notes-100401 - MATH 682 Notes Combinatorics and Graph...

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Unformatted text preview: MATH 682 Notes Combinatorics and Graph Theory II 1 Ramsey Theory 1.1 Classical Ramsey numbers Furthermore, there is a beautiful recurrence to give bounds on Ramsey numbers, but we will start with a simple but distinctly nontrivial example, to set the stage: Proposition 1. R (3 , 4) 10 . Proof. Consider a vertex v in a coloring of a K 10 . v has 9 incident edges, in some variety of colors. It must either have at least 6 incident blue edges, or 4 incident red edges (since otherwise it would have fewer than 8 incident edges total). If v is incident on 6 blue edges, consider the endpoints { v 1 ,...,v 6 } of those edges. Among these 6 vertices, a K 6 is colored red and blue; by our previous proposition, this K 6 contains either a red K 3 or a blue K 3 . A blue K 3 among vertices { v i ,v j ,v k } , together with the blue edges to v from each of these, will form a blue K 4 . Thus, if v has 6 blue edges incident on it, we are guaranteed either a blue K 4 or a red K 3 . On the other hand, if v is incident on 4 red edges, then either there is a red edge between the endpoints of these edges, producing a red K 3 , or the edges among all four of them are blue, producing a blue K 4 . We can actually improve this result slightly by maknig use of a parity argument. Corollary 1. R (3 , 4) 9 . Proof. 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 bound is in fact sharp, as can be seen by this example of a coloring of a K 8 without a red K 3 or blue K 4 . The above argument doesnt really use anything special about R (3 , 4), and can in fact be generalized to give a nice upper bound on R ( k, ): Page 1 of 6 April 1, 2010 MATH 682 Notes Combinatorics and Graph Theory II Theorem 1. R ( k, ) R ( k- 1 , ) + R ( k,- 1) . Furthermore, if R ( k- 1 , ) and R ( k,- 1) are both even, this bound can be reduced by 1. Proof. Let n = R ( k- 1 , ) + R ( k,- 1), for brevity. Consider a vertex v in a coloring of a K n . v has n- 1 incident edges, and must either have at least R ( k- 1 , ) incident red edges, or R ( k,- 1) incident blue edges (since otherwise it would have fewer than R ( k- 1 , )+ R ( k,- 1) incident edges total)....
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notes-100401 - MATH 682 Notes Combinatorics and Graph...

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