##Ramsey Theory

Example:
    Given 6 people,
    either there are 3 who all like each other,
    or there are 3 no two of whom like each other.

Abstract version:
    $K_n :=$ "complete graph on $n$ vertices"
    :	$= n$ points with an edge between each pair.

    Colour the edges of $K_6$ each either red or blue,
    then there's a red copy of $K_3$ or there's a blue copy of $K_3$;
    i.e. there is a monochromatic triangle.

    Denote this fact
    :	$K_6 --> K_3,K_3$

Proof:
    Pick a vertex $v_0$.
    Consider the 5 edges from it.
    3 of them are red or 3 of them are blue, since $5 > (3-1) + (3-1)$.
    Say 3 are red, and consider the 3 other vertices of these red edges.
    If the edges between them are all blue, they form a blue triangle and
    we're done.
    Else, some edge is red; but then it along with the edges from $v_0$ form a
    red triangle, and we're done.

Ramsey's Theorem for 2-coloured graphs:
    Given $n$ and $m$ positive integers,
    there exists $r$ such that for any red-blue colouring of the edges of $K_r$,
    there are $n$ vertices all edges between which are red
    or there are $m$ vertices all edges between which are blue.
Notation:
    We write
    :	$K_r --> K_n,K_m$
    to mean that $r$ has this property,
    and we let $r(m,n)$ ("the (m,n)th Ramsey number") be the least such $r$.
Remarks:
    We saw that $K_6 --> K_3,K_3$;
    it's easy to see that $K_5 -/> K_3,K_3$,
    so $r(3,3) = 6$.

    It has been shown that
    :	$r(3,4) = 9$
    :	$r(3,5) = 14$
    :	$r(4,4) = 18$

    $r(5,5)$ is unknown! All we know is
    :	$43 <= r(5,5) <= 49$.

    Erdös:
	"Suppose aliens invade the earth and threaten to obliterate it in a
	year's time unless human beings can find the Ramsey number for red
	five and blue five. We could marshal the world's best minds and
	fastest computers, and within a year we could probably calculate the
	value. If the aliens demanded the Ramsey number for red six and blue
	six, however, we would have no choice but to launch a preemptive
	attack."

Proof of Theorem:
    Suppose inductively that
    :	$K_b --> K_{n-1},K_m$
    and
    :	$K_c --> K_n,K_{m-1}$.

    We show that
    :	$K_{b+c} --> K_n,K_m$.

    So colour $K_{b+c}$, and suppose there's no red $K_n$ and no blue $K_m$.

    Pick a vertex $v_0$; consider the $b+c-1$ edges from it.
    Since $b+c-1 > (b-1)+(c-1)$,
    $b$ of the edges are red or $c$ of the edges are blue.

    Say $b$ are red.
	Consider the $K_b$ formed by the vertices these edges connect to $v_0$.
	By the inductive hypothesis, it contains a red $K_{n-1}$ or a blue $K_m$.
	If it contains a red $K_{n-1}$, adjoining $v_0$ yields a red $K_n$;
	contradiction.
	If it contains a blue $K_m$, then so does our original $K_{b+c}$;
	contradiction.

    A symmetrical argument applies in the case that $c$ of the edges from
    $v_0$ are blue.

Remark:
    This proof yields a recursive upper bound on the Ramsey numbers:
    :	$r(m,n) <= r(n-1,m) + r(n,m-1)$
    (but this is far from sharp).
