1 Chapter 15-2: Dynamic Programming II

2 Matrix Multiplication Let A be a matrix of dimension p x q and B be a matrix of dimension q x r Then, if we multiply matrices A and B, we obtain a resulting matrix C = AB whose dimension is p x r We can obtain each entry in C using q operations  in total, pqr operations

3 Matrix Multiplication Example : How to obtain c 1,2 ?

4 Matrix Multiplication In fact, ((A 1 A 2 )A 3 ) = (A 1 (A 2 A 3 )) so that matrix multiplication is associative  Any way to write down the parentheses gives the same result E.g., (((A 1 A 2 )A 3 )A 4 ) = ((A 1 A 2 )(A 3 A 4 )) = (A 1 ((A 2 A 3 )A 4 )) = ((A 1 (A 2 A 3 ))A 4 ) = (A 1 (A 2 (A 3 A 4 )))

5 Matrix Multiplication Question: Why do we bother this? Because different computation sequence may use different number of operations! E.g., Let the dimensions of A 1, A 2, A 3 be: 1 x 100, 100 x 1, 1 x 100, respectively #operations to get ((A 1 A 2 )A 3 ) = ?? #operations to get (A 1 (A 2 A 3 )) = ??

6 Lemma: Suppose that to multiply B 1,B 2,…,B n, the way with minimum #operations is to: (i) first, obtain B 1 B 2 … B x (ii) then, obtain B x+1 … B n (iii) finally, multiply the matrices of part (i) and part (ii) Then, the matrices in part (i) and part (ii) must be obtained with min #operations Optimal Substructure (allows recursion)

7 Let f i,j denote the min #operations to obtain the product A i A i+1 … A j  f i,i = 0 Let r k and c k denote #rows and #cols of A k Then, we have: Optimal Substructure Lemma: For any j > i, f i,j = min x { f i,x + f x+1,j + r i c x c j }

8 Define a function Compute_F(i,j) as follows: Compute_F(i, j) /* Finding f i,j */ 1. if (i == j) return 0; 2. m =  ; 3. for (x = i, i+1, …, j-1) { g = Compute_F(i,x) + Compute_F(x+1,j) + r i c x c j ; if (g  m) m = g; } 4. return m ; Recursive-Matrix-Chain

9 The recursion tree for the computation of RECURSUVE-MATRIX-CHAIN(P, 1, 4)

10 Time Complexity Question: Time to get Compute_F(1,n)? By substituion method, we can show that Running time =  (3 n )

11 Counting the number of parenthesizations Remark: On the other hand, #operations for each possible way of writing parentheses are computed at most once  Running time = O( C(2n-2,n-1)/n )  Catalan Number

12 Overlapping Subproblems Here, we can see that : To Compute_F(i,j) and Compute_F(i,j+1), both have many COMMON subproblems: Compute_F(i,i+1), …, Compute_F(i,j-1) So, in our recursive algorithm, there are many redundant computations ! Question: Can we avoid it ?

13 Bottom-Up Approach We notice that f i,j depends only on f x,y with y-x  j-i Let us create a 2D table F to store all f i,j values once they are computed Then, compute f i,j for j-i = 1,2,…,n-1

14 BottomUp_F( ) /* Finding min #operations */ 1. for j = 1, 2, …, n, set F[j, j] = 0 ; 2. for (length = 1,2,…, n-1) { Compute F[i,i+length] for all i; // Based on F[x,y] with |x-y| < length } 3. return F[1,n] ; Running Time =  (n 3 ) Bottom-Up Approach

15 Example:

16 The m and s table computed by MATRIX-CHAIN-ORDER for n = 6 Optimal Solution:

17 Example m[2,5]= Min { m[2,2]+m[3,5]+p 1 p 2 p 5 =  15  20=13000, m[2,3]+m[4,5]+p 1 p 3 p 5 =  5  20=7125, m[2,4]+m[5,5]+p 1 p 4 p 5 =  10  20=11374 } =7125

18 Remarks Again, a slight change in the algorithm allows us to get the exact sequence of steps (or the parentheses) that achieves the minimum number of operations Also, we can make minor changes to the recursive algorithm and obtain a memoized version (whose running time is O(n 3 ))

19 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

20 When should we apply DP? Optimal structure: an optimal solution to the problem contains optimal solutions to subproblems. –Example: Matrix-multiplication problem Overlapping subproblems: a recursive algorithm revisits the same subproblem over and over again.

21 Common pattern in discovering optimal substructure 1. Show that a solution to a problem consists of making a choice, which leaves one or subproblems to solve. 2. Suppose that for a given problem, you are given the choice that leads to an optimal solution. You do not concern how to determine this choice. You just assume that it has been given to you. 3. Given this choice, determine which subproblems ensure and how to best characterize the resulting (solution) space of subproblems.

22 Common pattern in discovering optimal substructure 4. Show that the solutions to the subproblems used within the optimal solution must themselves be optimal. Usually use “ cut-and- paste ” technique.

23 Optimal substructure Optimal substructure varies across problem domains in two ways: 1. how many subproblems are used in an optimal solution to the original problem, and 2. how many choices we have in determining which subproblem(s) to use in an optimal solution. Informally, running time depends on (# of subproblems overall) x (# of choices).

24 Refinement 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.

25 Shortest path Shortest path has optimal substructure. Suppose p is shortest path u -> v. Let w be any vertex on p. Let p 1 be the portion of p going u -> w. Then p 1 is a shortest path u -> w.

26 Longest simple path Does longest path have optimal substructure? Consider q -> r -> t = longest path q -> t. Are its subpaths longest paths? No! Longest simple path q -> r is q -> s -> t -> r. Longest simple path r -> t is r -> q ->s -> t. Not only isn’t there optimal substructure, but we can’t even assemble a legal solution from solutions to subproblems.

27 Overlapping subproblems These occur when a recursive algorithm revisits the same problem over and over. Solutions 1. Bottom up 2. Memorization (memorize the natural, but inefficient)

28 Top down approach RECURSIVE_MATRIX_CHAIN(p, i, j) 1 if i = j 2 then return 0 3 m[i, j]   4 for k  i to j – 1 5 do q  RMC(p,i,k) + RMC(p,k+1,j) + p i-1 p k p j 6 if q < m[i, j] 7 then m[i, j]  q 8 return m[i, j]

29 The recursion tree for the computation of RECURSUVE-MATRIX-CHAIN(P, 1, 4)

30 Memoization Alternative approach to dynamic programming:  “Store, don’t recompute.”  Make a table indexed by subproblem.  When solving a subproblem: Lookup in table.  If answer is there, use it.  Else, compute answer, then store it.  In bottom-up dynamic programming, we go one step further. We determine in what order we’d want to access the table, and fill it in that way.

31 MEMORIZED_MATRIX_CHAIN MEMORIZED_MATRIX_CHAIN(p) 1 n  length[p] –1 2 for i  1 to n 3 do for j  1 to n 4 do m[i, j]   5 return LC (m,p,1,n)

32 LOOKUP_CHAIN LOOKUP_CHAIN(m, p,i,j) 1if m[i, j] <  2then return m[i, j] 3if i = j 4 then m[i, j]  0 5 else for k  i to j – 1 6 do q  LC(m,p,i,k) +LC(m,p,k+1,j)+p i-1 p k p j 7 if q < m[i, j] 8 then m[i, j]  q 9return m[i, j] Time Complexity :

33 Homework Problem: 15.2 Practice at home: , , ,