Design and Analysis of Algorithms

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Design and Analysis of Algorithms

ผศ. ดร. สมชาย ประสิทธิ์จูตระกูล ภาควิชาวิศวกรรมคอมพิวเตอร

จุฬาลงกรณมหาวิทยาลัย 2542

http://www.cp.eng.chula.ac.th/faculty/spj

คําเตือน

เนื้อหาอันรวมถึงขอความ ตัวเลข สัญลักษณ รูปภาพ และ คําบรรยาย อาจมีขอผิดพลาดแฝงอยู ผูจัดทําจะไมรับผิด ชอบตอความเสียหายทั้งทางดานผลการเรียน สุขภาพกาย และสุขภาพจิตใดๆ อันเนื่องมาจากการใชสื่อการเรียนนี้

Dynamic Programming

Matrix-Chain Multiplication

Outline

• Definition

• Optimal Substructures

• Recursive Solution to Subproblems

• Bottom-Up Dynamic Programming

• Examples

• Top-Down + Memoization

Matrix Multiplication

x x

x x x

x x x

x x

x x 2 x 3 x 4 scalar multiplications [a x b] x [b x c] costs a b c scalar multiplications

Matrix-Chain Multiplication

• A₁ A₂ A₃ A₄

• (A₁(A₂(A₃A₄)))

• (A₁((A₂A₃)A₄))

• ((A₁A₂)(A₃A₄))

• ((A₁(A₂A₃)A₄))

• (((A₁A₂)A₃)A₄)

Choose a full parenthesization which minimizes the number of

scalar multiplications

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( ) ( ) ( )

Example

• A₁ = a 10 × 100 matrix

• A₂ = a 100 × 5 matrix

• A₃ = a 5 × 50 matrix A₁ A₂ A₃

10× 5 10× 100× 5 + 10× 5× 50 A₁( ) A₂ A₃

100× 50 100× 5× 50 + 10× 100× 50

( )( )

Brute Force

• How many full parenthesizations are there in a matrix chain of length n ?

( X ... X X ... X )

n

k n - k

P( n ) P( k ) × P( n-k ) P( n ) = P( k ) P( n-k ) _{k = 1}Σ^n-1 = Ω( 4 ⁿ / n^1.5 )

Notation

• A₁ has dimension p₀× p₁

• A₂ has dimension p₁× p₂

• A_i has dimension p_i-1× p_i

• A_i... A_j has dimension p_i-1× p_j

A₁A₂... A_i ... A_j ... A_n p₀× p₁× p₂... p_i-1× p_i... p_j-1× p_j... × p_n

Notation

• m[ i, j ] = the min. number of scalar mults needed to compute the chain A_i... A_j

• m[ 1, n ] : solution

Optimal Substructure

( A₁ A₂ A₃ A₄ A₅ A₆ ) ( A₁ A₂ A₃)( A₄ A₅ A₆ )

m[1, 3 ] + m[ 4, 6 ] + p₀p₃p₆ p₀×p₃ p₃×p₆ m[ 1, 3 ] m[ 4, 6 ]

Optimal Substructure

( A₁ A₂ A₃ A₄ A₅ A₆ ) m[1,6] = ?

( ( A₁)( A₂ A₃ A₄ A₅ A₆ ) ) m[1,1] + m[2,6] + p₀p₁p₆ ( ( A₁ A₂)( A₃ A₄ A₅ A₆ ) ) m[1,2] + m[3,6] + p₀p₂p₆ ( ( A₁ A₂ A₃)( A₄ A₅ A₆ ) ) m[1,3] + m[4,6] + p₀p₃p₆ ( ( A₁ A₂ A₃ A₄)(A₅ A₆ ) ) m[1,4] + m[5,6] + p₀p₄p₆ ( ( A₁ A₂ A₃ A₄A₅)( A₆ ) ) m[1,5] + m[6,6] + p₀p₅p₆

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Recursive Solution

m[ i, j ] = m[ i, k ] + m[ k +1, j ] + p_i-1p_kp_j ( ( A_i .... A_k )(A_k+1 ... A_j ) )

min { }

i ≤ k < j

m[ i, j ] = 0 if i = j if i < j

R_Matrix_Chain( p, i, j ) {

if i = j then return 0 m = INFINITY

for k = i to j-1

q = R_Matrix_Chain( p, i, k ) + R_matrix_Chain( p, k+1, j ) + p[i-1]*p[k]*p[j]

if q < m then m = q return m

}

m[ i, j ] = min { }m[ i, k ] + m[ k +1, j ] + p_i-1p_kp_j

i ≤ k ^< j

Recursive Algorithm

Recursive Algorithm : Analysis

i ≤ k < j

T( n ) = ( T( k ) + T( n-k ) + O(1) ) _{k = 1}Σ^n-1

= Ω( 2 ⁿ ) [ CLR p.311 ]

Observation

• There are only n² distinct m[ i, j ]’s

• Q : Why does it take Ω( 2ⁿ) time ?

