between 2 k and3 k −
1. This leads to the following definition of C k − ( m ). V ( C k − ( m )) = { ( v, i ) | v ∈ Z k − , ≤ i ≤ m } ( v, i ) ∼ ( u, i ) if d ( v, u ) ∈ { k − , k + 1 , k + 2 , ..., k − } ( v, i ) ∼ ( u, j ) if i = j and d ( v, u ) ∈ { k, k + 1 , k + 2 , ..., k − } By the above discussion C k − (4) = C k +1 , and C k − ( m ) is 3ECTF for m ≥
4, since any threevertices belong to an isomorphic copy of C k +1 . For the same reason µ ( C k − ( m )) ≥ (cid:0) kk (cid:1) , andfor constant k and large m the graph is Θ( n )-regular.To introduce the second variation, consider the following graph. C ′ k = (cid:10) Z k , { , , , ..., k − , k } (cid:11) . By considering the two matchings on odd-parity and on even-parity vectors we see that this is anAlbert graph. In fact, each odd distance d can be separately replaced by 4 k − d to yield anotherAlbert graph. 6 k +1 and C ′ k are the first and last members of a simple sequence of 3ECTF graphs, C k,j for j ∈ { , ..., k } . These graphs are also defined in terms of the Hamming metric on the hypercube.We describe these graphs without proving their properties, since this is not needed henceforth. C k,j = D Z k + j , { k + 1 , k + 3 , ..., k + (2 j − } ∪ { k + j ) , k + j ) + 1 , ..., k + j } E . Having seen 3ECTF graphs with unbounded µ , we move to the next construction in search ofmany such graphs.We start with variation of a 3ECTF construction from Section 13.2 of [12]. Given positiveintegers m , m , m and m , we define the twisted graph G ( m , m , m , m ) as follows. V ( G ( m , m , m , m )) = { ( i, j, x ) | i ∈ { , , , } , ≤ j ≤ m i , x ∈ Z } ( i, j, x ) ∼ ( i, j, x + 1) for all i, j, x ( i, j, x ) ∼ ( i, j ′ , x + 2) for all i, x, j = j ′ ( i, j, x ) ∼ ( i ′ , j ′ , x + 3) for all x, j, j ′ , ( i, i ′ ) ∈ { (0 , , (0 , , (0 , , (1 , , (2 , , (3 , } . Remark: the graph G (1 , m , m , m ) differs from G ( m , m , m ) in Example 13.3 of [12], unlesswe switch two edges inside V . We do not know how to place both graphs on a common ground.The following proposition reveals some of the structure of these graphs. Although it is a specialcase of Proposition 12, we believe that it is easier to follow and would make the more complicatedProposition 12 more transparent. Proposition 10.
For m , m , m , m ≥ , the graph G ( m , m , m , m ) is 3ECTF.Proof. For i ∈ { , , , } let V i = { ( i, j, x ) ∈ V ( G ) | x ∈ Z , ≤ j ≤ m i } . Note that the restriction of G to each V i is isomorphic to A ( m i ) (see Section 3). Moreover, thesubgraph induced on V i ∪ V i ′ is A ( m i + m i ′ ). To see this, ”twist” the m i ′ last 4-cycles in A ( m i + m i ′ ),by sending x x + 1 mod 4. This affects only cross edges with x + 2 → x + 3. Therefore, we onlyneed to consider triplets from distinct V i ’s.We first verify triangle-freeness. Along an edge of such a triangle, the Z -coordinate changesfrom x to x ±
