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

Complexity of Connected Components in Evolving Graphs 265a321, 4,7cd65bFig. 4. Overlapping SCC’s.node in the network disappears for any reason, then upon rejoining the network,it will still have all the information that it had received before its disappearance.3.2 Verification of Strong Connectivity in FSDN’sGiven an FSDN network, we must determine if it is strongly connected. It isequivalent to the following proposition over the corresponding evolving digraph.Proposition 1. Given an evolving digraph G with N nodes and M links overa sequence of length T , it is possible to determine if it is strongly connected ornot in O(NM(logT + logN )) time steps.Proof sketch: The transitive closure of G is defined as the digraph R G =(V,E R ), where E R = {(v i ,v j ) : ∃ a journey J (v i ,v j )}. Hence, G is stronglyconnected if the underlying graph of R G is a complete graph. The verificationis executed simply and efficiently by forming the shortest journeys tree for eachnode in the network using the algorithm proposed in [2]. For N nodes, thealgorithm is repeated N times, for an overall time of O(NM(logT + logN )).3.3 Decomposition into SCC’sTarjan’s algorithm [4], based on the concept of forefathers in a depth-first searchtree over a digraph, is used to decompose standard digraphs into SCC’s. HoweverSCC’s in evolving digraphs have the following unique properties, which make itimpossible to use Tarjan’s algorithm.Property 1. Two different SCC’s can have common vertices.For example, consider the digraph given in Figure 4, where arcs are labeled withthe respective arc schedule times. From the definition of SCC’s we see that thereare two such components a, c, d and b, c, d which have the common vertices c, dbetween them.Property 2. For any two vertices in the SCC (respectively, o-SCC) there maybe journeys connecting them which use vertices outside the SCC (respectively,o-SCC).