Data Structures and Algorithms in Java[1].pdf - Fulvio Frisone

once, and each edge is traversed exactly once (from its origin). Hence, if ns vertices
and ms edges are reachable from vertex s, a directed DFS starting at s runs in O(n s
+ m s ) time, provided the digraph is represented with a data structure that supports
constant-time vertex and edge methods. The adjacency list structure satisfies this
requirement, for example.
By Proposition 13.16, we can use DFS to find all the vertices reachable from a
given vertex, and hence to find the transitive closure of . That is, we can perform
a DFS, starting from each vertex v of , to see which vertices w are reachable
from v, adding an edge (v, w) to the transitive closure for each such w. Likewise, by
repeatedly traversing digraph with a DFS, starting in turn at each vertex, we can
easily test whether is strongly connected. Namely, is strongly connected if
each DFS visits all the vertices of
Thus, we may immediately derive the proposition that follows.
Proposition 13.17: Let be a digraph with n vertices and m edges. The
following problems can be solved by an algorithm that traverses n times using
DFS, runs in O (n(n+m)) time, and uses O(n) auxiliary space:
• Computing, for each vertex v of , the subgraph reachable from v
• Testing whether is strongly connected
• Computing the transitive closure of .
Testing for Strong Connectivity
Actually, we can determine if a directed graph is strongly connected much
faster than this, just using two depth-first searches. We begin by performing a
DFS of our directed graph starting at an arbitrary vertex s. If there is any
vertex of that is not visited by this DFS, and is not reachable from s, then the
graph is not strongly connected. So, if this first DFS visits each vertex of , then
we reverse all the edges of (using the reverse Direction method) and perform
another DFS starting at s in this "reverse" graph. If every vertex of is visited
by this second DFS, then the graph is strongly connected, for each of the vertices
visited in this DFS can reach s. Since this algorithm makes just two DFS
traversals of , it runs in O(n + m) time.
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