On the algorithmic complexity of determining the AVD and NSD chromatic indices of graphs

Abstract

A proper k-edge-colouring ϕ of a graph G is an assignment of colours from {1,…,k} to the edges of G such that no two adjacent edges receive the same colour. If, additionally, ϕ guarantees that no two adjacent vertices of G are incident to the same sets or sums of colours, then ϕ is called an AVD or NSD edge-colouring, respectively (the abbreviations AVD and NSD standing for “adjacent vertex distinguishing” and “neighbour sum distinguishing”). The chromatic index χ′(G) of G is the smallest k such that proper k-edge-colourings of G exist. Similarly, the AVD and NSD chromatic indices χAVD′(G) and χNSD′(G) of G are the smallest k such that AVD and NSD k-edge-colourings of G exist, respectively. These chromatic parameters are quite related, as we always have χNSD′(G)≥χAVD′(G)≥χ′(G).By a well-known result of Vizing, we know that, for any graph G, we must have χ′(G)∈{Δ(G),Δ(G)+1}. Still, determining χ′(G) is NP-hard in general. Regarding χAVD′(G) and χNSD′(G), it is conjectured that, in general, they should always lie in {Δ(G),Δ(G)+1,Δ(G)+2}.In this work, we prove that determining whether a given graph G has AVD or NSD chromatic index Δ(G) is NP-hard for every Δ(G)≥3. We also prove that, for a given graph G, determining whether the AVD or NSD chromatic index is Δ(G)+1 is NP-hard for every Δ(G)≥3. Through other NP-hardness results, we also establish that there are infinitely many graphs for which the AVD and NSD chromatic indices are different. We actually come up, for every k≥4, with infinitely many graphs with maximum degree k, AVD chromatic index k, and NSD chromatic index k+1, and similarly, for every k≥3, with infinitely many graphs with maximum degree k, AVD chromatic index k+1, and NSD chromatic index k+2. In both cases, recognising graphs having those properties is actually NP-hard.