Ramsey theory for graphs By Zhixuan Li

1. Ramsey theory for graphsBy Zhixuan Li Combinatorics Ramsey theory

2. Frank Plumpton Ramsey • 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?

3. Theorem on friends and strangers • 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.

4. 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 Kn

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

6. Proof of Theorem on friends and strangers • First we choose any point, and note that five lines come out of it - one to each of the other five points

7. 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

8. 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):

9. Ramsey number • the Ramsey number, R( s, t ) , is the order of the smallest complete graph which, when 2-coloured, must contain a red Ks or a blue Kt • 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

10. Ramsey’s Theorem • 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 Ks or a blue Kt

11. Proof of R(3,4) = 9

12. R(4,3) ≤ 10 • 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

13. Case 1 Case 2

14. R(4,3) ≤ 9 • 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 K3or a blue K4. In order to prevent this from happening in a coloring of K9, 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 K9contains a red K3or a blue K4

15. R(4,3) ≥ 8 • This coloring shows K8may be 2-coloured such that it does not contain a red K4 or a blue K3 as a subgraph

17. Reference • http://aleph.math.louisville.edu/teaching/2010SP-682/notes-100401.pdf • http://en.wikipedia.org/wiki/Theorem_on_friends_and_strangers • http://www.mathdb.org/notes_download/elementary/combinatorics/de_D7.pdf • http://math.mit.edu/~fox/MAT307-lecture05.pdf • http://highered.mcgraw-hill.com/sites/dl/free/0073383090/299355/Apps_Ch8.pdf