Page 2 Lecture Notes in Computer Science 2865 Edited by G. Goos ...

266 S. Bhadra and A. FerreiraThis stands directly from Property 1. As an example, take in Figure 4 the journeyfrom d to c, which uses vertex a that lies outside the SCC {b, c, d}.The main problem calls for decomposing the evolving digraph into all possibleSCC’s. Consider a subproblem COMPONENT defined as follows.COMPONENT: Given an evolving digraph G =(V G ,E G ) and an integer k,is there a SCC of size k?We shall subsequently demonstrate that COMPONENT is NP-Complete,thereby precluding a polynomial time algorithm for the decomposition problem,unless P=NP.Theorem 1. COMPONENT is in NP.Proof sketch: Given a subset V G ′ of V G and the integer k, we must have a meansof verifying in polynomial time if V G ′ is indeed a SCC of size k. First, verify that|V G ′| = k. Verifying that the subdigraph G ′ induced by V G ′ on G is stronglyconnected and maximum is possible in polynomial time from Proposition 1.We now define a strong reachability digraph for an evolving digraph G as anundirected graph S G =(V G ,E S ), where E S = {(v i ,v j )} if and only if (v i ,v j ) ∪(v j ,v i ) ∈ R G , the transitive closure digraph of G.To prove the NP-Completeness of COMPONENT we reduce the CLIQUEproblem to COMPONENT. CLIQUE is formally defined as follows: Given adigraph G =(V,E), and an integer k, is there a clique of size k in G?Lemma 1. Finding an SCC in G is equivalent to finding a maximal clique inS G , the strong connectivity graph of G.Proof: Directly from the definitions of strong reachability, SCC and maximalclique, we see that the SCC in G is equivalent to finding the maximal clique inS G .Theorem 2. CLIQUE can be reduced to COMPONENT in polynomial time.Proof sketch: Given an undirected graph G =(V,E) and the integer k, weconstruct an evolving digraph G =(V G ,E G ) as follows (cf. Figure 5):1. For each node u i ∈ V create a node v i ∈ V G ,anodeh ii ∈ V G , and arcs(v i ,h ii ), (h ii ,v i ) with arc schedule time 2;2. For each edge {u i ,u j }∈E, do(a) create nodes h ij ,h ji ∈ V G ,(b) create arcs (v i ,h ij ),(h ij ,v i ), and arcs (v j ,h ji ), (h ji ,v j ) with arc scheduletime 2,(c) create arcs (h ij ,v j ), (v j ,h ij ) and arcs (h ji ,v i ), (v i ,h ji ) with arc scheduletime 3.3. Create an SCC connecting all h-nodes. Label these arcs with schedule times1 and 4.By construction, S G contains a clique of size n ′ = |{(h ij ,h ii ):1≤ i, j ≤|V G |}| formed of the h-nodes alone. We can then prove that finding an SCC in Gis the same as finding a clique in G, since a clique of size k in G will correspondto a clique of size n ′ + k in S G , corresponding, in turn, to an SCC of size n ′ + kin G (via Lemma 1).