1. Lecture 36
CSE 331
Nov 30, 2012

2. HW 9 due today I will not take any HW after 1:15pm
Q1, Q 2 and Q 3 in separate piles
I will not take any HW after 1:15pm

3. Other HW related stuff Solutions to HW 9 at the end of the lecture
HW 8 should be available for pickup from Monday
HW 10 has been posted on the blog (Due in last lecture)

4. Sample final exam Posted on blog/piazza Final exam from last year
Can submit Q1 and Q2 as Q4 on HW 10 next Friday

5. Weighted Interval Scheduling
Input: n jobs (si,ti,vi)
Output: A schedule S s.t. no two jobs in S have a conflict
Goal: max Σi in S vj
Assume: jobs are sorted by their finish time

6. Couple more definitions
p(j) = largest i<j s.t. i does not conflict with j
p(j) < j
= 0 if no such i exists
OPT(j) = optimal value on instance 1,..,j

7. HW 9 due today I will not take any HW after 1:15pm
Q1, Q 2 and Q 3 in separate piles
I will not take any HW after 1:15pm

8. Property of OPT OPT(j) = max { vj + OPT( p(j) ), OPT(j-1) }
j not in OPT(j)
j in OPT(j)
OPT(j) = max { vj + OPT( p(j) ), OPT(j-1) }
Given OPT(1),…., OPT(j-1), how can one figure out if j in optimal solution or not?

10. Proof of correctness by induction on j
A recursive algorithm
Proof of correctness by induction on j
Correct for j=0
Compute-Opt(j)
If j = 0 then return 0
return max { vj + Compute-Opt( p(j) ), Compute-Opt( j-1 ) }
= OPT( p(j) )
= OPT( j-1 )
OPT(j) = max { vj + OPT( p(j) ), OPT(j-1) }

11. Exponential Running Time1 2 3 4 5
p(j) = j-2
Only 5 OPT values!
OPT(5)
OPT(3)
OPT(4)
Formal proof: Ex.
OPT(2)
OPT(3)
OPT(1)
OPT(2)
OPT(1)
OPT(2)
OPT(1)

13. How many distinct OPT values?

14. A recursive algorithm M-Compute-Opt(j) = OPT(j) M-Compute-Opt(j)
If j = 0 then return 0
If M[j] is not null then return M[j]
M[j] = max { vj + M-Compute-Opt( p(j) ), M-Compute-Opt( j-1 ) }
return M[j]
Run time = O(# recursive calls)

15. Bounding # recursions M-Compute-Opt(j) If j = 0 then return 0
O(n) overall
If j = 0 then return 0
If M[j] is not null then return M[j]
M[j] = max { vj + M-Compute-Opt( p(j) ), M-Compute-Opt( j-1 ) }
return M[j]
Whenever a recursive call is made an M value of assigned
At most n values of M can be assigned

17. Property of OPT Given OPT(1), …, OPT(j-1), one can compute OPT(j)
OPT(j) = max { vj + OPT( p(j) ), OPT(j-1) }
Given OPT(1), …, OPT(j-1),
one can compute OPT(j)

18. Recursion+ memory = Iteration
Iteratively compute the OPT(j) values
Iterative-Compute-Opt
M[0] = 0
M[j] = max { vj + M[p(j)], M[j-1] }
For j=1,…,n
M[j] = OPT(j)
O(n) run time

20. Reading Assignment
Sec 6.1, 6.2 of [KT]

21. When to use Dynamic Programming
There are polynomially many sub-problems
Richard Bellman
Optimal solution can be computed from solutions to sub-problems
There is an ordering among sub-problem that allows for iterative solution

22. Shortest Path Problem
Input: (Directed) Graph G=(V,E) and for every edge e has a cost ce (can be <0)
t in V
Output: Shortest path from every s to t
Assume that G has no negative cycle 1
100
-1000
899 s t
Shortest path has cost negative infinity

23. Rest of today’s agenda
Dynamic Program for shortest path

24. May the Bellman force be with you