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Dynamic Programming

- An algorithm design technique for optimization problems (similar to divide and conquer)
- Divide and conquer
- Partition the problem into independentsubproblems
- Solve the subproblems recursively
- Combine the solutions to solve the original problem

CS 477/677 - Lecture 12

Dynamic Programming

- Used for optimization problems
- Find a solution with the optimal value (minimum or maximum)
- A set of choices must be made to get an optimal solution
- There may be many solutions that return the optimal value: we want to find one of them
- Applicable when subproblems are not independent
- Subproblems share subsubproblems

CS 477/677 - Lecture 12

Dynamic Programming Algorithm

- Characterize the structure of an optimal solution
- Recursively define the value of an optimal solution
- Compute the value of an optimal solution in a bottom-up fashion
- Construct an optimal solution from computed information

CS 477/677 - Lecture 12

Elements of Dynamic Programming

- Optimal Substructure
- An optimal solution to a problem contains within it an optimal solution to subproblems
- Optimal solution to the entire problem is built in a bottom-up manner from optimal solutions to subproblems
- Overlapping Subproblems
- If a recursive algorithm revisits the same subproblems again and again the problem has overlapping subproblems

CS 477/677 - Lecture 12

Assembly Line Scheduling

- Automobile factory with two assembly lines
- Each line has n stations: S1,1, . . . , S1,n and S2,1, . . . , S2,n
- Corresponding stations S1, j and S2, j perform the same function but can take different amounts of time a1, j and a2, j
- Times to enter are e1 and e2and times to exit are x1 and x2

CS 477/677 - Lecture 12

Assembly Line

- After going through a station, the car can either:
- stay on same line at no cost, or
- transfer to other line: cost after Si,j is ti,j , i = 1, 2, j = 1, . . . , n-1

CS 477/677 - Lecture 12

Assembly Line Scheduling

- Problem:

What stations should be chosen from line 1 and what from line 2 in order to minimize the total time through the factory for one car?

CS 477/677 - Lecture 12

S1,j

a1,j-1

a1,j

t2,j-1

a2,j-1

S2,j-1

1. Structure of the Optimal Solution- Let’s consider all possible ways to get from the starting point through station S1,j
- We have two choices of how to get to S1, j:
- Through S1, j - 1, then directly to S1, j
- Through S2, j - 1, then transfer over to S1, j

CS 477/677 - Lecture 12

2. A Recursive Solution

- f* = the fastest time to get through the entire factory
- fi[j] = the fastest time to get from the starting point through station Si,j

f* = min (f1[n] + x1, f2[n] + x2)

CS 477/677 - Lecture 12

2. A Recursive Solution

- fi[j] = the fastest time to get from the starting point through station Si,j
- j = 1 (getting through station 1)

f1[1] = e1 + a1,1

f2[1] = e2 + a2,1

CS 477/677 - Lecture 12

S1,j

a1,j-1

a1,j

t2,j-1

a2,j-1

S2,j-1

2. A Recursive Solution- Compute fi[j] for j = 2, 3, …,n, and i = 1, 2
- Fastest way through S1, j is either:
- the way through S1, j - 1 then directly through S1, j, or

f1[j - 1] + a1,j

- the way through S2, j - 1, transfer from line 2 to line 1, then through S1, j

f2[j -1] + t2,j-1 + a1,j

f1[j] = min(f1[j - 1] + a1,j ,f2[j -1] + t2,j-1 + a1,j)

CS 477/677 - Lecture 12

3. Computing the Optimal Solution

- For j ≥ 2, each value fi[j] depends only on the values of f1[j – 1] and f2[j - 1]
- Compute the values of fi[j]
- in increasing order of j
- Bottom-up approach
- First find optimal solutions to subproblems
- Find an optimal solution to the problem from the subproblems

1

2

3

4

5

f1[j]

f2[j]

CS 477/677 - Lecture 12

4. Construct the Optimal Solution

- We need the information about which line has been used at each station:
- li[j] – the line number (1, 2) whose station (j - 1) has been used to get in fastest time through Si,j, j = 2, 3, …, n
- l* – the line number (1, 2) whose station n has been used to get in fastest time through the exit point

2

3

4

5

l1[j]

l2[j]

CS 477/677 - Lecture 12

Compute initial values of f1 and f2

FASTEST-WAY(a, t, e, x, n)- f1[1] ← e1 + a1,1
- f2[1] ← e2 + a2,1
- for j ← 2to n
- do if f1[j - 1] + a1,j ≤ f2[j - 1] + t2, j-1 + a1, j
- then f1[j] ← f1[j - 1] + a1, j
- l1[j] ← 1
- else f1[j] ← f2[j - 1] + t2, j-1 + a1, j
- l1[j] ← 2
- if f2[j - 1] + a2, j ≤ f1[j - 1] + t1, j-1 + a2, j
- then f2[j] ← f2[j - 1] + a2, j
- l2[j] ← 2
- else f2[j] ← f1[j - 1] + t1, j-1 + a2, j
- l2[j] ← 1

Compute the values of

f1[j] and l1[j]

Compute the values of

f2[j] and l2[j]

CS 477/677 - Lecture 12

FASTEST-WAY(a, t, e, x, n) (cont.)

