Combinatorics Ramsey theory

 a precocious British mathematician, philosopher and economist  a problem in mathematical logic: can one always find order in chaos? If so, how much? Just how large a slice of chaos do we need to be sure to find a particular amount of order in it?

 Among any six people, there are three any two of whom are friends, or there are three such that no two of them are friends.

 a complete graph is a simple undirected graph in which every pair of distinct vertices is connected by a unique edge. The complete graph on n vertices is denoted by K n

 A 2-coloured graph is a complete graph whose edges have been colored with 2 different colors

 First we choose any point, and note that five lines come out of it - one to each of the other five points

 With five lines, there must be at least three lines of one color. Let's suppose there are three red lines.So we have four points like this

 For the three remaining lines, if al least one of them is red, then it makes a red triangle together with point A (left). And if not, the three together make a blue triangle (right):

 the Ramsey number, R( s, t ), is the order of the smallest complete graph which, when 2- coloured, must contain a red K s or a blue K t  2 properties: ◦ For any positive integers s,t≥2, we have R(s,t) = R(t, s) ◦ For any positive integer s≥2, we have R(2,s ) = s

 Lemma: R(r, s) ≤ R(r − 1, s) + R(r, s − 1)  Theorem :For any two natural numbers, s and t, there exists a natural number, R ( s, t ) = n, such that any 2-coloured complete graph of order at least n, colored red and blue, must contain a monochromatic red K s or a blue K t

 Suppose we have 10 points, joined with red and blue lines in the usual way. Choose a point, and note that 9 lines come out of it. There must either be at least 6 red ones, or at least 4 blue ones

Case 1Case 2

 The above argument demonstrated that if any vertex is incident on 6 blue edges or 4 red edges, then the graph must contain a red K 3 or a blue K 4. In order to prevent this from happening in a coloring of K 9, in which each vertex is incident on 8 edges, every vertex must be incident specifically on 5 blue edges and 3 red edges. However, the total number of pairs (v,e) of vertices and incident blue edges would then be 9 *5 = 45, but the number of such pairs must be even, since for each blue edge e, there are exactly two vertices which are its endpoints. Thus, a coloring with the incidence properties described above is impossible, and some vertex has either 6 incident blue edges or 4 incident blue edges; thus any coloring of K 9 contains a red K 3 or a blue K 4

 This coloring shows K 8 may be 2-coloured such that it does not contain a red K 4 or a blue K 3 as a subgraph

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