Analysis of Algorithms
CS 477/677
Instructor: Monica Nicolescu
Lecture 14
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
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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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One Solution
• Brute force
– Enumerate all possibilities of selecting stations
– Compute how long it takes in each case and
choose the best one
1
2
3
4
n
1
0
0
1
1
0 if choosing line 2
at step j (= 3)
1 if choosing line 1
at step j (= n)
– There are 2n possible ways to choose stations
– Infeasible when n is large
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1. Structure of the Optimal Solution
• How do we compute the minimum time of going
through the station?
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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
S1,j-1
S1,j
a1,j-1
a1,j
t2,j-1
a2,j-1
S2,j-1
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1. Structure of the Optimal Solution
• Suppose that the fastest way through S1, j is
through S1, j – 1
– We must have taken the fastest way from entry through S1, j – 1
– If there were a faster way through S1, j - 1, we would use it
instead
• Similarly for S2, j – 1
S1,j-1
S1,j
a1,j-1
a1,j
t2,j-1
a2,j-1
S2,j-1
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Optimal Substructure
• Generalization: an optimal solution to the
problem find the fastest way through S1, j contains
within it an optimal solution to subproblems:
find the fastest way through S1, j - 1 or S2, j - 1.
• This is referred to as the optimal substructure
property
• We use this property to construct an optimal
solution to a problem from optimal solutions
to subproblems
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2. A Recursive Solution
• Define the value of an optimal solution in terms of the
optimal solution to subproblems
• Assembly line subproblems
– Finding the fastest way through each station j on each line i
(i = 1,2, j = 1, 2, …, n)
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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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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
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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
S1,j-1
S1,j
a1,j-1
a1,j
t2,j-1
f1[j] = min(f1[j - 1] + a1,j ,f2[j -1] + t2,j-1 + a1,j)
a2,j-1
S2,j-1
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2. A Recursive Solution
f1[j] =
f2[j] =
e1 + a1,1
if j = 1
min(f1[j - 1] + a1,j ,f2[j -1] + t2,j-1 + a1,j) if j ≥ 2
e2 + a2,1
if j = 1
min(f2[j - 1] + a2,j ,f1[j -1] + t1,j-1 + a2,j) if j ≥ 2
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3. Computing the Optimal Value
f* = min (f1[n] + x1, f2[n] + x2)
f1[j] = min(f1[j - 1] + a1,j ,f2[j -1] + t2,j-1 + a1,j)
1
2
3
4
5
f1[j]
f1(1)
f1(2)
f1(3)
f1(4)
f1(5)
f2[j]
f2(1)
f2(2)
f2(3)
f2(4)
f2(5)
4 times
2 times
• Solving top-down would result in
exponential running time
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3. Computing the Optimal Value
• 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
increasing j
1
2
3
4
5
f1[j]
f2[j]
• Bottom-up approach
– First find optimal solutions to subproblems
– Find an optimal solution to the problem from the subproblems
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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
increasing j
2
3
4
5
l1[j]
l2[j]
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FASTEST-WAY(a, t, e, x, n)
1. f1[1] ← e1 + a1,1
2. f2[1] ← e2 + a2,1
Compute initial values of f1 and f2
3. for j ← 2 to n
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
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
Compute the values of
f1[j] and l1[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
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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.
16.
17.
18.
then f* = f1[n] + x1
l* = 1
else f* = f2[n] + x2
Compute the values of
the fastest time through the
entire factory
l* = 2
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Example
e1 + a1,1,
f1[j] = min(f1[j - 1] + a1,j ,f2[j -1] + t2,j-1 + a1,j)
1
2
3
4
5
f1[j]
9
18[1]
20[2]
24[1]
32[1]
f2[j]
12
16[1]
22[2]
25[1]
30[2]
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if j = 1
if j ≥ 2
f* = 35[1]
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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
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]
f2[j] l2[j]
12
16[1]
22[2]
25[1]
30[2]
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l* = 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 bottomup 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∙B
colA = rowB
rowC = rowA
colC = colB
A1 ∙ A2 ∙ Ai ∙ Ai+1 ∙∙∙ An
coli = rowi+1
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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
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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]
rows[A] ∙ cols[A] ∙ cols[B]
multiplications
do C[i, j] = 0
for k ← 1 to columns[A]
do C[i, j] ←C[i, j] + A[i, k] B[k, j]
k
j
i
rows[A]
*
A
j
cols[B]
cols[B]
= i
k
B
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rows[A]
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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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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
p0 x p1
p1 x p2
∙∙∙ Ai ∙ Ai+1
pi-1 x pi
∙∙∙ An
pi x pi+1
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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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2. A Recursive Solution
• 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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2. A Recursive Solution
Consider the subproblem of parenthesizing
Ai…j = Ai Ai+1 ∙∙∙Aj
= Ai…k Ak+1…j
m[i, k]
for 1 ≤ i ≤ j ≤ n
pi-1pkpj
for i ≤ k < j
m[k+1,j]
• Assume that the optimal parenthesization
splits the product Ai Ai+1 ∙∙∙ Aj at k (i ≤ k < j)
m[i, j] = m[i, k]
min # of multiplications
to compute Ai…k
+
m[k+1, j]
+
pi-1pkpj
min # of multiplications # of multiplications
to compute Ak+1…j
to compute
Ai…kAk…j
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Readings
• Chapters 14, 15
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