1 The number of orientations having no fixed tournament Noga Alon Raphael Yuster

2 Definitions Let T be a tournament. Let G be an undirected graph. Define: D(G,T)= # orientations of G that are T-free D(n,T)= max D(G,T) where G has n vertices Example: GT=C3T=C3 D(G,C 3 )=18 (out of 32 possible) Observations If T has k vertices and G has no K k then D(G,T)=2 e(G) Let t k-1 (n) denote the maximum number of edges in a K k -free n-vertex graph. Then, D(n,T) ≥ 2 t k-1 (n) The value of t k-1 (n) is well-known and given by Turan’s Theorem. In particular, D(n, C 3 ) ≥ 2  n 2 /4  In some cases the lower bound given by Turan’s theorem is not the correct answer. If T has a cycle then, D(n,T) ≥ n! Hence: D(7,C 3 ) ≥ 5040 > 4096=2 12

3 Main result Theorem: If T is a k-vertex tournament then for n sufficiently (very) large, D(n,T) = 2 t k-1 (n) In other words: the “race” between adding edges to the Turan graph (which doubles the total number of orientations) and the disqualified “bad” orientations that result, is decided in favor of the Turan graph for n sufficiently large Conjecture D(n,C 3 )= 2  n 2 /4  for all n ≥ 8. Known for n=8 (computer) and n ≥ The only tournament for which all is known D(n,TT 3 )= 2  n 2 /4  for all n ≥ 1 Not so easy to prove directly. Relies on an older result on two-edge colorings with no monochromatic triangle TT 3

4 Tools used The Szemeredi regularity lemma for directed graphs The stability theorem of Simonovits The first two tools lead us to “the edge of the river” (an approximate result). We then use counting and inductive ideas to cross the river (the exact result). The stability theorem For every α > 0 and positive integer k there exists β = β(α,k) > 0 such that the following holds: If G has m vertices, has no K k+1 and has at least t k (m)- βm 2 edges then we can delete from G at most αm 2 edges and make it k-partite

5 The regularity lemma for directed graphs Let G=(V,E) be a directed graph. Let A and B be disjoint subsets of V. If A and B are nonempty we define the density from A to B as For ε > 0 the pair (A,B) is ε-regular if: for every X  A and Y  B with |X| > ε|A| and |Y| > ε|B| we have |d(X,Y) - d(A,B)| < ε |d(Y,X) - d(B,A)| < ε An equitable partition of V into V 1,…,V m is called ε-regular if all parts have size at most ε|V| and all but at most εm 2 pairs (V i,V j ) are ε-regular The directed regularity lemma states that: For every ε > 0 there exists M=M(ε) such that for every graph G with n > M vertices, there is an ε-regular partition of the vertex set of G into m parts where 1/ε < m < M

6 The embedding lemma We can use the ε-regularity and show that if C(  ) contains K k+1 then the original directed graph G contains any (k+1)-vertex tournament. More precisely, we have the following embedding lemma: Let T be a fixed tournament with k+1 vertices. Let  > 0 and suppose that ε < (  /2) k /k. Let G be a directed graph with an ε-regular partition into m parts.  If C(  ) contains K k+1 then G contains T  If C(  ) does not contain K k+1 and d(V s,V t ) is ε-regular but dense only in one direction and the addition of (s,t) to C(  ) forms a K k+1 then G contains T A useful notion associated with an ε-regular partition is the undirected cluster graph C(  ) (where  is a small parameter but much larger than ε):  The vertex set of C(  ) is {1,…,m}.  (i,j) is an edge if (V i,V j ) is an ε-regular pair with density at least  in both directions

7 Phase 1: An approximate result Fix a tournament T with k+1 vertices. We prove the following lemma: For all  > 0 there exists n 0 =n 0 (k,  ) such that if G is a graph with n > n 0 vertices which has at least 2 t k (n) distinct T-free orientations then we can delete at most  n 2 edges from G and make it k-partite Sketch of proof: –Let  > 0 –Put α <  / (4k+7) –Put β= β(α,k) as in the stability theorem –Let  < β (  is chosen sufficiently small during the proof to guarantee some inequality) –Let ε < (  /2) k / k –Let M=M(ε) be as in the regularity lemma –Let G be an undirected graph with n vertices and at least 2 t k (n) distinct T-free orientations. For every T-free orientation of G we can apply the directed regularity lemma and by the embedding lemma, the resulting cluster graph C(  ) has no K k+1 and hence has at most t k (m) edges

