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Graph Traversals

Is G connected?. Is there a path from vertex a to vertex b ? . Does G have a cycle? . Will removing a single edge disconnect G ?. If G is directed, what are the strongly connected components?. If G is the WWW, follow links to locate information. .

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Graph Traversals

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  1. Is G connected? Is there a path from vertex a to vertex b? Does G have a cycle? Will removing a single edge disconnect G? If G is directed, what are the strongly connected components? If G is the WWW, follow links to locate information. Most graph algorithms need to examine or process each vertex and each edge. Graph Traversals Visit vertices of a graph G to determine some property:

  2. the distance of each vertex from s. a tree of shortest paths called thebreadth-first tree. Breadth-First Search Find the shortest path from a source node s to all other nodes. Outputs Idea: Find nodes at distance 0, then at distance 1, then at distance 2, …

  3. L =  s  0 A BFS Example b s e c g f a d

  4. L =  s  0 L =  a, c frontier of exploration 1 b s e c g f a d Visualized as many simultaneous explorations starting from s and spreading out independently.

  5. L =  s  0 L =  d, e, f  2 b s e c g f a d Dashed edges were explored but had resulted in the discovery of no new vertices. L =  a, c 1

  6. L =  s  0 L =  d, e, f  L =  b, g  2 3 b s e c g f a d L =  a, c 1

  7. L =  s  0 L =  d, e, f  L =  b, g  3 2 The Finish b s e c g f a d L =  a, c 1

  8. b d(b): shortest distance from s to b s e c g f a d Breadth-First Tree 3 0 2 3 1 2 1 2 # visited vertices = 1 + # tree edges  1 + # explored edges.

  9. The BFS Algorithm BFS(G, s) for eachv V(G) – sdo d(v)   // shortest distance from s d(s)  0 // initialization L  s T   i  0 whileL ≠  do // all nodes at distance i from s L   for eachu  Ldo for eachv  Adj(u) do ifd(v) =  then // not yet visited d(v)  d(u) + 1 insert v into L T  T  {(u, v)} i  i + 1 0 i i+1 i i+1

  10. Basis: L = s includes the only node at distance 0 from s. 0 Inductive step: Suppose L consists of all nodes at distance k≥ 0 from s. k k … s u v By hypothesis u L . Thus v  L . k k+1 Correctness LemmaFor each i, the set L includes all nodes at distance i from s. i Proof By induction on the distance k from s. Every node v at distance k+1 from s must be adjacent to a node u at distance k. Corollary At the finish of BFS, d(v) is the length of the shortest path from s to v.

  11. Every vertex is inserted at most once into some list L . i # edge scans u e ≤ |E| for a directed graph v ≤ 2 |E| for an undirected graph e v u 2 O(|V| )if represented as an adjacency matrix. Running Time O(|E|)if G is represented using adjacency lists. # inserted vertices  1 + # scanned edges  |E| + 1. Courtesy of Dr. Fernandez-Baca

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