Not-All-Equal and 1-in-Degree Decompositions: Algorithmic Complexity and Applications

Abstract

A Not-All-Equal (NAE) decomposition of a graph G is a decomposition of the vertices of G into two parts such that each vertex in G has at least one neighbor in each part. Also, a 1-in-Degree decomposition of a graph G is a decomposition of the vertices of G into two parts A and B such that each vertex in the graph G has exactly one neighbor in part A. Among our results, we show that for a given graph G, if G does not have any cycle of length congruent to 2 mod 4, then there is a polynomial time algorithm to decide whether G has a 1-in-Degree decomposition. In sharp contrast, we prove that for every r, r ≥ 3, for a given r-regular bipartite graph G determining whether G has a 1-in-Degree decomposition is NP-complete. These complexity results have been especially useful in proving NP-completeness of various graph related problems for restricted classes of graphs. In consequence of these results we show that for a given bipartite 3-regular graph G determining whether there is a vector in the null-space of the 0,1-adjacency matrix of G such that its entries belong to {±1, ±2} is NP-complete. Among other results, we introduce a new version of Planar 1-in-3 SAT and we prove that this version is also NP-complete. In consequence of this result, we show that for a given planar (3, 4)-semiregular graph G determining whether there is a vector in the null-space of the 0,1-incidence matrix of G such that its entries belong to {±1, ±2} is NP-complete.

$(2/2/3)$-SAT problem and its applications in dominating set problems

“…The satisfiability problem is known to be NP-complete in general and for many restricted cases, for example see [5,6,7,8,11,19]. Finding the strongest possible restrictions under which the satisfiability problem remains NP-complete is important since this can make it easier to prove the NP-completeness of new problems by allowing easier reductions.…”