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Graph Traversals Visit vertices of a graph G to determine some property: 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.

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

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A BFS Example s d b g f e c a L = s 0

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

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s d b g f e c a L = s 0 L = a, c 1 L = d, e, f 2 Dashed edges were explored but had previously been discovered.

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s d b g f e c a L = s 0 L = a, c 1 L = d, e, f 2 L = b, g 3

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

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s d b g f e c a Breadth-First Tree 0 1 1 2 2 2 3 3 d(b): shortest distance from s to b The number of visited vertices is one more than the number of tree edges and hence at most one more than the number of explored edges.

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The BFS Algorithm BFS(G, s) for each v V(G) – s do d(v) // shortest distance from s d(s) 0 // initialization L s T i 0 while L ≠ do // all nodes at distance i from s L for each u L do for each v Adj(u) do if d(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

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Correctness Lemma For each i, the set L includes all nodes at distance i from s. i Proof By induction on the distance k from s. s … uv k 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. Every node v at distance k+1 from s must be adjacent to a node u at distance k. By hypothesis u L. Thus v L. k k k+1 Corollary At the finish of BFS, d(v) is the length of the shortest path from s to v.

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Running Time O(|E|) if G is represented using adjacency lists. The total number of edge scans is ≤ |E| for a directed graph ≤ 2 |E| for an undirected graph u v e u v e O(|V| ) if represented as an adjacency matrix. 2 Every vertex is inserted at most once into some list L. i The number of inserted vertices is at most one more than the number of scanned edges; hence ≤ |E| + 1.

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Depth-First Search Idea: Keep going forward as long as there are unseen nodes to be visited. Backtrack when stuck. v G G G 1 2 3 G is completely traversed before exploring G and G. 1 23 From Computer Algorithms by S. Baase and A. van Gelder

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The DFS Algorithm DFS(G) time 0 // global variable for each v V(G) do disc(v) unseen for each v V(G) do if disc(v) = unseen then DFS-visit(v) time time + 1 disc(v) time for each u Adj(v) do if disc(u) = unseen then DFS-visit(u)

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A DFS Example a l g f b ced j k ih time = 1

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a l g f b ced j k ih 1 2

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a l g f b c e d j k ih 1 2 3 4 5

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a l g f b ced j k ih 1 2 3 4 5 6 7

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a l g f b ced j k ih 1 2 3 4 5 6 7 8 9

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a l g f b ced j k ih 1 2 3 4 5 6 7 8 9 10 11 12

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