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CS161, Winter 2011 Handout #
Notes on Strongly Connected Components
Recall from Section 3.5 of the Kleinberg-Tardos book that the strongly connected components of a directed graph G are the equivalence classes of the following equivalence relation: u ∼ v if and only if there is a directed u v path and also there is a directed v u path. (Check that this is indeed an equivalence relation.) For example, in the directed graph in Figure 1, the strongly connected components are identified by the dashed circles.
Figure 1: The strongly connected components of a directed graph.
1 The Algorithm
Goal of Lecture: to give a linear-time (i.e., O(m+n)-time) algorithm that computes the strongly connected components of a directed graph. The algorithm we present is essentially two passes of depth-first search, plus some extremely clever additional book-keeping. The algorithm is described in a top-down fashion in Figures 2–4.
Input: a directed graph G = (V, E), in adjacency list representation. Assume that the vertices V are labeled 1 , 2 , 3 ,... , n.
- Let Grev^ denote the graph G after the orientation of all arcs have been reversed.
- Run the DFS-Loop subroutine on Grev^ , processing vertices according to the given order, to obtain a finishing time f (v) for each vertex v ∈ V.
- Run the DFS-Loop subroutine on G, processing vertices in decreasing order of f (v), to assign a leader to each vertex v ∈ V.
- The strongly connected components of G correspond to vertices of G that share a common leader.
Figure 2: The top level of our SCC algorithm. The f -values and leaders are computed in the first and second calls to DFS-Loop, respectively (see below).
Input: a directed graph G = (V, E), in adjacency list representation.
- Initialize a global variable t to 0. [This keeps track of the number of vertices that have been fully explored.]
- Initialize a global variable s to NULL. [This keeps track of the vertex from which the last DFS call was invoked.]
- For i = n downto 1:
[In the first call, vertices are labeled 1, 2 ,... , n arbitrarily. In the second call, vertices are labeled by their f (v)-values from the first call.]
(a) if i not yet explored: i. set s := i ii. DFS(G, i)
Figure 3: The DFS-Loop subroutine.
Input: a directed graph G = (V, E), in adjacency list representation, and a source vertex i ∈ V.
- Mark i as explored. [It remains explored for the entire duration of the DFS-Loop call.]
- Set leader(i) := s
- For each arc (i, j) ∈ G:
(a) if j not yet explored: i. DFS(G, j)
- t + +
- Set f (i) := t
Figure 4: The DFS subroutine. The f -values only need to be computed during the first call to DFS-Loop, and the leader values only need to be computed during the second call to DFS-Loop.
C 1
C 2
C 3
C 4
(a) SCC graph for Figure 1
C 3 C 2 C 1
(b) SCC graph for Figure 5(b)
Figure 6: The DAGs of the SCCs of the graphs in Figures 1 and 5(b), respectively.
Key Lemma: Consider two “adjacent” strongly connected components of a graph G: components C 1 and C 2 such that there is an arc (i, j) of G with i ∈ C 1 and j ∈ C 2. Let f (v) denote the finishing time of vertex v in some execution of DFS-Loop on the reversed graph Grev^. Then
max v∈C 1
f (v) < max v∈C 2
f (v).
Proof of Key Lemma: Consider two adjacent SCCs C 1 and C 2 , as they appear in the reversed graph Grev — where there is an arc (j, i), with j ∈ C 2 and i ∈ C 1 (Figure 7). Because the equivalence relation defining the SCCs is symmetric, G and Grev^ have the same SCCs; thus C 1 and C 2 are also SCCs of Grev^. Let v denote the first vertex of C 1 ∪ C 2 visited by DFS-Loop in Grev^. There are now two cases. First, suppose that v ∈ C 1 (Figure 7(a)). Since there is no non-trivial cycle of SCCs (Section 3.1), there is no directed path from v to C 2 in Grev^. Since DFS discovers everything reachable and nothing more, it will finish exploring all vertices in C 1 without reaching any vertices in C 2. Thus, every finishing time in C 1 will be smaller that every finishing time in C 2 , and this is even stronger than the assertion of the lemma. (Cf., the left and middle SCCs in Figure 5.) Second, suppose that v ∈ C 2 (Figure 7(b)). Since DFS discovers everything reachable and nothing more, the call to DFS at v will finish exploring all of the vertices in C 1 ∪ C 2 before ending. Thus, the finishing time of v is the largest amongst vertices in C 1 ∪ C 2 , and in particular is larger than all finishing times in C 1. (Cf., the middle and right SCCs in Figure 5.) This completes the proof.
C 1 C 2 i (^) j v
(a) All f -values in C 1 smaller than in C 2
C 1 C 2 i (^) j v
(b) v has the largest f -value in C 1 ∪ C 2
Figure 7: Proof of Key Lemma. Vertex v is the first in C 1 ∪ C 2 visited during the execution of DFS-Loop on Grev^.
3.3 The Final Argument
The Key Lemma says that traversing an arc from one SCC to another (in the original, unreversed graph) strictly increases the maximum f -value of the current SCC. For example, if fi denotes the largest f -value of a vertex in Ci in Figure 6(a), then we must have f 1 < f 2 , f 3 < f 4. Intuitively, when DFS-Loop is invoked
on G, processing vertices in decreasing order of finishing times, the successive calls to DFS peel off the SCCs of the graph one at a time, like layers of an onion. We now formally prove correctness of our algorithm for computing strongly connected components. Consider the execution of DFS-Loop on G. We claim that whenever DFS is called on a vertex v, the vertices explored — and assigned a common leader — by this call are precisely those in v’s SCC in G. Since DFS-Loop eventually explores every vertex, this claim implies that the SCCs of G are precisely the groups of vertices that are assigned a common leader. We proceed by induction. Let S denote the vertices already explored by previous calls to DFS (initially empty). Inductively, the set S is the union of zero or more SCCs of G. Suppose DFS is called on a vertex v and let C denote v’s SCC in G. Since the SCCs of a graph are disjoint, S is the union of SCCs of G, and v /∈ S, no vertices of C lie in S. Thus, this call to DFS will explore, at the least, all vertices of C. By the Key Lemma, every outgoing arc (i, j) from C leads to some SCC C′^ that contains a vertex w with a finishing time larger than f (v). Since vertices are processed in decreasing order of finishing time, w has already been explored and belongs to S; since S is the union of SCCs, it must contain all of C′. Summarizing, every outgoing arc from C leads directly to a vertex that has already been explored. Thus this call to DFS explores the vertices of C and nothing else. This completes the inductive step and the proof of correctness.