3, but three such numbers do not sum to zero mod 4.We next find a common neighbor for three independent vertices ( i, j, x ), ( i ′ , j ′ , x ′ ), and ( i ′′ , j ′′ , x ′′ )for distinct i , i ′ , and i ′′ . Suppose first that the parity of x differs from those of x ′ and x ′′ .Which is the neighbor of ( i ′ , j ′ , x ′ ) in the 4-cycle ( i, j, · )? From parity considerations, and since( i, j, x ) ( i ′ , j ′ , x ′ ), it must be ( i, j, x + 2). For similar reasons ( i, j, x + 2) is the neighbor of( i ′′ , j ′′ , x ′′ ) in that 4-cycle. This implies that ( i, j ′′′ , x +2) is a common neighbor of ( i, j, x ), ( i ′ , j ′ , x ′ ),and ( i ′′ , j ′′ , x ′′ ) for any j ′′′ = j .Now if x , x ′ and x ′′ have equal parities, then each of ( i ′ , j ′ , x ′ ) and ( i ′′ , j ′′ , x ′′ ) is adjacent toone of the elements of the set { ( i, j, x − , ( i, j, x + 1) } . If both are adjacent to the same one,it is the common neighbor and we are done. A similar solution may exist in different roles ofthe three vertices. If all fails, then either ( i, i ′ , i ′′ ) = (1 , ,
3) and x = x ′ = x ′′ or ( i, i ′ , i ′′ ) ∈{ (0 , , , (0 , , , (0 , , } and x = x ′′ = x ′ + 2. In this case, the common neighbor is in the fourthpart V i ′′′ , being ( i ′′′ , j ′′′ , x + 1) and ( i ′′′ , j ′′′ , x + 3), respectively, for any j ′′′ .A tournament is an orientation of a complete graph. Edges ( i, i ′ ) ∈ E ( T ) are denoted i → i ′ .Note that the set of ordered pairs { (0 , , (0 , , (0 , , (1 , , (2 , , (3 , } , that appears in the abovedefinition of G ( m , m , m , m ), can be encoded by the following tournament that we call T .
102 3 T ′ is obtained from T by reversing all edges. These two four-vertex tournamentsare characterized by the property that every two vertices are connected by exactly one path oflength two. In other words, for each pair of vertices one remaining vertex is attached to them inthe same way, while the other one is attached in opposite ways. It is exactly this property thatallowed us to find the common neighbor in the previous argument. A tournament is shattered ifevery three of its vertices extend to a copy of either T or T ′ .We next associate a graph G T ( m ) to a shattered tournament T and a positive integer m . V ( G T ( m )) = { ( i, j, x ) | i ∈ V ( T ) , ≤ j ≤ m, x ∈ Z } ( i, j, x ) ∼ ( i, j, x + 1) for all i, j, x ( i, j, x ) ∼ ( i, j ′ , x + 2) for all i, x, j = j ′ ( i, j, x ) ∼ ( i ′ , j ′ , x + 3) for all x, j, j ′ , i → i ′ . It is easy to apply the proof of Proposition 10 to general shattered tournaments and conclude thefollowing broader statement.
Corollary 11.
The graph G T ( m ) is 3ECTF for every shattered tournament T and every m ≥ . The graphs that were constructed so far in this section have µ ( G ) = 2 and µ ( G ) = 1. Thenext example integrates them with C k − ( m ), thus increasing the 2-multiplicity. To this end, wefirst define a similar twist function in the hypercube. For x ∈ Z n , define τ (( x , x , x , x , ..., x n )) = ( x , x + 1 , x , x , ..., x n ) . Note that τ ( x ) = x for all x ∈ Z n .We turn to extend this construction and associate a graph G T ( m, k ) with every shatteredtournament T and positive numbers m, k . V ( G T ( m, k )) = { ( i, j, x ) | i ∈ V ( T ) , ≤ j ≤ m, x ∈ Z k − } ( i, j, x ) ∼ ( i, j, x ′ ) if d ( x, x ′ ) ∈ { k − , k + 1 , k + 2 , ..., k − } ( i, j, x ) ∼ ( i, j ′ , x ′ ) if j = j ′ and d ( x, x ′ ) ∈ { k, k + 1 , ..., k − } ( i, j, x ) ∼ ( i ′ , j ′ , x ′ ) if i → i ′ and d ( x, τ ( x ′ )) ∈ { k, k + 1 , ..., k − } . Note that G T ( m ) is isomorphic to G T ( m, Proposition 12.
