Ramsey's theorem


In combinatorics, Ramsey's theorem, in one of its graph-theoretic forms, states that one will find monochromatic cliques in any edge labelling of a sufficiently large complete graph.
As the simplest example, consider two colours. Let and be any two positive integers. Ramsey's theorem states that there exists a least positive integer for which every blue-red edge colouring of the complete graph on vertices contains a blue clique on vertices or a red clique on vertices.
Ramsey's theorem is a foundational result in combinatorics. The first version of this result was proved by Frank Ramsey. This initiated the combinatorial theory now called Ramsey 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,, and any given integers, there is a number,, such that if the edges of a complete graph of order are coloured with different colours, then for some between 1 and, it must contain a complete subgraph of order whose edges are all colour. The special case above has .

Examples

''R''(3, 3) = 6

Suppose the edges of a complete graph on 6 vertices are coloured red and blue. Pick a vertex,. There are 5 edges incident to and so at least 3 of them must be the same colour. Without loss of generality we can assume at least 3 of these edges, connecting the vertex,, to vertices,, and, are blue. If any of the edges,,,, 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 contains a monochromatic, and therefore. The popular version of this is called the theorem on friends and strangers.
An alternative proof works by double counting. It goes as follows: Count the number of ordered triples of vertices,,,, such that the edge,, is red and the edge,, is blue. Firstly, any given vertex will be the middle of either , , or such triples. Therefore, there are at most such triples. Secondly, for any non-monochromatic triangle, 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 are monochromatic.
Conversely, it is possible to 2-colour a without creating any monochromatic, showing that. The unique colouring is shown to the right. Thus.
The task of proving that was one of the problems of William Lowell Putnam Mathematical Competition in 1953, as well as in the Hungarian Math Olympiad in 1947.

A multicolour example: ''R''(3, 3, 3) = 17

A multicolour Ramsey number is a Ramsey number using 3 or more colours. There are only two non-trivial multicolour Ramsey numbers for which the exact value is known, namely and.
Suppose that we have an edge colouring of a complete graph using 3 colours, red, green and blue. Suppose further that the edge colouring has no monochromatic triangles. Select a vertex. Consider the set of vertices that have a red edge to the vertex. This is called the red neighbourhood of. The red neighbourhood of cannot contain any red edges, since otherwise there would be a red triangle consisting of the two endpoints of that red edge and the vertex. Thus, the induced edge colouring on the red neighbourhood of has edges coloured with only two colours, namely green and blue. Since, the red neighbourhood of can contain at most 5 vertices. Similarly, the green and blue neighbourhoods of can contain at most 5 vertices each. Since every vertex, except for itself, is in one of the red, green or blue neighbourhoods of, the entire complete graph can have at most vertices. Thus, we have.
To see that, it suffices to draw an edge colouring on the complete graph on 16 vertices with 3 colours that avoids monochromatic triangles. It turns out that there are exactly two such colourings on, the so-called untwisted and twisted colourings. Both colourings are shown in the figures to the right, with the untwisted colouring on the left, and the twisted colouring on the right.
If we select any colour of either the untwisted or twisted colouring on, and consider the graph whose edges are precisely those edges that have the specified colour, we will get the Clebsch graph.
It is known that there are exactly two edge colourings with 3 colours on that avoid monochromatic triangles, which can be constructed by deleting any vertex from the untwisted and twisted colourings on, respectively.
It is also known that there are exactly 115 edge colourings with 3 colours on that avoid monochromatic triangles, provided that we consider edge colourings that differ by a permutation of the colours as being the same.
It is of interest to find the sequence of the multicolor Ramsey numbers, where there are 3's. The sequence currently is only known up to, with the bounds for values as early as being relatively loose:.

Proof

2-colour case

The theorem for the 2-colour case can be proved by induction on. It is clear from the definition that for all,. This starts the induction. We prove that exists by finding an explicit bound for it. By the inductive hypothesis and exist.
Proof. Consider a complete graph on vertices whose edges are coloured with two colours. Pick a vertex from the graph, and partition the remaining vertices into two sets and, such that for every vertex, is in if edge is blue, and is in if is red. Because the graph has vertices, it follows that either or In the former case, if has a red then so does the original graph and we are finished. Otherwise has a blue and so has a blue by the definition of. The latter case is analogous. Thus the claim is true and we have completed the proof for 2 colours.
In this 2-colour case, if and are both even, the induction inequality can be strengthened to:
Proof. Suppose and are both even. Let and consider a two-coloured graph of vertices. If is the degree of the -th vertex in the blue subgraph, then by the Handshaking lemma, is even. Given that is odd, there must be an even. Assume without loss of generality that is even, and that and are the vertices incident to vertex 1 in the blue and red subgraphs, respectively. Then both and are even. By the Pigeonhole principle, either or Since is even and is odd, the first inequality can be strengthened, so either or Suppose Then either the subgraph has a red and the proof is complete, or it has a blue which along with vertex 1 makes a blue. The case is treated similarly.

