Complexity of the improper twin edge coloring of graphs

Let G be a graph whose each component has order at least 3. Let s:E(G)→Zks:E(G)→Zk for some integer k≥2k≥2 be an improper edge coloring of G (where adjacent edges may be assigned the same color). If the induced vertex coloring c:V(G)→Zkc:V(G)→Zk defined by c(v)=∑e∈Evs(e) in Zk,c(v)=∑e∈Evs(e) in Zk, (where the indicated sum is computed in ZkZk and EvEv denotes the set of all edges incident to v) results in a proper vertex coloring of G, then we refer to such a coloring as an improper twin k-edge coloring. The minimum k for which G has an improper twin k-edge coloring is called the improper twin chromatic index of G and is denoted by χ′it(G)χit′(G) . It is known that χ′it(G)=χ(G)χit′(G)=χ(G) , unless χ(G)≡2(mod4)χ(G)≡2(mod4) and in this case χ′it(G)∈{χ(G),χ(G)+1}χit′(G)∈{χ(G),χ(G)+1} . In this paper, we first give a short proof of this result and we show that if G admits an improper twin k-edge coloring for some positive integer k, then G admits an improper twin t-edge coloring for all t≥kt≥k ; we call this the monotonicity property. In addition, we provide a linear time algorithm to construct an improper twin edge coloring using at most k+1k+1 colors, whenever a k-vertex coloring is given. Then we investigate, to the best of our knowledge the first time in literature, the complexity of deciding whether χ′it(G)=χ(G)χit′(G)=χ(G) or χ′it(G)=χ(G)+1χit′(G)=χ(G)+1 , and we show that, just like in case of the edge chromatic index, it is NP-hard even in some restricted cases. Lastly, we exhibit several classes of graphs for which the problem is polynomial.