15.Dynamic Programming

Transcription

1 15.Dynamic Programming

2 Dynamic Programming is an algorithm design technique for optimization problems: often minimizing or maximizing. Like divide and conquer, DP solves problems by combining solutions to subproblems. Unlike divide and conquer, subproblems are not independent. Subproblems may share subsubproblems, However, solution to one subproblem may not affect the solutions to other subproblems of the same problem. (More on this later.) DP reduces computation by Solving subproblems in a bottom-up fashion. Storing solution to a subproblem the first time it is solved. Looking up the solution when subproblem is encountered again. Key: determine structure of optimal solutions Chapter 15 P.2

3 The development of a dynamic programming algorithm can be broken into a sequence of four steps: 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 fashion. 4. Construct an optimal solution from computed information. Chapter 15 P.3

4 Rod cutting [New in the third edition of the CLRS] How to cut steel rods into pieces in order to maximize the revenue you can get? Each cut is free. Rod lengths are always an integral number of inches. Input: A length n and table of prices p i, for i =1, 2,, n. Output: The maximum revenue obtainable for rods whose lengths sum to n, computed as the sum of the prices for the individual rods. If p n is large enough, an optimal solution might require no cuts, i.e., just leave the rod as n inches long. Chapter 15 P.4

5 Chapter 15 P.5 IVPL

6 Chapter 15 P.6 IVPL

7 Chapter 15 P.7 IVPL

8 Chapter 15 P.8 IVPL

9 Recursive top-down solution Chapter 15 P.9

10 Chapter 15 P.10 IVPL

11 Dynamic-programming solution IVPL Chapter 15 P.11

12 Chapter 15 P.12 IVPL

13 Chapter 15 P.13 IVPL

14 Subproblem graphs IVPL Chapter 15 P.14

15 Reconstructing a solution IVPL Chapter 15 P.15

16 Chapter 15 P.16 IVPL

17 Matrix-chain multiplication 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. A 1 A 2... How to compute A n where A i is a matrix for every i. Example: A1 A2 A3 A4 ( A ( A ( A A ))) ( A (( A A ) A )) (( A A )( A A )) (( A ( A A )) A ) ((( A A ) A ) A ) Chapter 15 P.17

18 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 do for j 1 to columns[b] 5 do c[ i, j] 0 6 for k 1 to columns[a] 7 do 8 return C c[ i, j ] c[ i, j ] A[ i, k ] B[ k, j ] Chapter 15 P.18

19 Complexity IVPL Let A be a q r p q r matrix, and B be a matrix. Then the complexity is. p q Example: A 1 is a matrix, A 2 is a matrix, and A 3 is a 5 50 matrix. Then (( A A ) A ) takes time. However ( A ( A A )) takes time. Chapter 15 P.19

20 The matrix-chain multiplication problem: IVPL Given a chain A, A,..., A 1 2 n of n matrices, where for i=0,1,,n, matrix A i has dimension p i-1 p i, fully parenthesize the product A 1 A 2... A n in a way that minimizes the number of scalar multiplications. Chapter 15 P.20

21 Step 1: The structure of an optimal parenthesization IVPL Optimal (( A A... A )( A A... A )) 1 2 k k1 k2 n Combine Optimal L R 可能的 k 位置 L 的維度 = p 0 p k R 的維度 = p k p n Chapter 15 P.21

22 Step 2: A recursive solution Define m[i, j]= minimum number of scalar multiplications needed to compute the matrix A A A A goal m[1, n] m[ i, j ] min ik j 0 { m[ i, k] m[ k 1, i.. j i i1.. j j] p i1 p k p j } i i j j Chapter 15 P.22

23 Step 3: Computing the optimal costs IVPL 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 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]+ 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: O( n ) Chapter 15 P.23 3