Case of more colours

Lemma 2. If, then
Proof. Consider a complete graph of vertices and colour its edges with colours. Now 'go colour-blind' and pretend that and are the same colour. Thus the graph is now -coloured. Due to the definition of such a graph contains either a mono-chromatically coloured with colour for some or a -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 we must have either a -monochrome or a -monochrome. In either case the proof is complete.
Lemma 1 implies that any is finite. The right hand side of the inequality in Lemma 2 expresses a Ramsey number for colours in terms of Ramsey numbers for fewer colours. Therefore, any is finite for any number of colours. This proves the theorem.

Ramsey numbers

The numbers in Ramsey's theorem are known as Ramsey numbers. The Ramsey number gives the solution to the party problem, which asks the minimum number of guests,, that must be invited so that at least will know each other or at least will not know each other. In the language of graph theory, the Ramsey number is the minimum number of vertices,, such that all undirected simple graphs of order, contain a clique of order, or an independent set of order. Ramsey's theorem states that such a number exists for all and.
By symmetry, it is true that. An upper bound for can be extracted from the proof of the theorem, and other arguments give lower bounds. However, there is a vast gap between the tightest lower bounds and the tightest upper bounds. There are also very few numbers and for which we know the exact value of.
Computing a lower bound for usually requires exhibiting a blue/red colouring of the graph with no blue subgraph and no red subgraph. Such a counterexample is called a Ramsey graph. Brendan McKay maintains a list of known Ramsey graphs. Upper bounds are often considerably more difficult to establish: one either has to check all possible colourings to confirm the absence of a counterexample, or to present a mathematical argument for its absence.

Computational complexity

A sophisticated computer program does not need to look at all colourings individually in order to eliminate all of them; nevertheless it is a very difficult computational task that existing software can only manage on small sizes. Each complete graph has edges, so there would be a total of graphs to search through if brute force is used. Therefore, the complexity for searching all possible graphs is for colourings and at most nodes.
The situation is unlikely to improve with the advent of quantum computers. One of the best-known searching algorithms for unstructured datasets exhibits only a quadratic speedup relative to classical computers, so that the computation time is still exponential in the number of nodes.

Known values

As described above,. It is easy to prove that, and, more generally, that for all : a graph on nodes with all edges coloured red serves as a counterexample and proves that ; among colourings of a graph on nodes, the colouring with all edges coloured red contains a -node red subgraph, and all other colourings contain a 2-node blue subgraph
Using induction inequalities and the handshaking lemma, it can be concluded that, and therefore. There are only two graphs among different 2-colourings of 16-node graphs, and only one graph among colourings. It follows that.
The fact that was first established by Brendan McKay and Stanisław Radziszowski in 1995.
The exact value of is unknown, although it is known to lie between 43 and 46, inclusive.
In 1997, McKay, Radziszowski and Exoo employed computer-assisted graph generation methods to conjecture that. They were able to construct exactly 656 graphs, arriving at the same set of graphs through different routes. None of the 656 graphs can be extended to a graph.
For with, only weak bounds are available. Lower bounds for and have not been improved since 1965 and 1972, respectively.
with are shown in the table below. Where the exact value is unknown, the table lists the best known bounds. with are given by and for all values of.
The standard survey on the development of Ramsey number research is the Dynamic Survey 1 of the Electronic Journal of Combinatorics, by Stanisław Radziszowski, which is periodically updated. Where not cited otherwise, entries in the table below are taken from the June 2024 edition.
12345678910
1
2
3
4
5
6
7
8
9
10

It is also interesting that Erdos showed
R =. + 1,
for a path and a complete graph with n and m vertices respectively. Also Chvatal showed
R =. + 1,
for a tree and a complete graph with n and m vertices respectively. These two theorems are the best examples of formulating Ramsey numbers for some special graphs.