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ITCS 6114

ITCS 6114. Graph Algorithms. Depth-First Search. Depth-first search is another strategy for exploring a graph Explore “deeper” in the graph whenever possible Edges are explored out of the most recently discovered vertex v that still has unexplored edges

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ITCS 6114

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  1. ITCS 6114 Graph Algorithms 110/23/2019

  2. Depth-First Search • Depth-first search is another strategy for exploring a graph • Explore “deeper” in the graph whenever possible • Edges are explored out of the most recently discovered vertex v that still has unexplored edges • When all of v’s edges have been explored, backtrack to the vertex from which v was discovered 210/23/2019

  3. Depth-First Search • Vertices initially colored white • Then colored gray when discovered • Then black when finished 310/23/2019

  4. DFS(G) { for each vertex u  G->V { u->color = WHITE; } time = 0; for each vertex u  G->V { if (u->color == WHITE) DFS_Visit(u); } } DFS_Visit(u) { u->color = GREY; time = time+1; u->d = time; for each v  u->Adj[] { if (v->color == WHITE) DFS_Visit(v); } u->color = BLACK; time = time+1; u->f = time; } Depth-First Search: The Code 410/23/2019

  5. DFS(G) { for each vertex u  G->V { u->color = WHITE; } time = 0; for each vertex u  G->V { if (u->color == WHITE) DFS_Visit(u); } } DFS_Visit(u) { u->color = GREY; time = time+1; u->d = time; for each v  u->Adj[] { if (v->color == WHITE) DFS_Visit(v); } u->color = BLACK; time = time+1; u->f = time; } Depth-First Search: The Code What does u->d represent? 510/23/2019

  6. DFS(G) { for each vertex u  G->V { u->color = WHITE; } time = 0; for each vertex u  G->V { if (u->color == WHITE) DFS_Visit(u); } } DFS_Visit(u) { u->color = GREY; time = time+1; u->d = time; for each v  u->Adj[] { if (v->color == WHITE) DFS_Visit(v); } u->color = BLACK; time = time+1; u->f = time; } Depth-First Search: The Code What does u->f represent? 610/23/2019

  7. DFS(G) { for each vertex u  G->V { u->color = WHITE; } time = 0; for each vertex u  G->V { if (u->color == WHITE) DFS_Visit(u); } } DFS_Visit(u) { u->color = GREY; time = time+1; u->d = time; for each v  u->Adj[] { if (v->color == WHITE) DFS_Visit(v); } u->color = BLACK; time = time+1; u->f = time; } Depth-First Search: The Code Will all vertices eventually be colored black? 710/23/2019

  8. DFS(G) { for each vertex u  G->V { u->color = WHITE; } time = 0; for each vertex u  G->V { if (u->color == WHITE) DFS_Visit(u); } } DFS_Visit(u) { u->color = GREY; time = time+1; u->d = time; for each v  u->Adj[] { if (v->color == WHITE) DFS_Visit(v); } u->color = BLACK; time = time+1; u->f = time; } Depth-First Search: The Code What will be the running time? 810/23/2019

  9. DFS(G) { for each vertex u  G->V { u->color = WHITE; } time = 0; for each vertex u  G->V { if (u->color == WHITE) DFS_Visit(u); } } DFS_Visit(u) { u->color = GREY; time = time+1; u->d = time; for each v  u->Adj[] { if (v->color == WHITE) DFS_Visit(v); } u->color = BLACK; time = time+1; u->f = time; } Depth-First Search: The Code Running time: O(n2) because call DFS_Visit on each vertex, and the loop over Adj[] can run as many as |V| times 910/23/2019

  10. DFS(G) { for each vertex u  G->V { u->color = WHITE; } time = 0; for each vertex u  G->V { if (u->color == WHITE) DFS_Visit(u); } } DFS_Visit(u) { u->color = GREY; time = time+1; u->d = time; for each v  u->Adj[] { if (v->color == WHITE) DFS_Visit(v); } u->color = BLACK; time = time+1; u->f = time; } Depth-First Search: The Code BUT, there is actually a tighter bound. How many times will DFS_Visit() actually be called? 1010/23/2019

