Combinatorics (Fall 2010)/Existence, the probabilistic method

Counting arguments

Circuit complexity

This is a fundamental problem in in Computer Science.

A boolean function is a function is the form [math]\displaystyle{ f:\{0,1\}^n\rightarrow \{0,1\} }[/math].

Formally, a boolean circuit is a directed acyclic graph. Nodes with indegree zero are input nodes, labeled [math]\displaystyle{ x_1, x_2, \ldots , x_n }[/math]. A circuit has a unique node with outdegree zero, called the output node. Every other node is a gate. There are three types of gates: AND, OR (both with indegree two), and NOT (with indegree one).

Computations in Turing machines can be simulated by circuits, and any boolean function in P can be computed by a circuit with polynomially many gates. Thus, if we can find a function in NP that cannot be computed by any circuit with polynomially many gates, then NP[math]\displaystyle{ \neq }[/math]P.

The following theorem due to Shannon says that functions with exponentially large circuit complexity do exist.

Theorem (Shannon 1949)

There is a boolean function [math]\displaystyle{ f:\{0,1\}^n\rightarrow \{0,1\} }[/math] with circuit complexity greater than [math]\displaystyle{ \frac{2^n}{3n} }[/math].

Proof. We first count the number of boolean functions [math]\displaystyle{ f:\{0,1\}^n\rightarrow \{0,1\} }[/math]. There are [math]\displaystyle{ 2^{2^n} }[/math] boolean functions [math]\displaystyle{ f:\{0,1\}^n\rightarrow \{0,1\} }[/math]. Then we count the number of boolean circuit with fixed number of gates. Fix an integer [math]\displaystyle{ t }[/math], we count the number of circuits with [math]\displaystyle{ t }[/math] gates. By the De Morgan's laws, we can assume that all NOTs are pushed back to the inputs. Each gate has one of the two types (AND or OR), and has two inputs. Each of the inputs to a gate is either a constant 0 or 1, an input variable [math]\displaystyle{ x_i }[/math], an inverted input variable [math]\displaystyle{ \neg x_i }[/math], or the output of another gate; thus, there are at most [math]\displaystyle{ 2+2n+t-1 }[/math] possible gate inputs. It follows that the number of circuits with [math]\displaystyle{ t }[/math] gates is at most [math]\displaystyle{ 2^t(t+2n+1)^{2t} }[/math]. If [math]\displaystyle{ t=2^n/3n }[/math], then [math]\displaystyle{ \frac{2^t(t+2n+1)^{2t}}{2^{2^n}}=o(1)\lt 1, }[/math] thus, [math]\displaystyle{ 2^t(t+2n+1)^{2t} \lt 2^{2^n}. }[/math] Each boolean circuit computes one boolean function. Therefore, there must exist a boolean function [math]\displaystyle{ f }[/math] which cannot be computed by any circuits with [math]\displaystyle{ 2^n/3n }[/math] gates. [math]\displaystyle{ \square }[/math]

Note that by Shannon's theorem, not only there exists a boolean function with exponentially large circuit complexity, but almost all boolean functions have exponentially large circuit complexity.

Double counting

Handshaking lemma

Handshaking Lemma

At a party, the number of guests who shake hands an odd number of times is even.

We model this scenario as an undirected graph [math]\displaystyle{ G(V,E) }[/math] with [math]\displaystyle{ |V|=n }[/math] standing for the [math]\displaystyle{ n }[/math] guests. There is an edge [math]\displaystyle{ uv\in E }[/math] if [math]\displaystyle{ u }[/math] and [math]\displaystyle{ v }[/math] shake hands. Let [math]\displaystyle{ d(v) }[/math] be the degree of vertex [math]\displaystyle{ v }[/math], which represents the number of times that [math]\displaystyle{ v }[/math] shakes hand. The handshaking lemma states that in any undirected graph, the sum of odd degrees is even.

The handshaking lemma is a direct consequence of the following lemma, which is proved by Euler in a 1736 paper that began the study of graph theory.

