Graphs omitting a finite set of cycles

Abstract

We prove that for C a finite set of cycles, there is a universal C-free graph if and only if C consists precisely of all the odd cycles of order less than same specified bound.The sufficiency of this condition was proved by Komjath, Mekler, and Pach (Israel J.

A Combinatorial Classic — Sparse Graphs with High Chromatic Number

“…On the other hand, such a universal graph exists for the class of all triangle-free graphs. (The same is true, more generally, for the class of all graphs not containing short odd cycles [18].) In fact one can "forbid" homomorphisms from any finite set of graphs, see [17].…”

Research Trends in Combinatorial Optimization

“…(Komjáth 1999;Komjáth and Mekler 1988;Cherlin and Shi 1996;Cameron 1998). (Komjáth 1999;Komjáth and Mekler 1988;Cherlin and Shi 1996;Cameron 1998).…”

Universal graphs with a forbidden subtree

“…1 First author supported by NSF Grant DMS 0100794. 2 We deal here with the problem of determining the finite connected constraint graphs C for which there is a countable universal C-free graph. We introduce a new inductive method and use it to settle the case in which C is a tree, confirming a long-standing conjecture of Tallgren.…”

“…In the other direction, the nonexistence of universal graphs has been treated in the following cases: (1) arrows, which are trees consisting of a path with two more edges adjoined to either endpoint, a case treated in [8]; (2) trees with a unique vertex of maximal degree d 4, which is moreover adjacent to a leaf, treated in [7]; and (3) "bushy" trees, that is trees with no vertex of degree 2, treated in [5]. Of these, the case treated in [7] now seems the most suggestive.…”

“…These phenomena remain invisible when one considers only single constraints, and a number of the more general examples are exceedingly natural (universal graphs omitting all cycles of odd order up to some fixed bound [2]; or universal graphs omitting a path and any further set of constraints).…”