Probabilistic Methods in Combinatorics

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PROBABILISTIC METHODS IN COMBINATORICS

YUFEI ZHAO

Abstract. This is a talk I gave for the Cambridge Part III seminar at the end of Lent term 2011.

The probabilistic method, originally popularized by Paul Erdos, is a powerful technique in

combinatorics. In this talk, we provide an introduction to the probabilistic method using examples

from Ramsey theory and set systems.

1. Introduction

The probabilistic method is a tool for tackling combinatorics problem by introducing randomness.For instance, to prove that a combinatorial object with certain property exists, we could constructsome object randomly, and then show that it has the desired property with positive probability.

Here is a simple example illustrating this idea.

Theorem 1.1. Every graph G = (V, E ) contains a bipartite subgraph with at least |E |

2edges.

Proof. Randomly assign each vertex of G with black or white, independently with uniform prob-ability. Consider the set of edges E with different colours on its endpoints. Then (V, E ) is abipartite subgraph of G. Note that every edge belongs to E with probability 1

2, so by linearity of

expectation, E[|E |] = 1

2|E |. Thus there is some colouring for which |E | ≥ 1

2|E |, and this gives

the desired bipartite subgraph.

2. Ramsey numbers

The Ramsey number R(k, ) is defined to be the smallest n such that if the edges of the completegraph K n are coloured red or blue, then it contains either a copy of a red K k or a copy of a blueK . It was shown by Ramsey that every R(k, ) is finite. However, very few exact values of R(k, ).It is an active problem in combinatorics to study the growth behavior of these Ramsey numbers.

In this section, we give several lower bounds to the diagonal Ramsey numbers R(k, k), and witheach bound we introduce a new idea in the probabilistic method.

2.1. Lower bound to Ramsey numbers. The following lower bound to Ramsey numbers isgiven by Erdo in a 1947 paper that “started” the probabilistic method.

Theorem 2.1. If nk

21−(k

2) < 1, then R(k, k) > n.

By optimizing n, this theorem gives us

R(k, k) >1

e√

2(1 + o(1))k2k/2

Proof. We need to show that there exists a colouring of the edges of K n with two colours containingno monochromatic K k. Let us colour the edges of K n randomly. The probability that a particular

subgraph K k is monochromatic is exactly 2

2(k2)

= 21−(k2). By consider all

nk

copies of K k in K n, we

find that the probability that there is some monochromatic K k is at mostnk

21−(k

2) < 1. Therefore,

with positive probability, the colouring has no monochromatic K k, thereby proving the existenceof such a colouring.

Date : 17 April 2011.

1



2 YUFEI ZHAO

2.2. Alterations. Next we give a slightly better lower bound to R(k, k) using the idea of alter-

ations . Our approach in the previous proof is that to randomly pick an edge-colouring of K n andthen hope that it contains no monochromatic K k. Alternatively, we can first pick a random edge-colouring of K k, and then modify the graph to get rid of the bad parts, namely the monochromaticK k.

Theorem 2.2. For any k, n, we have R(k, k) > n − nk

21−(k2).

By optimizing the choice of n, this theorem gives us

R(k, k) >1

e(1 + o(1))k2k/2,

which improves the previous bound by a constant factor of √

2.

Proof. Randomly colour the edges of K n with two colours. Whenever we see a monochromaticK k, delete one of its vertices. As in the previous proof, the probability that a particular subgraph

K k is monochromatic is exactly 21−(k2), so the expected number of monochromatic K k’s is exactly

nk21−(k2). Since we delete at most one vertex per every monochromatic K k, we remove at most

nk

21−(k2) vertices on expectation. Hence with some positive probability, the remaining graph has

at least n − nk

21−(k

2) vertices, and it has no monochromatic K k. This proves the desired lower

bound to R(k, k).

2.3. Lovasz local lemma. We give one more improvement to the lower bound, using the Lovaszlocal lemma, which we state without proof.

Theorem 2.3 (Lovasz local lemma). Let E 1, . . . , E n be events, with Pr[E i] ≤ p for all i. Suppose

that each E i is mutually independent of all other E j except for at most d of them. If

ep(d + 1) < 1,

then with some positive probability, none of the events E i occur.

Here is some intuition about the local lemma. We can view the events E i as “bad events” that wewant to avoid. If they are all independent, then we know easily there is some positive probabilitythat none of the bad events occur as long as each bad event has probability less than 1. On theother hand, if the probability of each E i is very small, say smaller than 1

n , then we can apply theunion bound to see that there is some probability that none of them occur. The situation reflectedinn the local lemma is between these two extremes. We know that p is small, but not as small as1

n . We know that most of the events mutually independent, but not all are. The local lemma tellsus that even in this situation, we can conclude that there is some positive probability that none of the bad events occur.

