Lecture 8. Dynamic Programming

Transcription

1 Lecture 8. Dynamic Programming T. H. Cormen, C. E. Leiserson and R. L. Rivest Introduction to Algorithms, 3rd Edition, MIT Press, 2009 Sungkyunkwan University Hyunseung Choo Copyright Networking Laboratory

2 Dynamic Programming Dynamic programming solves optimization problems by combining solutions to subproblems Programming refers to a tabular method with a series of choices, not coding Recall the divide-and-conquer approach Partition the problem into independent subproblems Solve the subproblems recursively Combine solutions of subproblems Dynamic programming is applicable when subproblems are not independent i.e., subproblems share subsubproblems Solve every subsubproblem only once and store the answer for use when it reappears A divide-and-conquer approach will do more work than necessary Algorithms Networking Laboratory 2/36

3 Dynamic Programming Solution 4 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 Algorithms Networking Laboratory 3/36

4 Matrix Multiplication Matrix multiplication Two matrices A and B can be multiplied The number of columns of A must equal the number of rows of B A(pq) x B(qr) C(pr) The number of scalar multiplication is pqr For an pq matrix A and a qr matrix B, the product AB is the pr matrix AB C [ c i, j ] pr q k1 a i, k b k, j pr Algorithms Networking Laboratory 4/36

5 Matrix Multiplication Algorithms Networking Laboratory 5/36

6 Matrix Multiplication Matrix multiplication Algorithms Networking Laboratory 6/36

7 Matrix-Chain Multiplication Matrix multiplication A(pq) x B(qr) C(pr) The number of scalar multiplication is pqr e.g. < A 1 (10100), A 2 (1005), A 3 (550) > ((A 1 A 2 )A 3 ) 10x100x5 + 10x5x50 = = 7,500 (A 1 (A 2 A 3 )) 100x5x x100x50 = = 75, times faster Given a sequence (chain) <A 1, A 2,, A n > of n matrices to be multiplied, where i=1,2,, 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 Determine an order for multiplying matrices that has the lowest cost Algorithms Networking Laboratory 7/36

8 Matrix-Chain Multiplication Determine an order for multiplying matrices that has the lowest cost Counting the number of parenthesizations A 1 A 2 A k A k+1 A n-1 A n 1 if n 1 P( n) n1 ( 2 P( k) P( n k) if n 2 n ) k 1 Exercise on page 338 Impractical to check all possible parenthesizations Algorithms Networking Laboratory 8/36

9 Step 1 The structure of an optimal parenthesization Notation: A i..j Result from evaluating A i A i+1 A j (i < j) Any parenthesization of A i A i+1 A j must split the product between A k and A k+1 for some integer k in the range i k < j The cost of this parenthesization cost of computing A i..k + cost of computing A k+1..j + cost of multiplying A i..k and A k+1..j together Algorithms Networking Laboratory 9/36

10 Step 1 Suppose that an optimal parenthesization of A i A i+1 A j splits the product between A k and A k+1 The parenthesization of the prefix sub-chain A i A i+1 A k must be an optimal parenthesization of A i A i+1 A j The parenthesization of the prefix sub-chain A k+1 A i+1 A j must be an optimal parenthesization of A i A i+1 A j That is, the optimal solution to the problem contains within it the optimal solution to subproblems Algorithms Networking Laboratory 10/36

11 Step 1 Minimal Cost_A Cost_A p 0 p 6 p 9 A 1 A 2 A 3 A 4 A 5 A 6 A 7 A 8 A 9 Suppose ((A 1 A 2 )(A 3 ((A 4 A 5 )A 6 ))) ((A 7 A 8 )A 9 ) is optimal Then (A 1 A 2 ) (A 3 ((A 4 A 5 )A 6 )) must be optimal for A 1 A 2 A 3 A 4 A 5 A 6 Otherwise, if (A 1 (A 2 A 3 )) ((A 4 A 5 )A 6 ) is optimal for A 1 A 2 A 3 A 4 A 5 A 6 Then ((A 1 (A 2 A 3 )) ((A 4 A 5 )A 6 )) ((A 7 A 8 )A 9 ) will be better than ((A 1 A 2 )(A 3 ((A 4 A 5 )A 6 ))) ((A 7 A 8 )A 9 ) Contradiction! Algorithms Networking Laboratory 11/36