For m ≥ , k ≥ , and a shattered tournament T , the graph G T ( m, k ) is 3ECTF.Proof. We adapt the proof of Proposition 10 to k >
1. For i ∈ V ( T ), we partition the vertices asfollows: V i = { ( i, j, x ) | x ∈ Z k − , ≤ j ≤ m } . The induced graph on each V i is isomorphic to C k − ( m ). Since τ is an isometry, also V i ∪ V i ′ induces a copy of C k − (2 m ). By our previous comments these subgraphs are 3ECTF.We next show that three vertices from distinct parts never form a triangle. For each edgebetween parts ( i, j, x ) ∼ ( i ′ , j ′ , x ′ ) we have d ( x, τ ( x ′ )) ≥ k . Therefore at least 2 k − k − x and x ′ differ. Suppose, toward a contradiction, that a triangle of such threeedges exists, and consider two cases.1. Suppose that along some of the edges the Hamming distance of the last 3 k − k −
1. Then the three differences cannot add up to 0, because at least one of thelast 3 k − k −
2, then in each such edge d ( x, τ ( x ′ )) = 2 k , andhence parity( x ) = parity( τ ( x ′ )). But τ switches parity, so the parity changes three timesaround the triangle – a contradiction.Next, we seek a common neighbor for the independent set { ( i , j , x ) , ( i , j , x ) , ( i , j , x ) } ,with i , i , i distinct. Define τ = τ if i → i , and τ = τ − otherwise. Correspondingly define8 and so on. Also, define the following two sets of three distances. d = d ( x , τ ( x )) d ∗ = d ( τ ( x ) , τ ( x )) d = d ( x , τ ( x )) d ∗ = d ( τ ( x ) , τ ( x )) d = d ( x , τ ( x )) d ∗ = d ( τ ( x ) , τ ( x ))By the independence assumption, d , d , d ≤ k −
1. Note that d and d ∗ are defined by thesame two vectors up to three applications of τ or τ − , so that d ∗ = d ± d ∗ = d ±
1, and d ∗ = d ± d ∗ ≤ k −
2. Apply Lemma 7 to the vectors x , τ ( x ), and τ ( x ), with a = k and b = c = k − v ∈ Z k − with d ( v, x ) ≤ k, d ( v, τ ( x )) ≤ k − , d ( v, τ ( x )) ≤ k − . If d ( v, x ) = k then ( i , j , ¯ v ) is a common neighbor, while for d ( v, x ) = k − i , j, ¯ v ) for any j = j . For d ( v, x ) ≤ k − j would do.By applying the same argument to d ∗ and d ∗ , we may assume that d ∗ , d ∗ , d ∗ ≥ k −
1. Inparticular, they exceed d , d and d by one, respectively.The restricted parity of x ∈ Z n , is defined as the parity of its first two coordinates. We claimthat, under the above assumption, x , x and x all have the same restricted parity. Otherwise,say x is the exception. This implies several further properties.1. The vectors τ ( x ) and τ ( x ) have different restricted parity.2. Since d ∗ = d + 1, at least one of the first two coordinates of τ ( x ) and τ ( x ) thatappear in the definition of d ∗ must differ. But then, from restricted parity considerations,they both differ.3. Also by restricted parity and by the previous property, x agrees either with τ ( x ) orwith τ ( x ) on the first two coordinates, and disagrees on them with the other one. Say itdisagrees with τ ( x ).Properties 1 and 3 imply that d ∗ < d , an already settled case. Therefore, the three vectors musthave the same restricted parity. Now, by the reasoning of property 2, each of the three followingvector pairs differs in both of the first two coordinates. τ ( x ) , τ ( x ) τ ( x ) , τ ( x ) τ ( x ) , τ ( x )Here the shattered tournament comes in. There is an i for which { i , i , i , i } induces either a T or a T ′ tournament. Therefore exactly one of ( τ , τ ) differs from its counterpart in ( τ , τ ).Consequently, τ ( x ) and τ ( x ) agree on their first two coordinates. Denoting their distance by d +12 , we have d +12 = d ∗ − d − ≤ k − , and likewise d +23 , d +31 ≤ k −
2. Apply Lemma 7 to τ ( x ), τ ( x ), and τ ( x ) with a = b = c = k −
1, to obtain v ∈ Z k − at distance at most k − i , j, ¯ v ) for any possible j .As every two vertices of G T (2 , k ) are covered by some embedded copy of C k +1 , we have µ ( G T (2 , k )) ≥ (cid:0) kk (cid:1) by Proposition 9. One can verify that this is in fact an equality. Since thereare 2 Ω( n / k ) nonisomorphic tournaments on n/ k vertices, and for large n almost all of them areshattered, Proposition 12 yields Corollary 13.
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