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CSCE 411 Design and Analysis of Algorithms

CSCE 411 Design and Analysis of Algorithms. Set 5: Dynamic Programming Slides by Prof. Jennifer Welch Spring 2014. Drawback of Divide & Conquer. Sometimes can be inefficient Fibonacci numbers: F 0 = 0 F 1 = 1 F n = F n-1 + F n-2 for n > 1 Sequence is 0, 1, 1, 2, 3, 5, 8, 13, ….

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CSCE 411 Design and Analysis of Algorithms

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  1. CSCE 411Design and Analysis of Algorithms Set 5: Dynamic Programming Slides by Prof. Jennifer Welch Spring 2014 CSCE 411, Spring 2014: Set 5

  2. Drawback of Divide & Conquer • Sometimes can be inefficient • Fibonacci numbers: • F0 = 0 • F1 = 1 • Fn = Fn-1 + Fn-2 for n > 1 • Sequence is 0, 1, 1, 2, 3, 5, 8, 13, … CSCE 411, Spring 2014: Set 5

  3. Computing Fibonacci Numbers • Obvious recursive algorithm: • Fib(n): • if n = 0 or 1 then return n • else return (F(n-1) + Fib(n-2)) CSCE 411, Spring 2014: Set 5

  4. Recursion Tree for Fib(5) Fib(5) Fib(4) Fib(3) Fib(3) Fib(2) Fib(2) Fib(1) Fib(2) Fib(1) Fib(1) Fib(0) Fib(1) Fib(0) Fib(1) Fib(0) CSCE 411, Spring 2014: Set 5

  5. How Many Recursive Calls? • If all leaves had the same depth, then there would be about 2n recursive calls. • But this is over-counting. • However with more careful counting it can be shown that it is Ω((1.6)n) • Still exponential! CSCE 411, Spring 2014: Set 5

  6. Save Work • Wasteful approach - repeat work unnecessarily • Fib(2) is computed three times • Instead, compute Fib(2) once, store result in a table, and access it when needed CSCE 411, Spring 2014: Set 5

  7. More Efficient Recursive Alg • F[0] := 0; F[1] := 1; F[n] := Fib(n); • Fib(n): • if n = 0 or 1 then return F[n] • if F[n-1] = NIL then F[n-1] := Fib(n-1) • if F[n-2] = NIL then F[n-2] := Fib(n-2) • return (F[n-1] + F[n-2]) • computes each F[i] only once called memoization CSCE 411, Spring 2014: Set 5

  8. fills in F[4] with 3, returns 3+2 = 5 fills in F[2] with 1, returns 1+1 = 2 fills in F[3] with 2, returns 2+1 = 3 returns 0+1 = 1 Example of Memoized Fib F 0 1 2 3 4 5 Fib(5) 0 1 1 NIL Fib(4) 2 NIL 3 NIL Fib(3) 5 NIL Fib(2) CSCE 411, Spring 2014: Set 5

  9. Get Rid of the Recursion • Recursion adds overhead • extra time for function calls • extra space to store information on the runtime stack about each currently active function call • Avoid the recursion overhead by filling in the table entries bottom up, instead of top down. CSCE 411, Spring 2014: Set 5

  10. Subproblem Dependencies • Figure out which subproblems rely on which other subproblems • Example: F0 F1 F2 F3 … Fn-2 Fn-1 Fn CSCE 411, Spring 2014: Set 5

  11. Order for Computing Subproblems • Then figure out an order for computing the subproblems that respects the dependencies: • when you are solving a subproblem, you have already solved all the subproblems on which it depends • Example: Just solve them in the order F0 , F1 , F2 , F3 , … called Dynamic Programming CSCE 411, Spring 2014: Set 5

  12. DP Solution for Fibonacci • Fib(n): • F[0] := 0; F[1] := 1; • for i := 2 to n do • F[i] := F[i-1] + F[i-2] • return F[n] • Can perform application-specific optimizations • e.g., save space by only keeping last two numbers computed time reduced from exponential to linear! CSCE 411, Spring 2014: Set 5

