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Algorithms

Algorithms. Dynamic Programming. Dynamic Programming. General approach – combine solutions to subproblems to get the solution to an problem Unlike Divide and Conquer in that subproblems are dependent rather than independent bottom-up approach

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Algorithms

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  1. Algorithms Dynamic Programming

  2. Dynamic Programming • General approach – combine solutions to subproblems to get the solution to an problem • Unlike Divide and Conquer in that • subproblems are dependent rather than independent • bottom-up approach • save values of subproblems in a table and use more than one time

  3. Usual Approach • Start with the smallest, simplest subproblems • Combine “appropriate” subproblem solutions to get a solution to the bigger problem • If you have a D&C algorithm that does a lot of duplicate computation, you can often find a DP solution that is more efficient

  4. A Simple Example • Calculating binomial coefficients • n choose k is the number of different combinations of n things taken k at a time • These are also the coefficients of the binomial expansion (x+y)n

  5. Two Definitions

  6. Algorithm for Recursive Definition function C(n,k) if k = 0 or k = n then return 1 else return C(n-1, k-1) + C(n-1, k)

  7. C(5,3) C(4,2) C(4,3) C(3,1) C(3,2) C(3,2) C(3,3)

  8. Complexity of D & C Algorithm • Time complexity is (n!) • But we did a lot of duplicate computation • Dynamic programming approach • Store solutions to sub-problems in a table • Bottom-up approach

  9. k n k 0 1 2 3 0 1 2 3 4 5 n

  10. Analysis of DP Version • Time complexity O(n k) • Storage requirements • full table O(n k) • better solution

  11. Typical Problem • Dynamic Programming is often used for optimization problems that satisfy the principle of optimality • Principle of optimality • In an optimal sequence of decisions or choices, each subsequence must be optimal

  12. Optimization Problems • Problem has many possible solutions • Each solution has a value • Goal is to find the solution with the optimal value

  13. Steps in DP Solutions to Optimization Problems 1 Characterize the structure of an optimal solution 2 Recursively define the value of an optimal solution 3 Compute the value of an optimal solution in a bottom-up manner 4 Construct an optimal solution from computed information

  14. Matrix Chain Multiplication A1A2A3An • Matrix multiplication is associative • All ways that a sequence can be parenthesized give the same answer • But some are much less expensive to compute

  15. Matrix Chain Multiplication Problem • Given a chain < A1,A2, . . .,An> of n matrices, where i = 1, 2, . . ., n and matrix Ai has dimension pi-1pi, fully parenthesize the product A1A2An in a way that minimizes the number of scalar multiplications

  16. Example Matrix Dimensions A 13 x 5 B 5 X 89 C 89 X 3 D 3 X 34 M = A B C D 13 x 34

  17. Parenthesization Scalar multiplications 1 ((A B) C) D 10,582 2 (A B) (C D) 54,201 3 (A (B C)) D 2, 856 4 A ((B C) D) 4, 055 5 A (B (C D)) 26,418 1 13 x 5 x 89 to get (A B) 13 x 89 result 13 x 89 x 3 to get ((AB)C) 13 x 3 result 13 x 3 x 34 to get (((AB)C)D) 13 x 34

  18. Number of Parenthesizations

  19. T(n) ways to parenthesize n 1 2 3 4 5 10 15 T(n) 1 1 2 5 14 4,862 2,674,440

  20. Steps in DP Solutions to Optimization Problems 1 Characterize the structure of an optimal solution 2 Recursively define the value of an optimal solution 3 Compute the value of an optimal solution in a bottom-up manner 4 Construct an optimal solution from computed information

  21. Step 1 • Show that the principle of optimality applies • An optimal solution to the problem contains within it optimal solutions to sub-problems • Let Ai..j be the optimal way to parenthesize AiAi+1. . .Aj • Suppose the optimal solution has the first split at position k A1..k Ak+1..n • Each of these sub-problems must be optimally parenthesized

  22. Step 2 • Define value of the optimal solution recursively in terms of optimal solutions to sub-problems. • Consider the problem Ai..j • Let m[i,j] be the minimum number of scalar multiplications needed to compute matrix Ai..j • The cost of the cheapest way to compute A1..n is m[1,n]

  23. Step 2 continued • Define m[i,j] If i = j the chain has one matrix and the cost m[i,i] = 0 for all i  n If i < j Assume the optimal split is at position k where i k < j m[i,j] = m[i,k] + m[k+1,j] + pi-1 pkpj to compute Ai..k and Ak+1..j

  24. Step 2 continued • Problem: we don’t know what the value for k is, but there are only j-i possibilities

  25. Step 3 • Compute optimal cost by using a bottom-up approach • Example: 4 matrix chain Matrix Dimensions A1 100 x 1 A2 1 x 100 A3 100 x 1 A4 1 x 100 • po p1 p2 p3 p4 100 1 100 1 100

  26. j j 1 2 3 4 1 2 3 4 i i 1 2 3 4 1 2 3 4 s m po p1 p2 p3 p4 100 1 100 1 100

  27. MATRIX-CHAIN-ORDER(p) 1 n length[p] - 1 2 for i 1 to n 3 do m[i,i] 0 4 for l 2 to n 5 do for i 1 to n - l + 1 6 do ji + l - 1 7 m[i,j]  8 for k i to j - 1 9 do q m[i,k]+m[k+1,j]+ pi-1 pkpj 10 if q < m[i,j] 11 then m[i,j] = q 12 s[i,j] = k 13 return m and s

  28. Time and Space Complexity • Time complexity • Triply nested loops • O(n3) • Space complexity • n x n two dimensional table • O(n2)

