Chromatic Number and Orientations of Graphs and Signed Graphs

Type: Article

Publication Date: 2018-10-22

Citations: 3

DOI: https://doi.org/10.11650/tjm/181005

Abstract

Assume $D$ is a digraph, and $D'$ is a spanning sub-digraph of $D$. We say $D'$ is a modulo-$k$ Eulerian sub-digraph of $D$ if for each vertex $v$ of $D'$, $d_{D'}^+(v) \equiv d_{D'}^-(v) \pmod{k}$. A modulo-$k$ Eulerian sub-digraph $D'$ of $D$ is special if for every vertex $v$, $d_D^+(v) = 0$ implies $d_{D'}^-(v) = 0$ and $d_{D'}^+(v) = d_D^+(v) > 0$ implies $d_{D'}^-(v) > 0$. We denote by $\operatorname{OE}_k(D)$ or $\operatorname{EE}_k(D)$ (respectively, $\operatorname{OE}_k^s(D)$ or $\operatorname{EE}_k^s(D)$) the sets of spanning modulo-$k$ Eulerian sub-digraphs (respectively, the sets of spanning special modulo-$k$ Eulerian sub-digraphs) of $D$ with an odd number or even number of edges. Matiyasevich [A criterion for vertex colorability of a graph stated in terms of edge orientations, (in Russia), Diskretnyi Analiz, issue 26, 65--71 (1974)] proved that a graph $G$ is $k$-colourable if and only if $G$ has an orientation $D$ such that $|\operatorname{OE}_k(D)| \neq |\operatorname{EE}_k(D)|$. In this paper, we give another characterization of $k$-colourable graphs: a graph $G$ is $k$-colourable if and only if $G$ has an orientation $D$ such that $|\operatorname{OE}_{k-1}^s(D)| \neq |\operatorname{EE}_{k-1}^s(D)|$. We extend the characterizations of $k$-colourable graphs to $k$-colourable signed graphs: If $k$ is an even integer, then a signed graph $(G,\sigma)$ is $k$-colourable if and only if $G$ has an orientation $D$ such that $|\operatorname{OE}_k(D)| \neq |\operatorname{EE}_k(D)|$; if $k$ is an odd integer, then $(G,\sigma)$ is $k$-colourable if and only if $G$ has an orientation $D$ such that $|\operatorname{OE}_{k-1}^s(D)| \neq |\operatorname{EE}_{k-1}^s(D)|$, where a (special) modulo-$k$ Eulerian sub-digraph is even or odd if it has an even or odd number of positive edges. The characterization of $k$-colourable signed graphs for even $k$ (respectively, for odd $k$) fails for odd $k$ (respectively, for even $k$).

Locations

  • Taiwanese Journal of Mathematics - View - PDF

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