- This article goes into technical details quite quickly. For a slightly gentler introduction see Ramsey theory.
In combinatorics, Ramsey's theorem states that in
colouring a large complete graph, one will find complete subgraphs all
of the same colour. In a precise statement, for any pair of positive integers (r,s), there exists an integer
R(r,s) such that for any complete 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.
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 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 homogeneous subsets, that is, subsets
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 colors c, and any given integers n1,...,nc, there
is a number, R(n1, ..., nc; c), such that if the edges of a complete
graph of order R(n1, ..., nc; c) are colored with c different
colors, then for some i between 1 and c, it must contain a complete subgraph of order ni
whose edges are all color i. The special case above has c = 2 (and n1 = r and
n2 = s).
Example: R(3,3;2) is 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 with an entirely red triangle. Since this argument works for any colouring any K6 contains a
monchromatic K3 and therefore that R(3,3;2) ≤ 6.
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
exists precisely two such triples. Therefore there are at most 18 non-monochromatic triangles. Therefore there are at least 2
monochromatic triangles.
Conversely, it is possible to 2-colour a K5 without creating any monochromatic K3,
showing that R(3,3;2) > 5. The unique coloring is shown to the right. Thus R(3,3;2) = 6.
Proof of the theorem
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 consider two subgraphs M
and N where a vertex w is in M if and only
if (v, w) is blue and is in N otherwise.
Now either |M| ≥ R(r −1,s) or |N| ≥
R(r,s −1), again by the pigeonhole principle. In the former case if M has a red
Ks then so does the original graph and we are finished. Otherwise M has a blue
Kr-1 and so M union {v} has blue Kr 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(n1,...,nc;c) ≤
R(n1,...,nc−2,R(nc-1,nc;2);c−1)
Proof: The right-hand side of the inequality exists by inductive hypothesis. Consider a graph on this many
vertices and colour it 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
Kni monochromatically coloured with colour i for some 1 ≤ i ≤ (c-2) or
a KR(nc−1,nc;2)-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(nc−1,nc;2) we must have either a
(c−1)-monochrome Knc−1 or a c-monochrome
Knc. 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. 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 KL-1 with no blue Kr subgraph and no red
Ks subgraph. Searching all colourings of a graph Kn becomes computationally
extremely difficult as n increases; the number of colourings grows super-exponentially.
At the time of writing, even the exact value of R(5,5) is unknown, although it is known to lie between 43 and 49
(inclusive), and, barring a breakthrough in theory, it is probably the case that the exact value of R(6,6) will remain
unknown forever.
- "Imagine an alien force, vastly more powerful than us landing on Earth and demanding the value of R(5, 5) or they will
destroy our planet. In that case, we should marshal all our computers and all our mathematicians and attempt to find the value.
But suppose, instead, that they asked for R(6, 6), we should attempt to destroy the aliens". - Paul Erdös
Extensions of the theorem
The theorem can also be extended to hypergraphs. 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
n1,...,nc, there is an integer
R(n1,...,nc;c,m) such that if the hyperedges of a
complete m-hypergraph of order
R(n1,...,nc;c,m) are colored with c
different colors, then for some i between 1 and c, the hypergraph must contain a complete
sub-m-hypergraph of order ni whose hyperedges are all color 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 pictoral 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 infinite set X, choose element a0 of X (note
that in the statement of the theorem we insisted that X was countably infinite, if X were a general
infinite set we would require the axiom of choice to make this
choice) and let Y = X\a0. We then induce a 2-colouring of the r-element subsets of
Y, by just adding a0 to each r-element subset (to get an (r+1)-element subset of
X. By the induction hypothesis, there exists an infinite subset Y1 within Y such that every
r-element subset of Y is coloured the same colour in the induced colouring. Therefore we have chosen an element
a0 and a subset Y1 such that every (r+1)-element subset of X consisting
of a0 and r elements of Y1 has the same colour. Continuing in this we can choose
a1 from Y1 and subset Y2 of Y1 with the same
properties. We end with a sequence {a0,a1,a2,...} such that the
colour of each (r+1)-element subset
(ai(1),ai(2),...,ai(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
ai(n)'s to get the desired monochromatic set.
Infinite version implies the finite
Since we can only colour finite objects a finite number of times, considering all possible colourings of X(n), we see that infinitely many agree on a given finite subset Y(n) where Y is a finite subset of X. This observation enables us to
deduce the finite Ramsey theorem from the infinite one using a proof by contradiction. For suppose the finite Ramsey theorem is false. Then we can colour
Y(n) for every finite Y and never have a monochromatic set of size
N, say. Call such a colouring a "bad" colouring. Let (Yn) be an
increasing sequence of finite sets whose union is the infinite set X and colour each with a "bad" colouring. Then since
infinitely many of the colourings agree on any given finite subset, we can use this to define a colouring of X(n) with no monochromatic set of size N, contradicting the infinite Ramsey
theorem.
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.
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