CSE331 Lecture 36

CSE331 Lecture 36 - Lecture 36 CSE 331 New Office 319 Davis...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Lecture 36 CSE 331 Nov 28, 2011 New Office: 319 Davis CSEd Week celebrations http://csedweek.wordpress.com/ Volunteers needed! Warning on blog posts/scribes AT MOST 3 blog posts (and scribe)/lecture Need volunteers for today A new experiment today 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 Couple more definitions p(j) = largest i<j s.t. i does not conflict with j = 0 if no such i exists OPT(j) = optimal value on instance 1,..,j p(j) < j Property of OPT j in OPT(j) j not in OPT(j) OPT(j) = max { vj + OPT( p(j) ), OPT(j-1) } Given OPT(1),…., OPT(j1), how can one figure out if j in optimal solution or not? A recursive algorithm Compute-Opt(j) Correct for j=0 Proof of correctness by induction on 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) } Exponential Running Time 1 2 p(j) = j-2 3 4 Only 5 OPT values! 5 OPT( 5) OPT( 3) Formal proof: Ex. OPT( 1) OPT( 4) OPT( 2) OPT( 2) OPT( 1) OPT( 1) OPT( 3) OPT( 1) OPT( 2) OPT( 1) A recursive algorithm M-Compute-Opt(j) M-Compute-Opt(j) = 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) Bounding # recursions 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] Whenever a recursive call is made an M value of assigned At most n values of M can be assigned O(n) overall Property of OPT OPT(j) = max { vj + OPT( p(j) ), OPT(j-1) } Given OPT(1), …, OPT(j-1), one can compute OPT(j) Recursion+ memory = Iteration Iteratively compute the OPT(j) values Iterative-Compute-Opt M[0] = 0 For j=1,…,n M[j] = max { vj + M[p(j)], M[j-1] } M[j] = OPT(j) O(n) run time Reading Assignment Sec 6.1, 6.2 of [KT] 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 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 s Shortest path has cost negative infinity 1 1 899 100 -1000 t Assume that G has no negative cycle Today’s agenda Dynamic Program for shortest path May the Bellman force be with you ...
View Full Document

This note was uploaded on 12/11/2011 for the course CSE 331 taught by Professor Rudra during the Fall '11 term at SUNY Buffalo.

Ask a homework question - tutors are online