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1 Strongly connected components. 2 Definition: the strongly connected components (SCC) C 1, …, C k of a directed graph G = (V,E) are the largest disjoint.

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Presentation on theme: "1 Strongly connected components. 2 Definition: the strongly connected components (SCC) C 1, …, C k of a directed graph G = (V,E) are the largest disjoint."— Presentation transcript:

1 1 Strongly connected components

2 2 Definition: the strongly connected components (SCC) C 1, …, C k 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 C i, 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|).

3 3 Example: strongly connected components d bfeac g h

4 4 d bfeac g h

5 5 Strongly connected components graph Definition: the SCC graph G ~ = (V ~,E ~ ) of the graph G = (V,E) is as follows: n V ~ = {C 1, …, C k }. Each SCC is a vertex. n E ~ = {(C i,C j )| i≠j and there is (x,y)  E, where x  C i and y  C j }. 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,E R ) of the graph G = (V,E) is G with its edge directions reversed: E R = {(u,v)| (v,u)  E}.

6 6 Example: reverse graph G R d bfeac g h d bfeac g h G G R

7 7 Example: SCC graph C1C1 C2C2 C4C4 C3C3 d bfe a c g h

8 8 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

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

10 10 DFS numbering DFS(G) 0. time  1 1Para todo v em G 2Se v não visitado então 3DFS-Visit(G, v) DFS-Visit(G, v) 1Marque v como visitado 1.5pre(v)  time; time++ 2 Para todo w em Adj(v) 3 Se w não visitado então 4 Insira aresta (v, w) na árvore 5DFS-Visit(G, w) 6Post(v)  time; time++

11 11 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 DFS(c) visit all nodes in D (they are reachable from c due to the existing edge) n Thus, post(c) > post(x) for every x in D

12 12 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) { "@context": "http://schema.org", "@type": "ImageObject", "contentUrl": "http://images.slideplayer.com/13/4047149/slides/slide_12.jpg", "name": "12 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.", "description": "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)

13 13 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 )~

14 14 SCC algorithm Idea: compute the SCC graph G ~ = (V ~,E ~ ) with two DFS, one for G and one for its reverse G R, 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.

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

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

17 17 Example: computing SCC d b f eac g h 8 1/6 d b f e a c g h 3/4 10/15 11/12 9/16 7/8 13/14 2/5 16 C1C1 C4C4 C2C2 C3C3 1 2 3 4


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