- Ramsey's theorem
:"This article goes into technical details quite quickly. For a slightly gentler introduction see

Ramsey theory ."In

combinatorics ,**Ramsey's theorem**states that in any colouring of the edges of a sufficiently largecomplete graph (that is, asimple graph in which an edge connects every pair of vertices), one will find monochromatic complete subgraphs. For 2 colours, Ramsey's theorem states that for any pair of positive integers ("r","s"), there exists a least positive integer "R"("r","s") such that for anycomplete graph on "R"("r","s") vertices, whose edges are coloured "red" or "blue", there exists either a complete subgraph on "r" vertices which is entirely blue, or a complete subgraph on "s" vertices which is entirely red. Here "R(r,s)" signifies an integer that depends on both "r" and "s". It is understood to represent the smallest integer for which the theorem holds.Ramsey's theorem is a foundational result in combinatorics. The first version of this result was proved by

F. P. Ramsey . This initiated the combinatorial theory, now calledRamsey theory , that seeks regularity amid disorder: general conditions for the existence of substructures with regular properties. In this application it is a question of the existence of**"monochromatic subsets**", that is, subsets of connected edges of just one colour.An extension of this theorem applies to any finite number of colours, rather than just two. More precisely, the theorem states that for any given number of colours "c", and any given integers "n"

_{1},...,"n_{c}", there is a number, "R"("n"_{1}, ..., "n_{c}"), such that if the edges of a complete graph oforder "R"("n"_{1}, ..., "n_{c}") are coloured with "c" different colours, then for some "i" between 1 and "c", it must contain a complete subgraph of order "n_{i}" whose edges are all colour "i". The special case above has "c" = 2 (and "n"_{1}= "r" and "n"_{2}= "s").

=Example: "R"(3,3)=6=Suppose the edges of a complete graph on 6 vertices are coloured red and blue. Pick a vertex "v". There are 5 edges incident to "v" and so (by the

pigeonhole principle ) at least 3 of them must be the same colour.Without loss of generality we can assume at least 3 of these edges, connecting to vertices "r", "s" and "t", are blue. (If not, exchange red and blue in what follows.) If any of the edges ("r", "s"), ("r", "t"), ("s", "t") are also blue then we have an entirely blue triangle. If not, then those three edges are all red and we have an entirely red triangle. Since this argument works for any colouring, "any" "K"_{6}contains a monochromatic "K"_{3}, and therefore that "R"(3,3) ≤ 6. The popular version of this is called thetheorem on friends and strangers .An alternate proof works by double counting. It goes as follows: Count the number of ordered triples of vertices "x", "y", "z" such that the edge ("xy") is red and the edge ("yz") is blue. Firstly, any given vertex will be the middle of either 0×5=0, 1×4=4 or 2 × 3 = 6 such triples. Therefore there are at most 6×6=36 such triples. Secondly, for any non-monochromatic triangle (

**xyz**), there exist precisely two such triples. Therefore there are at most 18 non-monochromatic triangles. Therefore at least 2 of the 20 triangles in the "K"_{6}are monochromatic.Conversely, it is possible to 2-colour a "K"

_{5}without creating any monochromatic "K"_{3}, showing that "R"(3,3) > 5. The unique colouring is shown to the right. Thus "R"(3,3) = 6.**Proof of the theorem**First we prove the theorem for the 2-colour case, by induction on "r" + "s".It is clear from the definition that for all "n", "R"("n", 1) = "R"(1, "n") = "1". This starts the induction.We prove that "R"("r", "s") exists by finding an explicit bound for it. By the inductive hypothesis "R"("r" − 1, "s") and "R"("r", "s" − 1) exist.

__Claim: "R"("r", "s") ≤ "R"("r" − 1, "s") + "R"("r", "s" − 1):__Consider a complete graph on "R"("r" − 1, "s") + "R"("r", "s" − 1) vertices.Pick a vertex "v" from the graph, and partition the remaining vertices into two sets "M" and "N", such that for every vertex "w", "w" is in "M" if ("v", "w") is blue, and "w" is in "N" if ("v", "w") is red.Because the graph has "R"("r" - 1, "s") + "R"("r", "s" - 1) = |"M"| + |"N"| + 1 vertices, it follows that either |"M"| ≥ "R"("r" − 1, "s") or |"N"| ≥ "R"("r", "s" − 1). In the former case, if "M" has a red "K

_{s}" then so does the original graph and we are finished. Otherwise "M" has a blue "K"_{"r"−1}and so $M\; cup\; \{v\}$ has blue "K_{r}" by definition of "M". The latter case is analogous.Thus the claim is true and we have completed the proof for 2 colours. We now prove the result for the general case of "c" colours. The proof is again by induction, this time on the number of colours "c". We have the result for "c" = 1 (trivially) and for "c" = 2 (above). Now let "c" > 2.