• A : Overlapping subproblems

m[1,6] --> m[1,1], m[2,6], m[1,2], m[3,6], m[1,3], m[4,6], m[1,4], m[5,6], m[1,5], m[6,6]

m[2,6] --> m[2,2], m[3,6], m[2,3], m[4,6], m[2,4], m[5,6], m[2,5], m[6,6]

Computing Optimal Cost : Bottom Up

i ≤ k ^< j

1 2 3 4 5 6 1

2 3 4 5 6

0 0

Solution m[1, 6]

x

x x

Computing Optimal Cost : Bottom Up

i ≤ k ^< j

1 2 3 4 5 6 1

2 3 4 5 6

0 0

Solution m[1, 6]

x

x x

x x x

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Computing Optimal Cost : Bottom Up

i ≤ k < j

1 2 3 4 5 6 1

2 3 4 5 6

0 0

Solution m[1, 6]

x

x x

x x x

x

Computing Optimal Cost : Bottom Up

i ≤ k < j

1 2 3 4 5 6 1

2 3 4 5 6

0 0

Solution m[1, 6]

x

x x

x x x

x x

Computing Optimal Cost : Bottom Up

i ≤ k ^< j

1 2 3 4 5 6 1

2 3 4 5 6

0 0

Solution m[1, 6]

x

x x

x x x

x x

x

Computing Optimal Cost : Bottom Up

i ≤ k ^< j

1 2 3 4 5 6 1

2 3 4 5 6

0 0

Solution m[1, 6]

x

x x

x x x

x x

Computing Optimal Cost : Bottom Up

i ≤ k ^< j

1 2 3 4 5 6 1

2 3 4 5 6

0 0

Solution m[1, 6]

x

x x

x x x

x x

x

Computing Optimal Cost : Bottom Up

i ≤ k ^< j

1 2 3 4 5 6 1

2 3 4 5 6

0 0

Solution m[1, 6]

x

x x

x

x x

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Computing Optimal Cost : Bottom Up

i ≤ k < j

1 2 3 4 5 6 1

2 3 4 5 6

0 0

Solution m[1, 6]

x

x x

x x x

x x

x

Computing Optimal Cost : Bottom Up

i ≤ k < j

1 2 3 4 5 6 1

2 3 4 5 6

0 0

Solution m[1, 6]

x

x x

x

x x

Computing Optimal Cost : Bottom Up

i ≤ k ^< j

1 2 3 4 5 6 1

2 3 4 5 6

0 0

Solution m[1, 6]

x

x x

x x x

x x

x x x

Example

1 2 3 4 5 1

2 3 4 5

i ≤ k ^< j 0

0 0

A₁A₂A₃A₄A₅ 10 × 5× 1 × 5 × 10 × 2

Example

A₁A₂A₃A₄A₅ 10 × 5× 1 × 5 × 10 × 2

1 2 3 4 5 1

2 3 4 5

i ≤ k < j 0

0 0

m[1,2]

= m[1,1] + m[2,2] + 10x5x1 = 0 + 0 + 50

= 50

Example

A₁A₂A₃A₄A₅ 10 × 5× 1 × 5 × 10 × 2

1 2 3 4 5 1

2 3 4 5

i ≤ k < j 0

0 0

m[2,3]

= m[2,2] + m[3,3] + 5x1x5 = 0 + 0 + 25

= 25

150

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Example

A₁A₂A₃A₄A₅ 10 × 5× 1 × 5 × 10 × 2

1 2 3 4 5 1

2 3 4 5

i ≤ k ^< j 0

0 0

m[3,4]

= m[3,3] + m[4,4] + 1x5x10 = 0 + 0 + 50

= 50

50 25 1

2

Example

A₁A₂A₃A₄A₅ 10 × 5× 1 × 5 × 10 × 2

1 2 3 4 5 1

2 3 4 5

i ≤ k ^< j 0

0 0

m[4,5]

= m[4,4] + m[5,5] + 5x10x2 = 0 + 0 + 100

= 100

50 25

50 1

2 3

Example

A₁A₂A₃A₄A₅ 10 × 5× 1 × 5 × 10 × 2

1 2 3 4 5 1

2 3 4 5

i ≤ k ^< j 0

0 0

m[1,3]

m[1,1] + m[2,3] + 10x5x5 = 0 + 25 + 250 = 275 m[1,2] + m[3,3] + 10x1x5 = 50 + 0 + 50 = 100

50 25

50 100 1

2 3

4

Example

A₁A₂A₃A₄A₅ 10 × 5× 1 × 5 × 10 × 2

1 2 3 4 5 1

2 3 4 5

i ≤ k ^< j 0

0 0

m[2,4]

m[2,2] + m[3,4] + 5x1x10 = 0 + 50 + 50 = 100 m[2,3] + m[4,4] + 5x5x10 = 25 + 0 + 250 = 275

