Slides15 - Dynamic Programming Introblerner/cs312/Slides/Slides15.pdf · Slides15 - Dynamic...

CS Lunch

Wednesday, 12:15 PMKendade 307

RABBIT TRACKSAn Animated film by Luke Jaegar!

A journey through a mortality-infused landscape populated by mysterious chickens, inconvenient

frogs, and other cartoonish creatures.

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Midterm 2!

Monday, April 8

In class

Covers Greedy Algorithms and Dynamic Programming

Closed book

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Dynamic Programming “Recipe”

Solve subproblems, remember the solutions in an array

Gradually build up solution to bigger problems based on subproblem solutions

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Slides15 - Dynamic Programming Intro.key - March 25, 2019

Interval Scheduling (Yes, this is an old problem!)

Job j starts at sj and finishes at fj.Two jobs compatible if they don't overlap.Goal: find maximum subset of mutually compatible jobs.

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Interval Scheduling: Greedy Solution

Sort jobs by earliest finish time.

Take each job provided it's compatible with the ones already taken.

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Interval Scheduling: Greedy Solution

Sort jobs by earliest finish time.

Take each job provided it's compatible with the ones already taken.

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Weighted Interval SchedulingJob j starts at sj, finishes at fj, and has weight or value vj Two jobs compatible if they don't overlap.Goal: find maximum weight subset of mutually compatible jobs.

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Weighted Interval SchedulingLabel jobs by finishing time: f1 ≤ f2 ≤ . . . ≤ fn .p(j) = largest index i < j such that job i is compatible with j.Ex: p(8) = 5, p(7) = 5, p(2) = 0.

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Dynamic Programming: Binary Choice

OPT(j) = value of optimal solution to the problem consisting of job requests 1, 2, ..., j.

Case 1: OPT selects job j.Case 2: OPT does not select job j.

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If OPT selects job j...can't use incompatible jobs { p(j) + 1, p(j) + 2, ..., j - 1 }must include optimal solution to problem consisting of remaining compatible jobs 1, 2, ..., p(j)

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If OPT selects job j...can't use incompatible jobs { p(j) + 1, p(j) + 2, ..., j - 1 }must include optimal solution to problem consisting of remaining compatible jobs 1, 2, ..., p(j)

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p(j) + 1, p(j) + 2, ..., j - 1

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If OPT does not select job j...must include optimal solution to problem consisting of remaining compatible jobs 1, 2, ..., j-1

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If OPT does not select job j...must include optimal solution to problem consisting of remaining compatible jobs 1, 2, ..., j-1

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Optimal Substructure

€

OPT( j) =0 if j = 0

max v j + OPT( p( j)), OPT( j −1){ } otherwise# $ %

OPT(j) = value of optimal solution to the problem consisting of job requests 1, 2, ..., j.

Case 1: OPT selects job j.Case 2: OPT does not select job j.

Case 1 Case 2

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Input: n, s1,…,sn , f1,…,fn , v1,…,vn

Sort jobs by finish times so that f1 ≤ f2 ≤ ... ≤ fn.

Compute p(1), p(2), …, p(n)

Compute-Opt(j) { if (j = 0) return 0 else return max(vj + Compute-Opt(p(j)), Compute-Opt(j-1))}

Straightforward Recursive Algorithm12


Worst Case

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p(1) = 0, p(j) = j-2

13-1

Worst Case

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p(1) = 0, p(j) = j-2

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13-2

Worst Case

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p(1) = 0, p(j) = j-2

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13-3


Worst Case

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p(1) = 0, p(j) = j-2

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3 2

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Worst Case

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p(1) = 0, p(j) = j-2

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3 2 2 1

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Worst Case

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p(1) = 0, p(j) = j-2

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3 2 2 1

2 1

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Worst Case

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p(1) = 0, p(j) = j-2

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3 2 2 1

2 1 1 0

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Worst Case

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p(1) = 0, p(j) = j-2

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3 2 2 1

2 1 1 0 1 0

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Worst Case

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p(1) = 0, p(j) = j-2

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3 2 2 1

2 1

1 0

1 0 1 0

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Sort jobs by finish times so that f1 ≤ f2 ≤ ... ≤ fn.Compute p(1), p(2), …, p(n)

for j = 1 to n M[j] = emptyM[0] = 0

M-Compute-Opt(j) { if (M[j] is empty) M[j] = max(wj + M-Compute-Opt(p(j)), M-Compute-Opt(j-1)) return M[j]}

MemoizationMemoization. Store results of each sub-problem in a cache; lookup as needed.

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Memoization

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M

M[8] = max(1 + M-Compute-Opt(5), M-Compute-Opt(7)

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Memoization

3 3 4 4 6 8 9 9M

M[8] = max(1 + 6, 9)

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Memoization

3 3 4 4 6 8 9 9M

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Running TimeMemoized version of algorithm takes O(n log n) time.

Sort by finish time: O(n log n).Computing p(⋅) : O(n log n) due to sorting by start time.M-Compute-Opt(j): each invocation takes O(1) time and either(i) returns an existing value M[j] (ii) fills in one new entry M[j] and makes two recursive callsProgress measure Φ = # filled entries of M[].

initially Φ = 0, throughout Φ ≤ n. (ii) increases Φ by 1, making 2 recursive callsTo fill n squares, we make at 2n recursive calls.

Overall running time of M-Compute-Opt(n) is O(n).

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Iterative SolutionBottom-up dynamic programming. Unwind recursion.

Sort jobs by finish times so that f1 ≤ f2 ≤ ... ≤ fn.

Compute p(1), p(2), …, p(n)

Iterative-Compute-Opt { M[0] = 0 for j = 1 to n M[j] = max(vj + M[p(j)], M[j-1])}

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Dynamic Programming Formula

Divide a problem into a polynomial number of smaller subproblems

Solve subproblem, recording its answer

Build up answer to bigger problem by using stored answers of smaller problems

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http://imgs.xkcd.com/comics/travelling_salesman_problem.png

Traveling Salesman (sic) Problem

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