组合数学 (Fall 2019)/Ramsey theory

Ramsey's Theorem

Ramsey's theorem for graph

Ramsey's Theorem

Let [math]\displaystyle{ k,\ell }[/math] be positive integers. Then there exists an integer [math]\displaystyle{ R(k,\ell) }[/math] satisfying: If [math]\displaystyle{ n\ge R(k,\ell) }[/math], for any coloring of edges of [math]\displaystyle{ K_n }[/math] with two colors red and blue, there exists a red [math]\displaystyle{ K_k }[/math] or a blue [math]\displaystyle{ K_\ell }[/math].

Proof. We show that [math]\displaystyle{ R(k,\ell) }[/math] is finite by induction on [math]\displaystyle{ k+\ell }[/math]. For the base case, it is easy to verify that [math]\displaystyle{ R(k,1)=R(1,\ell)=1 }[/math]. For general [math]\displaystyle{ k }[/math] and [math]\displaystyle{ \ell }[/math], we will show that [math]\displaystyle{ R(k,\ell)\le R(k,\ell-1)+R(k-1,\ell) }[/math]. Suppose we have a two coloring of [math]\displaystyle{ K_n }[/math], where [math]\displaystyle{ n=R(k,\ell-1)+R(k-1,\ell) }[/math]. Take an arbitrary vertex [math]\displaystyle{ v }[/math], and split [math]\displaystyle{ V\setminus\{v\} }[/math] into two subsets [math]\displaystyle{ S }[/math] and [math]\displaystyle{ T }[/math], where [math]\displaystyle{ \begin{align} S&=\{u\in V\setminus\{v\}\mid uv \text{ is blue }\}\\ T&=\{u\in V\setminus\{v\}\mid uv \text{ is red }\} \end{align} }[/math] Since [math]\displaystyle{ |S|+|T|+1=n=R(k,\ell-1)+R(k-1,\ell) }[/math], we have either [math]\displaystyle{ |S|\ge R(k,\ell-1) }[/math] or [math]\displaystyle{ |T|\ge R(k-1,\ell) }[/math]. By symmetry, suppose [math]\displaystyle{ |S|\ge R(k,\ell-1) }[/math]. By induction hypothesis, the complete subgraph defined on [math]\displaystyle{ S }[/math] has either a red [math]\displaystyle{ K_k }[/math], in which case we are done; or a blue [math]\displaystyle{ K_{\ell-1} }[/math], in which case the complete subgraph defined on [math]\displaystyle{ S\cup{v} }[/math] must have a blue [math]\displaystyle{ K_\ell }[/math] since all edges from [math]\displaystyle{ v }[/math] to vertices in [math]\displaystyle{ S }[/math] are blue. [math]\displaystyle{ \square }[/math]

Ramsey's Theorem (graph, multicolor)

Let [math]\displaystyle{ r, k_1,k_2,\ldots,k_r }[/math] be positive integers. Then there exists an integer [math]\displaystyle{ R(r;k_1,k_2,\ldots,k_r) }[/math] satisfying: For any [math]\displaystyle{ r }[/math]-coloring of a complete graph of [math]\displaystyle{ n\ge R(r;k_1,k_2,\ldots,k_r) }[/math] vertices, there exists a monochromatic [math]\displaystyle{ k_i }[/math]-clique with the [math]\displaystyle{ i }[/math]th color for some [math]\displaystyle{ i\in\{1,2,\ldots,r\} }[/math].

