Dynamic Programming: The Matrix Chain Algorithm

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Dynamic Programming: The Matrix Chain Algorithm Andreas Klappenecker [partially based on slides by Prof. Welch]

Matrix Chain Problem Suppose that we want to multiply a sequence of rectangular matrices. In which order should we multiply? A x (BxC) or (AxB) x C

Matrices An n x m matrix A over the real numbers is a rectangular array of nm real numbers that are arranged in n rows and m columns. For example, a 3 x 2 matrix A has 6 entries a 11 a 12! A = a 21 a 22 a 31 a 32 where each of the entries a ij is e.g. a real number.

Matrix Multiplication Let A be an n x m matrix B an m x p matrix The product of A and B is n x p matrix AB whose (i,j)-th entry is! k=1 m a ik b kj In other words, we multiply the entries of the i-th row of A with the entries of the j-th column of B and add them up.

Matrix Multiplication

Complexity of Matrix Multiplication Let A be an n x m matrix, B an m x p matrix. Thus, AB is an n x p matrix. Computing the product AB takes nmp scalar multiplications n(m-1)p scalar additions for the standard matrix multiplication algorithm.

Matrix Chain Order Problem Matrix multiplication is associative, meaning that (AB)C = A(BC). Therefore, we have a choice in forming the product of several matrices. What is the least expensive way to form the product of several matrices if the naïve matrix multiplication algorithm is used? [We use the number of scalar multiplications as cost.]

Why Order Matters Suppose we have 4 matrices: A: 30 x 1 B: 1 x 40 C: 40 x 10 D: 10 x 25 ((AB)(CD)) : requires 41,200 scalar multiplications (A((BC)D)) : requires 1400 scalar multiplications

Matrix Chain Order Problem Given matrices A 1, A 2,, A n, where A i is a d i-1 x d i matrix. [1] What is minimum number of scalar multiplications required to compute the product A 1 A 2 A n? [2] What order of matrix multiplications achieves this minimum? We focus on question [1], and sketch an answer to [2].

A Possible Solution Try all possibilities and choose the best one. Drawback: There are too many of them (exponential in the number of matrices to be multiplied) We need to be smarter: Let s try dynamic programming!

Step 1: Develop a Recursive Solution Define M(i,j) to be the minimum number of multiplications needed to compute!! A i A i+1 A j Goal: Find M(1,n). Basis: M(i,i) = 0. Recursion: How can one define M(i,j) recursively?

Defining M(i,j) Recursively Consider all possible ways to split A i through A j into two pieces. Compare the costs of all these splits: best case cost for computing the product of the two pieces plus the cost of multiplying the two products Take the best one M(i,j) = min k (M(i,k) + M(k+1,j) + d i-1 d k d j )

Defining M(i,j) Recursively (A i A k ) (A k+1 A j ) P 1 P 2 minimum cost to compute P 1 is M(i,k) minimum cost to compute P 2 is M(k+1,j) cost to compute P 1 P 2 is d i-1 d k d j

Step 2: Find Dependencies Among Subproblems M: 1 2 3 4 5 1 0 2 n/a 0 3 n/a n/a 0 4 n/a n/a n/a 0 5 n/a n/a n/a n/a 0

Step 2: Find Dependencies Among Subproblems M: 1 2 3 4 5 1 0 GOAL! 2 n/a 0 3 n/a n/a 0 4 n/a n/a n/a 0 5 n/a n/a n/a n/a 0

Step 2: Find Dependencies Among Subproblems M: 1 2 3 4 5 1 0 GOAL! 2 n/a 0 3 n/a n/a 0 4 n/a n/a n/a 0 5 n/a n/a n/a n/a 0

Step 2: Find Dependencies Among Subproblems M: 1 2 3 4 5 1 0 GOAL! 2 n/a 0 3 n/a n/a 0 4 n/a n/a n/a 0 5 n/a n/a n/a n/a 0

Step 2: Find Dependencies Among Subproblems M: 1 2 3 4 5 1 0 2 n/a 0 3 n/a n/a 0 4 n/a n/a n/a 0 5 n/a n/a n/a n/a 0 GOAL! computing the pink square requires the purple ones: to the left and below.