24 Chapter 15 P.24 Example: p p A p p A p p A p p A p p A p p A

25 the m and s table computed by MATRIX-CHAIN-ORDER for n=6 Chapter 15 P.25

26 m[2,5]= min{ m[2,2]+m[3,5]+p 1 p 2 p 5 = =13000, m[2,3]+m[4,5]+p 1 p 3 p 5 = =7125, m[2,4]+m[5,5]+p 1 p 4 p 5 = =11374 } =7125 Chapter 15 P.26

27 Step 4: Constructing an optimal solution MATRIX_CHAIN_MULTIPLY(A, s, i, j) 1 if j > i 2 then 3 y x 4 return MATRIX-MULTIPLY(X,Y) 5 else return A i MATRIX_CHA IN_MULTIPLY( MATRIX_CHA IN_MULTIPLY( A, s, s[ i, A, s, i, s[ i, j] 1, j]) j) example: (( A1 ( A2 A3 ))(( A4 A5 ) A6 )) Chapter 15 P.27

28 Elements of Dynamic Programming Optimal substructure Overlapping subproblems Chapter 15 P.28

29 Optimal substructure: We say that a problem exhibits optimal substructure if an optimal solution to the problem contains within its optimal solution to subproblems. Example: Matrix-multiplication problem Chapter 15 P.29

30 Optimal Substructure IVPL Show that a solution to a problem consists of making a choice, which leaves one or more subproblems to solve. Suppose that you are given this last choice that leads to an optimal solution. Given this choice, determine which subproblems arise and how to characterize the resulting space of subproblems. Show that the solutions to the subproblems used within the optimal solution must themselves be optimal. Usually use cut-and-paste. Need to ensure that a wide enough range of choices and subproblems are considered. Chapter 15 P.30

31 Optimal Substructure IVPL Optimal substructure varies across problem domains: 1. How many subproblems are used in an optimal solution. 2. How many choices in determining which subproblem(s) to use. Informally, running time depends on (# of subproblems overall) (# of choices). How many subproblems and choices do the examples considered contain? Dynamic programming uses optimal substructure bottom up. First find optimal solutions to subproblems. Then choose which to use in optimal solution to the problem. Chapter 15 P.31

32 Subtleties( 細微的差異 ) One should be careful not to assume that optimal substructure applies when it does not. consider the following two problems in which we are given a directed graph G = (V, E) and vertices u, v V. Unweighted shortest path: Find a path from u to v consisting of the fewest edges. Good for Dynamic programming. Unweighted longest simple path: Find a simple path from u to v consisting of the most edges. Not good for Dynamic programming. Chapter 15 P.32

33 Unweighted undirected graph Unweighted directed graph Weighted directed graph Chapter 15 P.33

34 Chapter 15 P.34 IVPL

35 Chapter 15 P.35 IVPL

36 Overlapping Subproblems The space of subproblems must be small. The total number of distinct subproblems is a polynomial in the input size. A recursive algorithm is exponential because it solves the same problems repeatedly. If divide-and-conquer is applicable, then each problem solved will be brand new. Chapter 15 P.36

37 Longest Common Subsequence X = < A, B, C, B, D, A, B > Y = < B, D, C, A, B, A > < B, C, A > is a common subsequence of both X and Y. < B, C, B, A > or < B, C, A, B > is the longest common subsequence of X and Y. Chapter 15 P.37

38 Problem: Given 2 sequences, X = x 1,...,x m and Y = y 1,...,y n, find a common subsequence whose length is maximum. IVPL Subsequence need not be consecutive, but must be in order. Chapter 15 P.38

39 Longest-common-subsequence problem: We are given two sequences X = <x 1,x 2,...,x m > and Y = <y 1,y 2,...,y n > and wish to find a maximum length common subsequence of X and Y. Chapter 15 P.39

40 Optimal substructure of LCS X m = x 1 x 2... x m-2 x m-1 x m X m-1 = or Y n = y 1 y 2... y n-2 y n-1 y n Y n-1 LCS Z k = z 1 z 2... z k-2 z k-1 z k Z k-1 Chapter 15 P.40