Lemma (the "mixing color" trick)

[math]\displaystyle{ R(r;k_1,k_2,\ldots,k_r)\le R(r-1;k_1,k_2,\ldots,k_{r-2},R(2;k_{r-1},k_r)) }[/math]

Proof. We transfer the [math]\displaystyle{ r }[/math]-coloring to [math]\displaystyle{ (r-1) }[/math]-coloring by identifying the [math]\displaystyle{ (r-1) }[/math]th and the [math]\displaystyle{ r }[/math]th colors. If [math]\displaystyle{ n\ge R(r-1;k_1,k_2,\ldots,k_{r-2},R(2;k_{r-1},k_r)) }[/math], then for any [math]\displaystyle{ r }[/math]-coloring of [math]\displaystyle{ K_n }[/math], there either exist an [math]\displaystyle{ i\in\{1,2,\ldots,r-2\} }[/math] and a [math]\displaystyle{ k_i }[/math]-clique which is monochromatically colored with the [math]\displaystyle{ i }[/math]th color; or exists clique of [math]\displaystyle{ R(2;k_{r-1},k_r) }[/math] vertices which is monochromatically colored with the mixed color of the original [math]\displaystyle{ (r-1) }[/math]th and [math]\displaystyle{ r }[/math]th colors, which again implies that there exists either a [math]\displaystyle{ k }[/math]-clique which is monochromatically colored with the original [math]\displaystyle{ (r-1) }[/math]th color, or a [math]\displaystyle{ \ell }[/math]-clique which is monochromatically colored with the original [math]\displaystyle{ r }[/math]th color. This implies the recursion. [math]\displaystyle{ \square }[/math]

Ramsey number

The smallest number [math]\displaystyle{ R(k,\ell) }[/math] satisfying the condition in the Ramsey theory is called the Ramsey number.

Alternatively, we can define [math]\displaystyle{ R(k,\ell) }[/math] as the smallest [math]\displaystyle{ N }[/math] such that if [math]\displaystyle{ n\ge N }[/math], for any 2-coloring of [math]\displaystyle{ K_n }[/math] in red and blue, there is either a red [math]\displaystyle{ K_k }[/math] or a blue [math]\displaystyle{ K_\ell }[/math]. The Ramsey theorem is stated as:

"[math]\displaystyle{ R(k,\ell) }[/math] is finite for any positive integers [math]\displaystyle{ k }[/math] and [math]\displaystyle{ \ell }[/math]."

The core of the inductive proof of the Ramsey theorem is the following recursion

[math]\displaystyle{ \begin{align} R(k,1) &=R(1,\ell)=1\\ R(k,\ell) &\le R(k,\ell-1)+R(k-1,\ell). \end{align} }[/math]

From this recursion, we can deduce an upper bound for the Ramsey number.

Theorem

[math]\displaystyle{ R(k,\ell)\le{k+\ell-2\choose k-1} }[/math].

Proof. It is easy to verify the bound by induction. [math]\displaystyle{ \square }[/math]

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The following theorem is due to Spencer in 1975, which is the best known lower bound for diagonal Ramsey number.

Theorem (Spencer 1975)

[math]\displaystyle{ R(k,k)\ge Ck2^{k/2} }[/math] for some constant [math]\displaystyle{ C\gt 0 }[/math].

Its proof uses the Lovász local lemma in the probabilistic method.

Lovász Local Lemma (symmetric case)

Let [math]\displaystyle{ A_1,A_2,\ldots,A_n }[/math] be a set of events, and assume that the following hold: for all [math]\displaystyle{ 1\le i\le n }[/math], [math]\displaystyle{ \Pr[A_i]\le p }[/math]; each event [math]\displaystyle{ A_i }[/math] is independent of all but at most [math]\displaystyle{ d }[/math] other events, and [math]\displaystyle{ ep(d+1)\le 1 }[/math]. Then [math]\displaystyle{ \Pr\left[\bigwedge_{i=1}^n\overline{A_i}\right]\gt 0 }[/math].

We can use the local lemma to prove a lower bound for the diagonal Ramsey number.

