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Complexity of planar signed graph homomorphisms to cycles
François Dross, Florent Foucaud, Valia Mitsou, Pascal Ochem, Thé · 2019-07-07 · via cs.DS updates on arXiv.org

We study homomorphism problems of signed graphs. A signed graph is an undirected graph where each edge is given a sign, positive or negative. An important concept for signed graphs is the operation of switching at a vertex, which is to change the sign of each incident edge. A homomorphism of a graph is a vertex-mapping that preserves the adjacencies; in the case of signed graphs, we also preserve the edge-signs. Special homomorphisms of signed graphs, called s-homomorphisms, have been studied. In an s-homomorphism, we allow, before the mapping, to perform any number of switchings on the source signed graph. This concept has been extensively studied, and a full complexity classification (polynomial or NP-complete) for s-homomorphism to a fixed target signed graph has recently been obtained. Such a dichotomy is not known when we restrict the input graph to be planar (not even for non-signed graph homomorphisms). We show that deciding whether a (non-signed) planar graph admits a homomorphism to the square $C_t^2$ of a cycle with $t\ge 6$, or to the circular clique $K_{4t/(2t-1)}$ with $t\ge2$, are NP-complete problems. We use these results to show that deciding whether a planar signed graph admits an s-homomorphism to an unbalanced even cycle is NP-complete. (A cycle is unbalanced if it has an odd number of negative edges). We deduce a complete complexity dichotomy for the planar s-homomorphism problem with any signed cycle as a target. We also study further restrictions involving the maximum degree and the girth of the input signed graph. We prove that planar s-homomorphism problems to signed cycles remain NP-complete even for inputs of maximum degree~$3$ (except for the case of unbalanced $4$-cycles, for which we show this for maximum degree~$4$). We also show that for a given integer $g$, the problem for signed bipartite planar inputs of girth $g$ is either trivial or NP-complete.