Combinatorics (Fall 2010)/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].

Ramsey number

Theorem

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

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, and let [math]\displaystyle{ k_{r-1}=R_t(2;k_{r-1},k_r) }[/math]. If [math]\displaystyle{ n\ge R_t(r-1;k_1,k_2,\ldots,k_{r-2},R_t(2;k_{r-1},k_r)) }[/math], then for any [math]\displaystyle{ r }[/math]-coloring of [math]\displaystyle{ {[n]\choose t} }[/math], there either exist an [math]\displaystyle{ i\in\{1,2,\ldots,r-2\} }[/math] and [math]\displaystyle{ S\subseteq[n] }[/math] with [math]\displaystyle{ |S|=k_i }[/math] such that [math]\displaystyle{ {S\choose t} }[/math] is monochromatically colored with the [math]\displaystyle{ i }[/math]th color; or exists a [math]\displaystyle{ T\subseteq[n] }[/math] with [math]\displaystyle{ |T|=R_t(2;k_{r-1},k_r) }[/math] such that [math]\displaystyle{ {T\choose t} }[/math] 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 an [math]\displaystyle{ X\subseteq T }[/math] with [math]\displaystyle{ |X|=k_{r-1} }[/math] such that [math]\displaystyle{ {X\choose t} }[/math] is monochromatically colored with the original [math]\displaystyle{ (r-1) }[/math]th color, or a [math]\displaystyle{ Y\subseteq T }[/math] with [math]\displaystyle{ |Y|=k_{r-1} }[/math] such that [math]\displaystyle{ {Y\choose t} }[/math] is monochromatically colored with the original [math]\displaystyle{ r }[/math]th color. This implies that [math]\displaystyle{ R_t(r;k_1,k_2,\ldots,k_r)\le n\le R_t(r-1;k_1,k_2,\ldots,k_{r-2},R_t(2;k_{r-1},k_r)) }[/math]. [math]\displaystyle{ \square }[/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{ n\ge 3 }[/math], there is an [math]\displaystyle{ N(n) }[/math] such that any set of at least [math]\displaystyle{ N(n) }[/math] points in general position in the plane (i.e., no three of the points are on a line) contains [math]\displaystyle{ n }[/math] points that are the vertices of a convex [math]\displaystyle{ n }[/math]-gon.

Yao's lower bound on implicit data structures

Lemma

Let [math]\displaystyle{ n\ge 2 }[/math] and [math]\displaystyle{ N\ge 2n-1 }[/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{ \lceil\log(n+1)\rceil }[/math] accesses to the data structure in the worst case.

Proof. We will show by an adversarial argument that [math]\displaystyle{ \lceil\log(n+1)\rceil }[/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_0\gt 2 }[/math]. Assume the induction hypothesis to be true for all [math]\displaystyle{ n\lt n_0 }[/math]; we will prove it for [math]\displaystyle{ n=n_0, m\ge 2n_0-1 }[/math] and [math]\displaystyle{ x=n_0 }[/math]. By symmetry, assume that the first access position [math]\displaystyle{ \ell }[/math] satisfies [math]\displaystyle{ \ell\le\lceil n_0/2\rceil }[/math]. The adversary answers [math]\displaystyle{ T[\ell]=\ell }[/math]. Then the key [math]\displaystyle{ x=n_0 }[/math] may be in any position [math]\displaystyle{ i }[/math], where [math]\displaystyle{ \lceil n_0/2\rceil+1\le i\le n_0 }[/math]. In fact, [math]\displaystyle{ T[\lceil n_0/2\rceil+1] }[/math] through [math]\displaystyle{ T[n_0] }[/math] is a sorted table of size [math]\displaystyle{ n'=\lfloor n_0/2\rfloor }[/math] which may contain any [math]\displaystyle{ n' }[/math]-subset of [math]\displaystyle{ \{\lceil n_0/2\rceil+1,\lceil n_0/2\rceil+2,\ldots,N\} }[/math], and hence, in particular, any subset of the universe [math]\displaystyle{ U'=\{\lceil n_0/2\rceil+1,\lceil n_0/2\rceil+2,\ldots,N-\lceil n_0/2\rceil\} }[/math]. The size [math]\displaystyle{ N' }[/math] of [math]\displaystyle{ U' }[/math] satisfies [math]\displaystyle{ N'=N-2\lceil n_0/2\rceil\ge (2n_0-1)-2\lceil n_0/2\rceil\ge 2\lfloor n_0/2\rfloor-1=2n'-1 }[/math], and the desired key [math]\displaystyle{ n_0 }[/math] has the relative value [math]\displaystyle{ x'=n_0-\lceil n_0/2\rceil=n' }[/math] in the universe [math]\displaystyle{ U' }[/math]. By the induction hypothesis, [math]\displaystyle{ \lceil\log(n'+1)\rceil }[/math] more accesses will be required. Hence the total number of accesses is at least [math]\displaystyle{ 1+\lceil\log(n'+1)\rceil=1+\lceil\log(\lfloor n_0/2\rfloor+1)\rceil\ge\lceil\log(n_0+1)\rceil }[/math]. [math]\displaystyle{ \square }[/math]

Ramsey-like Theorems

Van der Waerden's Theorem

Theorem (Van der Waerden 1927)

For every choice of positive integers [math]\displaystyle{ r }[/math] and [math]\displaystyle{ t }[/math], there exists an integer [math]\displaystyle{ W(r,t) }[/math] such that for every [math]\displaystyle{ r }[/math]-coloring of [math]\displaystyle{ [n] }[/math] where [math]\displaystyle{ n\ge W(r,t) }[/math], there exists a monochromatic arithmetic progression of length [math]\displaystyle{ t }[/math].

Hales–Jewett Theorem

Theorem (Hales-Jewett 1963)

Let [math]\displaystyle{ A }[/math] be a finte alphabet of [math]\displaystyle{ t }[/math] symbols and let [math]\displaystyle{ r }[/math] be a positive integer. Then there exists an integer [math]\displaystyle{ \mathrm{HJ}(r,t) }[/math] such that for every [math]\displaystyle{ r }[/math]-coloring of the cube [math]\displaystyle{ A^n }[/math] where [math]\displaystyle{ n\ge \mathrm{HJ}(r,t) }[/math], there exists a combinatorial line, which is monochromatic.

Theorem (Hales-Jewett 1963)

Let [math]\displaystyle{ A }[/math] be a finte alphabet of [math]\displaystyle{ t }[/math] symbols and let [math]\displaystyle{ m,r }[/math] be positive integers. Then there exists an integer [math]\displaystyle{ \mathrm{HJ}(m,r,t) }[/math] such that for every [math]\displaystyle{ r }[/math]-coloring of the cube [math]\displaystyle{ A^n }[/math] where [math]\displaystyle{ n\ge \mathrm{HJ}(r,t) }[/math], there exists a combinatorial [math]\displaystyle{ m }[/math]-space, which is monochromatic.