Proof. To prove a lower bound [math]\displaystyle{ R(k,k)\gt n }[/math], it is sufficient to show that there exists a 2-coloring of [math]\displaystyle{ K_n }[/math] without a monochromatic [math]\displaystyle{ K_k }[/math]. We prove this by the probabilistic method. Pick a random 2-coloring of [math]\displaystyle{ K_n }[/math] by coloring each edge uniformly and independently with one of the two colors. For any set [math]\displaystyle{ S }[/math] of [math]\displaystyle{ k }[/math] vertices, let [math]\displaystyle{ A_S }[/math] denote the event that [math]\displaystyle{ S }[/math] forms a monochromatic [math]\displaystyle{ K_k }[/math]. It is easy to see that [math]\displaystyle{ \Pr[A_s]=2^{1-{k\choose 2}}=p }[/math]. For any [math]\displaystyle{ k }[/math]-subset [math]\displaystyle{ T }[/math] of vertices, [math]\displaystyle{ A_S }[/math] and [math]\displaystyle{ A_T }[/math] are dependent if and only if [math]\displaystyle{ |S\cap T|\ge 2 }[/math]. For each [math]\displaystyle{ S }[/math], the number of [math]\displaystyle{ T }[/math] that [math]\displaystyle{ |S\cap T|\ge 2 }[/math] is at most [math]\displaystyle{ {k\choose 2}{n\choose k-2} }[/math], so the max degree of the dependency graph is [math]\displaystyle{ d\le{k\choose 2}{n\choose k-2} }[/math]. Take [math]\displaystyle{ n=Ck2^{k/2} }[/math] for some appropriate constant [math]\displaystyle{ C\gt 0 }[/math]. [math]\displaystyle{ \begin{align} \mathrm{e}p(d+1) &\le \mathrm{e}2^{1-{k\choose 2}}\left({k\choose 2}{n\choose k-2}+1\right)\\ &\le 2^{3-{k\choose 2}}{k\choose 2}{n\choose k-2}\\ &\le 1 \end{align} }[/math] Applying the local lemma, the probability that there is no monochromatic [math]\displaystyle{ K_k }[/math] is [math]\displaystyle{ \Pr\left[\bigwedge_{S\in{[n]\choose k}}\overline{A_S}\right]\gt 0 }[/math]. Therefore, there exists a 2-coloring of [math]\displaystyle{ K_n }[/math] which has no monochromatic [math]\displaystyle{ K_k }[/math], which means [math]\displaystyle{ R(k,k)\gt n=Ck2^{k/2} }[/math]. [math]\displaystyle{ \square }[/math]

Theorem

[math]\displaystyle{ \Omega\left(k2^{k/2}\right)\le R(k,k)\le{2k-2\choose k-1}=O\left(k^{-1/2}4^{k}\right) }[/math].

[math]\displaystyle{ k }[/math],[math]\displaystyle{ l }[/math]	1	2	3	4	5	6	7	8	9	10
1	1	1	1	1	1	1	1	1	1	1
2	1	2	3	4	5	6	7	8	9	10
3	1	3	6	9	14	18	23	28	36	40–43
4	1	4	9	18	25	35–41	49–61	56–84	73–115	92–149
5	1	5	14	25	43–49	58–87	80–143	101–216	125–316	143–442
6	1	6	18	35–41	58–87	102–165	113–298	127–495	169–780	179–1171
7	1	7	23	49–61	80–143	113–298	205–540	216–1031	233–1713	289–2826
8	1	8	28	56–84	101–216	127–495	216–1031	282–1870	317–3583	317-6090
9	1	9	36	73–115	125–316	169–780	233–1713	317–3583	565–6588	580–12677
10	1	10	40–43	92–149	143–442	179–1171	289–2826	317-6090	580–12677	798–23556

Ramsey's theorem for hypergraph

Ramsey's Theorem (hypergraph, multicolor)

Let [math]\displaystyle{ r, t, k_1,k_2,\ldots,k_r }[/math] be positive integers. Then there exists an integer [math]\displaystyle{ R_t(r;k_1,k_2,\ldots,k_r) }[/math] satisfying: For any [math]\displaystyle{ r }[/math]-coloring of [math]\displaystyle{ {[n]\choose t} }[/math] with [math]\displaystyle{ n\ge R_t(r;k_1,k_2,\ldots,k_r) }[/math], there exist an [math]\displaystyle{ i\in\{1,2,\ldots,r\} }[/math] and a subset [math]\displaystyle{ X\subseteq [n] }[/math] with [math]\displaystyle{ |X|\ge k_i }[/math] such that all members of [math]\displaystyle{ {X\choose t} }[/math] are colored with the [math]\displaystyle{ i }[/math]th color.

