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Dynamic Programming 2012/11/20 1 Dynamic Programming (DP) Dynamic programming is typically applied to optimization problems. Problems that can be solved by dynamic programming satisfy the principle of optimality. P.2 Principle of optimality Suppose that in solving a problem, we have to make a sequence of decisions D1, D2, …, Dn-1, Dn If this sequence of decisions D1, D2, …, Dn-1, Dn is optimal, then the last k, 1 k n, decisions must be optimal under the condition caused by the first n-k decisions. 3 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. 4 The Greedy Method E.g. Find a shortest path from v0 to v3. The greedy method can solve this problem. The shortest path: 1 + 2 + 4 = 7. 5 The Greedy Method E.g. Find a shortest path from v0 to v3 in the multi-stage graph. Greedy method: v0v1,2v2,1v3 = 23 Optimal: v0v1,1v2,2v3 = 7 The greedy method does not work for this problem. This is because decisions at different stages influence one another. 6 Multistage graph A multistage graph G=(V,E) is a directed graph in which the vertices are partitioned into k2 disjoint sets Vi, 1i k In addition, if <u,v> is an edge in E then uVi and vVi+i for some i, 1i<k The set V1 and Vk are such that V1 =Vk=1 The multistage graph problem is to find a minimum cost path from s in V1 to t in Vk Each set Vi defines a stage in the graph 7 Greedy Method vs. Multistage graph E.g. 4 A D 1 11 S 2 5 B 18 9 E 16 13 T 2 5 C 2 F 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. 8 Dynamic Programming Dynamic programming approach: 1 S 2 B 5 A C d(A, T) d(B, T) T d(C, T) d(S, T) = min{1+d(A, T), 2+d(B, T), 5+d(C, T)} 9 Dynamic Programming 4 A D 1 A 4 11 11 D E d(D, T) S T d(E, T) 2 5 B 18 9 E 16 13 T 2 5 C 2 F d(A, T) = min{4+d(D, T), 11+d(E, T)} = min{4+18, 11+13} = 22. 10 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. 4 A D 1 11 S 2 9 5 B E 16 13 T B D d (D , T ) d (E , T ) 5 E T 2 5 C 9 18 2 F 16 d (F , T ) F 11 4 A D 1 11 S Save computation 2 5 B 18 9 E 16 13 T 2 5 C 2 F For example, we never calculate (as a whole) the length of the path SBDT ( 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… 12 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. 13 Comment If a problem can be described by a multistage graph then it can be solved by dynamic programming. 14 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. 15 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. P.16 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) 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. How similar are two paths in Freebies island of Second Life? 19 LCSS-DC－path transfers sequence SeqA:C60.C61.C62.C63.C55.C47.C39.C31.C32 20 LCSS-DC － similar path thresholds SeqA :C60.C61.C62.C63.C55.C47.C39.C31.C32 SeqB :C60.C61.C62.C54.C62.C63.C64 LCSSAB :C60.C61.C62. C63 21 Longest-common-subsequence problem: We are given two sequences X = <x1,x2,...,xm> and Y = <y1,y2,...,yn> and wish to find a maximum length common subsequence of X and Y. We define Xi = < x1,x2,...,xi > and Yj = <y1,y2,...,yj>. P.22 Brute Force Solution m * 2n = O(2n ) or n * 2m = O(2m) 23 A recursive solution to subproblem