__Claim:__"R"("n"_{1}, ..., "n"_{"c"}) ≤ "R"("n"_{1}, ..., "n"_{"c"−2}, "R"("n"_{"c"−1}, "n"_{"c"}))Note, that the right hand side only contains Ramsey numbers for "c" − 1 colours and 2 colours, and therefore exists and is the finite number "t", by the inductive hypothesis. Thus, proving the claim will prove the theorem.

**Proof of claim**: Consider a graph on "t" vertices and colour its edges with "c" colours. Now 'go colour-blind' and pretend that "c" − 1 and "c" are the same colour. Thus the graph is now ("c" − 1)-coloured. By the inductive hypothesis, it contains either a "K_{n}_{i}" monochromatically coloured with colour "i" for some 1 ≤ "i" ≤ ("c" − 2) or a "K"_{"R"("n"}_{c−1,"n""c")}-coloured in the 'blurred colour'. In the former case we are finished. In the latter case, we recover our sight again and see from the definition of "R"("n"_{"c"−1}, "n"_{"c"}) we must have either a ("c" − 1)-monochrome "K"_{"n"}_{"c"−1}or a "c"-monochrome "K"_{"n"}_{"c"}. In either case the proof is complete.**Ramsey numbers**The numbers "R"("r","s") in Ramsey's theorem (and their extensions to more than two colours) are known as

**Ramsey numbers**. An upper bound for "R"("r","s") can be extracted from the proof of the theorem, and other arguments give lower bounds. (The first lower bound was obtained byPaul Erdős using theprobabilistic method .) However, there is a vast gap between the tightest lower bounds and the tightest upper bounds. Consequently, there are very few numbers "r" and "s" for which we know the exact value of "R"("r","s"). Computing a lower bound "L" for "R"("r","s") usually requires exhibiting a blue/red colouring of the graph "K"_{"L"−1}with no blue "K_{r}" subgraph and no red "K"_{"s"}subgraph. Searching all colourings of a graph "K_{n}" becomes computationally extremely difficult as "n" increases; the number of colourings grows super-exponentially.The complexity for searching all possible graphs is O(2

^{(n-1)(n-2)/2}) for an upper bound of "n" nodes. [*http://www.learner.org/channel/courses/mathilluminated/units/2/textbook/06.php*]At the time of writing, even the exact value of "R"(5,5) is unknown, although it is known to lie between 43 (Geoffrey Exoo) and 49(

Brendan McKay andStanisław Radziszowski ) (inclusive);barring a breakthrough in theory, it is probable that the exact value of "R"(6,6) will remain unknown forever."R"("r","s") for values of "r" and "s" up to 10 are shown in the table below. Where the exact value is unknown, the table lists the best known bounds. "R"("r","s") for values of "r" and "s" less than 3 are given by "R"(1,"s") = 1 and "R"(2,"s") = "s" for all values of "s". The standard survey on the development of Ramsey number research has been written by

Stanisław Radziszowski , who also found the exact value of R(4,5) (withBrendan McKay ).There is a trivial symmetry across the diagonal.

This table is extracted from a larger table compiled by

Stanisław Radziszowski [*http://www.combinatorics.org/Surveys/index.html*] .

=A Multicolour Example: "R"(3,3,3) = 17=A multicolour Ramsey number is a Ramsey number using 3 or more colours. There is only one nontrivial multicolour Ramsey number for which the exact value is known, namely "R"(3,3,3) = 17.

Suppose that you have an edge colouring of a complete graph using 3 colours, red, yellow and green. Suppose further that the edge colouring has no monochromatic triangles. Select a vertex "v". Consider the set of vertices which have a green edge to the vertex "v". This is called the green neighborhood of "v". The green neighborhood of "v" cannot contain any green edges, since otherwise there would be a green triangle consisting of the two endpoints of that green edge and the vertex "v". Thus, the induced edge colouring on the green neighborhood of "v" has edges coloured with only two colours, namely yellow and red. Since "R"(3,3) = 6, the green neighborhood of "v" can contain at most 5 vertices. Similarly, the red and yellow neighborhoods of "v" can contain at most 5 vertices each. Since every vertex, except for "v" itself, is in one of the green, red or yellow neighborhoods of "v", the entire complete graph can have at most 1 + 5 + 5 + 5 = 16 vertices. Thus, we have "R"(3,3,3) ≤ 17.

