CS Lunch. Dynamic Programming Recipe. 2 Midterm 2! Slides17 - Segmented Least Squares.key - November 16, 2016

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1 CS Lunch 1 Michelle Oraa Ali Tien Dao Vladislava Paskova Nan Zhuang Surabhi Sharma Wednesday, 12:15 PM Kendade Midterm 2! Monday, November 21 In class Covers Greedy Algorithms Closed book Dynamic Programming Recipe 3 Solve subproblems, remember the solutions in an array Gradually build up solution to bigger problems based on subproblem solutions

2 Weighted Interval Scheduling 4 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 Time Optimal Substructure 5 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. # 0 if j = 0 OPT( j) = $ % max { v j + OPT( p( j)), OPT( j 1) } otherwise Case 1 Case 2 Straightforward Recursive Algorithm 6 Input: n, s 1,,s n, f 1,,f n, v 1,,v n Sort jobs by finish times so that f 1 f 2... f n. Compute p(1), p(2),, p(n) Compute-Opt(j) { if (j = 0) return 0 else return max(v j + Compute-Opt(p(j)), Compute-Opt(j-1)) }

3 Iterative Solution Bottom-up dynamic programming. Unwind recursion. 7 Sort jobs by finish times so that f 1 f 2... f n. Compute p(1), p(2),, p(n) Iterative-Compute-Opt { M[0] = 0 for j = 1 to n M[j] = max(v j + M[p(j)], M[j-1]) } Dynamic Programming Formula 8 Divide a problem into a polynomial number of smaller subproblems We often think recursively to identify the subproblems Solve subproblem, recording its answer Build up answer to bigger problem by using stored answers of smaller problems We develop algorithm that builds up the answer iteratively Iterative Solution Bottom-up dynamic programming. Unwind recursion. 9 Sort jobs by finish times so that f 1 f 2... f n. Compute p(1), p(2),, p(n) Iterative-Compute-Opt { Subproblems M[0] = 0 for j = 1 to n M[j] = max(v j + M[p(j)], M[j-1]) }

4 Iterative Solution Bottom-up dynamic programming. Unwind recursion. 10 Sort jobs by finish times so that f 1 f 2... f n. Compute p(1), p(2),, p(n) Record subproblem Iterative-Compute-Opt { M[0] = 0 answers for j = 1 to n M[j] = max(v j + M[p(j)], M[j-1]) } Iterative Solution Bottom-up dynamic programming. Unwind recursion. 11 Sort jobs by finish times so that f 1 f 2... f n. Compute p(1), p(2),, p(n) Combine subproblem Iterative-Compute-Opt { M[0] = 0 answers for j = 1 to n M[j] = max(v j + M[p(j)], M[j-1]) } 12 Traveling Salesman (sic) Problem

5 Least Squares Foundational problem in statistic and numerical analysis. 13 Given n points in the plane: (x 1, y 1 ), (x 2, y 2 ),..., (x n, y n ). Find a line y = ax + b that minimizes the sum of the squared error: n SSE = ( y i ax i b) 2 i=1 y x Least Squares Solution 14 Result from calculus, least squares achieved when: a = n x i i y i ( i x i ) ( i y i ) i, b = n x 2 i ( x i ) 2 i i y i a i x i n Least Squares 15 Sometimes a single line does not work very well. y x

6 Segmented Least Squares Points lie roughly on a sequence of several line segments. Given n points in the plane (x 1, y 1 ), (x 2, y 2 ),..., (x n, y n ) with x 1 < x 2 <... < x n, find a sequence of lines that fits well. 16 y x Segmented Least Squares 17-1 Segmented Least Squares 17-2 Segmented least squares for points i..j

7 Segmented Least Squares 17-3 Segmented least squares for points i..j Different values for the starting point of the line segment that ends at point j Segmented Least Squares 17-4 Segmented least squares for points i..j Different values for the starting point of the line segment that ends at point j e(i, j) + c + OPT(i-1) Segmented Least Squares 17-5 Segmented least squares for points i..j Different values for the starting point of the line segment that ends at point j e(i, j) + c + OPT(i-1) min over all values of i from 1 to j-1

8 Calculating Line Segment Error 18 First calculate slope and y-intercept for least squares line: a = n x i i y i ( i x i ) ( i y i ) i, b = n x 2 i ( x i ) 2 i i y i a i x i n Then calculate the error associated with that line n SSE = ( y i ax i b) 2 i=1 The Price is Right! (or shopping with somebody else s money) Spend as much money as possible without going over $100. You can buy at most 1 of each. 19 CD $11 Jeans $33 DVD $20 Dinner $15 Book $13 Ice cream $5 Shoes $70 Pizza $7 Subset sum 20-1

9 Subset sum 20-2 The most we can spend considering items 1..j Subset sum 20-3 The most we can spend considering items 1..j Case 1: Do not buy item j Case 2: Buy item j Subset sum 20-4 The most we can spend considering items 1..j Case 1: Do not buy item j Case 2: Buy item j Case 1: OPT(j-1) Case 2:???

10 Subset sum 20-5 The most we can spend considering items 1..j Case 1: Do not buy item j Case 2: Buy item j Case 1: OPT(j-1) Case 2:??? max( ) Dynamic Programming Formula 21 Divide a problem into a polynomial number of smaller subproblems We often think recursively to identify the subproblems Solve subproblem, recording its answer Build up answer to bigger problem by using stored answers of smaller problems We develop algorithm that builds up the answer iteratively Knapsack Problem 22 Knapsack problem. Given n objects and a "knapsack." Item i weighs w i > 0 pounds and has value v i > 0. Knapsack has capacity of W pounds. Goal: fill knapsack so as to maximize total value.