Strongly connected components 1 Strongly connected components Definition

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Strongly connected components 1

Strongly connected components 1

Strongly connected components Definition: the strongly connected components (SCC) C 1, …, Ck of

Strongly connected components Definition: the strongly connected components (SCC) C 1, …, Ck of a directed graph G = (V, E) are the largest disjoint sub-graphs (no common vertices or edges) such that for any two vertices u and v in Ci, there is a path from u to v and from v to u. Equivalence classes of the binary path(u, v) relation, denoted by u ~ v. Applications: networking, communications. Problem: compute the strongly connected components of G in linear time Θ(|V|+|E|). 2

Example: strongly connected components b a e d c h g f 3

Example: strongly connected components b a e d c h g f 3

Example: strongly connected components b a e d c h g f 4

Example: strongly connected components b a e d c h g f 4

Strongly connected components graph Definition: the SCC graph G~ = (V~, E~) of the

Strongly connected components graph Definition: the SCC graph G~ = (V~, E~) of the graph G = (V, E) is as follows: n n V~ = {C 1, …, Ck}. Each SCC is a vertex. E~ = {(Ci, Cj)| i≠j and there is (x, y) E, where x Ci and y Cj}. A directed edge between components corresponds to a directed edge between them from any of their vertices. G~ is a directed acyclic graph (no directed cycles)! Definition: the reverse graph G R = (V, ER) of the graph G = (V, E) is G with its edge directions reversed: E R = {(u, v)| (v, u) E}. 5

Example: reverse graph G R G b a e d c g h GR

Example: reverse graph G R G b a e d c g h GR f b a e d c h g f 6

Example: SCC graph b a e d c C 2 h C 1 C

Example: SCC graph b a e d c C 2 h C 1 C 4 g f C 3 7

SCC algorithm: Approach H G~ For i=k, . . . , 1 v node

SCC algorithm: Approach H G~ For i=k, . . . , 1 v node of G that lies in a sink node of H (*) C connected component retuned by DFS(v) H H-C End n If we manage to execute (*) we are done 8

DFS numbering Useful information n n pre(v): “time” when a node is visited in

DFS numbering Useful information n n pre(v): “time” when a node is visited in a DFS post(v): “time” when DFS(v) finishes 2/15 1/16 3/14 b a e 8/11 d c 9/10 g 7/12 h f 5/6 4/13 9

DFS numbering DFS(G) 0. time 1 1 Para todo v em G 2 Se

DFS numbering DFS(G) 0. time 1 1 Para todo v em G 2 Se v não visitado então 3 DFS-Visit(G, v) 1 Marque v como visitado 1. 5 pre(v) time; time++ 2 Para todo w em Adj(v) 3 Se w não visitado então 4 Insira aresta (v, w) na árvore 5 DFS-Visit(G, w) 6 Post(v) time; time++ 10

DFS numbering Property If C an D are strongly connected components and there is

DFS numbering Property If C an D are strongly connected components and there is an edge from a node in C to a node in D, then the highest post() in C is bigger than the highest post() number in D Proof. Let c be the first node visited in C and d be the first node visited in D. Case i) c is visited before d. n n DFS(c) visit all nodes in D (they are reachable from c due to the existing edge) Thus, post(c) > post(x) for every x in D 11

DFS numbering Property If C an D are strongly connected components and there is

DFS numbering Property If C an D are strongly connected components and there is an edge from a node in C to a node in D, then the highest post() in C is bigger than the highest post() number in D Proof. Let c be the first node visited in C and d be the first node visited in D Case 2) d is visited before c. n DFS(d) visit all nodes in D because all of them are reachable from d n DFS(d) does not visit nodes from C since they are not reachable from d. n Thus , post(x) <post(y) for every pair of nodes x, y, with x in D and y in C 12

DFS numbering Corollary. The node of G with highest post number lies in a

DFS numbering Corollary. The node of G with highest post number lies in a source node in G~ Observation 1. The strongly connected components are the same in G and G R Observation 2. If a node lies in a source node of G~ then it lies in a sink node in (G R)~ 13

SCC algorithm Idea: compute the SCC graph G~ = (V~, E~) with two DFS,

SCC algorithm Idea: compute the SCC graph G~ = (V~, E~) with two DFS, one for G and one for its reverse GR, visiting the vertices in reverse order. SCC(G) 1. DFS(G) to compute post[v], ∀v V 2. Compute G R 3. DFS(G R) in the order of decreasing post[v] 4. Output the vertices of each tree in the DFS forest as a separate SCC. 14

2/5 1/6 b a Example: computing SCC 3/4 e 9/16 d c 7/8 g

2/5 1/6 b a Example: computing SCC 3/4 e 9/16 d c 7/8 g 11/12 h f 13/14 10/15 15

2/5 1/6 b a Example: computing SCC 3/4 e 9/16 d c 7/8 g

2/5 1/6 b a Example: computing SCC 3/4 e 9/16 d c 7/8 g 11/12 h f 13/14 10/15 b e 16 a d c 8 g h f 16

2/5 1/6 b a Example: computing SCC 3/4 e 9/16 d c 7/8 g

2/5 1/6 b a Example: computing SCC 3/4 e 9/16 d c 7/8 g 11/12 h f C 1 13/14 10/15 b e d c 8 3 C 4 4 C 2 C 3 1 16 a 2 g h f 17