Combinatorics (Fall 2010)/Ramsey theory
Ramsey's Theorem
Ramsey's Theorem - Let [math]\displaystyle{ k,\ell }[/math] be positive integers. Then there exists an integer [math]\displaystyle{ R(k,\ell) }[/math] satisfying:
- In any graph of [math]\displaystyle{ n\ge R(k,\ell) }[/math] vertices, there exists a clique of [math]\displaystyle{ k }[/math] vertices or an independent set of [math]\displaystyle{ \ell }[/math] vertices.
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'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 - [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]
Ramsey number
| [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 |
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.