1. 1 Dynamic Programming 2012/11/20

2. P.2 Dynamic Programming (DP) Dynamic programming Dynamic programming is typically applied to optimization problems. Problems that can be solved by dynamic programming satisfy the principle of optimality.

3. 3 Principle of optimality Suppose that in solving a problem, we have to make a sequence of decisions D 1, D 2, …, D n-1, D n If this sequence of decisions D 1, D 2, …, D n-1, D n is optimal, then the last k, 1  k  n, decisions must be optimal under the condition caused by the first n-k decisions.

4. 4 Dynamic method v.s. Greedy method Comparison: In the greedy method, any decision is locally optimal. These locally optimal solutions will finally add up to be a globally optimal solution.

5. 5 The Greedy Method E.g. Find a shortest path from v 0 to v 3. The greedy method can solve this problem. The shortest path: 1 + 2 + 4 = 7.

6. 6 The Greedy Method E.g. Find a shortest path from v 0 to v 3 in the multi-stage graph. Greedy method: v 0 v 1,2 v 2,1 v 3 = 23 Optimal: v 0 v 1,1 v 2,2 v 3 = 7 The greedy method does not work for this problem. This is because decisions at different stages influence one another.

7. 7 Multistage graph A multistage graph G=(V,E) is a directed graph in which the vertices are partitioned into k  2 disjoint sets V i, 1  i  k In addition, if is an edge in E then u  V i and v  V i+i for some i, 1  i<k The set V 1 and V k are such that  V 1  =  V k  =1 The multistage graph problem is to find a minimum cost path from s in V 1 to t in V k Each set V i defines a stage in the graph

8. 8 Greedy Method vs. Multistage graph E.g. The greedy method cannot be applied to this case: S A D T 1+4+18 = 23. The shortest path is: S C F T 5+2+2 = 9.

9. 9 Dynamic Programming Dynamic programming approach: d(S, T) = min{1+d(A, T), 2+d(B, T), 5+d(C, T)}

10. 10 Dynamic Programming d(A, T) = min{4+d(D, T), 11+d(E, T)} = min{4+18, 11+13} = 22.

11. 11 Dynamic Programming d(B, T) = min{9+d(D, T), 5+d(E, T), 16+d(F, T)} = min{9+18, 5+13, 16+2} = 18. d(C, T) = min{ 2+d(F, T) } = 2+2 = 4 d(S, T) = min{1+d(A, T), 2+d(B, T), 5+d(C, T)} = min{1+22, 2+18, 5+4} = 9.

12. 12 Save computation For example, we never calculate (as a whole) the length of the path S  B  D  T ( namely, d(S,B)+d(B,D)+d(D,T) ) because we have found d(B, E)+d(E, T)<d(B,D)+d(D,T) There are some more examples … Compare with the brute-force method …

13. 13 The advantages of dynamic programming approach To avoid exhaustively searching the entire solution space (to eliminate some impossible solutions and save computation). To solve the problem stage by stage systematically. To store intermediate solutions in a table (array) so that they can be retrieved from the table in later stages of computation.

14. Comment If a problem can be described by a multistage graph then it can be solved by dynamic programming. If a problem can be described by a multistage graph then it can be solved by dynamic programming. 14

15. 15 The longest common subsequence (LCS or LCSS) problem A sequence of symbols A = b a c a d A subsequence of A: deleting 0 or more symbols (not necessarily consecutive) from A. E.g., ad, ac, bac, acad, bacad, bcd. Common subsequences of A = b a c a d and B = a c c b a d c b : ad, ac, bac, acad. The longest common subsequence of A and B: a c a d.

16. P.16 DNA Matching DNA = {A|C|G|T}* S1=ACCGGTCGAGTGCGGCCGAAGCCGGCCGAA S2=GTCGTTCGGAATGCCGTTGCTGTAAA Are S1 and S2 similar DNAs? The question can be answered by figuring out the longest common subsequence.

17. Networked virtual environments (NVEs) virtual worlds full of numerous virtual objects to simulate a variety of real world scenes allowing multiple geographically distributed users to assume avatars to concurrently interact with each other via network connections. E.G., MMOGs: World of Warcraft (WoW), Second Life (SL)

18. Avatar Path Clustering Because of similar personalities, interests, or habits, users may possess similar behavior patterns, which in turn lead to similar avatar paths within the virtual world. We would like to group similar avatar paths as a cluster and find a representative path (RP) for them.

19. How similar are two paths in Freebies island of Second Life? 19

20. SeqA:C60.C61.C62.C63.C55.C47.C39.C31.C32 LCSS-DC － path transfers sequence 20

21. SeqA :C60.C61.C62.C63.C55.C47.C39.C31.C32 SeqB :C60.C61.C62.C54.C62.C63.C64 LCSS AB :C60.C61.C62. C63 LCSS-DC － similar path thresholds 21

22. P.22 Longest-common-subsequence problem: We are given two sequences X = and Y = and wish to find a maximum length common subsequence of X and Y. We define X i = and Y j =.

