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Lecture 16: DFS, DAG, and Strongly Connected Components

Lecture 16: DFS, DAG, and Strongly Connected Components. Shang-Hua Teng. Directed Acyclic Graphs. A directed acyclic graph or DAG is a directed graph with no directed cycles:. DFS and DAGs. Theorem: a directed graph G is acyclic iff a DFS of G yields no back edges:

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Lecture 16: DFS, DAG, and Strongly Connected Components

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  1. Lecture 16:DFS, DAG, and Strongly Connected Components Shang-Hua Teng

  2. Directed Acyclic Graphs • A directed acyclic graph or DAG is a directed graph with no directed cycles:

  3. DFS and DAGs • Theorem: a directed graph G is acyclic iff a DFS of G yields no back edges: • => if G is acyclic, will be no back edges • Trivial: a back edge implies a cycle • <= if no back edges, G is acyclic • Proof by contradiction: G has a cycle   a back edge • Let v be the vertex on the cycle first discovered, and u be the predecessor of v on the cycle • When v discovered, whole cycle is white • Must visit everything reachable from v before returning from DFS-Visit() • So path from uv is graygray, thus (u, v) is a back edge

  4. Topological Sort • Topological sort of a DAG: • Linear ordering of all vertices in graph G such that vertex u comes before vertex v if edge (u, v)  G • Real-world application: Scheduling a dependent graph, find a feasible course plan for university studies

  5. A Topological Sort Algorithm Topological-Sort() { • Call DFS to compute finish time f[v] for each vertex • As each vertex is finished, insert it onto the front of a linked list • Return the linked list of vertices } • Time: O(V+E) • Correctness: need to prove that (u,v)  G  f[u]>f[v]

  6. Correctness of Topological Sort • Lemma: (u,v)  G  f[u] > f[v] • When (u,v) is explored, u is gray, consider the following cases: • v is gray  (u,v) is back edge. Can’t happen, if G is a DAG. • v if white  v becomes descendent of u  f[v] < f[u] (since must finish v before backtracking and finishing u) • v is black  v already finished  f[v] < f[u]

  7. Our Algorithm for Topological Sorting is correct

  8. Strongly Connected Directed graphs • Every pair of vertices are reachable from each other a g c d e f b

  9. Strongly-Connected Graph G is strongly connected if, for every u and v in V, there is some path from u to v and some path from v to u. Not Strongly Connected Strongly Connected

  10. Strongly-Connected Components A strongly connected component of a graph is a maximal subset of nodes (along with their associated edges) that is strongly connected. Nodes share a strongly connected component if they are inter-reachable.

  11. Strongly Connected Components a { a , c , g } g c { f , d , e , b } d e f b

  12. a { a , c , g } g c { f , d , e , b } d e f b Reduced Component Graph of Strongly Connected Components • Component graph GSCC=(VSCC, ESCC): one vertex for each component • (u, v) ESCC if there exists at least one directed edge from the corresponding components

  13. Strongly Connected Components

  14. Graph of Strongly Connected Components • Theorem: the Component graph GSCC=(VSCC, ESCC) is a DAG • Each component is maximal in the sense that no other vertices can be added to it. If GSCC=(VSCC, ESCC) is not a DAG, then one can merge components on along a circle of GSCC • Therefore, GSCC has a topological ordering

  15. Finding Strongly-Connected Components • Input: A directed graph G = (V,E) • Output: a partition of V into disjoint sets so that each set defines a strongly connected component of G • How should we compute the partition?

  16. Graph of Strongly Connected Components • Recall: Theorem: the Component graph GSCC=(VSCC, ESCC) is a DAG • Each component is maximal in the sense that no other vertices can be added to it. If GSCC=(VSCC, ESCC) is not a DAG, then one can merge components on along a circle of GSCC • Therefore, GSCC has a topological ordering

  17. DFS on G Topological Sort GSCC=(VSCC, ESCC) • Let U be a subset of V • If we output U in VSCC in the decreasing order of f[U], then we topologically sort GSCC • Lemma: Let U and U’ be distinct strongly connected component, suppose there is an edge (u,v) in E where u in U and v in U’. Then f[U] > f[U’]

  18. Proof of the Lemma Lemma: Let U and U’ be distinct strongly connected component, suppose there is an edge (u,v) in E where u in U and v in U’. Then f[U] > f[U’] Proof: Two cases • d[U] < d[U’], say x in U is the first vertex • d[U’] < d[U], say y is the first, but U is not reachable from y

  19. Transpose of a Digraph Transpose of G = (V,E): GT=(V, ET), where ET={(u, v): (v, u) E} If G is a DAG then GT is also a DAG If we print the topological order of G in the reverse order, then it is a topological order of GT

  20. Strongly-Connected Components Strongly-Connected-Components(G) • call DFS(G) to compute finishing times f[u] for each vertex u. • compute GT • call DFS(GT), but in the main loop of DFS, consider the vertices in order of decreasing f[u] • output the vertices of each tree in the depth-first forest of step 3 as a separate strongly connected component. The graph GT is the transpose of G, which is visualized by reversing the arrows on the digraph.

  21. Strong Components: example a b a b a4 b3 d c d c after step 1 c2 a4 d1 a4 b3 d1 c2 d1 c2 b3 df spanning forest for Gr Graph Gr

  22. Runtime Lines 1 and 3 are (E+V) due to DFS Line 2 involves creating an adjacency list or matrix, and it is also O(E+V) Line 4 is constant time So, SCC(G) is (E+V)

  23. node a d=13 f=14 node b d=11 f=16 node c d=1 f=10 node d d=8 f=9 node e d=12 f=15 node f d=3 f=4 node g d=2 f=7 node h d=5 f=6 Strongly-Connected Components DFS on G, starting at c.

  24. node a d=13 f=14 node b d=11 f=16 node c d=1 f=10 node d d=8 f=9 node a d=2 f=5 =b node b d=1 f=6 =NIL node c d=7 f=10 =NIL node d d=8 f=9 =c node e d=12 f=15 node f d=3 f=4 node g d=2 f=7 node h d=5 f=6 node e d=3 f=4 =a node f d=12 f=13 =g node g d=11 f=14 =NIL node h d=15 f=16 =NIL DFS on GT

  25. node a d=2 f=5 =b node b d=1 f=6 =NIL node c d=7 f=10 =NIL node d d=8 f=9 =c node e d=3 f=4 =a node f d=12 f=13 =g node g d=11 f=14 =NIL node h d=15 f=16 =NIL DFS on GT This is GT, labeled after running DFS(GT). In order of decreasing finishing time, process the nodes in this order: b e a c d g h f. GT:

  26. b c g h a b c d e f g h a d f abe cd e fg h Strongly-Connected Components These are the 4 trees that result, yielding the strongly connected components. Finally, merge the nodes of any given tree into a super-node, and draw links between them, showing the resultant acyclic component graph. Component Graph

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