Corrigendum to "On pairwise compatibility graphs having Dilworth number two" [Theoret. Comput. Science (2014) 34-40]

In [4] we put in relation graphs with Dilworth number at most two and the two classes LPG and mLPG. In order to prove this relation, we have heavily exploited a result we deduced from [1]. We have now realized that this result is not always true, so in this corrigendum we correctly restate the result concerning the relation between graphs with Dilworth number at most two and the two classes of LPG and mLPG. A graph G = (V, E) is a leaf power graph, LPG (respectively minimum leaf power graph, mLPG) if there exists a tree T , a positive edge-weight function w on T , and a non-negative real number dmax (respectively dmin) such that V coincides with the set of leaves of T , and there is an edge (u, v) ∈ E if and only if dT,w(u, v) ≤ dmax (respectively dT,w(u, v) ≥ dmin), where dT,w(u, v) is the sum of the weights of the edges on the unique path from u to v in T . Property 1. [2] The complement of every graph in LPG is in mLPG and, conversely, the complement of every graph in mLPG is in LPG. The Dilworth number of a graph is the size of the largest subset of its nodes in which the close neighborhood of no node contains the open neighborhood of another. Graphs with Dilworth number at most two (also called threshold signed graphs) can be defined in several ways. Among them, we use the following [1]: a graph G = (V, E) has Dilworth number at most two if there are positive real numbers S ,T and for every node v there is a real weight a(v) < min(S ,T ) such that (v,w) is an edge if and only if either |a(v) + a(w)| ≥ S or |a(v) − a(w)| ≥ T . Notice that if S = T then the threshold signed graph is simply a threshold graph and their Dilworth number is exactly one. For any node v, if a(v) = 0 then v is an isolated node, so if (u, v) is an edge in a connected graph G, then a(u) · a(v) , 0. Consider an edge (u, v) of a threshold Preprint submitted to Elsevier August 24, 2015 signed graph. It is not difficult to see that only one of the two conditions concerning the thresholds can be satisfied; so, if a(u) · a(v) > 0, meaning that a(u) and a(v) have the same signs, then it must be that |a(u) + a(v)| ≥ S and the edge (u, v) is called S-edge; if, on the contrary, a(u) · a(v) < 0, that is a(u) and a(v) have different sign, then it must be that |a(u) − a(v)| ≥ T and the edge is called T-edge. We can hence consider the partition of the nodes of a connected graph with Dilworth number at most two G into two sets X and Y such that X = {x ∈ V s.t. a(x) < 0} and Y = {y ∈ V s.t. a(y) > 0}. As consequence, (u, v) is an S-edge (T-edge) if u and v are in the same (different) set X or Y . In the following we denote a graph with Dilworth number at most two by G = (X ∪ Y, a, S ,T ) when we want to highlight that it is given by means of its weight function a (determining the partition of V into sets X and Y) and the two thresholds S and T . We will call D1 the class of threshold graphs and D2 the wider class of graphs with Dilworth number at most two. The class D2 is self-complemented; given a graph G = (X ∪ Y, a, S ,T ) in D2, we indicate its complement by Ḡ = (X̄ ∪ Ȳ , ā, S̄ , T̄ ). The class of threshold graphs is in LPG ∩mLPG [5]. In [4], the class of graphs inD2\D1 with S > T is considered in relation to the two classes LPG and mLPG, so deducing the following results: Lemma 1. Let G = (X ∪ Y, a, S ,T ) be inD2. If S ≥ T then G is in mLPG. Lemma 2. Let G = (X ∪ Y, a, S ,T ) and Ḡ = (X̄ ∪ Ȳ , ā, S̄ , T̄ ) be a graph inD2 and its complement. If S̄ ≥ T̄ then G is in LPG. Arguing from [1] that if G is in D2 \ D1 then it can always be found a weight function a and two values S and T such that S > T , we deduced from the previous two lemmas that the whole classD2 was in LPG ∩ mLPG. Instead, the condition S > T cannot be always assumed, as shown by the following theorem. Theorem 1. Let be given a graph G = (X ∪ Y, a, S ,T ) in D2 \ D1. If there exist two S-edges (x1, x2) and (y1, y2) such that xi ∈ X, yi ∈ Y, i = 1, 2 and they induce a 2K2, then it cannot be S > T. Proof. First, notice that there exists a graph (shown in Figure 1) having the pair of edges (x1, x2), (y1, y2) as described in the statement. We prove the claim by contradiction, and hence we suppose S > T in this graph. W.l.o.g., assume that −a(x1) ≥ −a(x2) ≥ 0 and a(y1) ≥ a(y2) ≥ 0. The fact that (x1, x2) and (y1, y2) are S-edges in G and the two previous inequalities imply