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The number of orientations having no fixed tournament

Noga Alon

Raphael Yuster

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Definitions• Let T be a tournament. Let G be an undirected graph. Define:

D(G,T) = # orientations of G that are T-freeD(n,T) = max D(G,T) where G has n vertices

Example:

G T=C3 D(G,C3)=18 (out of 32 possible)

Observations• If T has k vertices and G has no Kk then D(G,T)=2e(G)

• Let tk-1(n) denote the maximum number of edges in a Kk-free n-vertex graph. Then, D(n,T) ≥ 2tk-1(n)

• The value of tk-1(n) is well-known and given by Turan’s Theorem.

In particular, D(n, C3) ≥ 2 n2/4

• 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,C3) ≥ 5040 > 4096=212

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Main result• Theorem: If T is a k-vertex tournament then for n sufficiently (very) large,

D(n,T) = 2tk-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,C3)= 2n2/4 for all n ≥ 8. Known for n=8 (computer) and n ≥ 600000

The only tournament for which all is known• D(n,TT3)= 2n2/4 for all n ≥ 1

Not so easy to prove directly. Relies on an older result on two-edge colorings with no monochromatic triangle

TT3

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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 Kk+1 and has at least tk(m)- βm2 edges then we can delete from G at most αm2 edges and make it k-partite

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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

|A||B|

A,BeA,Bd

)()(

• 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 V1,…,Vm is called ε-regular if all parts have size at most ε|V| and all but at most εm2 pairs (Vi,Vj) 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

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The embedding lemma• We can use the ε-regularity and show that if C() contains Kk+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 thatε < (/2)k/k. Let G be a directed graph with an ε-regular partition into m parts.

If C() contains Kk+1 then G contains T

If C() does not contain Kk+1 and d(Vs,Vt) is ε-regular but dense only in one direction and the addition of (s,t) to C() forms a Kk+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 (Vi,Vj) is an ε-regular pair with density at least

in both directions

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Phase 1: An approximate result• Fix a tournament T with k+1 vertices. We prove the following lemma:

For all > 0 there exists n0=n0(k, ) such that if G is a graph with n > n0 vertices which has at least 2tk(n) distinct T-free orientations then we can delete at most n2 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 2tk(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 Kk+1 and hence has at most tk(m) edges

G

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– We show that for some T-free orientation of G the resulting cluster graph has more than tk(m) - βm2 edges. Assume this is false:

– The number of ways to partition n vertices into at most M equitable parts is at most nM+1

– For each such partition there are 2M2/2 choices for C() and at most2M2/2 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 2tk(n)/ (nM+1 2M2), 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

tk(m)- βm2 such pairs– We may now fix a for which C() has at least tk(m)- βm2 edges

By the stability theorem we can delete at most αm2 more edges and obtain a k-partite spanning subgraph of C(). The number of edges is at least tk(m)-(β+α)m2 = tk(m)-m2

G

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– We cannot have more than (2k+1)m2 one-sided dense regular pairs, since otherwise, some one-sided dense pair causes the formation of aKk+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 Vi. There are at most εn2

• (b) edges belonging to non-regular pairs. There are at most εn2

• (c) the edges belonging to non-dense pairs or one-sided dense pairs. There are at most (2+(2k+1))n2

• (d) the edges corresponding to the αm2 “deleted” pairs. There are at most αn2

– 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 n2 edges

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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 W1,…,Wk be subsets of vertices of G such that for any pair of subsets Xi Wi with |Xi| > 10-k|Wi|,Xj Wj with |Xj| > 10-k|Wj|,there are at least 0.1|Xi||Xj| edges in the “right” direction, then G contains a copy of S where the role of vertex i is played by a vertex from Wi

• Sketch of proof of the main result:

– We use n0 large enough for the approximation lemma with = 10-8k

– Suppose G has n > n02 vertices and at least 2tk(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 2tk(n-1)+m+1

distinct T-free orientations. Iterating downwards until n0 we obtain a graph with at least

2tk(n0)+m+n-n0 > 2n02 distinct T-free orientations, a contradiction

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– 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 V1,…,Vk of G that minimizes the “inside edges”.By the choice of n0 the number of “inside edges” is at most 10-8kn2

– 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 2tk(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 2tk(n-2)+m+2

distinct T-free orientations