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Lecture 13

Lecture 13. CSE 331 Sep 24, 2013. Reminders. HW 3 due on Friday. Today ’ s agenda. Every edge in is between consecutive layers. Computing Connected component. BFS Tree. BFS naturally defines a tree rooted at s. Add non-tree edges. L j forms the j th “ level ” in the tree.

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Lecture 13

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  1. Lecture 13 CSE 331 Sep 24, 2013

  2. Reminders HW 3 due on Friday

  3. Today’s agenda Every edge in is between consecutive layers Computing Connected component

  4. BFS Tree BFS naturally defines a tree rooted at s Add non-tree edges Lj forms the jth “level” in the tree u in Lj+1is child of v in Lj from which it was “discovered” L0 1 7 1 2 3 L1 2 3 8 4 7 8 L2 5 5 4 6 6 L3

  5. Computing Connected Component Explore(s) Start with R = {s} While exists (u,v) edge v not in R and u in R Add v to R Output R

  6. Questions?

  7. BFS all

  8. Depth First Search (DFS) http://xkcd.com/761/

  9. DFS(u) Mark u as explored and add u to R For each edge (u,v) If v is not explored then DFS(v)

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