Lecture12

Analysis of AlgorithmsCS 477/677

Instructor: Monica Nicolescu

Lecture 12

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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 independent subproblems

– Solve the subproblems recursively

– Combine the solutions to solve the original problem

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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

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Dynamic Programming Algorithm

1. Characterize the structure of an optimal

solution

2. Recursively define the value of an optimal

solution

3. Compute the value of an optimal solution in a

bottom-up fashion

4. Construct an optimal solution from computed

information

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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

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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 e2 and times to exit are x1 and x2

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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

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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?

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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

a1,ja1,j-1

a2,j-1

t2,j-1

S1,jS1,j-1

S2,j-1

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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)

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• 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

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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)

a1,ja1,j-1

a2,j-1

t2,j-1

S1,jS1,j-1

S2,j-1

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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

f1[j]

f2[j]

1 2 3 4 5

increasing j

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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

l1[j]

l2[j]

2 3 4 5

increasing j

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FASTEST-WAY(a, t, e, x, n)1. f1[1] ← e1 + a1,1

2. f2[1] ← e2 + a2,1

3. for j ← 2 to n

4. do if f1[j - 1] + a1,j ≤ f2[j - 1] + t2, j-1 + a1, j

5. then f1[j] ← f1[j - 1] + a1, j

6. l1[j] ← 1

7. else f1[j] ← f2[j - 1] + t2, j-1 + a1, j

8. l1[j] ← 2

9. if f2[j - 1] + a2, j ≤ f1[j - 1] + t1, j-1 + a2, j

10. then f2[j] ← f2[j - 1] + a2, j

11. l2[j] ← 2

12. else f2[j] ← f1[j - 1] + t1, j-1 + a2, j

13. l2[j] ← 1

Compute initial values of f1 and f2

Compute the values of f1[j] and l1[j]

Compute the values of f2[j] and l2[j]

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

14. if f1[n] + x1 ≤ f2[n] + x2

15. then f* = f1[n] + x1

16. l* = 1

17. else f* = f2[n] + x2

18. l* = 2

Compute the values of the fastest time through theentire factory

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4. Construct an Optimal Solution

Alg.: PRINT-STATIONS(l, n)

i ← l* print “line ” i “, station ” n for j ← n downto 2

do i ←li[j]

print “line ” i “, station ” j - 1

f1[j] l1[j]

f2[j] l2[j]

1 2 3 4 5

9

12 16[1]

18[1] 20[2]

22[2]

24[1]

25[1]

32[1]

30[2]l* = 1

line 1, station 5

line 1, station 4

line 1, station 3

line 2, station 2

line 1, station 1

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Dynamic Programming Algorithm

1. Characterize the structure of an optimal solution– Fastest time through a station depends on the fastest time on

previous stations

2. Recursively define the value of an optimal solution– f1[j] = min(f1[j - 1] + a1,j ,f2[j -1] + t2,j-1 + a1,j)

3. Compute the value of an optimal solution in a bottom-up

fashion– Fill in the fastest time table in increasing order of j (station #)

4. Construct an optimal solution from computed information– Use an additional table to help reconstruct the optimal solution

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Matrix-Chain Multiplication

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

A1 A2 An

• Matrix compatibility:

C = A BcolA = rowB

rowC = rowA

colC = colB

A1 A2 Ai Ai+1 An

coli = rowi+1

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• 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

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MATRIX-MULTIPLY(A, B)

if columns[A] rows[B]then error “incompatible dimensions”else for i 1 to rows[A]

do for j 1 to columns[B] do C[i, j] = 0

for k 1 to columns[A]do C[i, j] C[i, j] + A[i, k] B[k,

j]

rows[A]rows[A]

cols[B]cols[B]

i

jj

i

A B C

* =

k

k

rows[A] cols[A] cols[B]multiplications

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Example

A1 A2 A3

• A1: 10 x 100

• A2: 100 x 5

• A3: 5 x 50

1. ((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!!

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• 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 x pn

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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

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Optimal Substructure

Ai…j = Ai…k Ak+1…j

• The parenthesization of the “prefix” Ai…k must 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

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• Subproblem:

determine the minimum cost of parenthesizing

Ai…j = Ai Ai+1 Aj for 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

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• Consider the subproblem of parenthesizing

Ai…j = Ai Ai+1 Aj for 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] =min # of multiplications to compute Ai…k

# of multiplications to compute Ai…kAk…j

min # of multiplications to compute Ak+1…j

m[i, k] m[k+1,j]

pi-1pkpj

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

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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+1 Aj becomes:

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

ik<j

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3. Computing the Optimal Costs

0 if i = jm[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

(n2)1

1

2 3 n

2

3

n

j

i

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3. Computing the Optimal Costs 0 if i = j

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

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3. Computing the Optimal Costs

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

ik<j

• Length = 1: i = j, i = 1, 2, …, n• Length = 2: j = i + 1, i = 1, 2, …, n-11

1

2 3 n

2

3

n

firstse

cond

Compute elements on each diagonal, starting with the longest diagonal.In a similar matrix s we keep the optimal values of k.

m[1, n] gives the optimalsolution to the problem

i

j

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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

1

1

2 3 6

2

3

6

i

j

4 5

4

5

m[2, 5] = min

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

k = 2

k = 3

k = 4

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

1pkpj}

Compute A1 A2 A3

• A1: 10 x 100 (p0 x p1)

• A2: 100 x 5 (p1 x p2)

• A3: 5 x 50 (p2 x 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)

0

0

0

1

1

2

2

3

3

50001

250002

75002

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• 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

k

1

1

2 3 n

2

3

n

j

i

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• 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

1

2 3 n

2

3

n

j

i

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

3 3 3 5 5 -

3 3 3 4 -

3 3 3 -

1 2 -

1 -

-

1

1

2 3 6

2

3

6

i

j

4 5

4

5• 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

CS 477/677 - Lecture 12 3737

Readings

• Chapter 15

Lecture12

Art & Photos

Transcript of Lecture12