41 Theorem 15.1 (Optimal substructure of LCS) IVPL Let X = <x 1,x 2,...,x m > and Y = <y 1,y 2,...,y n > be the sequences, and let Z = <z 1,z 2,...,z k > be any LCS of X and Y. 1. If x m = y n then z k = x m = y n and Z k-1 is an LCS of X m-1 and Y n If x m y n then z k x m implies Z is an LCS of X m-1 and Y. 3. If x m y n then z k y n implies Z is an LCS of X and Y n-1. Chapter 15 P.41

42 A recursive solution to subproblem We Define X i = < x 1,x 2,...,x i >. Define c [i, j] is the length of the LCS of X i and Y j. c[ i, j] 0 c[ i 1, j 1] 1 max{ c[ i, j 1], c[ i 1, j]} if i= 0 or j= 0 if i,j> 0 and if i,j> 0 and x x=y i i j y j Chapter 15 P.42

43 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 7 for i 1 to m 8 do for j 1 to n 9 do 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 Chapter 15 P.43

44 Complexity: O(mn) Chapter 15 P.44 IVPL

45 PRINT_LCS PRINT_LCS(b, X, c, j ) 1 if i = 0 or j = 0 2 then return Complexity: O(m+n) 3 if b[i, j] = 4 then PRINT_LCS(b, X, i-1, j-1) 5 print x i 6 else if b[i, j] = 7 then PRINT_LCS(b, X, i-1, j) 8 then PRINT_LCS(b, X, i, j-1) Chapter 15 P.45

46 Optimal Binary search trees cost:2.80 cost:2.75 optimal!! Chapter 15 P.46

47 Chapter 15 P.47 Expected cost the expected cost of a search in T is n i n i p i q i n i i n i i T i i T i n i n i i T i i T q d p k q d p k ) ( depth ) ( depth 1 1) ) ( (depth 1) ) ( (depth E[search cost in T]

48 For a given set of probabilities, our goal is to construct a binary search tree whose expected search is smallest. We call such a tree an optimal binary search tree. Chapter 15 P.48

49 Step 1: The structure of an optimal binary search tree Consider any subtree of a binary search tree. It must contain keys in a contiguous range k 1,...,k j, for some 1 i j n. In addition, a subtree that contains keys k i,..., k j must also have as its leaves the dummy keys d i-1,..., d j. If an optimal binary search tree T has a subtree T' containing keys k i,..., k j, then this subtree T' must be optimal as well for the subproblem with keys k i,..., k j and dummy keys d i-1,..., d j. Chapter 15 P.49

50 Chapter 15 P.50 Step 2: A recursive solution. if )}, ( ] 1, [ 1], [ min{ -1, i j if 1 ], [ ), ( 1 j i j w i j r e r i e q j i e q p j w i j r i i j i l l j i l l

51 Step 3:computing the expected search cost of an optimal binary search tree OPTIMAL-BST(p,q,n) 1 for i 1 to n do e[i, i 1] q i-1 3 w[i, i 1] q i-1 4 for l 1 to n 5 do for i 1 to n l do j i + l 1 7 e[i, j] 8 w[i, j] w[i, j 1] + p j +q j IVPL Chapter 15 P.51

52 9 for r i to j 10 do t e[i, r 1]+e[r +1, j]+w[i, j] 11 if t < e[i, j] 12 then e[i, j] t 13 root[i, j] r 14 return e and root the OPTIMAL-BST procedure takes (n 3 ), just like MATRIX-CHAIN-ORDER Chapter 15 P.52

53 The table e[i,j], w[i,j], and root[i,j] computer by OPTIMAL-BST on the key distribution. Chapter 15 P.53

54 Knuth has shown that there are always roots of optimal subtrees such that root[i, j 1] root[i+1, j] for all 1 i j n. We can use this fact to modify Optimal-BST procedure to run in (n 2 ) time. Chapter 15 P.54