Abstract

A graph is Hamiltonian if it contains a cycle passing through every vertex. One of the cornerstone results in the theory of random graphs asserts that for edge probability p≫log⁡nnp \gg \frac{\log n}{n}, the random graph G(n,p)G(n,p) is asymptotically almost surely Hamiltonian. We obtain the following strengthening of this result. Given a graph G=(V,E)G=(V,E), an {\em incompatibility system} F\mathcal{F} over GG is a family F={Fv}v∈V\mathcal{F}=\{F_v\}_{v\in V} where for every v∈Vv\in V, the set FvF_v is a set of unordered pairs Fv⊆{{e,e′}:e≠e′∈E,e∩e′={v}}F_v \subseteq \{\{e,e'\}: e\ne e'\in E, e\cap e'=\{v\}\}. An incompatibility system is {\em ΔΔ-bounded} if for every vertex vv and an edge ee incident to vv, there are at most ΔΔ pairs in FvF_v containing ee. We say that a cycle CC in GG is {\em compatible} with F\mathcal{F} if every pair of incident edges e,e′e,e' of CC satisfies {e,e′}∉Fv\{e,e'\} \notin F_v. This notion is partly motivated by a concept of transition systems defined by Kotzig in 1968, and can be used as a quantitative measure of robustness of graph properties. We prove that there is a constant μ>0μ>0 such that the random graph G=G(n,p)G=G(n,p) with p(n)≫log⁡nnp(n) \gg \frac{\log n}{n} is asymptotically almost surely such that for any μnpμnp-bounded incompatibility system F\mathcal{F} over GG, there is a Hamilton cycle in GG compatible with F\mathcal{F}. We also prove that for larger edge probabilities p(n)≫log⁡8nnp(n)\gg \frac{\log^8n}{n}, the parameter μμ can be taken to be any constant smaller than 1−121-\frac{1}{\sqrt 2}. These results imply in particular that typically in G(n,p)G(n,p) for p≫log⁡nnp \gg \frac{\log n}{n}, for any edge-coloring in which each color appears at most μnpμnp times at each vertex, there exists a properly colored Hamilton cycle.

Explore this paper’s authors, topics and related work.

Building this map preview

BZPEER is loading the nearby papers, people, topics and institutions for this page.

Reviews 0

Write a reviewWrite

No reviews yet.

Discussion 0

Add a commentWrite

No comments yet.