[math]\displaystyle{ n\rightarrow(k_1,k_2,\ldots,k_r)^t }[/math]

Lemma (the "mixing color" trick)

[math]\displaystyle{ R_t(r;k_1,k_2,\ldots,k_r)\le R_t(r-1;k_1,k_2,\ldots,k_{r-2},R_t(2;k_{r-1},k_r)) }[/math]

It is then sufficient to prove the Ramsey's theorem for the two-coloring of a hypergraph, that is, to prove [math]\displaystyle{ R_t(k,\ell)=R_t(2;k,\ell) }[/math] is finite.

Lemma

[math]\displaystyle{ R_t(k,\ell)\le R_{t-1}(R_t(k-1,\ell),R_t(k,\ell-1))+1 }[/math]

Proof. Let [math]\displaystyle{ n=R_{t-1}(R_t(k-1,\ell),R_t(k,\ell-1))+1 }[/math]. Denote [math]\displaystyle{ [n]=\{1,2,\ldots,n\} }[/math]. Let [math]\displaystyle{ f:{[n]\choose t}\rightarrow\{{\color{red}\text{red}},{\color{blue}\text{blue}}\} }[/math] be an arbitrary 2-coloring of [math]\displaystyle{ {[n]\choose t} }[/math]. It is then sufficient to show that there either exists an [math]\displaystyle{ X\subseteq[n] }[/math] with [math]\displaystyle{ |X|=k }[/math] such that all members of [math]\displaystyle{ {X\choose t} }[/math] are colored red by [math]\displaystyle{ f }[/math]; or exists an [math]\displaystyle{ X\subseteq[n] }[/math] with [math]\displaystyle{ |X|=\ell }[/math] such that all members of [math]\displaystyle{ {X\choose t} }[/math] are colored blue by [math]\displaystyle{ f }[/math]. We remove [math]\displaystyle{ n }[/math] from [math]\displaystyle{ [n] }[/math] and define a new coloring [math]\displaystyle{ f' }[/math] of [math]\displaystyle{ {[n-1]\choose t-1} }[/math] by [math]\displaystyle{ f'(A)=f(A\cup\{n\}) }[/math] for any [math]\displaystyle{ A\in{[n-1]\choose t-1} }[/math]. By the choice of [math]\displaystyle{ n }[/math] and by symmetry, there exists a subset [math]\displaystyle{ S\subseteq[n-1] }[/math] with [math]\displaystyle{ |X|=R_t(k-1,\ell) }[/math] such that all members of [math]\displaystyle{ {S\choose t-1} }[/math] are colored with red by [math]\displaystyle{ f' }[/math]. Then there either exists an [math]\displaystyle{ X\subseteq S }[/math] with [math]\displaystyle{ |X|=\ell }[/math] such that [math]\displaystyle{ {X\choose t} }[/math] is colored all blue by [math]\displaystyle{ f }[/math], in which case we are done; or exists an [math]\displaystyle{ X\subseteq S }[/math] with [math]\displaystyle{ |X|=k-1 }[/math] such that [math]\displaystyle{ {X\choose t} }[/math] is colored all red by [math]\displaystyle{ f }[/math]. Next we prove that in the later case [math]\displaystyle{ {X\cup{n}\choose t} }[/math] is all red, which will close our proof. Since all [math]\displaystyle{ A\in{S\choose t-1} }[/math] are colored with red by [math]\displaystyle{ f' }[/math], then by our definition of [math]\displaystyle{ f' }[/math], [math]\displaystyle{ f(A\cup\{n\})={\color{red}\text{red}} }[/math] for all [math]\displaystyle{ A\in {X\choose t-1}\subseteq{S\choose t-1} }[/math]. Recalling that [math]\displaystyle{ {X\choose t} }[/math] is colored all red by [math]\displaystyle{ f }[/math], [math]\displaystyle{ {X\cup\{n\}\choose t} }[/math] is colored all red by [math]\displaystyle{ f }[/math] and we are done. [math]\displaystyle{ \square }[/math]

Applications of Ramsey Theorem

The "Happy Ending" problem

The happy ending problem

Any set of 5 points in the plane, no three on a line, has a subset of 4 points that form the vertices of a convex quadrilateral.

