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Intermediate Information Structures LECTURE 15

Chain Matrix Multiplication

Jon Rokne

Computer Science

University of Calgary

Canada

CPSC 335

Mostly from: https://www.cse.ust.hk/~dekai/271/notes/L12/L12.pdf‎

Outline of this Lecture

Recalling matrix multiplication.

The chain matrix multiplication problem.

A dynamic programming algorithm for chain matrix multiplication.

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Recalling Matrix Multiplication

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is a two-

.

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n  Example: consider the chain A1, A2, A3, A4 of 4 matrices

n  Let us compute the product A1A2A3A4

•  5 different orderings = 5 different parenthesizations

1.  (A1(A2(A3A4)))

2.  (A1((A2A3)A4))

3.  ((A1A2)(A3A4))

4.  ((A1(A2A3))A4)

5.  (((A1A2)A3)A4)

Matrix Chain Multiplication cont..

•  Matrix multiplication is associative , e.g.,

so parenthenization does not change result.

•  To compute the number of scalar multiplications necessary, we must know: •  Algorithm to multiply two matrices •  Matrix dimensions

Algorithm..

Matrix-Multiply(A,B)

1.If columns[A]!=rows[B]

2. then error “incomplete 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 C[i,j]=C[i,j]+A[i,k]*B[k,j]

8. Return C

• The time to compute C is dominated by the number of scalar multiplications. • To illustrate the different costs incurred by different paranthesization of a

matrix product. Example: Consider three matrices A10×100, B100×5, and C5×50 There are 2 ways to parenthesize

((AB)C) = D10×5 · C5×50

AB ⇒ 10·100·5=5,000 scalar multiplications

DC ⇒ 10·5·50 =2,500 scalar multiplications

(A(BC)) = A10×100 · E100×50

BC ⇒ 100·5·50=25,000 scalar multiplications Total: 75,000

AE ⇒ 10·100·50 =50,000 scalar multiplications

Total: 7,500

•  Matrix-chain multiplication problem ü  Given a chain A1, A2, …, An of n matrices, where for i=1, 2, …, n,

matrix Ai has dimension pi-1×pi ü  Parenthesize the product A1A2…An such that the total number of

scalar multiplications is minimized

•  Before solving by Dynamic programming exhaustively check all

paranthesizations. •  P(n) : paranthesization of a sequence of n matrices

Counting the Number of Parenthesizations

• We obtain the recurrence

P nif n

P k p n k if nk

n( ) ( ) ( )==

−∑ ≥

⎧

⎨⎪

⎩⎪ =

−1 1

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1.The Structure of an Optimal Parenthesization Step 1: Characterize the structure of an optimal solution • Ai..j : matrix that results from evaluating the product Ai Ai+1 Ai+2 ... Aj

• An optimal parenthesization of the product A1A2 ... An

– Splits the product between Ak and Ak+1, for some 1≤k<n

(A1A2A3 ... Ak) · (Ak+1Ak+2 ... An)

– i.e., first compute A1..k and Ak+1..n and then multiply these two •  The cost of this optimal parenthesization Cost of computing A1..k + Cost of computing Ak+1..n + Cost of multiplying A1..k · Ak+1..n

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n  Idea #1: repeatedly select the product that uses the fewest operations.

n  Counter-example: n  A is 101 × 11 n  B is 11 × 9 n  C is 9 × 100 n  D is 100 × 99 n  Greedy idea #1 gives A*((B*C)*D)), which takes

109989+9900+108900=228789 ops n  (A*B)*(C*D) takes 9999+89991+89100=189090 ops

n  The greedy approach is not giving us the optimal value.

Greedy Approach

Is there a better approach? Yes – Dynamic Programming

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We store the subproblems in an array.

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1..4

1..1 2..4 1..2 3..4 1..3 4..4

2..2 3..4 2..3 4..4 3..3 4..4 1..1 2..2

3..3 4..4 2..2 3..3

...

Subproblem Overlap

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7125}

1137520*10*3504375,712520*5*3510002625,1300020*15*3525000

min{

5414,43,1

5314,32,1

5214,21,1

4,1

=

=++=++

=++=++

=++=++

=

dddNNdddNNdddNN

N

n  A0: 30 X 35; A1: 35 X15; A2: 15X5; A3: 5X10; A4: 10X20; A5: 20 X 25

}{min 11,1,, +++<≤

++= jkijkkijkiji dddNNN

Algorithm Visualization – second example

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n  A0: 30 X 35; A1: 35 X15; A2: 15X5; A3: 5X10; A4: 10X20; A5: 20 X 25

Algorithm Visualization – second example

(A0*(A1*A2))*((A3*A4)*A5)

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