50 25

50 100 1 100

2 3

4 1

Example

A₁A₂A₃A₄A₅ 10 × 5× 1 × 5 × 10 × 2

1 2 3 4 5 1

2 3 4 5

i ≤ k < j 0

0 0

m[3,5]

m[3,3] + m[4,5] + 1x5x2 = 0 + 100 + 10 = 110 m[3,4] + m[5,5] + 1x10x2 = 50 + 0 + 20 = 70

50 25

50 100 100

100 1

2 3

4 1

2

Example

A₁A₂A₃A₄A₅ 10 × 5× 1 × 5 × 10 × 2

1 2 3 4 5 1

2 3 4 5

i ≤ k < j 0

0 0

m[1,4]

m[1,1] + m[2,4] + 10x5x10 = 0 + 100 + 500 = 600 m[1,2] + m[3,4] + 10x1x10 = 50 + 50 + 100 = 200 m[1,3] + m[4,4] + 10x5x10 = 100 + 0 + 500 = 600

50 25

50 100 100

100 70 1

2 3

4 1

2 4

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Example

A₁A₂A₃A₄A₅ 10 × 5× 1 × 5 × 10 × 2

1 2 3 4 5 1

2 3 4 5

i ≤ k ^< j 0

0 0

m[2,5]

m[2,2] + m[3,5] + 5x1x2 = 0 + 70 + 10 = 80 m[2,3] + m[4,5] + 5x5x2 = 25 + 100 + 50 = 175 m[2,4] + m[5,5] + 5x10x2 = 100 + 0 + 100 = 200

50 25

50 100 100

100 70 1 200

2 3

4 1

2 4 2

Example

A₁A₂A₃A₄A₅ 10 × 5× 1 × 5 × 10 × 2

1 2 3 4 5 1

2 3 4 5

i ≤ k ^< j 0

0 0

m[1,1] + m[2,5] + 10x5x2 = 0 + 80 + 100 = 180 m[1,2] + m[3,5] + 10x1x2 = 50 + 70 + 20 = 140 m[1,3] + m[4,5] + 10x5x2 = 100 + 100 + 100 = 300 m[1,4] + m[5,5] + 10x10x2 = 200 + 0 + 200 = 400

50 25

50 100 100

100 70 200

80 1

2 3

4 1

2 4 2

2

Example

A₁A₂A₃A₄A₅ 10 × 5× 1 × 5 × 10 × 2

1 2 3 4 5 1

2 3 4 5

i ≤ k ^< j 0

0 0

0 0 50

25 50

100 100

100 70 200

80 1 140

2 3

4 1

2 4 2

2

2 A₁A₂A₃A₄A₅ (A₁A₂)(A₃A₄A₅) (A₁A₂)((A₃A₄)A₅)

Matrix-Chain Mult. Dynamic Prog.

Matrix-Chain-Order( p, n ) {

for i = 1 to n m[i,i] = 0 for len = 2 to n

for i = 1 to n - len + 1 j = i + len - 1

return m[1,n]

}

m[ i, j ] =min { }m[ i, k ] + m[ k +1, j ] + p_i-1p_kp_j

i ≤ k < j

Matrix-Chain Mult. Dynamic Prog.

Matrix-Chain-Order( p, n ) {

for i = 1 to n m[i,i] = 0 for len = 2 to n

for i = 1 to n - len + 1 j = i + len - 1 m[i,j] = ∞ for k = i to j-1

q = m[i,k] + m[k+1,j] + p[i-1]*p[k]*p[j]

if q < m[i,j] then m[i,j] = q s[i,j] = k return s

}

Θ( n³ )

Matrix-Chain Multiplication

Matrix-Chain-Multiply( A, p, n ) {

s = Matrix-Chain-Order( p, n ) Matrix-Chain-Mult( A, s, 1, n ) }

Matrix-Chain-Mult( A, s, i, j ) {

if i < j

X = Matrix-Chain-Mult( A, s, i, s[i,j] ) Y = Matrix-Chain-Mult( A, s, s[i,j]+1, j ) return Matrix-Multiply( X, Y )

else

return A[i]

}

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R_Matrix_Chain( p, i, j ) {

for k = i to j-1

q = R_Matrix_Chain( p, i, k ) + R_matrix_Chain( p, k+1, j ) + p[i-1]*p[k]*p[j]

}

Top-Down Recursive Alg. : Revisited

Ω( 2ⁿ )

Lookup_Chain( p, i, j ) {

for k = i to j-1

q = Lookup_Chain( p, i, k ) + Lookup_Chain( p, k+1, j ) + p[i-1]*p[k]*p[j]

}

Top-Down Recursive Alg. : Revisited

if i = j then return 0 m[i,j] = INFINITY for k = i to j-1

if q < m[i,j] then m[i,j] = q return m[i,j]

}

Top-Down Recursive Alg. : Revisited

if i = j then return 0 m[i,j] = INFINITY for k = i to j-1

}

Top-Down + Memoization

if m[i,j] < INFINITY then return m[i,j]

if i = j then return 0 for k = i to j-1

}

Top-Down + Memoization

Θ( n³ )