See the article [1] for the proof.

We say a set of points in the plane in general positions if no three of the points are on the same line.

Theorem (Erdős-Szekeres 1935)

For any positive integer [math]\displaystyle{ m\ge 3 }[/math], there is an [math]\displaystyle{ N(m) }[/math] such that any set of at least [math]\displaystyle{ N(m) }[/math] points in general position in the plane (i.e., no three of the points are on a line) contains [math]\displaystyle{ m }[/math] points that are the vertices of a convex [math]\displaystyle{ m }[/math]-gon.

Proof. Let [math]\displaystyle{ N(m)=R_3(m,m) }[/math]. For [math]\displaystyle{ n\ge N(m) }[/math], let [math]\displaystyle{ X }[/math] be an arbitrary set of [math]\displaystyle{ n }[/math] points in the plane, no three of which are on a line. Define a 2-coloring of the 3-subsets of points [math]\displaystyle{ f:{X\choose 3}\rightarrow\{0,1\} }[/math] as follows: for any [math]\displaystyle{ \{a,b,c\}\in{X\choose 3} }[/math], let [math]\displaystyle{ \triangle_{abc}\subset X }[/math] be the set of points covered by the triangle [math]\displaystyle{ abc }[/math]; and [math]\displaystyle{ f(\{a,b,c\})=|\triangle_{abc}|\bmod 2 }[/math], that is, [math]\displaystyle{ f(\{a,b,c\}) }[/math] indicates the oddness of the number of points covered by the triangle [math]\displaystyle{ abc }[/math]. Since [math]\displaystyle{ |X|\ge R_3(m,m) }[/math], there exists a [math]\displaystyle{ Y\subseteq X }[/math] such that [math]\displaystyle{ |Y|=m }[/math] and all members of [math]\displaystyle{ {Y\choose 3} }[/math] are colored with the same value by [math]\displaystyle{ f }[/math]. We claim that the [math]\displaystyle{ m }[/math] points in [math]\displaystyle{ Y }[/math] are the vertices of a convex [math]\displaystyle{ m }[/math]-gon. If otherwise, by the definition of convexity, there exist [math]\displaystyle{ \{a,b,c,d\}\subseteq Y }[/math] such that [math]\displaystyle{ d\in\triangle_{abc} }[/math]. Since no three points are in the same line, [math]\displaystyle{ \triangle_{abc}=\triangle_{abd}\cup\triangle_{acd}\cup\triangle_{bcd}\cup\{d\} }[/math], where all unions are disjoint. Then [math]\displaystyle{ |\triangle_{abc}|=|\triangle_{abd}|+|\triangle_{acd}|+|\triangle_{bcd}|+1 }[/math], which implies that [math]\displaystyle{ f(\{a,b,c\}), f(\{a,b,d\}), f(\{a,c,d\}), f(\{b,c,d\})\, }[/math] cannot be equal, contradicting that all members of [math]\displaystyle{ {Y\choose 3} }[/math] have the same color. [math]\displaystyle{ \square }[/math]

Yao's lower bound on implicit data structures

Lemma

Let [math]\displaystyle{ n\ge 2 }[/math] be a power of 2 and [math]\displaystyle{ N\ge 2n }[/math]. Suppose the universe is [math]\displaystyle{ [N] }[/math] and the size of the data set is [math]\displaystyle{ n }[/math]. If the data structure is a sorted table, any search algorithm requires at least [math]\displaystyle{ \log n }[/math] accesses to the data structure in the worst case.