8 –We show that for some T-free orientation of G the resulting cluster graph has more than t k (m) - βm 2 edges. Assume this is false: –The number of ways to partition n vertices into at most M equitable parts is at most n M+1 –For each such partition there are 2 M 2 /2 choices for C(  ) and at most 2 M 2 /2 choices for non-regular pairs –We show that for each possible partition, cluster graph, and non-regular pairs, the number of T-free orientations that give rise to them is less than 2 t k (n) / (n M+1 2 M 2 ), giving the contradiction. We show this by counting the possible number of T-free orientations of edges that are: (a) inside a vertex class (b) in a non-regular pair (c) in one-way dense or no-way dense regular pairs (this defines  ) (d) in two-way dense pairs (recall that we assume there are at most t k (m)- βm 2 such pairs –We may now fix a for which C(  ) has at least t k (m)- βm 2 edges By the stability theorem we can delete at most αm 2 more edges and obtain a k-partite spanning subgraph of C(  ). The number of edges is at least t k (m)-(β+α)m 2 = t k (m)-  m 2

9 –We cannot have more than (2k+1)  m 2 one-sided dense regular pairs, since otherwise, some one-sided dense pair causes the formation of a K k+1 and by the embedding lemma, this implies the orientation is not T- free –We delete from G the following edges: (a) edges with both endpoints in the same V i. There are at most εn 2 (b) edges belonging to non-regular pairs. There are at most εn 2 (c) the edges belonging to non-dense pairs or one-sided dense pairs. There are at most (2  +(2k+1)  )n 2 (d) the edges corresponding to the αm 2 “deleted” pairs. There are at most αn 2 –In other words, we keep only edges belonging to pairs in the k-partite spanning subgraph of C(  ). Hence, the resulting spanning subgraph of G is also k-partite and we deleted at most  n 2 edges

10 Phase 2: The exact result We need the following technical lemma: Let S be a tournament with the vertices {1,…,k}. Let G be a directed graph and let W 1,…,W k be subsets of vertices of G such that for any pair of subsets X i  W i with |X i | > 10 -k |W i |, X j  W j with |X j | > 10 -k |W j |, there are at least 0.1|X i ||X j | edges in the “right” direction, then G contains a copy of S where the role of vertex i is played by a vertex from W i Sketch of proof of the main result: –We use n 0 large enough for the approximation lemma with  = 10 -8k –Suppose G has n > n 0 2 vertices and at least 2 t k (n)+m T-free orientations for some m ≥ 0. We use induction on n with an improvement in every step –We show that if G is not the Turan graph then we can find a vertex x such that G – x has at least 2 t k (n-1)+m+1 distinct T-free orientations. Iterating downwards until n 0 we obtain a graph with at least 2 t k (n 0 )+m+n-n 0 > 2 n 0 2 distinct T-free orientations, a contradiction

11 –We may assume that all the vertices have degree at least as large as the minimum degree of the Turan graph, otherwise we are done (using some easy properties of the Turan graphs) –We consider a partition V 1,…,V k of G that minimizes the “inside edges”. By the choice of n 0 the number of “inside edges” is at most 10 -8k n 2 –We consider two cases:  Some vertex x has “many” neighbors in its own class of the partition, say more than n/(400k) (and hence also this number of neighbors in every other part). We show, using the previous technical lemma, that in this case G - x has at least 2 t k (n-1)+m+1 distinct T-free orientations. This is done by showing that there are not too many ways to “extend” a T-free orientation of G - x to a T-free orientation of G by orienting the edges incident with x. The computations are rather involved  Every vertex has degree at most n/(400k) in its own class. We may assume G is not the Turan graph (otherwise we are done). Hence, there is some edge {x,y} inside a vertex class. We show that in this case G - {x,y} completes two induction steps. Namely, has at least 2 t k (n-2)+m+2 distinct T-free orientations