Lemma (Euler 1736)

[math]\displaystyle{ \sum_{v\in V}d(v)=2|E| }[/math]

Proof. We count the number of directed edges. A directed edge is an ordered pair [math]\displaystyle{ (u,v) }[/math] such that [math]\displaystyle{ \{u,v\}\in E }[/math]. There are two ways to count the directed edges. First, we can enumerate by edges. Pick every edge [math]\displaystyle{ uv\in E }[/math] and apply two directions [math]\displaystyle{ (u,v) }[/math] and [math]\displaystyle{ (v,u) }[/math] to the edge. This gives us [math]\displaystyle{ 2|E| }[/math] directed edges. On the other hand, we can enumerate by vertices. Pick every vertex [math]\displaystyle{ v\in V }[/math] and for each of its [math]\displaystyle{ d(v) }[/math] neighbors, say [math]\displaystyle{ u }[/math], generate a directed edge [math]\displaystyle{ (v,u) }[/math]. This gives us [math]\displaystyle{ \sum_{v\in V}d(v) }[/math] directed edges. It is obvious that the two terms are equal, since we just count the same thing twice with different methods. The lemma follows. [math]\displaystyle{ \square }[/math]

The handshaking lemma is implied directly by the above lemma, since the sum of even degrees is even.

Cayley's formula

Caylay's formula for trees

There are [math]\displaystyle{ n^{n-2} }[/math] different trees on [math]\displaystyle{ n }[/math] distinct vertices.

The Pigeonhole Principle

Monotonic subsequences

Theorem (Erdős-Szekeres 1935)

A sequence of more than [math]\displaystyle{ mn }[/math] different real numbers must contain either an increasing subsequence of length [math]\displaystyle{ m+1 }[/math], or a decreasing subsequence of length [math]\displaystyle{ n+1 }[/math].

Proof. (due to Seidenberg 1959) [math]\displaystyle{ \square }[/math]

Dirichlet's theorem

Theorem (Dirichlet 1879)

Let [math]\displaystyle{ x }[/math] be a real number. For any natural number [math]\displaystyle{ n }[/math], there is a rational number [math]\displaystyle{ \frac{p}{q} }[/math] such that [math]\displaystyle{ 1\le q\le n }[/math] and [math]\displaystyle{ \left|x-\frac{p}{q}\right|\lt \frac{1}{nq} }[/math].

The Probabilistic Method

The probabilistic method provides another way of proving the existence of objects: instead of explicitly constructing an object, we define a probability space of objects in which the probability is positive that a randomly selected object has the required property.

The basic principle of the probabilistic method is very simple, and can be stated in intuitive ways:

• If an object chosen randomly from a universe satisfies a property with positive probability, then there must be an object in the universe that satisfies that property.

For example, for a ball(the object) randomly chosen from a box(the universe) of balls, if the probability that the chosen ball is blue(the property) is >0, then there must be a blue ball in the box.

• Any random variable assumes at least one value that is no smaller than its expectation, and at least one value that is no greater than the expectation.

For example, if we know the average height of the students in the class is [math]\displaystyle{ \ell }[/math], then we know there is a students whose height is at least [math]\displaystyle{ \ell }[/math], and there is a student whose height is at most [math]\displaystyle{ \ell }[/math].

Although the idea of the probabilistic method is simple, it provides us a powerful tool for existential proof.

Sampling

Ramsey number

Recall the Ramsey theorem which states that in a meeting of at least six people, there are either three people knowing each other or three people not knowing each other. In graph theoretical terms, this means that no matter how we color the edges of [math]\displaystyle{ K_6 }[/math] (the complete graph on six vertices), there must be a monochromatic [math]\displaystyle{ K_3 }[/math] (a triangle whose edges have the same color).

Generally, the Ramsey number [math]\displaystyle{ R(k,\ell) }[/math] is the smallest integer [math]\displaystyle{ n }[/math] such that in any two-coloring of the edges of a complete graph on [math]\displaystyle{ n }[/math] vertices [math]\displaystyle{ K_n }[/math] by red and blue, either there is a red [math]\displaystyle{ K_k }[/math] or there is a blue [math]\displaystyle{ K_\ell }[/math].