12 Step 2 A Recursive Solution Subproblem m[i, j] Determine the minimum cost of a parenthesization of A i A i+1 A j (1 i j n) the minimum number of scalar multiplications needed to compute the matrix A i..j m[i, j] = m[i, k] + m[k+1, j] + p i-1 p k p j However, we do not know the value of k (=s[i,j]), so we have to try all j-i possibilities 0 m[ i, j] min { m[ i, k] m[ k 1, j]} pi pk p 1 ik j A recursive solution takes exponential time j if if i j i j Algorithms Networking Laboratory 12/36

13 Step 3 Computing the optimal costs How much subproblems in total? One for each choice of i and j satisfying 1 i j n (n 2 ) MATRIX-CHAIN-ORDER(p) Input: a sequence p = < p 0, p 1, p 2,, p n > (length[p] = n+1) Try to fill in the table m in a manner that corresponds to solving the parenthesization problem on matrix chains of increasing length Lines 4-12: compute m[i, i+1], m[i, i+2], each time Algorithms Networking Laboratory 13/36

14 Step 3 e.g. < A 1 (83), A 2 (35), A 3 (510) > m 12 = 8x3x5 = 120 m 23 = 3x5x10 = 150 m 13 = 390 m 11 + m 23 + p 0 p 1 p 3 = x3x10 = 390 m 12 + m 33 + p 0 p 2 p 3 = x5x10 = 520 j i 2 3 m s j 1 2 i Algorithms Networking Laboratory 14/36

15 Step 3 O(n 3 ), (n 3 ) (n 3 ) running time (n 2 ) space Algorithms Networking Laboratory 15/36

16 l =3 l = 2 35x15x5= x20x25 =5000 m[3,5] = min m[3,4]+m[5,5] + 15x10x20 = = 3750 m[3,3]+m[4,5] + 15x5x20 = = 2500 Algorithms Networking Laboratory 16/30

17 Step 4 Constructing an optimal solution Each entry s[i, j] records the value of k such that the optimal parenthesization of A i A i+1 A j splits the product between A k and A k+1 A 1..n A 1..s[1..n] A s[1..n]+1..n A 1..s[1..n] A 1..s[1, s[1..n]] A s[1, s[1..n]]+1..s[1..n] Recursive Algorithms Networking Laboratory 17/36

18 Step 4 Constructing an optimal solution ( A 1 ( A 2 A 3 ) ) ( ( A 4 A 5 ) A 6 ) Algorithms Networking Laboratory 18/36

19 Elements of Dynamic Programming Optimal substructure If an optimal solution contains within it optimal solutions to subproblems Build an optimal solution from optimal solutions to subproblems Example Matrix-chain multiplication An optimal parenthesization of A i A i+1 A j that splits the product between A k and A k+1 contains within it optimal solutions to the problem of parenthesizing A i A i+1 A k and A k+1 A k+2 A j Algorithms Networking Laboratory 19/36

20 Elements of Dynamic Programming Overlapping subproblems The space of subproblems must be small in the sense that a recursive algorithm for the problem solves the same subproblems over and over, rather than always generating new subproblems Typically, the total number of distinct subproblems is a polynomial in the input size Divide-and-Conquer is suitable usually generate brand-new problems at each step of the recursion Algorithms Networking Laboratory 20/36

21 Characteristics of Optimal Substructure How many subproblems are used in an optimal solution to the original problem? Matrix-Chain scheduling 2 (A 1 A 2 A k and A k+1 A k+2 A j ) How may choice we have in determining which subproblems to use in an optimal solution? Matrix-chain scheduling: j - i (choice for k) Informally, the running time of a dynamic-programming algorithm relies on: the number of subproblems overall and how many choices we look at for each subproblem Matrix-Chain scheduling: (n 2 ) * O(n) = O(n 3 ) Algorithms Networking Laboratory 21/36

