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Sharp concentration of the equitable chromatic number of dense random graphs
Combinatorics, Probability and Computing ( IF 0.9 ) Pub Date : 2019-11-26 , DOI: 10.1017/s0963548319000397
Annika Heckel

An equitable colouring of a graph G is a vertex colouring where no two adjacent vertices are coloured the same and, additionally, the colour class sizes differ by at most 1. The equitable chromatic number χ=(G) is the minimum number of colours required for this. We study the equitable chromatic number of the dense random graph ${\mathcal{G}(n,m)}$ where $m = \left\lfloor {p\left( \matrix{ n \cr 2 \cr} \right)} \right\rfloor $ and 0 < p < 0.86 is constant. It is a well-known question of Bollobás [3] whether for p = 1/2 there is a function f(n) → ∞ such that, for any sequence of intervals of length f(n), the normal chromatic number of ${\mathcal{G}(n,m)}$ lies outside the intervals with probability at least 1/2 if n is large enough. Bollobás proposes that this is likely to hold for f(n) = log n. We show that for the equitable chromatic number, the answer to the analogous question is negative. In fact, there is a subsequence ${({n_j})_j}_{ \in {\mathbb {N}}}$ of the integers where $\chi_=({\mathcal{G}(n_j,m_j)})$ is concentrated on exactly one explicitly known value. This constitutes surprisingly narrow concentration since in this range the equitable chromatic number, like the normal chromatic number, is rather large in absolute value, namely asymptotically equal to n/(2logbn) where b = 1/(1 − p).

中文翻译:

密集随机图等色数的急剧集中

一个公平着色图的G是一种顶点着色,其中没有两个相邻顶点的颜色相同,此外,颜色类大小最多相差 1。等式色数 χ=(G) 是为此所需的最少颜色数。我们研究了稠密随机图的等色数${\mathcal{G}(n,m)}$在哪里$m = \left\lfloor {p\left( \matrix{ n \cr 2 \cr} \right)} \right\rfloor $和 0 <p< 0.86 是常数。Bollobás [3] 的一个众所周知的问题是:p= 1/2 有一个函数F(n) → ∞ 使得,对于任何长度的区间序列F(n), 的正常色数${\mathcal{G}(n,m)}$位于区间之外的概率至少为 1/2,如果n足够大。Bollobás 认为这可能会持续到F(n) = 日志n. 我们表明,对于公平色数,类似问题的答案是否定的。其实还有一个子序列${({n_j})_j}_{ \in {\mathbb {N}}}$的整数,其中$\chi_=({\mathcal{G}(n_j,m_j)})$集中在一个明确的已知值上。这构成了令人惊讶的窄浓度,因为在这个范围内,公平的色数,就像正常的色数一样,在绝对值上相当大,即渐近地等于n/(2logbn) 在哪里b= 1/(1 -p)。
更新日期:2019-11-26
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