  11. DFS(G) { for each vertex u  G->V { u->color = WHITE; } time = 0; for each vertex u  G->V { if (u->color == WHITE) DFS_Visit(u); } } DFS_Visit(u) { u->color = GREY; time = time+1; u->d = time; for each v  u->Adj[] { if (v->color == WHITE) DFS_Visit(v); } u->color = BLACK; time = time+1; u->f = time; } Depth-First Search: The Code So, running time of DFS = O(V+E) 1110/23/2019

  12. Depth-First Sort Analysis • This running time argument is an informal example of amortized analysis • “Charge” the exploration of edge to the edge: • Each loop in DFS_Visit can be attributed to an edge in the graph • Runs once/edge if directed graph, twice if undirected • Thus loop will run in O(E) time, algorithm O(V+E) • Considered linear for graph, b/c adj list requires O(V+E) storage • Important to be comfortable with this kind of reasoning and analysis 1210/23/2019

  13. DFS Example sourcevertex 1310/23/2019

  14. DFS Example sourcevertex d f 1 | | | | | | | | 1410/23/2019

  15. DFS Example sourcevertex d f 1 | | | 2 | | | | | 1510/23/2019

  16. DFS Example sourcevertex d f 1 | | | 2 | | 3 | | | 1610/23/2019

  17. DFS Example sourcevertex d f 1 | | | 2 | | 3 | 4 | | 1710/23/2019

  18. DFS Example sourcevertex d f 1 | | | 2 | | 3 | 4 5 | | 1810/23/2019

  19. DFS Example sourcevertex d f 1 | | | 2 | | 3 | 4 5 | 6 | 1910/23/2019

  20. DFS Example sourcevertex d f 1 | 8 | | 2 | 7 | 3 | 4 5 | 6 | 2010/23/2019

  21. DFS Example sourcevertex d f 1 | 8 | | 2 | 7 | 3 | 4 5 | 6 | 2110/23/2019

  22. DFS Example sourcevertex d f 1 | 8 | | 2 | 7 9 | 3 | 4 5 | 6 | What is the structure of the grey vertices? What do they represent? 2210/23/2019

  23. DFS Example sourcevertex d f 1 | 8 | | 2 | 7 9 |10 3 | 4 5 | 6 | 2310/23/2019

  24. DFS Example sourcevertex d f 1 | 8 |11 | 2 | 7 9 |10 3 | 4 5 | 6 | 2410/23/2019

  25. DFS Example sourcevertex d f 1 |12 8 |11 | 2 | 7 9 |10 3 | 4 5 | 6 | 2510/23/2019

  26. DFS Example sourcevertex d f 1 |12 8 |11 13| 2 | 7 9 |10 3 | 4 5 | 6 | 2610/23/2019

  27. DFS Example sourcevertex d f 1 |12 8 |11 13| 2 | 7 9 |10 3 | 4 5 | 6 14| 2710/23/2019

  28. DFS Example sourcevertex d f 1 |12 8 |11 13| 2 | 7 9 |10 3 | 4 5 | 6 14|15 2810/23/2019

  29. DFS Example sourcevertex d f 1 |12 8 |11 13|16 2 | 7 9 |10 3 | 4 5 | 6 14|15 2910/23/2019

  30. DFS: Kinds of edges • DFS introduces an important distinction among edges in the original graph: • Tree edge: encounter new (white) vertex • The tree edges form a spanning forest • Can tree edges form cycles? Why or why not? 3010/23/2019

  31. DFS Example sourcevertex d f 1 |12 8 |11 13|16 2 | 7 9 |10 3 | 4 5 | 6 14|15 Tree edges 3110/23/2019

  32. DFS: Kinds of edges • DFS introduces an important distinction among edges in the original graph: • Tree edge: encounter new (white) vertex • Back edge: from descendent to ancestor • Encounter a grey vertex (grey to grey) 3210/23/2019

