Lecture 13 CSE 331 Sep 24, 2013.

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Presentation transcript:

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 Lj forms the jth “level” in the tree u in Lj+1 is child of v in Lj from which it was “discovered” 1 7 1 L0 2 3 L1 2 3 8 4 5 4 5 7 8 L2 6 6 L3

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

Questions?

BFS all

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

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