1. Lecture 22
CSE 331
Oct 22, 2010

2. A non-CSE 331 poll
Vs.
iphone
Andriod

3. Announcements Graded HW 5 next week HW 6 has been posted
Jeff will be out of town starting next week
See the blog for more details

4. Mid Term stuff Mid term post up on the blog
Read it before asking for regrade requests
A temp grade assigned by early next week
I’ll ask some of you to meet me in person

5. Student blog posts on new blog

6. Interval Scheduling Problem
Input: n intervals [s(i), f(i)) for 1≤ i ≤ n
Output: A schedule S of the n intervals
No two intervals in S conflict
|S| is maximized

7. Analyzing the algorithm
R: set of requests
A has no conflicts
Set A to be the empty set
While R is not empty
Choose i in R with the earliest finish time
A is an optimal solution
Add i to A
Remove all requests that conflict with i from R
Return A

8. Today’s agenda Analyze run-time of the greedy algorithm
Prove correctness of the greedy algorithm
Analyze run-time of the greedy algorithm

9. Greedy “stays ahead”
Greedy
OPT

10. What can you say about f(i1) and f(j1)?
Greedy “stays ahead”
A = i1,…,ik
O = j1,…,jm
k = m
What do we need to prove?
What can you say about f(i1) and f(j1)?

11. A formal claim For every r ≤ k, f(ir) ≤ f(jr) A = i1,…,ik O = j1,…,jm
The greedy algorithm outputs an optimal A
Proof by contradiction: A is not optimal
A = i1,…,ik
O = j1,…,jm
m > k
After ik, R was non-empty! ik
No conflict! jk
jk+1

12. Greedy can always pick jr
Proof of claim
For every r ≤ k, f(ir) ≤ f(jr)
Why is r=1 OK?
By induction on r
Assume true up to r-1
Greedy can always pick jr ?
ir-1 ir
jr-1 jr