23. Brute Force Solution m * 2 n = O(2 n ) or n * 2 m = O(2 m ) 23

24. P.24 A recursive solution to subproblem Define c [i, j] is the length of the LCS of X i and Y j.        j i j i y x i,j>jicjic =yx i,j> jic j=i= jic and 0 if]},1[],1,[max{ and 0if 1]1,1[ 0or 0 if 0 ],[

25. P.25 Computing the length of an LCS LCS_LENGTH(X,Y) 1 m  length[X] 2 n  length[Y] 3 for i  1 to m 4 do c[i, 0]  0 5 for j  1 to n 6 do c[0, j]  0

26. P.26 7 for i  1 to m 8 for j  1 to n 9 if x i = y j 10 then c[i, j]  c[i-1, j-1]+1 11 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

27. P.27 Complexity: O(mn) rather than O(2 m ) or O(2 n ) of Brute force method

28. P.28 PRINT_LCS PRINT_LCS(b, X, i, j ) 1if i = 0 or j = 0 2then return 3if b[i, j] = “  ” 4then PRINT_LCS(b, X, i-1, j-1) 5 print x i 6else if b[i, j] = “  ” 7then PRINT_LCS(b, X, i-1, j) 8else PRINT_LCS(b, X, i, j-1) Complexity: O(m+n) By calling PRINT_LCS(b, X, length[X], length[Y]) to print LCS

29. Chapter 15P.29 Matrix-chain multiplication How to compute where is a matrix for every i. Example:

30. Chapter 15P.30 MATRIX MULTIPLY MATRIX MULTIPLY(A,B) 1if columns[A] rows[B] 2 then error “ incompatible dimensions ” 3 else for to rows[A] 4 for to columns[B] 5 6 for to columns[A] 7 8return C

31. Chapter 15P.31 Complexity: Let A be a matrix, and B be a matrix. Then the complexity of A xB is.

32. Chapter 15P.32 Example: is a matrix, is a matrix, and is a matrix. Then takes time. However takes time.

33. Chapter 15P.33 The matrix-chain multiplication problem: Given a chain of n matrices, where for i=0,1, …,n, matrix Ai has dimension p i-1  p i, fully parenthesize the product in a way that minimizes the number of scalar multiplications. A product of matrices is fully parenthesized if it is either a single matrix, or a product of two fully parenthesized matrix product, surrounded by parentheses.

34. Chapter 15P.34 Counting the number of parenthesizations: [Catalan number]

35. Chapter 15P.35 Step 1: The structure of an optimal parenthesization

36. Chapter 15P.36 Step 2: A recursive solution Define m[i, j]= minimum number of scalar multiplications needed to compute the matrix goal m[1, n]

37. Chapter 15P.37 Step 3: Computing the optimal costs Instead of computing the solution to the recurrence recursively, we compute the optimal cost by using a tabular, bottom-up approach. The procedure uses an auxiliary table m[1..n, 1..n] for storing the m[i, j] costs and an auxiliary table s[1..n, 1..n] that records which index of k achieved the optimal cost in computing m[i, j].

38. Chapter 15P.38 MATRIX_CHAIN_ORDER MATRIX_CHAIN_ORDER(p) 1 n  length[p] –1 2 for i  1 to n 3do m[i, i]  0 4 for l  2 to n 5do for i  1 to n – l + 1 6do j  i + l – 1 7 m[i, j]   8 for k  i to j – 1 9do q  m[i, k] + m[k+1, j]+ p i-1 p k p j 10 if q < m[i, j] 11 then m[i, j]  q 12 s[i, j]  k 13 return m and s Complexity:

39. Chapter 15P.39 Example:

40. Chapter 15P.40 the m and s table computed by MATRIX-CHAIN-ORDER for n=6

41. Chapter 15P.41 m[2,5]= min{ m[2,2]+m[3,5]+p 1 p 2 p 5 =0+2500+35  15  20=13000, m[2,3]+m[4,5]+p 1 p 3 p 5 =2625+1000+35  5  20=7125, m[2,4]+m[5,5]+p 1 p 4 p 5 =4375+0+35  10  20=11374 } =7125

42. Chapter 15P.42 MATRIX_CHAIN_MULTIPLY PRINT_OPTIMAL_PARENS(s, i, j) 1 if i=j 2 then print “A”i 3 else print “(“ 4 PRINT_OPTIMAL_PARENS(s, i, s[i,j]) 5 PRINT_OPTIMAL_PARENS(s, s[i,j]+1, j) 6 print “)” Example:

43. Q&A 43