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Lecture 32 CSE 331 Nov 16, 2016
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Talk today
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Three general techniques
High level view of CSE 331 Problem Statement Problem Definition Three general techniques Algorithm “Implementation” Data Structures Analysis Correctness+Runtime Analysis
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Greedy Algorithms Natural algorithms
Reduced exponential running time to polynomial
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Divide and Conquer Recursive algorithmic paradigm
Reduced large polynomial time to smaller polynomial time
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A new algorithmic technique
Dynamic Programming
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Dynamic programming vs. Divide & Conquer
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Same same because Both design recursive algorithms
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Different because Dynamic programming is smarter about solving recursive sub-problems
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End of Semester blues Can only do one thing at any day: what is the optimal schedule to obtain maximum value? Write up a term paper (30) (3) (2) (5) (10) Party! Exam study 331 HW Project Monday Tuesday Wednesday Thursday Friday
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Previous Greedy algorithm
Order by end time and pick jobs greedily Greedy value = 5+2+3= 10 Write up a term paper (10) OPT = 30 Party! (2) Exam study (5) 331 HW (3) Project (30) Monday Tuesday Wednesday Thursday Friday
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Today’s agenda Formal definition of the problem
Start designing a recursive algorithm for the problem
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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?
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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) }
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Exponential Running Time
1 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)
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How many distinct OPT values?
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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)
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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
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Reading Assignment Sec 6.1, 6.2 of [KT]
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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
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