On judicious partitions of graphs

Springer Science and Business Media LLC - Tập 31 - Trang 1383-1398 - 2015
Muhuo Liu1,2, Baogang Xu1
1Institute of Mathematics, School of Mathematical Science, Nanjing Normal University, Nanjing, China
2Department of Applied Mathematics, South China Agricultural University, Guangzhou, China

Tóm tắt

Let $$k, m$$ be positive integers, let $$G$$ be a graph with $$m$$ edges, and let $$h(m)=\sqrt{2m+\frac{1}{4}}-\frac{1}{2}$$ . Bollobás and Scott asked whether $$G$$ admits a $$k$$ -partition $$V_{1}, V_{2}, \ldots , V_{k}$$ such that $$\max _{1\le i\le k} \{e(V_{i})\}\le \frac{m}{k^2}+\frac{k-1}{2k^2}h(m)$$ and $$e(V_1, \ldots , V_k)\ge {k-1\over k} m +{k-1\over 2k}h(m) -\frac{(k-2)^{2}}{8k}$$ . In this paper, we present a positive answer to this problem on the graphs with large number of edges and small number of vertices with degrees being multiples of $$k$$ . Particularly, if $$d$$ is not a multiple of $$k$$ and $$G$$ is $$d$$ -regular with $$m\ge {9\over 128}k^4(k-2)^2$$ , then $$G$$ admits a $$k$$ -partition as desired. We also improve an earlier result by showing that $$G$$ admits a partition $$V_{1}, V_{2}, \ldots , V_{k}$$ such that $$e(V_{1},V_{2},\ldots ,V_{k})\ge \frac{k-1}{k}m+\frac{k-1}{2k}h(m)-\frac{(k-2)^{2}}{2(k-1)}$$ and $$\max _{1\le i\le k}\{e(V_{i})\}\le \frac{m}{k^{2}}+\frac{k-1}{2k^{2}}h(m)$$ .

Tài liệu tham khảo

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