Strongly Connected Components

Transcription

1 Strongly Connected Components Let G = (V, E) be a directed graph Write if there is a path from to in G Write if and is an equivalence relation: implies and implies s equivalence classes are called the strongly connected components (SCC s) of G For V, C( ) := s SCC 1

2 The component graph Idea: collapse each SCC s into a single node Formally: component graph G scc = (V scc, E scc ) V scc = the SCC s C 1,..., C k of G E scc = {(C, C j ) : = j, (, ) E for some C, C j } G : G scc : a b c d a b c d e f g h e f g h 2

3 Lemma 1. in G C( ) C( ) in G scc 3

4 Lemma 2. G scc is acyclic. Suppose there is a cycle. By definition, no self loops in G scc, so the cycle must contain two distinct nodes, say C( ) and C( ) Then we have C( ) C( ) and C( ) C( ) in G scc By Lemma 1, and in G Thus, C( ) = C( ) QED 4

5 An application Scheduling with constraints: We want to schedule a set of tasks Each task is represented by a node in a directed graph G Edges in G represent scheduling constraints: if and are distinct tasks, there is a path from to in G, and both and are performed, then must be performed before Each task has a profit associated with it: we want to schedule tasks to maximize profit 5

6 A solution: 1. Compute component graph, topologically sorted 2. For each SCC, select the task with maximum profit 3. Output the tasks from Step 2 in topological order G : G scc : a b c d a b c d e f g h e f g h Profits: = 1, b = 2, c = 3, etc. Optimal schedule: e, d, g, h 6

7 Special case: G is undirected (, ) E (, ) E SCC s are just called connected components The component graph consists of isolated nodes no edges between components Easy to compute: the trees in the DFS forest are the connected components 7

8 Computing SCC s: the Kosaraju/Sharir Algorithm For a graph G, let G T denote its transpose or reverse same as G but with all edges reversed G and G T have the same SCC s in fact, (G T ) scc = (G scc ) T Algorithm SCC(G): 1. call DFS(G), and order the nodes 1,..., n in order of decreasing finishing time (as in DFSTopSort) 2. compute G T 3. call DFS(G T ) but in the top-level loop, process in the order 1,..., n the trees in the DFS forest are the SCC s of G Running time: O( V + E ) 8

9 Example: G : 13/14 11/16 1/10 8/9 12/15 3/4 2/7 5/6 G T : 13/14 11/16 1/10 8/9 12/15 3/4 2/7 5/6 G T : 13/14 11/16 1/10 8/9 12/15 3/4 2/7 5/6 G T : 13/14 11/16 1/10 8/9 12/15 3/4 2/7 5/6 G T : 13/14 11/16 1/10 8/9 G T : 13/14 11/16 1/10 8/9 12/15 3/4 2/7 5/6 12/15 3/4 2/7 5/6 G scc : 13/14 11/16 1/10 8/9 12/15 3/4 2/7 5/6 9

10 Notation: let ƒ [ ] be the finish time in the first DFS, and let ƒ (U) := m x{ƒ [ ] : U} Lemma 3. Suppose (C, C ) E scc. Then ƒ (C) > ƒ (C ) Proof. In the first DFS, let be the first node discovered in C C Case 1: C C C By the White Path Theorem, all nodes in C C are descendents of in the DFS forest By the Parenthesis Theorem, ƒ [ ] = ƒ (C) > ƒ (C ) 10

11 Case 2: C C C By the White Path Theorem, all nodes in C are descendents of in the DFS forest By Lemma 2, there is no path from C to C in G scc, and so no node in C is reachable from so at time ƒ [ ], all nodes in C are still white ƒ (C) > ƒ [ ] = ƒ (C ). QED 11

12 Theorem. Algorithm SCC is correct. Proof. Let T 1,..., T l be the trees of the DFS forest created in step 3 Let C 1,..., C k be the SCC s, with ƒ (C ) > ƒ (C +1 ) G scc : C 1 C 2 C3 C 4 (G scc ) T : C 1 C 2 C3 C 4 12

13 (G scc ) T : C 1 C 2 C3 C 4 At step 3, we start with a vertex 1 in C 1 By White Path Theorem, all nodes in C 1 will be in T 1 By Lemma 3, in G T, there are no edges leaving C 1 the nodes of C 1 are exactly the nodes of T 1 13

14 (G scc ) T : C 1 C 2 C3 C 4 Next, we pick a node in C 2, and at this time, all nodes in C 1 are black, and all nodes in C 2,..., C k are white By White Path Theorem, T 2 contains all nodes in C 2, and by Lemma 3, T 2 contains no other nodes the nodes of C 2 are exactly the nodes of T 2 Proceeding by induction, we get T = C for = 1,..., l, and so k = l. QED 14

15 Representation of G scc Let C 1,..., C k be the SCC s Number the nodes 1.. k Standard adjacency list representation of G scc Also: An array mapping V to j {1,..., k}, where C j An array mapping j {1..., k} to a list representation of C j This can all be done in time O( V + E ), and we may assume that C 1,..., C k are already in topological order in fact Algorithm SCC outputs C 1,..., C k in topological order 15

16 Another application Problem: A graph G = (V, E) is called semi-connected if for all, V, or. Show how to test if G is semi-connected in time O( V + E ) 16

17 First consider the problem for DAG s Let 1,..., n be a topological sort of G Claim: G is semi-connected there is an edge +1 for each = 1.. n

18 Now consider a general graph Claim: G is semi-connected G scc is semi-connected (follows directly from Lemma 1) Algorithm: 1. Run algorithm SCC to get G scc (which is a DAG) 2. Test if G scc is semi-connected (as above) General principle: Try to reduce questions about graphs to questions about DAG s. 18

19 Another application Consider the gathering coins problem for a general directed graph Given a directed graph G = (V, E) On each node there are N[ ] coins Goal: find the max number of coins that can be gathered on any one path through G The path need not be simple, but once you pick up the coins on a node, they are gone 19

20 We already know how to solve this for a DAG For a general graph, we start by computing G scc For each SCC C, we assign to it C N[ ] coins Now run the DAG algorithm on G scc 20