Transcript The number of orientations having no fixed tournament Noga Alon Raphael Yuster Definitions • Let T be a tournament.

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-free
D(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.
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In particular, D(n, C3) ≥ 2 n /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)= 2n2/4 for all n ≥ 8. Known for n=8 (computer) and n ≥ 600000
The only tournament for which all is known
•
D(n,TT3)= 2n2/4 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
•
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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
d ( A,B ) 
e( A,B )
|A||B|
• 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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• 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
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 that
ε < (/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
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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 G 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
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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
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– For each such partition there are 2M /2 choices for C() and at most
2
2M /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
2
than 2tk(n)/ (nM+1 2M ), 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 G 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
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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 a
Kk+1 and by the embedding lemma, this implies the orientation is not Tfree
– 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
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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
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