1. Lecture 13
CSE 331
Sep 27, 2017

2. Mini Project Pitch due in a week

3. Today’s agenda
Computing Connected component (with DFS)

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

5. Why is DFS a special case of Explore?

6. Every non-tree edge is between a node and its ancestor
A DFS run
Every non-tree edge is between a node and its ancestor
DFS(u)
u is explored
For every unexplored
neighbor v of u
DFS(v) 1 1 7 2 2 3 8 4 4 5
Note that the tree is arbitrary based on which neighbor was visited first. 5
DFS tree 6 6 3 8 7

7. Questions?

8. Connected components are disjoint
Either Connected components of s and t are the same or are disjoint
Algorithm to compute ALL the connected components?
Run BFS on some node s. Then run BFS on t that is not connected to s

9. Reading Assignment
Sec 3.2 in [KT]

10. Rest of today’s agenda
Run-time analysis of BFS (DFS)

11. Stacks and Queues Last in First out First in First out
We will need to keep track of visited nodes during our traversals.
First in First out

12. But first… How do we represent graphs?
When using an algorithm to solve a problem, the runtime depends on how data is organized.

13. Graph representations1
Better for sparse graphs and traversals
Adjacency matrix
Adjacency List
(u,v) in E?
O(1)
O(n) [ O(nv) ]
All neighbors of u?
O(n)
O(nu)
O(n2)
Space?
O(m+n)

14. Questions?

15. 2m = Σ u in V nu Rest of the graph Give 2 pennies to each edge
2m = Σ u in V nu
Rest of the graph
Give 2 pennies to each edge
Total # of pennies = 2m
nv=3 u
nu=4 v
Each edges gives one penny to its end points
# of pennies u receives = nu

16. Breadth First Search (BFS)
Build layers of vertices connected to s
L0 = {s}
Assume L0,..,Lj have been constructed
Lj+1 set of vertices not chosen yet but are connected to Lj
Use CC[v] array
Stop when new layer is empty
Use linked lists

17. An illustration 1 2 3 4 5 7 8 6 1 7 2 3 8 4 5 6