Define c [i, j] is the length of the LCS of Xi and Yj . if i=0 or j=0 0 if i,j>0 and xi=y c[i, j] = c[i -1, j -1] +1 max{c[i, j -1], c[i -1, j]} if i,j>0 and x y j i P.24 j 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 P.25 7 for i 1 to m 8 for j 1 to n 9 if xi = yj 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 P.26 Complexity: O(mn) rather than O(2m) or O(2n) of Brute force method P.27 PRINT_LCS PRINT_LCS(b, X, i, j ) 1 if i = 0 or j = 0 Complexity: O(m+n) 2 then return 3 if b[i, j] = “” 4 then PRINT_LCS(b, X, i-1, j-1) 5 print xi 6 else if b[i, j] = “” 7 then PRINT_LCS(b, X, i-1, j) 8 else PRINT_LCS(b, X, i, j-1) By calling PRINT_LCS(b, X, length[X], length[Y]) to print LCSP.28 Matrix-chain multiplication How to compute A1 A2 ... An where Ai is a matrix for every i. Example: A1 A2 A3 A4 ( A1 ( A2 ( A3 A4 ))) ( A1 (( A2 A3 ) A4 )) (( A1 A2 )( A3 A4 )) (( A1 ( A2 A3 )) A4 ) ((( A1 A2 ) A3 ) A4 ) Chapter 15 P.29 MATRIX MULTIPLY MATRIX MULTIPLY(A,B) 1 if columns[A] rows[B] 2 then error “incompatible dimensions” 3 else for i 1 to rows[A] 4 for j 1 to columns[B] C [i , j ] 0 5 6 for k 1 to columns[A] C [ i , j ] C [ i , j ] + A[ i , k ] B [ k , j ] 7 8 return C Chapter 15 P.30 Complexity: Let A be a p q matrix, and B be a q r matrix. Then the complexity of A xB is p q r . Chapter 15 P.31 Example: is a 10 100 matrix, A2 is a 100 5 matrix, and A3 is a 5 50 matrix. Then (( A1 A2 ) A3 ) takes 10 100 5 + 10 5 50 = 7500 time. However ( A1 ( A2 A3 )) takes 100 5 50 + 10 100 50 = 75000 time. A1 Chapter 15 P.32 The matrix-chain multiplication problem: Given a chain A , A ,..., A of n matrices, where for i=0,1,…,n, matrix Ai has dimension pi-1pi, fully parenthesize the product A1 A2 ... An in a way that minimizes the number of scalar multiplications. 1 2 n 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. Chapter 15 P.33 Counting the number of parenthesizations: if n = 1 1 n -1 P( n ) = P ( k ) p( n - k ) k =1 if n 2 [Catalan number] P( n ) = C ( n - 1 ) n 2 n 4 = = ( 3 / 2 ) n +1 n n 1 Chapter 15 P.34 Step 1: The structure of an optimal parenthesization O ptim al (( A1 A2 ... A k )( A k + 1 A k + 2 ... A n )) C om bine Chapter 15 P.35 Step 2: A recursive solution Define m[i, j]= minimum number of scalar multiplications needed to compute the matrix Ai .. j = Ai Ai +1 .. A j goal m[1, n] m[ i, j ] = min 0 i k j { m [ i , k ] + m [ k + 1, j ] + p i -1 p k p j } i= j i j Chapter 15 P.36 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]. Chapter 15 P.37 MATRIX_CHAIN_ORDER 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-1pkpj 10 if q < m[i, j] 11 then m[i, j] q 12 s[i, j] k 13 return m and s 3 Complexity: O ( n ) Chapter 15 P.38 Example: A1 30 35 = p 0 p1 A2 35 15 = p1 p 2 A3 15 5 = p2 p3 A4 5 10 = p3 p4 A5 10 20 = p4 p5 A6 20 25 = p5 p6 Chapter 15 P.39 the m and s table computed by MATRIX-CHAIN-ORDER for n=6 Chapter 15 P.40 m[2,5]= min{ m[2,2]+m[3,5]+p1p2p5=0+2500+351520=13000, m[2,3]+m[4,5]+p1p3p5=2625+1000+35520=7125, m[2,4]+m[5,5]+p1p4p5=4375+0+351020=11374 } =7125 Chapter 15 P.41 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 “)” (( A1 ( A2 A3 ))(( A4 A5 ) A6 )) Example: Chapter 15 P.42 Q&A 43