Ramsey showed in 1929 that [math]\displaystyle{ R(k,\ell) }[/math] is finite for any [math]\displaystyle{ k }[/math] and [math]\displaystyle{ \ell }[/math]. It is extremely hard to compute the exact value of [math]\displaystyle{ R(k,\ell) }[/math]. Here we give a lower bound of [math]\displaystyle{ R(k,k) }[/math] by the probabilistic method.

Theorem (Erdős 1947)

If [math]\displaystyle{ {n\choose k}\cdot 2^{1-{k\choose 2}}\lt 1 }[/math] then it is possible to color the edges of [math]\displaystyle{ K_n }[/math] with two colors so that there is no monochromatic [math]\displaystyle{ K_k }[/math] subgraph.

Proof. Consider a random two-coloring of edges of [math]\displaystyle{ K_n }[/math] obtained as follows: For each edge of [math]\displaystyle{ K_n }[/math], independently flip a fair coin to decide the color of the edge. For any fixed set [math]\displaystyle{ S }[/math] of [math]\displaystyle{ k }[/math] vertices, let [math]\displaystyle{ \mathcal{E}_S }[/math] be the event that the [math]\displaystyle{ K_k }[/math] subgraph induced by [math]\displaystyle{ S }[/math] is monochromatic. There are [math]\displaystyle{ {k\choose 2} }[/math] many edges in [math]\displaystyle{ K_k }[/math], therefore [math]\displaystyle{ \Pr[\mathcal{E}_S]=2\cdot 2^{-{k\choose 2}}=2^{1-{k\choose 2}}. }[/math] Since there are [math]\displaystyle{ {n\choose k} }[/math] possible choices of [math]\displaystyle{ S }[/math], by the union bound [math]\displaystyle{ \Pr[\exists S, \mathcal{E}_S]\le {n\choose k}\cdot\Pr[\mathcal{E}_S]={n\choose k}\cdot 2^{1-{k\choose 2}}. }[/math] Due to the assumption, [math]\displaystyle{ {n\choose k}\cdot 2^{1-{k\choose 2}}\lt 1 }[/math], thus there exists a two coloring that none of [math]\displaystyle{ \mathcal{E}_S }[/math] occurs, which means there is no monochromatic [math]\displaystyle{ K_k }[/math] subgraph. [math]\displaystyle{ \square }[/math]

For [math]\displaystyle{ k\ge 3 }[/math] and we take [math]\displaystyle{ n=\lfloor2^{k/2}\rfloor }[/math], then

[math]\displaystyle{ \begin{align} {n\choose k}\cdot 2^{1-{k\choose 2}} &\lt \frac{n^k}{k!}\cdot\frac{2^{1+\frac{k}{2}}}{2^{k^2/2}}\\ &\le \frac{2^{k^2/2}}{k!}\cdot\frac{2^{1+\frac{k}{2}}}{2^{k^2/2}}\\ &= \frac{2^{1+\frac{k}{2}}}{k!}\\ &\lt 1. \end{align} }[/math]

By the above theorem, there exists a two-coloring of [math]\displaystyle{ K_n }[/math] that there is no monochromatic [math]\displaystyle{ K_k }[/math]. Therefore, the Ramsey number [math]\displaystyle{ R(k,k)\gt \lfloor2^{k/2}\rfloor }[/math] for all [math]\displaystyle{ k\ge 3 }[/math].

Note that for sufficiently large [math]\displaystyle{ k }[/math], if [math]\displaystyle{ n= \lfloor 2^{k/2}\rfloor }[/math], then the probability that there exists a monochromatic [math]\displaystyle{ K_k }[/math] is bounded by

[math]\displaystyle{ {n\choose k}\cdot 2^{1-{k\choose 2}} \lt \frac{2^{1+\frac{k}{2}}}{k!} \ll 1, }[/math]

which means that a random two-coloring of [math]\displaystyle{ K_n }[/math] is very likely not to contain a monochromatic [math]\displaystyle{ K_{2\log n} }[/math]. This gives us a very simple randomized algorithm for finding a two-coloring of [math]\displaystyle{ K_n }[/math] without monochromatic [math]\displaystyle{ K_{2\log n} }[/math].