- if f1[n] + x1 ≤ f2[n] + x2
- then f* = f1[n] + x1
- l* = 1
- else f* = f2[n] + x2
- l* = 2

Compute the values of

the fastest time through the

entire factory

CS 477/677 - Lecture 12

4. Construct an Optimal Solution

Alg.: PRINT-STATIONS(l, n)

i ← l*

print “line ” i “, station ” n

for j ← ndownto 2

do i ←li[j]

print “line ” i “, station ” j - 1

line 1, station 5

line 1, station 4

line 1, station 3

line 2, station 2

line 1, station 1

1

2

3

4

5

f1[j] l1[j]

9

18[1]

20[2]

24[1]

32[1]

l* = 1

f2[j] l2[j]

12

16[1]

22[2]

25[1]

30[2]

CS 477/677 - Lecture 12

Dynamic Programming Algorithm

- Characterize the structure of an optimal solution
- Fastest time through a station depends on the fastest time on previous stations
- Recursively define the value of an optimal solution
- f1[j] = min(f1[j - 1] + a1,j ,f2[j -1] + t2,j-1 + a1,j)
- Compute the value of an optimal solution in a bottom-up fashion
- Fill in the fastest time table in increasing order of j (station #)
- Construct an optimal solution from computed information
- Use an additional table to help reconstruct the optimal solution

CS 477/677 - Lecture 12

Matrix-Chain Multiplication

Problem: given a sequence A1, A2, …, An of matrices, compute the product:

A1 A2 An

- Matrix compatibility:

C = A B

colA = rowB

rowC = rowA

colC = colB

A1 A2 Ai Ai+1 An

coli = rowi+1

CS 477/677 - Lecture 12

Matrix-Chain Multiplication

- In what order should we multiply the matrices?

A1 A2 An

- Matrix multiplication is associative:
- E.g.:A1 A2 A3 = ((A1 A2) A3)

= (A1 (A2 A3))

- Which one of these orderings should we choose?
- The order in which we multiply the matrices has a significant impact on the overall cost of executing the entire chain of multiplications

CS 477/677 - Lecture 12

multiplications

k

k

MATRIX-MULTIPLY(A, B)ifcolumns[A] rows[B]

then error“incompatible dimensions”

else fori 1 to rows[A]

do forj 1 to columns[B]

doC[i, j] = 0

fork 1 to columns[A]

doC[i, j] C[i, j] + A[i, k] B[k, j]

j

cols[B]

j

cols[B]

i

i

=

*

B

A

C

rows[A]

rows[A]

CS 477/677 - Lecture 12

Example

A1 A2 A3

- A1: 10 x 100
- A2: 100 x 5
- A3: 5 x 50
- ((A1 A2) A3): A1 A2 takes 10 x 100 x 5 = 5,000

(its size is 10 x 5)

((A1 A2) A3) takes 10 x 5 x 50 = 2,500

Total: 7,500 scalar multiplications

2. (A1 (A2 A3)): A2 A3 takes 100 x 5 x 50 = 25,000

(its size is 100 x 50)

(A1 (A2 A3)) takes 10 x 100 x 50 = 50,000

Total: 75,000 scalar multiplications

one order of magnitude difference!!

CS 477/677 - Lecture 12

Matrix-Chain Multiplication

- Given a chain of matrices A1, A2, …, An, where for i = 1, 2, …, n matrix Ai has dimensions pi-1x pi, fully parenthesize the product A1 A2 An in a way that minimizes the number of scalar multiplications.

A1 A2 Ai Ai+1 An

p0 x p1 p1 x p2 pi-1 x pi pi x pi+1 pn-1 xpn

CS 477/677 - Lecture 12

1. The Structure of an Optimal Parenthesization

- Notation:

Ai…j = Ai Ai+1 Aj, i j

- For i < j:

Ai…j = Ai Ai+1 Aj

= Ai Ai+1 Ak Ak+1 Aj

= Ai…k Ak+1…j

- Suppose that an optimal parenthesization of Ai…j splits the product between Ak and Ak+1, where i k < j

CS 477/677 - Lecture 12

Optimal Substructure

Ai…j= Ai…k Ak+1…j

- The parenthesization of the “prefix” Ai…kmust be an optimal parentesization
- If there were a less costly way to parenthesize Ai…k, we could substitute that one in the parenthesization of Ai…j and produce a parenthesization with a lower cost than the optimum contradiction!
- An optimal solution to an instance of the matrix-chain multiplication contains within it optimal solutions to subproblems

CS 477/677 - Lecture 12

2. A Recursive Solution

- Subproblem:

determine the minimum cost of parenthesizing Ai…j = Ai Ai+1 Ajfor 1 i j n

- Let m[i, j] = the minimum number of multiplications needed to compute Ai…j
- Full problem (A1..n): m[1, n]
- i = j: Ai…i = Ai m[i, i] =

