Dynamic Programming: Interval Scheduling and Knapsack
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1 Dynamic Programming: Interval Scheduling and Knapsack
2 . Weighted Interval Scheduling
3 Weighted Interval Scheduling Weighted interval scheduling problem. Job j starts at s j, finishes at f j, and has weight or value v j. Two jobs compatible if they don't overlap. Goal: find maximum weight subset of mutually compatible jobs. a b c d How? Divide & Conquer? Greedy? e f g h Time 8
4 Unweighted Interval Scheduling Review Recall. Greedy algorithm works if all weights are. Consider jobs in ascending order of finish time. Add job to subset if it is compatible with previously chosen jobs. Observation. Greedy fails spectacularly with arbitrary weights. weight = weight = a b by finish Time weight = a weight = 999 a a a 4 a a a a 8 a 9 a b by weight Exercises: by density = weight per unit time? Other ideas? Time 9
5 Weighted Interval Scheduling Notation. Label jobs by finishing time: f f... f n. Def. p(j) = largest index i < j such that job i is compatible with j. p suggesting (last possible) predecessor Ex: p(8) =, p() =, p() =. j p(j) Time
6 principle of optimality Dynamic Programming: Binary Choice Notation. OPT(j) = value of optimal solution to the problem consisting of job requests,,..., j. key idea: binary choice Case : Optimum selects job j. can't use incompatible jobs { p(j) +, p(j) +,..., j - } must include optimal solution to problem consisting of remaining compatible jobs,,..., p(j) Case : Optimum does not select job j. must include optimal solution to problem consisting of remaining compatible jobs,,..., j- # if j = OPT( j) = $ % max v j + OPT( p( j)), OPT( j ) { } otherwise
7 Weighted Interval Scheduling: Brute Force Recursion Brute force recursive algorithm. Input: n, s,,s n, f,,f n, v,,v n Sort jobs by finish times so that f f... f n. Compute p(), p(),, p(n) Compute-Opt(j) { if (j = ) return else return max(v j + Compute-Opt(p(j)), Compute-Opt(j-)) }
8 Weighted Interval Scheduling: Brute Force Observation. Recursive algorithm is correct, but spectacularly slow because of redundant sub-problems exponential time. Ex. Number of recursive calls for family of "layered" instances grows like Fibonacci sequence. 4 4 p() = p() = ; p(j) = j-, j
9 Weighted Interval Scheduling: Bottom-Up Bottom-up dynamic programming. Unwind recursion. Input: n, s,,s n, f,,f n, v,,v n Sort jobs by finish times so that f f... f n. Compute p(), p(),, p(n) Iterative-Compute-Opt { OPT[] = for j = to n OPT[j] = max(v j + OPT[p(j)], OPT[j-]) } Output OPT[n] Claim: OPT[j] is value of optimal solution for jobs..j Timing: Easy. Main loop is O(n); sorting is O(n log n); what about p(j)?
10 Weighted Interval Scheduling Notation. Label jobs by finishing time: f f... f n. Def. p(j) = largest index i < j such that job i is compatible with j. Ex: p(8) =, p() =, p() =. j vj pj optj Time 8
11 Weighted Interval Scheduling Example Label jobs by finishing time: f f... f n. p(j) = largest i < j s.t. job i is compatible with j. Exercise: try other concrete examples: If all vj=: greedy by finish time,4,8 what if v > v?, but < v+v4? v>v+v4, but v+v < v+v, say? etc Time 8 j pj vj max(v j +opt[p(j)], opt[j-]) = opt[j] max(+, ) = max(+, ) = max(+, ) = 4 max(+, ) = 8 9 max(9+, 8) = 9 max(+, 9) = max(+, ) = 8? max(?+9, ) =? Exercise: What values of v8 cause it to be in/ex-cluded from opt? 9
12 Weighted Interval Scheduling: Finding a Solution Q. Dynamic programming algorithms computes optimal value. What if we want the solution itself? A. Do some post-processing traceback Run M-Compute-Opt(n) Run Find-Solution(n) Find-Solution(j) { if (j = ) output nothing else if (v j + OPT[p(j)] > OPT[j-]) print j Find-Solution(p(j)) else Find-Solution(j-) } the condition determining the max when computing OPT[ ] the relevant sub-problem # of recursive calls n O(n).
13 Sidebar: why does job ordering matter? It s Not for the same reason as in the greedy algorithm for unweighted interval scheduling. Instead, it s because it allows us to consider only a small number of subproblems (O(n)), vs the exponential number that seem to be needed if the jobs aren t ordered (seemingly, any of the n possible subsets might be relevant) Don t believe me? Think about the analogous problem for weighted rectangles instead of intervals (I.e., pick max weight non-overlapping subset of a set of axis-parallel rectangles.) Same problem for squares or circles also appears difficult.
14 .4 Knapsack Problem
15 Knapsack Problem Knapsack problem. Given n objects and a knapsack. Item i weighs w i > kilograms and has value v i >. Knapsack has capacity of W kilograms. Goal: maximize total value without overfilling knapsack Ex: {, 4 } has value 4. Item Value Weight V/W W = Greedy: repeatedly add item with maximum ratio v i / w i. Ex: {,, } achieves only value = greedy not optimal. [NB greedy is optimal for fractional knapsack : take # + 4/ of #4] 9
16 Dynamic Programming: False Start Def. OPT(i) = max profit subset of items,, i. Case : OPT does not select item i. OPT selects best of {,,, i- } Case : OPT selects item i. accepting item i does not immediately imply that we will have to reject other items without knowing what other items were selected before i, we don't even know if we have enough room for i Conclusion. Need more sub-problems!
17 Dynamic Programming: Adding a New Variable Def. OPT(i, w) = max profit subset of items,, i with weight limit w. Case : OPT does not select item i. OPT selects best of {,,, i- } using weight limit w Case : OPT selects item i. new weight limit = w w i OPT selects best of {,,, i } using this new weight limit # if i = % OPT(i, w) = $ OPT(i, w) if w i > w % & max{ OPT(i, w), v i + OPT(i, w w i )} otherwise
18 Knapsack Problem: Bottom-Up OPT(i, w) = max profit subset of items,, i with weight limit w. Input: n, w,,w N, v,,v N for w = to W OPT[, w] = for i = to n for w = to W if (w i > w) OPT[i, w] = OPT[i-, w] else OPT[i, w] = max {OPT[i-, w], v i + OPT[i-, w-w i ]} return OPT[n, W] (Correctness: prove it by induction on i & w.)
19 Knapsack Algorithm W φ { } n + {, } {,, } {,,, 4 } {,,, 4, } OPT: { 4, } value = + 8 = 4 W = if (w i > w) OPT[i, w] = OPT[i-, w] else OPT[i, w] = max{opt[i-,w],v i +OPT[i-,w-w i ]} Item 4 Value Weight 8 8
20 Knapsack Problem: Running Time Running time. Θ(n W). Not polynomial in input size! "Pseudo-polynomial. Knapsack is NP-hard. [Chapter 8] Knapsack approximation algorithm. There exists a polynomial time algorithm that produces a feasible solution (i.e., satisfies weight-limit constraint) that has value within.% (or any other desired factor) of optimum. [Section.8] 4
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