To see that "R"(3,3,3) ≥ 17, it suffices to draw an edge colouring on the complete graph on 16 vertices with 3 colours, which avoids monochromatic triangles. It turns out that there are exactly two such colourings on "K"

_{16}, the so-called untwisted and twisted colourings. Both colourings are shown in the figure to the right, with the untwisted colouring on the top, and the twisted colouring on the bottom. In both colourings in the figure, note that the vertices are labeled, and that the vertices "v"_{11}through "v"_{15}are drawn twice, on both the left and the right, in order to simplify the drawings.Thus, "R"(3,3,3) = 17.

If you select any colour of either the untwisted or twisted colouring on "K"

_{16}, and consider the graph whose edges are precisely those edges which have the specified colour, you will get the Clebsch Graph.It is known that there are exactly two edge colourings with 3 colours on "K"

_{15}which avoid monochromatic triangles, which can be constructed by deleting any vertex from the untwisted and twisted colourings on "K"_{16}, respectively.It is also known that there are exactly 115 edge colourings with 3 colours on "K"

_{14}which avoid monochromatic triangles, provided that we consider edge colourings which differ by a permutation of the colours as being the same.**Extensions of the theorem**The theorem can also be extended to

hypergraph s. An "m"-hypergraph is a graph whose "edges" are sets of "m" vertices - in a normal graph an edge is a set of 2 vertices. The full statement of Ramsey's theorem for hypergraphs is that for any integers "m" and "c",and any integers "n"_{1},...,"n"_{"c"},there is an integer "R"("n"_{1},...,"n"_{"c"};"c","m") such that if the hyperedges of a complete "m"-hypergraph of order "R"("n"_{1},...,"n"_{"c"};"c","m") are coloured with "c" different colours, then for some "i" between 1 and "c", the hypergraph must contain a complete sub-"m"-hypergraph of order "n"_{"i"}whose hyperedges are all colour "i". This theorem is usually proved by induction on "m", the 'hyper-ness' of the graph. The base case for the proof is "m"=2, which is exactly the theorem above.**Infinite Ramsey theory**A further result, also commonly called "Ramsey's theorem", applies to infinite graphs. In a context where finite graphs are also being discussed it is often called the "Infinite Ramsey theorem". As intuition provided by the pictorial representation of a graph is diminished when moving from finite to infinite graphs, theorems in this area are usually phrased in set-theoretic terminology.

**Theorem**: Let "X" be some countably infinite set and colour the elements of "X^{(n)}" (the subsets of "X" of size "n") in "c" different colours. Then there exists some infinite subset "M" of "X" such that the size "n" subsets of "M" all have the same colour.**Proof**: The proof is given for "c"=2. It is easy to prove the theorem for an arbitrary number of colours using a 'colour-blindness' argument as above. The proof is by induction on 'n', the size of the subsets. For "n" = 1,the statement is equivalent to saying that if you split an infinite set into two sets, one of them is infinite. This is evident. Assuming the theorem is true for "n" ≤ "r", we prove it for "n" = "r" + 1. Given a 2-colouring of the ("r" + 1)-element subsets of "X", let a_{0}be an element of "X" and let "Y" = "X""a"_{0}. We then induce a 2-colouring of the "r"-element subsets of "Y", by just adding "a"_{0}to each "r"-element subset (to get an ("r"+1)-element subset of "X"). By the induction hypothesis, there exists an infinite subset "Y"_{1}within "Y" such that every "r"-element subset of "Y" is coloured the same colour in the induced colouring. Thus there is an element "a"_{0}and an infinite subset "Y"_{1}such that all the ("r"+1)-element subsets of "X" consisting of "a"_{0}and "r" elements of "Y"_{1}have the same colour. By the same argument, there is an element "a"_{1}in "Y"_{1}and an infinite subset "Y"_{2}of "Y"_{1}with the same properties. Inductively, we obtain a sequence {"a"_{0},"a"_{1},"a"_{2},...} such that the colour of each ("r" + 1)-element subset ("a"_{"i"(1)},"a"_{"i"(2)},...,"a"_{"i"("r"+1)}) with "i"(1) < "i"(2) < ... < "i"("r" + 1) depends only on the value of "i"(1). Further, there are infinitely many values of "i"("n") such that this colour will be the same. Take these "a"_{"i"("n")}'s to get the desired monochromatic set.**Infinite version implies the finite**It is easy to deduce the finite Ramsey theorem from the infinite one using a