  13. Matrix Chain Order Problem • Multiplying non-square matrices: • A is n x m, B is m x p • AB is n x p whose (i,j) entry is ∑aik bkj • Computing AB takes nmp scalar multiplications and n(m-1)p scalar additions (using basic algorithm). • Suppose we have a sequence of matrices to multiply. What is the best order? must be equal CSCE 411, Spring 2014: Set 5

  14. Why Order Matters • Suppose we have 4 matrices: • A, 30 x 1 • B, 1 x 40 • C, 40 x 10 • D, 10 x 25 • ((AB)(CD)) : requires 41,200 mults. • (A((BC)D)) : requires 1400 mults. CSCE 411, Spring 2014: Set 5

  15. Matrix Chain Order Problem • Given matrices A1, A2, …, An, where Ai is di-1 x di: [1] What is minimum number of scalar mults required to compute A1·A2 ·… · An? [2] What order of matrix multiplications achieves this minimum? Focus on [1]; see textbook for how to do [2] CSCE 411, Spring 2014: Set 5

  16. A Possible Solution • Try all possibilities and choose the best one. • Drawback is there are too many of them (exponential in the number of matrices to be multiplied) • Need to be more clever - try dynamic programming! CSCE 411, Spring 2014: Set 5

  17. Step 1: Develop a Recursive Solution • Define M(i,j) to be the minimum number of mults. needed to compute Ai·Ai+1 ·… · Aj • Goal: Find M(1,n). • Basis: M(i,i) = 0. • Recursion: How to define M(i,j) recursively? CSCE 411, Spring 2014: Set 5

  18. Defining M(i,j) Recursively • Consider all possible ways to split Ai through Aj into two pieces. • Compare the costs of all these splits: • best case cost for computing the product of the two pieces • plus the cost of multiplying the two products • Take the best one • M(i,j) = mink(M(i,k) + M(k+1,j) + di-1dkdj) CSCE 411, Spring 2014: Set 5

  19. Defining M(i,j) Recursively (Ai ·…·Ak)·(Ak+1 ·… · Aj) P1 P2 • minimum cost to compute P1 is M(i,k) • minimum cost to compute P2 is M(k+1,j) • cost to compute P1· P2 is di-1dkdj CSCE 411, Spring 2014: Set 5

  20. Step 2: Find Dependencies Among Subproblems M: GOAL! computing the red square requires the blue ones: to the left and below. CSCE 411, Spring 2014: Set 5

  21. Defining the Dependencies • Computing M(i,j) uses • everything in same row to the left: M(i,i), M(i,i+1), …, M(i,j-1) • and everything in same column below: M(i,j), M(i+1,j),…,M(j,j) CSCE 411, Spring 2014: Set 5

  22. Step 3: Identify Order for Solving Subproblems • Recall the dependencies between subproblems just found • Solve the subproblems (i.e., fill in the table entries) this way: • go along the diagonal • start just above the main diagonal • end in the upper right corner (goal) CSCE 411, Spring 2014: Set 5

  23. Order for Solving Subproblems M: CSCE 411, Spring 2014: Set 5

  24. Pseudocode for i := 1 to n do M[i,i] := 0 for d := 2 to n-1 do // diagonals for i := 1 to n-d+1 to // rows w/ an entry on d-th diagonal j := i + d - 1 // column corresponding to row i on d-th diagonal M[i,j] := infinity for k := i to j-1 to M[i,j] := min(M[i,j], M[i,k]+M[k+1,j]+di-1dkdj) endfor endfor endfor pay attention here to remember actual sequence of mults. running time O(n3) CSCE 411, Spring 2014: Set 5

  25. Example M: 1: A is 30x1 2: B is 1x40 3: C is 40x10 4: D is 10x25 CSCE 411, Spring 2014: Set 5

  26. Keeping Track of the Order • It's fine to know the cost of the cheapest order, but what is that cheapest order? • Keep another array S and update it when computing the minimum cost in the inner loop • After M and S have been filled in, then call a recursive algorithm on S to print out the actual order CSCE 411, Spring 2014: Set 5

  27. Modified Pseudocode for i := 1 to n do M[i,i] := 0 for d := 1 to n-1 do // diagonals for i := 1 to n-d to // rows w/ an entry on d-th diagonal j := i + d // column corresponding to row i on d-th diagonal M[i,j] := infinity for k := 1 to j-1 to M[i,j] := min(M[i,j], M[i,k]+M[k+1,j]+di-1dkdj) if previous line changed value of M[i,j] then S[i,j] := k endfor endfor endfor keep track of cheapest split point found so far: between Ak and Ak+1 CSCE 411, Spring 2014: Set 5

