Knapsack Problem Items have weights and values The problem is to - PDF document

Knapsack Problem • Items have weights and values • The problem is to maximize total value subject to CSE 421 a bound on weght Algorithms g • Items {I 1 , I 2 , … I n } – Weights {w 1 , w 2 , …,w n } – Values {v 1 , v 2 , …, v n } Richard Anderson – Bound K Lecture 18 • Find set S of indices to: – Maximize Σ i ε S v i such that Σ i ε S w i <= K Dynamic Programming Midterm Probem Midterm Probem w i = 1 or w i = 2 w i = 1 or w i = 2 • Idea one: • Idea two: – sort items by v i /w i – pair up items of weight 1 – greedy packing – greedy packing K = 4 K = 6 7 12 11 1 7 12 6 6 11 3 Subset Sum Problem Adding a variable for Weight • Let w 1 ,…,w n = {6, 8, 9, 11, 13, 16, 18, 24} • Opt[ j, K ] the largest subset of {w 1 , …, w j } that sums to at most K • Find a subset that has as large a sum as possible, without exceeding 50 • {2, 4, 7, 10} – Opt[2, 7] = O t[2 7] – Opt[3, 7] = – Opt[3,12] = – Opt[4,12] = 1

Subset Sum Recurrence Subset Sum Grid Opt[ j, K] = max(Opt[ j – 1, K], Opt[ j – 1, K – w j ] + w j ) • Opt[ j, K ] the largest subset of {w 1 , …, w j } that sums to at most K 4 3 3 2 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 {2, 4, 7, 10} Subset Sum Code Knapsack Problem • Items have weights and values for j = 1 to n • The problem is to maximize total value subject to for k = 1 to W a bound on weght Opt[j, k] = max(Opt[j-1, k], Opt[j-1, k-w j ] + w j ) • Items {I 1 , I 2 , … I n } – Weights {w 1 , w 2 , …,w n } – Values {v 1 , v 2 , …, v n } – Bound K • Find set S of indices to: – Maximize Σ i ε S v i such that Σ i ε S w i <= K Knapsack Recurrence Knapsack Grid Opt[ j, K] = max(Opt[ j – 1, K], Opt[ j – 1, K – w j ] + v j ) Subset Sum Recurrence: Opt[ j, K] = max(Opt[ j – 1, K], Opt[ j – 1, K – w j ] + w j ) 4 3 Knapsack Recurrence: Knapsack Recurrence: 2 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Weights {2, 4, 7, 10} Values: {3, 5, 9, 16} 2

Dynamic Programming Billboard Placement Examples • Examples • Maximize income in placing billboards – Optimal Billboard Placement – b i = (p i , v i ), v i : value of placing billboard at position p i • Text, Solved Exercise, Pg 307 – Linebreaking with hyphenation – Linebreaking with hyphenation • Constraint: • Constraint: • Compare with HW problem 6, Pg 317 – At most one billboard every five miles – String approximation • Example • Text, Solved Exercise, Page 309 – {(6,5), (8,6), (12, 5), (14, 1)} Design a Dynamic Programming Opt[k] = fun(Opt[0],…,Opt[k-1]) Algorithm for Billboard Placement • Compute Opt[1], Opt[2], . . ., Opt[n] • How is the solution determined from sub problems? • What is Opt[k]? Input b 1 , …, b n , where b i = (p i , v i ), position and value of billboard i Input b 1 , …, b n , where bi = (p i , v i ), position and value of billboard i Optimal line breaking and hyphen- Solution ation • Problem: break lines and insert hyphens to j = 0; // j is five miles behind the current position make lines as balanced as possible // the last valid location for a billboard, if one placed at P[k] for k := 1 to n • Typographical considerations: while (P[ j ] < P[ k ] – 5) – Avoid excessive white space A id i hit j := j + 1; j j + 1 j := j – 1; – Limit number of hyphens Opt[ k] = Max(Opt[ k-1] , V[ k ] + Opt[ j ]); – Avoid widows and orphans – Etc. 3

Penalty Function String approximation • Pen(i, j) – penalty of starting a line a • Given a string S, and a library of strings B position i, and ending at position j = {b 1 , …b m }, construct an approximation of the string S by using copies of strings in B. Opt-i-mal line break-ing and hyph-en-a-tion is com-put-ed with dy-nam-ic pro-gram-ming B = {abab, bbbaaa, ccbb, ccaacc} S = abaccbbbaabbccbbccaabab • Key technical idea – Number the breaks between words/syllables Design a Dynamic Programming Formal Model Algorithm for String Approximation • Strings from B assigned to non- • Compute Opt[1], Opt[2], . . ., Opt[n] overlapping positions of S • What is Opt[k]? • Strings from B may be used multiple times • Cost of δ for unmatched character in S f δ f C t t h d h t i S • Cost of γ for mismatched character in S – MisMatch(i, j) – number of mismatched characters of b j , when aligned starting with position i in s. Target string S = s 1 s 2 …s n Library of strings B = {b 1, …,b m } MisMatch(i,j) = number of mismatched characters with b j when aligned starting at position i of S. Opt[k] = fun(Opt[0],…,Opt[k-1]) Solution • How is the solution determined from sub for i := 1 to n problems? Opt[k] = Opt[k-1] + δ ; for j := 1 to |B| p = i – len(b j ); Opt[k] = min(Opt[k], Opt[p-1] + γ MisMatch(p, j)); Target string S = s 1 s 2 …s n Library of strings B = {b 1, …,b m } MisMatch(i,j) = number of mismatched characters with b j when aligned starting at position i of S. 4