Theorem 2.4. If ek2

nk−2

+ 1

21−(nk) < 1, then R(k, k) > n.

By optimizing the choice of n, this theorem gives us

R(k, k) >

√ 2

e(1 + o(1))k2k/2,

once again improving the previous bound by a constant factor of √

2. This bound was given bySpencer in 1975. It is the best known lower bound to R(k, k) to date.

Proof. Consider a random colouring of the edges of K n. For each subset R of the vertices of K n with

k vertices, let E R denote teh event that R induces a monochromatic K k. Then Pr[E R] = 21−(k2).



PROBABILISTIC METHODS IN COMBINATORICS 3

The events E R and E S are independent when the subgraphs induced by R and S do not shareedges. Thus is E R and E S are dependent, then we necessarily have |R ∩ S | ≥ 2. For a fixed R,

there are at mostk

2

nk−2

choices of S with |S | = k and |R ∩ S | ≥ 2.

Applying the Lovasz local lemma to the events {E R : R ⊂ V (K n), |R| = k} and p = 21−(k2) and

d = k2

nk−2, we see that with positive probability none of the events E R occur, thereby giving a

colouring with no monochromatic K k’s.

3. Set systems

In this section we apply the probabilistic method to two extremal problems regarding families of subsets of {1, 2, . . . , n}.

3.1. Antichains. Let F be a collection of subsets of {1, 2, . . . , n} such that no set in F is containedin another set in F . Such a collection is called an antichain . We would like to know what is themaximum number of sets in an antichain.

If F is the collection of all size k subsets of {1, 2, . . . , n}, then F is an antichain, since no k-element set can contain another k-element set. This gives |F| =

nk

, which is maximized when

k = n2 or

n2 . The next result shows that we cannot do better.

Theorem 3.1 (Sperner). If F is an antichain of subsets of {1, 2, . . . , n}, then |F| ≤

n

n/2

.

Proof. Consider a random permutation σ of {1, 2, . . . , n}, and its associated chain of subsets

∅, {σ(1)} , {σ(1), σ(2)} , {σ(1), σ(2), σ(3)} , . . . , {σ(1), . . . , σ(n)}where the last set is always equal to {1, 2, . . . , n}. For each A ⊂ {1, 2, . . . , n}, let E A denote theevent that A is found in this chain. Then

Pr(E A) =|A|!(n − |A|)!

n!=

1n|A|

≥ 1n

n/2

.

Indeed, if A were to be found in the chain, it must occur in the (|A| + 1)-th position, and are |A|!ways to form a chain in front of it, and (n − |A|)! ways to continue the chain after it.Since F is an antichain, if A, B ∈ F are distinct, then E A and E B cannot both occur. Thus{E A : A ∈ F} is a set of disjoint events, each with probability at least 1

( nn/2)

. Thus |F| ≤ nn/2

.

3.2. Intersecting family. We say that family F of sets is intersecting if A ∩ B = ∅ for everyA, B ∈ F . Let F be an intersecting family of k-element subsets of {1, 2, . . . , n}. How large can |F|be?

If we take F to be the collection of all k-element subsets of {1, 2, . . . , n} containing the element1, then F is intersecting, and |F| ≤ n−1

k−1

. Now we show that this is the best we can do.

Theorem 3.2 (Erdos-Ko-Rado). If F is a intersecting family of k-element subsets of {1, 2, . . . , n},

then |F| ≤ n

−1

k − 1

.

Proof. Arrange 1, 2, . . . , n randomly around a circle. For each k-element subset of A of {1, 2, . . . , n},we say that A is contiguous if all the elements of A lie in a contiguous block on the circle. Theprobability that A forms a contiguous set on the circle is exactly n

(nk), since there are n possible

positions for a continuous block of length k on the circle, and for each such choice, the probabilitythat A falls exactly into those position is 1

(nk). It follows that the expected number of continguous

sets in F is exactly n|F|

(nk).



4 YUFEI ZHAO

Since F is intersecting, there are at most k continguous sets in F . Indeed, suppose that A ∈ F is contiguous. Then there are 2(k −1) other contingous sets (not necessarily in F ) that intersect A,but they can be paired off into disjoint pairs. Since F is intersecting, it follows that it contains at

most k continguous sets. Combining with result from the previous paragraph, we see thatn|F|

(nk)≤ k,

and hence |F| ≤ kn

nk =

n−1

k−1.

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Probabilistic Methods in Combinatorics