It is shown that if G and H are arbitrary fixed graphs and n is sufficiently large, then Also, we prove that r ( K 1 +F, K,) 5 (m+o(l))&(n -+ GO) for any forest Fwhose largest component has m edges. Thus r(Fe, K,) 5 (1 + o(l))&, where Fe = K1 + CK2. We conjecture that r(Fe, K,) -&(n + cm).
On Book-Complete Graph Ramsey Numbers
β Scribed by Yusheng Li; C.C. Rousseau
- Publisher
- Elsevier Science
- Year
- 1996
- Tongue
- English
- Weight
- 274 KB
- Volume
- 68
- Category
- Article
- ISSN
- 0095-8956
No coin nor oath required. For personal study only.
β¦ Synopsis
It is shown that a graph of order N and average degree d that does not contain the book B m =K 1 +K 1, m as a subgraph has independence number at least Nf (d ), where f (x)t(log xΓx) (x Γ ). From this result we find that the book-complete graph Ramsey number satisfies r(B m , K n ) mn 2 Γlog(nΓe). It is also shown that for every tree T m with m edges, r(K
where T m is an arbitrary tree with m edges. (If a conjecture of Erdo s and So s is true, the second result can be strengthened to r(K 1 +T m , K n ) mn 2 Γlog (nΓe).) For fixed m, these bounds are of the right order of article no.
π SIMILAR VOLUMES
## Abstract The ramsey number of any tree of order __m__ and the complete graph of order __n__ is 1 + (__m__ β 1)(__n__ β 1).
## Abstract A formula is presented for the ramsey number of any forest of order at least 3 versus any graph __G__ of order __n__ β₯ 4 having clique number __n__ β 1. In particular, if __T__ is a tree of order __m__ β₯ 3, then __r(T, G)__ = 1 + (__m__ β 1)(__n__ β 2).
The Ramsey numbers M,,, n,P,, ..., n,P,), p > 2, are calculated. ## 1. Introduction One class of generalized Ramsey numbers that are known exactly are those for the graphs nP2 which consist of n disjoint paths of length 2; E. J. Cockayne and the author proved in 111 that d r(nlp2, ..., n d P 2 ) =
This paper establishes that the local k-Ramsey number R(K m , k -loc) is identical with the mean k-Ramsey number R(K m , k -mean). This answers part of a question raised by Caro and Tuza.