  33. DFS Example sourcevertex d f 1 |12 8 |11 13|16 2 | 7 9 |10 3 | 4 5 | 6 14|15 Tree edges Back edges 3310/23/2019

  34. DFS: Kinds of edges • DFS introduces an important distinction among edges in the original graph: • Tree edge: encounter new (white) vertex • Back edge: from descendent to ancestor • Forward edge: from ancestor to descendent • Not a tree edge, though • From grey node to black node 3410/23/2019

  35. DFS Example sourcevertex d f 1 |12 8 |11 13|16 2 | 7 9 |10 3 | 4 5 | 6 14|15 Tree edges Back edges Forward edges 3510/23/2019

  36. DFS: Kinds of edges • DFS introduces an important distinction among edges in the original graph: • Tree edge: encounter new (white) vertex • Back edge: from descendent to ancestor • Forward edge: from ancestor to descendent • Cross edge: between a tree or subtrees • From a grey node to a black node 3610/23/2019

  37. DFS Example sourcevertex d f 1 |12 8 |11 13|16 2 | 7 9 |10 3 | 4 5 | 6 14|15 Tree edges Back edges Forward edges Cross edges 3710/23/2019

  38. DFS: Kinds of edges • DFS introduces an important distinction among edges in the original graph: • Tree edge: encounter new (white) vertex • Back edge: from descendent to ancestor • Forward edge: from ancestor to descendent • Cross edge: between a tree or subtrees • Note: tree & back edges are important; most algorithms don’t distinguish forward & cross 3810/23/2019

  39. DFS: Kinds Of Edges • Thm 23.9: If G is undirected, a DFS produces only tree and back edges • Proof by contradiction: • Assume there’s a forward edge • But F? edge must actually be a back edge (why?) source F? 3910/23/2019

  40. DFS: Kinds Of Edges • Thm 23.9: If G is undirected, a DFS produces only tree and back edges • Proof by contradiction: • Assume there’s a cross edge • But C? edge cannot be cross: • must be explored from one of the vertices it connects, becoming a treevertex, before other vertex is explored • So in fact the picture is wrong…bothlower tree edges cannot in fact betree edges source C? 4010/23/2019

  41. DFS And Graph Cycles • Thm: An undirected graph is acyclic iff a DFS yields no back edges • If acyclic, no back edges (because a back edge implies a cycle • If no back edges, acyclic • No back edges implies only tree edges (Why?) • Only tree edges implies we have a tree or a forest • Which by definition is acyclic • Thus, can run DFS to find whether a graph has a cycle 4110/23/2019

  42. DFS And Cycles • How would you modify the code to detect cycles? DFS(G) { for each vertex u  G->V { u->color = WHITE; } time = 0; for each vertex u  G->V { if (u->color == WHITE) DFS_Visit(u); } } DFS_Visit(u) { u->color = GREY; time = time+1; u->d = time; for each v  u->Adj[] { if (v->color == WHITE) DFS_Visit(v); } u->color = BLACK; time = time+1; u->f = time; } 4210/23/2019

  43. DFS And Cycles • What will be the running time? DFS(G) { for each vertex u  G->V { u->color = WHITE; } time = 0; for each vertex u  G->V { if (u->color == WHITE) DFS_Visit(u); } } DFS_Visit(u) { u->color = GREY; time = time+1; u->d = time; for each v  u->Adj[] { if (v->color == WHITE) DFS_Visit(v); } u->color = BLACK; time = time+1; u->f = time; } 4310/23/2019

  44. DFS And Cycles • What will be the running time? • A: O(V+E) • We can actually determine if cycles exist in O(V) time: • In an undirected acyclic forest, |E|  |V| - 1 • So count the edges: if ever see |V| distinct edges, must have seen a back edge along the way 4410/23/2019

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