  28. Example M: 1: A is 30x1 2: B is 1x40 3: C is 40x10 4: D is 10x25 S: 1 1 1 2 3 3 CSCE 411, Spring 2014: Set 5

  29. Using S to Print Best Ordering Call Print(S,1,n) to get the entire ordering. Print(S,i,j): if i =j then output "A" + i //+ is string concat else k := S[i,j] output "(" + Print(S,i,k) + Print(S,k+1,j) + ")" CSCE 411, Spring 2014: Set 5

  30. Example S: <<draw recursion tree on board>> CSCE 411, Spring 2014: Set 5

  31. Longest Common Subsequence • Another example of dynamic programming • Given two strings • X = [x1, x2 ,…, xm] and • Y = [y1, y2, …, yn] • find the longest common subsequence (LCS) of X and Y • not necessarily contiguous! • Example: • X = [A, B, C, B, D, A, B] • Y = [B, D, C, A, B, A] • one LCS is [B, C, B, A] • another LCS is [B, D, A, B] CSCE 411, Spring 2014: Set 5

  32. Brute Force Solution • Take one of the strings, say X • Consider each possible subsequence of X • Check if that subsequence appears in Y • Would take exponential time since there are 2m possible subsequences of X • for each of the m characters in X, there are two choices (it is in the subsequence or is not) CSCE 411, Spring 2014: Set 5

  33. Recursive Expression for Solution • Let C[i,j] be length of LCS of • Xi = [x1, …, xi] and • Yj = [y1, …, yj] • Goal: C[m,n] • Basis: • C[0,j] = 0 (Xi is empty) for all j • C[i,0] = 0 (Yj is empty) for all i • i ranges from 0 to m, j ranges from 0 to n CSCE 411, Spring 2014: Set 5

  34. Recursive Expression for Solution • Formula for C[i,j]: • Case 1: xi = yj. • Recursively find LCS of Xi-1 and Yj-1 and append xi • So C[i,j] = C[i-1,j-1] + 1 if i > 0, j > 0, and xi = yj Xi-1 xi yj equal Yj-1 CSCE 411, Spring 2014: Set 5

  35. Recursive Expression for Solution • Case 2: xi ≠ yj. • Recursively find LCS of Xi-1 and Yj • Recursively find LCS of Xi and Yj-1 • Take the longer one • So C[i,j] = max{C[i,j-1], C[i-1,j]} if i > 0, j > 0, and xi ≠ yj Xi-1 xi yj not equal Yj-1 CSCE 411, Spring 2014: Set 5

  36. Determine Dependencies Among Subproblems • C[i,j] depends on C[i-1,j-1], C[i-1,j], and C[i,j-1] j-1 j n 0 1 0 1 i-1 i m Goal CSCE 411, Spring 2014: Set 5

  37. Determine Order for Solving Subproblems • An order for solving the subproblems (i.e., filling in the array) that respects the dependencies is row major order: • do the rows from top to bottom • inside each row, go from left to right CSCE 411, Spring 2014: Set 5

  38. Pseudocode fill in row 0 with all 0’s fill in column 0 with all 0’s for each row 1 through m do for each column 1 through n do fill in C[i,j] according to formula CSCE 411, Spring 2014: Set 5

  39. Time Complexity • There are O(mn) subproblems (entries in the array). • Each subproblem takes O(1) time to compute. • Total time is O(mn). CSCE 411, Spring 2014: Set 5

  40. Computing the Actual Subsequence • Keep track of which neighboring table entry gives the optimal solution to a subproblem: • above, to the left, or diagonal • break ties arbitrarily • Use another 2-D array, b, for this information. • store  in b[i,j] if xi = yj (this character is part of a common subsequence) • store  in b[i,j] if LCS of Xi and Yj-1 was longer • store  in b[i,j] if LCS of Xi-1 and Yj was longer • Trace backwards from b[m,n] until reaching an edge • every time  is encountered, the corresponding character is added to the longest common subsequence being constructed CSCE 411, Spring 2014: Set 5

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