Tournament

A tournament (竞赛图) on a set [math]\displaystyle{ V }[/math] of [math]\displaystyle{ n }[/math] players is an orientation of the edges of the complete graph on the set of vertices [math]\displaystyle{ V }[/math]. Thus for every two distinct vertices [math]\displaystyle{ u,v }[/math] in [math]\displaystyle{ V }[/math], either [math]\displaystyle{ (u,v)\in E }[/math] or [math]\displaystyle{ (v,u)\in E }[/math], but not both.

We can think of the set [math]\displaystyle{ V }[/math] as a set of [math]\displaystyle{ n }[/math] players in which each pair participates in a single match, where [math]\displaystyle{ (u,v) }[/math] is in the tournament iff player [math]\displaystyle{ u }[/math] beats player [math]\displaystyle{ v }[/math].

Definition

We say that a tournament has property [math]\displaystyle{ S_k }[/math] if for every set of [math]\displaystyle{ k }[/math] players there is one who beats them all.

Is it true for every finite [math]\displaystyle{ k }[/math], there is a tournament (on more than [math]\displaystyle{ k }[/math] vertices, of course) with the property [math]\displaystyle{ S_k }[/math]? This problem was first raised by Schütte, and as shown by Erdős, can be solved almost trivially by the probabilistic method.

Theorem (Erdős 1963)

If [math]\displaystyle{ {n\choose k}\left(1-2^{-k}\right)^{n-k}\lt 1 }[/math] then there is a tournament on [math]\displaystyle{ n }[/math] vertices that has the property [math]\displaystyle{ S_k }[/math].

Linearity of expectation

Maximum cut

Given an undirected graph [math]\displaystyle{ G(V,E) }[/math], a set [math]\displaystyle{ C }[/math] of edges of [math]\displaystyle{ G }[/math] is called a cut if [math]\displaystyle{ G }[/math] is disconnected after removing the edges in [math]\displaystyle{ C }[/math]. We can represent a cut by [math]\displaystyle{ c(S,T) }[/math] where [math]\displaystyle{ (S,T) }[/math] is a bipartition of the vertex set [math]\displaystyle{ V }[/math], and [math]\displaystyle{ c(S,T)=\{uv\in E\mid u\in S,v\in T\} }[/math] is the set of edges crossing between [math]\displaystyle{ S }[/math] and [math]\displaystyle{ T }[/math].

We have seen how to compute min-cut: either by deterministic max-flow algorithm, or by Karger's randomized algorithm. On the other hand, max-cut is hard to compute, because it is NP-complete. Actually, the weighted version of max-cut is among the Karp's 21 NP-complete problems.

We now show by the probabilistic method that a max-cut always has at least half the edges.

Theorem

Given an undirected graph [math]\displaystyle{ G }[/math] with [math]\displaystyle{ n }[/math] vertices and [math]\displaystyle{ m }[/math] edges, there is a cut of size at least [math]\displaystyle{ \frac{m}{2} }[/math].

