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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?

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Presentation on theme: "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?"— Presentation transcript:

1 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.

2 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.

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

4 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

5 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.

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

7 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

8 s d b g f e c a Breadth-First Tree 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.

9 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 i i+1 i

10 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.

11 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.

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

13 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)

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

15 a l g f b ced j k ih 1 2

16 a l g f b c e d j k ih

17 a l g f b ced j k ih

18 a l g f b ced j k ih

19 a l g f b ced j k ih


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