A Study on the Existence of Null Labelling for 3-Hypergraphs

Abstract

A 3-uniform hypergraph H consists of a set V of vertices, and E ⊆ V 3 triples. Let a null labelling be an assignment of ±1 to the triples such that each vertex has signed degree equal to zero. Assumed as necessary condition the degree of every vertex of H to be even, the Null Labelling Problem consists in determining whether H has a null labelling. Although the problem is NP-complete, the subclasses where the problem turns out to be polynomially solvable are of interest. In this study we define the notion of 2-intersection graph related to a 3-uniform hypergraph, and we prove that the existence of a Hamiltonian cycle there, is sufficient to obtain a null labelling in the related hypergraph. The proof we propose provides an efficient way of computing the null labelling.

Structure and Complexity of 2-Intersection Graphs of 3-Hypergraphs

“…In [8], this notion, restricted to 3-hypergraphs, is studied through their intersection graph and a sufficient condition for its existence was provided. Finally, in [6], it was shown that as the number of edges of the related 3-hypergraph increases, the usefulness of the intersection graph decreases, and the stronger notion of 2-intersection graph was introduced. In particular, the presence of a Hamiltonian cycle in the 2-intersection graph guarantees the existence of, and allows one to compute, the null labelling of the related 3-hypergraph.…”

“…In the sequel, the study focuses on 3-hypergraphs. In [6] the notion of 2-intersection graph was introduced to study the null label problem. The 2-intersection graph of a 3-hypergraph H is denoted I 2 (H ) = (V 2H , E 2H ).…”

“…Property 1 [6] The edges of a 3-hypergraph H sharing the same pair of vertices determine a clique in the 2-intersection graph I 2 (H ).…”

“…Proposition 1[6] Let H be a 3-hypergraph. If the 2-intersection graph I 2 (H ) is Hamiltonian, then H admits a null labelling.We observe that, provided an even graph G with n connected components and such that each component has an even number of edges, we can find a null labelling of G by considering, for each component, an Eulerian cycle and assigning alternatively ±1 to its edges.…”