Proof. Enumerate the vertices in an arbitrary order. Partition the vertex set [math]\displaystyle{ V }[/math] into two disjoint sets [math]\displaystyle{ S }[/math] and [math]\displaystyle{ T }[/math] as follows. For each vertex [math]\displaystyle{ v\in V }[/math], independently choose one of [math]\displaystyle{ S }[/math] and [math]\displaystyle{ T }[/math] with equal probability, and let [math]\displaystyle{ v }[/math] join the chosen set. For each vertex [math]\displaystyle{ v\in V }[/math], let [math]\displaystyle{ X_v\in\{S,T\} }[/math] be the random variable which represents the set that [math]\displaystyle{ v }[/math] joins. For each edge [math]\displaystyle{ uv\in E }[/math], let [math]\displaystyle{ Y_{uv} }[/math] be the 0-1 random variable which indicates whether [math]\displaystyle{ uv }[/math] crosses between [math]\displaystyle{ S }[/math] and [math]\displaystyle{ T }[/math]. Clearly, [math]\displaystyle{ \Pr[Y_{uv}=1]=\Pr[X_u\neq X_v]=\frac{1}{2}. }[/math] The size of [math]\displaystyle{ c(S,T) }[/math] is given by [math]\displaystyle{ Y=\sum_{uv\in E}Y_{uv} }[/math]. By the linearity of expectation, [math]\displaystyle{ \mathbf{E}[Y]=\sum_{uv\in E}\mathbf{E}[Y_{uv}]=\sum_{uv\in E}\Pr[Y_{uv}=1]=\frac{m}{2}. }[/math] Therefore, there exist a bipartition [math]\displaystyle{ (S,T) }[/math] of [math]\displaystyle{ V }[/math] such that [math]\displaystyle{ |c(S,T)|\ge\frac{m}{2} }[/math], i.e. there exists a cut of [math]\displaystyle{ G }[/math] which contains at least [math]\displaystyle{ \frac{m}{2} }[/math] edges. [math]\displaystyle{ \square }[/math]

Maximum satisfiability

Suppose that we have a number of boolean variables [math]\displaystyle{ x_1,x_2,\ldots,\in\{\mathrm{true},\mathrm{false}\} }[/math]. A literal is either a variable [math]\displaystyle{ x_i }[/math] itself or its negation [math]\displaystyle{ \neg x_i }[/math]. A logic expression is a conjunctive normal form (CNF) if it is written as the conjunction(AND) of a set of clauses, where each clause is a disjunction(OR) of literals. For example:

[math]\displaystyle{ (x_1\vee \neg x_2 \vee \neg x_3)\wedge (\neg x_1\vee \neg x_3)\wedge (x_1\vee x_2\vee x_4)\wedge (x_4\vee \neg x_3)\wedge (x_4\vee \neg x_1). }[/math]

The satisfiability (SAT) problem ask whether the CNF is satisfiable, i.e. there exists an assignment of variables to the values of true and false so that all clauses are true. The maximum satisfiability (MAXSAT) is the optimization version of SAT, which ask for an assignment that the number of satisfied clauses is maximized.

SAT is the first problem known to be NP-complete (the Cook-Levin theorem). MAXSAT is also NP-complete. We then see that there always exists a roughly good truth assignment which satisfies half the clauses.

Theorem

For any set of [math]\displaystyle{ m }[/math] clauses, there is a truth assignment that satisfies at least [math]\displaystyle{ \frac{m}{2} }[/math] clauses.

Proof. For each variable, independently assign a random value in [math]\displaystyle{ \{\mathrm{true},\mathrm{false}\} }[/math] with equal probability. For the [math]\displaystyle{ i }[/math]th clause, let [math]\displaystyle{ X_i }[/math] be the random variable which indicates whether the [math]\displaystyle{ i }[/math]th clause is satisfied. Suppose that there are [math]\displaystyle{ k }[/math] literals in the clause. The probability that the clause is satisfied is [math]\displaystyle{ \Pr[X_k=1]\ge(1-2^{-k})\ge\frac{1}{2} }[/math]. Let [math]\displaystyle{ X=\sum_{i=1}^m X_i }[/math] be the number of satisfied clauses. By the linearity of expectation, [math]\displaystyle{ \mathbf{E}[X]=\sum_{i=1}^{m}\mathbf{E}[X_i]\ge \frac{m}{2}. }[/math] Therefore, there exists an assignment such that at least [math]\displaystyle{ \frac{m}{2} }[/math] clauses are satisfied. [math]\displaystyle{ \square }[/math]

Alterations

Independent sets

An independent set of a graph is a set of vertices with no edges between them. The following theorem gives a lower bound on the size of the largest independent set.

Theorem

Let [math]\displaystyle{ G(V,E) }[/math] be a graph on [math]\displaystyle{ n }[/math] vertices with [math]\displaystyle{ m }[/math] edges. Then [math]\displaystyle{ G }[/math] has an independent set with at least [math]\displaystyle{ \frac{n^2}{4m} }[/math] vertices.