proof by contradiction . Suppose the finite Ramsey Theorem is false. Then there exists $c,n,T$ such that for every integer $k$, there exists a $c$-colouring of $[k]\; ^\{(n)\}$, without a monochromatic set of size $T$. Let $C\_k$ denote the $c$-colourings of $[k]\; ^\{(n)\}$ without a monochromatic set of size $T$.For any integer "k", given any colouring in $C\_\{k+1\}$, if we restrict the colouring to $[k]\; ^\{(n)\}$ (by ignoring the colour of all sets containing $k+1$), then we get a colouring in $C\_k$. Define $C^\{1\}\_k$ to be the colourings in $C\_k$ which are restrictions of colourings in $C\_\{k+1\}$. Since $C\_\{k+1\}$ is not empty, nor is $C^\{1\}\_k$.

Similarly, the restriction of any colouring in $C^\{1\}\_\{k+1\}$ is in $C^\{1\}\_k$, allowing us to define $C^\{2\}\_k$ as the set of all such restrictions, which we can see is not empty. Continue doing so, defining $C^\{n\}\_k$ for all integers $n,k$.

Now, for any integer $k$, $C\_ksupseteq\; C^1\_ksupseteq\; C^2\_ksupseteq\; dots$, and each set is non-empty. Furthermore, $|C\_k|le\; c^\{frac\{k!\}\{n!(k-n)!$, and hence $C\_k$ is finite. It follows that the intersection of all of these sets must be non-empty. Let $D\_k=C\_kcap\; C^1\_kcap\; C^2\_kcap\; dots$. Then everything in $D\_k$ is the restriction of something in $D\_\{k+1\}$. Therefore we can start with something in $D\_n$, and unrestrict to something in $D\_\{n+1\}$, and continue doing so, to get a colouring of $mathbb\; N^\{(n)\}$ without a monochromatic set of size $T$.

If we take a suitable topological viewpoint, this argument becomes a standard compactness argument showing that the infinite version of the theorem implies the finite version.

**Directed graph Ramsey numbers**It is also possible to define Ramsey numbers for "directed" graphs.(These were introduced by P. Erdős & L. Moser.)Let R(n) be the smallest number Q such that any complete graph withsingly-directed arcs (also called a "tournament") and with ≥Q nodescontains an acyclic (also called "transitive") n-node subtournament.

This is the directed-graph analogue of what (above) has been called R(n,n;2), thesmallest number Z such that any 2-colouring of the edges of a complete"un"directed graph with ≥Z nodes, contains a monochromatic complete graph on n nodes.(The directed analogue of the two possible arc "colours" is the two "directions" of the arcs,the analogue of "monochromatic" is "all arc-arrows point the same way," i.e. "acyclic.")

Indeed many find the directed graph problem to actually be"more" elegant than the unidirected one.We have R(0)=0, R(1)=1, R(2)=2, R(3)=4, R(4)=8, R(5)=14, R(6)=28, 32≤R(7)≤55, andR(8) is again a problem you do not want powerful aliens to pose.

**ee also***

Paris–Harrington theorem

*Sim (pencil game)

*Infinite Ramsey theory **References***F. P. Ramsey: "On a problem of formal logic", Proc. London Math. Soc. series 2, vol.

**30**(1930), pp. 264-286

*R. Graham , B. Rothschild, J.H. Spencer, "Ramsey Theory", John Wiley and Sons, NY (1990).

*G. Exoo, "A Lower Bound for R(5,5)", J. Graph Theory, 13 (1989), 97-98.**External links*** [

*http://www.ramseyathome.com/ramsey/ Ramsey@Home*] is adistributed computing project designed to find new lower bounds for various Ramsey numbers using a host of different techniques.

* [*http://www.combinatorics.org/Surveys/ds1/sur.pdf Radziszowski's survey of small Ramsey numbers*]

* [*http://RangeVoting.org/PuzzRamsey.html Survey of directed-graph Ramsey numbers*]

* [*http://mathworld.wolfram.com/RamseyNumber.html Ramsey Number - from MathWorld*] (contains lower and upper bounds up to R(19,19))

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