- 0, for i = 1, 2, …, n

CS 477/677 - Lecture 12

m[i, k]

m[k+1,j]

2. A Recursive Solution- Consider the subproblem of parenthesizing Ai…j = Ai Ai+1 Ajfor 1 i j n

= Ai…k Ak+1…j for i k < j

- Assume that the optimal parenthesization splits the product Ai Ai+1 Aj at k (i k < j)

m[i, j] =

m[i, k] + m[k+1, j] + pi-1pkpj

min # of multiplications

to compute Ai…k

min # of multiplications

to compute Ak+1…j

# of multiplications

to computeAi…kAk…j

CS 477/677 - Lecture 12

2. A Recursive Solution

m[i, j] = m[i, k] + m[k+1, j] + pi-1pkpj

- We do not know the value of k
- There are j – i possible values for k: k = i, i+1, …, j-1
- Minimizing the cost of parenthesizing the product Ai Ai+1Aj becomes:

0if i = j

m[i, j]= min {m[i, k] + m[k+1, j] + pi-1pkpj}ifi < j

ik<j

CS 477/677 - Lecture 12

3. Computing the Optimal Costs

0if i = j

m[i, j]= min {m[i, k] + m[k+1, j] + pi-1pkpj}if i < j

ik<j

- How many subproblems do we have?
- Parenthesize Ai…j

for 1 i j n

- One subproblem for each

choice of i and j

1

2

3

n

(n2)

n

j

3

2

1

CS 477/677 - Lecture 12

i

3. Computing the Optimal Costs

0if i = j

m[i, j]= min {m[i, k] + m[k+1, j] + pi-1pkpj}ifi < j

ik<j

- How do we fill in table m[1..n, 1..n]?
- Determine which entries of the table are used in computing m[i, j]

Ai…j= Ai…k Ak+1…j

- Fill in m such that it corresponds to solving problems of increasing length

CS 477/677 - Lecture 12

first

m[1, n] gives the optimal

solution to the problem

3. Computing the Optimal Costs0if i = j

m[i, j]= min {m[i, k] + m[k+1, j] + pi-1pkpj}ifi < j

ik<j

- Length = 1: i = j, i = 1, 2, …, n
- Length = 2: j = i + 1, i = 1, 2, …, n-1

1

2

3

n

n

j

3

Compute elements on each diagonal, starting with the longest diagonal.

In a similar matrix s we keep the optimal values of k.

2

1

i

CS 477/677 - Lecture 12

Example: min {m[i, k] + m[k+1, j] + pi-1pkpj}

m[2, 2] + m[3, 5] + p1p2p5

m[2, 3] + m[4, 5] + p1p3p5

m[2, 4] + m[5, 5] + p1p4p5

k = 2

m[2, 5] = min

k = 3

k = 4

1

2

3

4

5

6

6

5

- Values m[i, j] depend only on values that have been previously computed

4

j

3

2

1

i

CS 477/677 - Lecture 12

0

25000

2

1

0

7500

5000

0

Example min {m[i, k] + m[k+1, j] + pi-1pkpj}1

2

3

Compute A1 A2 A3

- A1: 10 x 100 (p0x p1)
- A2: 100 x 5 (p1x p2)
- A3: 5 x 50 (p2x p3)

m[i, i] = 0 for i = 1, 2, 3

m[1, 2] = m[1, 1] + m[2, 2] + p0p1p2 (A1A2)

= 0 + 0 + 10 *100* 5 = 5,000

m[2, 3] = m[2, 2] + m[3, 3] + p1p2p3 (A2A3)

= 0 + 0 + 100 * 5 * 50 = 25,000

m[1, 3] = min m[1, 1] + m[2, 3] + p0p1p3 = 75,000 (A1(A2A3))

m[1, 2] + m[3, 3] + p0p2p3 = 7,500 ((A1A2)A3)

3

2

1

CS 477/677 - Lecture 12

4. Construct the Optimal Solution

- Store the optimal choice made at each subproblem
- s[i, j] = a value of k such that an optimal parenthesization of Ai..j splits the product between Ak and Ak+1

1

2

3

n

n

j

3

2

1

i

CS 477/677 - Lecture 12

4. Construct the Optimal Solution

- s[1, n] is associated with the entire product A1..n
- The final matrix multiplication will be split at k = s[1, n]

A1..n = A1..k Ak+1..n

A1..n = A1..s[1, n] As[1, n]+1..n

- For each subproduct recursively find the corresponding value of k that results in an optimal parenthesization

1

2

3

n

n

j

3

2

1

i

CS 477/677 - Lecture 12

4. Construct the Optimal Solution

- s[i, j] = value of k such that the optimal parenthesization of Ai Ai+1 Aj splits the product between Ak and Ak+1

1

2

3

4

5

6

6

- s[1, n] = 3 A1..6 = A1..3 A4..6
- s[1, 3] = 1 A1..3 = A1..1 A2..3
- s[4, 6] = 5 A4..6 = A4..5 A6..6

5

4

3

j

2

1

i

CS 477/677 - Lecture 12

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