22 Overlapping Subproblems Algorithms Networking Laboratory 22/36

23 Overlapping Subproblems m[3,4] is computed twice Algorithms Networking Laboratory 23/36

24 Algorithms Networking Laboratory 24/36 Recursive Procedure for Matrix-Chain Multiplication The time to compute m[1,n] is at least exponential in n Prove T(n) = (2 n ) using the substitution method Show that T(n) 2 n ) ( 2 ) ( 1) ) ( ) ( ( 1 ) ( 1 (1) n i n k n i T n T k n T k T n T T ) (2 1) 2( ) ( n n n n i i n i i n n n n n T

25 Memoization A variation of dynamic programming that often offers the efficiency of the usual dynamic programming approach while maintaining a top-down strategy Memoize the natural, but inefficient, recursive algorithm Maintain a table with subproblem solutions, but the control structure for filling in the table is more like the recursive algorithm Memoization for matrix-chain multiplication Calls in which m[i, j] = (n 2 ) calls Calls in which m[i, j] < O(n 3 ) calls Turns an (2 n )-time into an O(n 3 )-time algorithm Algorithms Networking Laboratory 25/36

26 Memoization Algorithms Networking Laboratory 26/36

27 Lookup-Chain(p, i, j) if m[i, j] < if i=j then return m[i, j] then m[i, j] 0 else for k i to j-1 return m[i, j] do qlookup-chain(p, i, k) + LOOKUP-CHAIN(p, k+1, j) + p i-1 p k p j if q < m[i, j] then m[i, j] q Algorithms Networking Laboratory 27/36

28 Lookup-Chain(p, i, j) Algorithms Networking Laboratory 28/36

29 Dynamic Programming vs. Memoization If all subproblems must be solved at least once, a bottom-up dynamic-programming algorithm usually outperforms a top-down memoized algorithm by a constant factor No overhead for recursion and less overhead for maintaining table There are some problems for which the regular pattern of table acces ses in the dynamic programming algorithm can be exploited to reduce the time or space requirements even further Algorithms Networking Laboratory 29/36

30 Self-Study Two more dynamic-programming problems Section 15.4 Longest Common Subsequence Section 15.5 Optimal Binary Search Trees Algorithms Networking Laboratory 30/36

31 Longest Common Subsequence (LCS) Problem: Given two sequences X=<x 1,,x m > and Y=<y 1,,y n >, find the longest subsequence Z=<z 1,,z k > that is common to X and Y. A subsequence is a subset of elements from the sequence with strictly increasing order (not necessarily contiguous) There are 2 m subsequences of X checking all subsequences is impractical for long sequences Example: X=<A,B,C,B,D,A,B> and Y=<B,D,C,A,B,A> Common subsequences: <A>; <B>; <C>; <D>; <A, A>; <B, B>; <B, C, A>; <B, C, B>; <C, B, A>; etc. The longest common subsequences: <B, C, B, A>; <B, D, A, B> Algorithms Networking Laboratory 31/36

32 Step 1: Optimal Structure of an LCS Let X=<x 1,,x m > and Y=<y 1,,y n > be sequences, and let Z=<z 1,,z k > be any LCS of X and Y. 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-1. If x m y n, then z k x m implies that Z is an LCS of X m-1 and Y. If x m y n, then z k y n implies that Z is an LCS of X and Y n-1. Algorithms Networking Laboratory 32/36

33 Step 2: Recursive Solution (1/2) Overlapping-subproblems Algorithms Networking Laboratory 33/36

34 Step 2: Recursive Solution (2/2) Define c[i, j] = length of LCS for X i and Y j Algorithms Networking Laboratory 34/36

35 Step 3: Computing the length of an LCS b[i, j] points to the table entry corresponding to the optimal subproblem solution chosen when computing c[i, j] LCS-LENGTH(X, Y) is O(mn) Algorithms Networking Laboratory 35/36

36 Step 4: Constructing an LCS PRINT-LCS(b, X, i, j) is O(m+n) Algorithms Networking Laboratory 36/36