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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 16/7/2014

  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 26/7/2014

  3. Depth-First Search • Vertices initially colored white • Then colored gray when discovered • Then black when finished 36/7/2014

  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 46/7/2014

  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? 56/7/2014

  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? 66/7/2014

  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? 76/7/2014

  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? 86/7/2014

  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 96/7/2014

  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? 106/7/2014

  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) 116/7/2014

  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 126/7/2014

  13. DFS Example sourcevertex 136/7/2014

  14. DFS Example sourcevertex d f 1 | | | | | | | | 146/7/2014

  15. DFS Example sourcevertex d f 1 | | | 2 | | | | | 156/7/2014

  16. DFS Example sourcevertex d f 1 | | | 2 | | 3 | | | 166/7/2014

  17. DFS Example sourcevertex d f 1 | | | 2 | | 3 | 4 | | 176/7/2014

  18. DFS Example sourcevertex d f 1 | | | 2 | | 3 | 4 5 | | 186/7/2014

  19. DFS Example sourcevertex d f 1 | | | 2 | | 3 | 4 5 | 6 | 196/7/2014

  20. DFS Example sourcevertex d f 1 | 8 | | 2 | 7 | 3 | 4 5 | 6 | 206/7/2014

  21. DFS Example sourcevertex d f 1 | 8 | | 2 | 7 | 3 | 4 5 | 6 | 216/7/2014

  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? 226/7/2014

  23. DFS Example sourcevertex d f 1 | 8 | | 2 | 7 9 |10 3 | 4 5 | 6 | 236/7/2014

  24. DFS Example sourcevertex d f 1 | 8 |11 | 2 | 7 9 |10 3 | 4 5 | 6 | 246/7/2014

  25. DFS Example sourcevertex d f 1 |12 8 |11 | 2 | 7 9 |10 3 | 4 5 | 6 | 256/7/2014

  26. DFS Example sourcevertex d f 1 |12 8 |11 13| 2 | 7 9 |10 3 | 4 5 | 6 | 266/7/2014

  27. DFS Example sourcevertex d f 1 |12 8 |11 13| 2 | 7 9 |10 3 | 4 5 | 6 14| 276/7/2014

  28. DFS Example sourcevertex d f 1 |12 8 |11 13| 2 | 7 9 |10 3 | 4 5 | 6 14|15 286/7/2014

  29. DFS Example sourcevertex d f 1 |12 8 |11 13|16 2 | 7 9 |10 3 | 4 5 | 6 14|15 296/7/2014

  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? 306/7/2014

  31. DFS Example sourcevertex d f 1 |12 8 |11 13|16 2 | 7 9 |10 3 | 4 5 | 6 14|15 Tree edges 316/7/2014

  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) 326/7/2014

  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 336/7/2014

  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 346/7/2014

  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 356/7/2014

  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 366/7/2014

  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 376/7/2014

  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 386/7/2014

  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? 396/7/2014

  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? 406/7/2014

  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 416/7/2014

  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; } 426/7/2014

  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; } 436/7/2014

  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 446/7/2014

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