  29. Step 4: Constructing the Optimal Solution • Matrix-Chain-Order determines the optimal number of scalar multiplications • Does not directly compute the product • Step 4 of the dynamic programming paradigm is to construct an optimal solution from computed information. • The s matrix contains the optimal split for every level

  30. MATRIX-CHAIN-MULTIPLY(A,s,i,j) 1 if j > i 2 then X MATRIX-CHAIN-MULTIPLY(A,s,i,s[i,j]) 3 Y MATRIX-CHAIN-MULTIPLY(A,s,s[i,j]+1,j) 4 return MATRIX-MULTIPLY(X,Y) 5 else return Ai Initial call MATRIX-CHAIN-MULTIPLY(A,s,1,n) where A = < A1,A2, . . .,An>

  31. Algorithms Dynamic Programming Continued

  32. Comments on Dynamic Programming • When the problem can be reduced to several “overlapping” sub-problems. • All possible sub-problems are computed • Computation is done by maintaining a large matrix • Usually has large space requirement • Running times are usually at least quadratic

  33. Memoization • Variation on dynamic programming • Idea–memoize the natural but inefficient recursive algorithm • As each sub-problem is solved, store values in table • Initialize table with values that indicate if the value has been computed

  34. MEMOIZED-MATRIX-CHAIN(p) 1 n  length(p) - 1 2 for i 1 to n 3 do for j i to n 4 do m[i,j]  5 return LOOKUP-CHAIN(p,1,n)

  35. LOOKUP-CHAIN(p,i,j) 1 if m[i,j]  2 then return m[i,j] 3 if i = j 4 then m[i,j]  5 else for k i to j - 1 6 do q LOOKUP-CHAIN(p,i,k) + LOOKUP-CHAIN(p,k+1,j) + pi-1pkpj 7 if q < m[i,j] 8 then m[i,j] q 9 return m[i,j]

  36. Space and Time Requirementsof Memoized Version • Running time (n3) • Storage (n2)

  37. Longest Common Subsequence • Definition 1: Subsequence Given a sequence X = < x1, x2, . . . , xm> then another sequence Z = < z1, z2, . . . , zk> is a subsequence of X if there exists a strictly increasing sequence <i1, i2, . . . , ik> of indices of x such that for all j = 1,2,...k we have xij= zj

  38. Example X = <A,B,D,F,M,Q> Z = <B, F, M> Z is a subsequence of X with index sequence <2,4,5>

  39. More Definitions • Definition 2: Common subsequence • Given 2 sequences X and Y, we say Z is a common subsequence of X and Y if Z is a subsequence of X and a subsequence of Y • Definition 3: Longest common subsequence problem • Given X = < x1, x2, . . . , xm> and Y = < y1, y2, . . . , yn> find a maximum length common subsequence of X and Y

  40. Example X = <A,B,C,B,D,A,B> Y = <B,D,C,A,B,A>

  41. Brute Force Algorithm 1 for every subsequence of X 2 Is there subsequence in Y? 3 If yes, is it longer than the longest subsequence found so far? Complexity?

  42. Yet More Definitions • Definition 4: Prefix of a subsequence If X = < x1, x2, . . . , xm> , the ith prefix of X for i = 0,1,...,m is Xi = < x1, x2, . . . , xi> • Example • if X = <A,B,C,D,E,F,H,I,J,L> then X4 = <A,B,C,D> and X0 = <>

  43. Optimal Substructure • Theorem 16.1 Optimal Substructure of LCS Let X = < x1, x2, . . . , xm> and Y = < y1, y2, . . . , yn> be sequences and let Z = < z1, z2, . . . ,zk> be any LCS of X and Y 1. if xm = yn then zk = xm = yn and zk-1 is an LCS of xm-1 and yn-1 2. if xm  yn and zk  x m Z is an LCS of Xm-1 and Y 3. if xm  yn and zk  yn Z is an LCS of Xm and Yn-1

  44. Sub-problem structure • Case 1 • if xm = yn then there is one sub-problem to solve find a LCS of Xm-1 and Yn-1 and append xm • Case 2 • if xm  yn then there are two sub-problems • find an LCS of Xmand Yn-1 • find an LCS of Xm-1 and Yn • pick the longer of the two

  45. Cost of Optimal Solution • Cost is length of the common subsequence • We want to pick the longest one • Let c[i,j] be the length of an LCS of the sequences Xi and Yj • Base case is an empty subsequence--then c[i,j] = 0 because there is no LCS

  46. The Recurrence

  47. Dynamic Programming Solution • Table c[0..m, 0..n] stores the length of an LCS of Xi and Yj c[i,j] • Table b[1..m, 1..n] stores pointers to optimal sub-problem solutions

  48. Y C B R F T S Q 0 1 2 3 4 5 6 7 0 A 1 B 2 D 3 F 4 M 5 Q 6 X Matrix C

  49. Y j C B R F T S Q 1 2 3 4 5 6 7 i A 1 B 2 D 3 F 4 M 5 Q 6 X Matrix B

  50. LCS-LENGTH(X,Y) 1 m length[X] 2 n length[Y] 3 for i  to m 4 do c[i,0]  5 for j  to n 6 do c[0,j]  7 for i  to m 8 do for j  to n 9 do if xi = yj then c[i,j] c[i-1,j-1] + 1 b[i,j] ”” 12 else if c[i-1,j] c[i,j-1] 13 then c[i,j] c[i-1,j] 14 b[i,j] ”” 15 else c[i,j] c[i,j-1] 16 b[i,j] ”” 17 return c and b

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