Proof. Let [math]\displaystyle{ S }[/math] be a set of vertices constructed as follows: For each vertex [math]\displaystyle{ v\in V }[/math]: [math]\displaystyle{ v }[/math] is included in [math]\displaystyle{ S }[/math] independently with probability [math]\displaystyle{ p }[/math], [math]\displaystyle{ p }[/math] to be determined. Let [math]\displaystyle{ X=|S| }[/math]. It is obvious that [math]\displaystyle{ \mathbf{E}[X]=np }[/math]. For each edge [math]\displaystyle{ e\in E }[/math], let [math]\displaystyle{ Y_{e} }[/math] be the random variable which indicates whether both endpoints of [math]\displaystyle{ }[/math] are in [math]\displaystyle{ S }[/math]. [math]\displaystyle{ \mathbf{E}[Y_{uv}]=\Pr[u\in S\wedge v\in S]=p^2. }[/math] Let [math]\displaystyle{ Y }[/math] be the number of edges in the subgraph of [math]\displaystyle{ G }[/math] induced by [math]\displaystyle{ S }[/math]. It holds that [math]\displaystyle{ Y=\sum_{e\in E}Y_e }[/math]. By linearity of expectation, [math]\displaystyle{ \mathbf{E}[Y]=\sum_{e\in E}\mathbf{E}[Y_e]=mp^2 }[/math]. Note that although [math]\displaystyle{ S }[/math] is not necessary an independent set, it can be modified to one if for each edge [math]\displaystyle{ e }[/math] of the induced subgraph [math]\displaystyle{ G(S) }[/math], we delete one of the endpoint of [math]\displaystyle{ e }[/math] from [math]\displaystyle{ S }[/math]. Let [math]\displaystyle{ S^* }[/math] be the resulting set. It is obvious that [math]\displaystyle{ S^* }[/math] is an independent set since there is no edge left in the induced subgraph [math]\displaystyle{ G(S^*) }[/math]. Since there are [math]\displaystyle{ Y }[/math] edges in [math]\displaystyle{ G(S) }[/math], there are at most [math]\displaystyle{ Y }[/math] vertices in [math]\displaystyle{ S }[/math] are deleted to make it become [math]\displaystyle{ S^* }[/math]. Therefore, [math]\displaystyle{ |S^*|\ge X-Y }[/math]. By linearity of expectation, [math]\displaystyle{ \mathbf{E}[|S^*|]\ge\mathbf{E}[X-Y]=\mathbf{E}[X]-\mathbf{E}[Y]=np-mp^2. }[/math] The expectation is maximized when [math]\displaystyle{ p=\frac{n}{2m} }[/math], thus [math]\displaystyle{ \mathbf{E}[|S^*|]\ge n\cdot\frac{n}{2m}-m\left(\frac{n}{2m}\right)^2=\frac{n^2}{4m}. }[/math] There exists an independent set which contains at least [math]\displaystyle{ \frac{n^2}{4m} }[/math] vertices. [math]\displaystyle{ \square }[/math]

The proof actually propose a randomized algorithm for constructing large independent set:

Algorithm

Given a graph on [math]\displaystyle{ n }[/math] vertices with [math]\displaystyle{ m }[/math] edges, let [math]\displaystyle{ d=\frac{2m}{n} }[/math] be the average degree. For each vertex [math]\displaystyle{ v\in V }[/math], [math]\displaystyle{ v }[/math] is included in [math]\displaystyle{ S }[/math] independently with probability [math]\displaystyle{ \frac{1}{d} }[/math]. For each remaining edge in the induced subgraph [math]\displaystyle{ G(S) }[/math], remove one of the endpoints from [math]\displaystyle{ S }[/math].

Let [math]\displaystyle{ S^* }[/math] be the resulting set. We have shown that [math]\displaystyle{ S^* }[/math] is an independent set and [math]\displaystyle{ \mathbf{E}[|S^*|]\ge\frac{n^2}{4m} }[/math].