Proof. We will show by an adversarial argument that [math]\displaystyle{ \log n }[/math] accesses are required to search for the key value [math]\displaystyle{ x=n }[/math] from the universe [math]\displaystyle{ [N]=\{1,2,\ldots,N\} }[/math]. The construction of the adversarial data set [math]\displaystyle{ S }[/math] is by induction on [math]\displaystyle{ n }[/math]. For [math]\displaystyle{ n=2 }[/math] and [math]\displaystyle{ N\ge 2n-1=3 }[/math] it is easy to see that two accesses are necessary. Let [math]\displaystyle{ n\gt 2 }[/math]. Assume the induction hypothesis to be true for all smaller [math]\displaystyle{ n }[/math]; we will prove it for the size of data set [math]\displaystyle{ n }[/math], size of universe [math]\displaystyle{ N\ge 2n }[/math] and the search key [math]\displaystyle{ x=n }[/math]. Suppose that the first access position is [math]\displaystyle{ k }[/math]. The adversary chooses the table content [math]\displaystyle{ T[k] }[/math]. The adversary's strategy is: [math]\displaystyle{ \begin{align} T[k]= \begin{cases} k & k\le \frac{n}{2},\\ N-(n-k) & k\gt \frac{n}{2}. \end{cases} \end{align} }[/math] By symmetry, suppose it is the first case that [math]\displaystyle{ k\le \frac{n}{2} }[/math]. Then the key [math]\displaystyle{ x=n }[/math] may be in any position [math]\displaystyle{ i }[/math], where [math]\displaystyle{ n/2+1\le i\le n }[/math]. In fact, [math]\displaystyle{ T[ n/2+1] }[/math] through [math]\displaystyle{ T[n] }[/math] is a sorted table of size [math]\displaystyle{ n'=n/2 }[/math] which may contain any [math]\displaystyle{ n' }[/math]-subset of [math]\displaystyle{ \{n/2+1, n/2+2,\ldots,N\} }[/math], and hence, in particular, any [math]\displaystyle{ n' }[/math]-subset of the universe [math]\displaystyle{ U'=\{n/2+1, n/2+2,\ldots,N-n/2\} }[/math]. The size [math]\displaystyle{ N' }[/math] of [math]\displaystyle{ U' }[/math] satisfies [math]\displaystyle{ N'=N-n/2-n/2\ge 2n-n\ge 2n' }[/math], and the desired key [math]\displaystyle{ n }[/math] has the relative value [math]\displaystyle{ x'=n- n/2=n' }[/math] in the universe [math]\displaystyle{ U' }[/math]. By the induction hypothesis, [math]\displaystyle{ \log n'=-1+\log n }[/math] more accesses will be required. Hence the total number of accesses is at least [math]\displaystyle{ 1+\log n'=\log n }[/math]. If the first access is [math]\displaystyle{ k\gt \frac{n}{2} }[/math], we symmetrically get that [math]\displaystyle{ T[1] }[/math] through [math]\displaystyle{ T[n/2] }[/math] is a sorted table of size [math]\displaystyle{ n'=n/2 }[/math] which may contain any [math]\displaystyle{ n' }[/math]-subset of the universe [math]\displaystyle{ U'=\{n/2+1, n/2+2,\ldots,N-n/2\} }[/math]. The rest is the same. [math]\displaystyle{ \square }[/math]

We have seen that on a sorted table, there is no search algorithm outperforming the binary search in the worst case. Our question is:

Is there any other order than the increasing order, on which there is a better search algorithm?

An implicit data structure use no extra space in addition to the original data set, thus a data structure can only be represented implicitly by the order of the data items in the table. That is, each data set is stored as a permutation of the set. Formally, an implicit data structure is a function

[math]\displaystyle{ f:{U\choose n}\rightarrow[n!] }[/math],

where each [math]\displaystyle{ \pi\in[n!] }[/math] specify a permutation of the sorted table. Thus, the sorted table is the simplest implicit data structure, in which [math]\displaystyle{ f(S) }[/math] is the identity for all [math]\displaystyle{ S\in{U\choose n} }[/math].