Defining the Dependencies Computing M(i,j) uses everything in same row to the left:! M(i,i), M(i,i+1),, M(i,j-1) and everything in same column below:! M(i,j), M(i+1,j),,M(j,j)

Step 3: Identify Order for Solving Subproblems Recall the dependencies between subproblems just found Solve the subproblems (i.e., fill in the table entries) this way: - go along the diagonal - start just above the main diagonal - end in the upper right corner (goal)

Order for Solving Subproblems M: 1 2 3 4 5 1 0 2 n/a 0 3 n/a n/a 0 4 n/a n/a n/a 0 5 n/a n/a n/a n/a 0

Order for Solving Subproblems M: 1 2 3 4 5 1 0 2 n/a 0 3 n/a n/a 0 1 4 n/a n/a n/a 0 5 n/a n/a n/a n/a 0

Order for Solving Subproblems M: 1 2 3 4 5 1 0 2 n/a 0 3 n/a n/a 0 1 2 4 n/a n/a n/a 0 5 n/a n/a n/a n/a 0

Order for Solving Subproblems M: 1 2 3 4 5 1 0 2 n/a 0 3 n/a n/a 0 1 2 3 4 n/a n/a n/a 0 5 n/a n/a n/a n/a 0

Order for Solving Subproblems M: 1 2 3 4 5 1 0 2 n/a 0 3 n/a n/a 0 1 2 3 4 4 n/a n/a n/a 0 5 n/a n/a n/a n/a 0

Pseudocode for i := 1 to n do M[i,i] := 0 for d := 1 to n-1 do // diagonals for i := 1 to n-d to // rows w/ an entry on d-th diagonal j := i + d // column corresp. to row i on d-th diagonal M[i,j] := infinity for k := i to j-1 to M[i,j] := min(m[i,j], M[i,k]+M[k+1,j]+d i-1 d k d j ) endfor endfor endfor

Pseudocode for i := 1 to n do M[i,i] := 0 for d := 1 to n-1 do // diagonals for i := 1 to n-d to // rows w/ an entry on d-th diagonal j := i + d // column corresp. to row i on d-th diagonal M[i,j] := infinity for k := i to j-1 to M[i,j] := min(m[i,j], M[i,k]+M[k+1,j]+d i-1 d k d j ) endfor endfor endfor running time O(n 3 )

Pseudocode for i := 1 to n do M[i,i] := 0 for d := 1 to n-1 do // diagonals for i := 1 to n-d to // rows w/ an entry on d-th diagonal j := i + d // column corresp. to row i on d-th diagonal M[i,j] := infinity for k := i to j-1 to M[i,j] := min(m[i,j], M[i,k]+M[k+1,j]+d i-1 d k d j ) pay attention here to remember actual sequence of mults. endfor endfor endfor running time O(n 3 )

Example M: 1 2 3 4 1 0 1200 700 1400 2 n/a 0 400 650 3 n/a n/a 0 10,000 4 n/a n/a n/a 0 1: A is 30x1 2: B is 1x40 3: C is 40x10 4: D is 10x25 BxC: 1x40x10 (BxC)xD: 400 + 1x10x25 Bx(CxD):... + 10,000

Keeping Track of the Order It's fine to know the cost of the cheapest order, but what is that cheapest order? Keep another array S and update it when computing the minimum cost in the inner loop After M and S have been filled in, then call a recursive algorithm on S to print out the actual order

Modified Pseudocode for i := 1 to n do M[i,i] := 0 for d := 1 to n-1 do // diagonals for i := 1 to n-d to // rows w/ an entry on d-th diagonal j := i + d // column corresponding to row i on d-th diagonal M[i,j] := infinity for k := i to j-1 to M[i,j] := min(m[i,j], M[i,k]+M[k+1,j]+d i-1 d k d j ) if previous line changed value of M[i,j] then S[i,j] := k endfor endfor endfor keep track of cheapest split point found so far: between A k and A k+1

Example M: 1 2 3 4 S: 1 0 1200 700 1400 2 n/a 0 400 650 1 3 n/a n/a 0 10,000 4 n/a n/a n/a 0 1 2 1 3 3 1: A is 30x1 2: B is 1x40 3: C is 40x10 4: D is 10x25 A x (BCD) A x ((BC) x D) A x ((BxC) x D)

Using S to Print Best Ordering Call Print(S,1,n) to get the entire ordering. Print(S,i,j): if i = j then output "A" + i //+ is string concat else k := S[i,j] output "(" + Print(S,i,k) + Print(S,k+1,j) + ")"

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