Chapter 3 Dynamic Programming
„The Binomial Coefficient
Floyd’s Algorithm for Shortest Paths
„Dynamic Programming and Optimization Problems
„Chained Matrix Multiplication
„The Traveling Salesperson Problem
„
Divide-and-Conquer
„
top-down approach
„
„
divide an instance of a problem into
smaller instances
both smaller instances are unrelated
„
„
e.g., in mergesort, instances are sorted
independently
not suitable for related smaller instances
„
„
e.g., Fibonacci F(n) = F(n-1)+F(n-2)
Computing F(4) and F(3) both require F(2)
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Dynamic Programming
„
bottom up approach
„
„
„
divide an instance into smaller instances
solve small instance first, store it, look it up later
not recomputing
using array (table) to store solution
„
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e.g., Fibonacci F[n] = F[n-1]+F[n-2]
steps:
„
„
1. Establish a recursive property to a problem
2. Solve an instance of the problem in a bottomup fashion; solving smaller instance first.
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3.1 The Binomial Coefficient
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3.1 The Binomial Coefficient
„
„
„
⎛n⎞
n!
⎜⎜ ⎟⎟ =
⎝ k ⎠ k!(n − k )!
for 0<=k<=n
⎛15 ⎞
⎜⎜ ⎟⎟ = ?
⎝7⎠
⎧⎛ n − 1⎞ ⎛ n − 1⎞
⎛ n ⎞ ⎪⎜⎜
⎟+⎜
⎟,
⎜⎜ ⎟⎟ = ⎨⎝ k − 1⎟⎠ ⎜⎝ k ⎟⎠
⎝ k ⎠ ⎪1,
⎩
0<k <n
k = 0 or k = n
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3.1 The Binomial Coefficient
„
Algorithm 3.1 binomial coefficient
„
„
„
using divide-and-conquer pp. 93
bin(n, k) = bin(n-1, k-1)+ bin(n-1, k)
inefficient
„
recomputing bin(n-2, k-1) when computing
both bin(n-1,k-1) and bin(n-1, k)
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3.1 The Binomial Coefficient
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3.1 The Binomial Coefficient
„
using dynamic programming
„
1. Establish a recursive property
⎧ B[i − 1][ j − 1] + B[i − 1][ j ] 0 < j < i
B[i ][ j ] = ⎨
j = 0 or j = i
⎩1
„
2. Solve an instance of the problem in a bottomup fashion; solving smaller instance first.
„
for (i=1, j=1; ....)
{computing B[i][j]; }
„
Fig. 3.1 pp.94
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3.1 The Binomial Coefficient
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3.1 The Binomial Coefficient
„
„
Example 3.1 pp. 94
Algorithm 3.2 pp. 95
„
„
Analysis
total number of passes (for-j loop) pp. 94
i= 0 1
2
k
k+1
n
1 + 2 + 3+....+ k + (k+1) + (k+1) + ....+ (k+1)
„
„
k(k+1)/2 + (n-k+1)(k+1) ∈ Θ(nk)
Improvements
„
„
using only one dimensional array
B[i-1][j-1] B[i-1][j]
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B[i][j]
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3.1 The Binomial Coefficient
⎛n⎞ ⎛ n ⎞
⎜⎜ ⎟⎟ = ⎜⎜
⎟⎟
⎝k ⎠ ⎝n − k ⎠
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3.2 Floyd’s Algorithm for
Shortest Paths
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3.2 Floyd’s Algorithm for Shortest Paths
„
graph G=(V,E)
„
„
„
„
„
vertices
edges
digraph
weights (weighted graph )
path (a sequence of vertices) [v1, v4, v3]
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3.2 Floyd’s Algorithm for Shortest Paths
„
cycle ([v1, v4, v5, v1])
„
„
„
simple path ( never pass through same
vertex twice)
„
„
„
cyclic (G contains a cycle)
Acyclic
length
length [v1, v4, v3] =1+2=3
shortest path
„
„
„
length [v1, v2, v3] = 1+3=4
length [v1, v4, v3] = 1+2=3 → shortest
length [v1, v2, v4, v3] = 1+2+2=5
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3.2 Floyd’s Algorithm for Shortest Paths
„
Shortest Path Problem
„
„
„
optimization problem (min, max)
solution may be not unique
solve: an obvious algorithm
„
„
„
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for each vertex, determining the lengths of all
paths from it to every other vertex, selecting
the minimum
worst than exponential-time
(n-2)*(n-3)*......*1= (n-2)!
2nd, 3rd, …
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3.2 Floyd’s Algorithm for Shortest Paths
„
„
An dynamic programming approach
with cubic-time
adjacency matrix W[i][j]
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3.2 Floyd’s Algorithm for Shortest Paths
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3.2 Floyd’s Algorithm for Shortest Paths
„
shortest path lengths matrix D[i][j]
„
„
„
e.g., in Fig. 3.2, D[3][5]=7
find a way to calculate D from W
D (k) [i][j]
„
„
length of a shortest path from vi to vj
using only the vertices in set {v1, v2, ...,
vk}
e.g., Example 3.2 pp.99
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3.2 Floyd’s Algorithm for Shortest Paths
„
compute D(k)[i][j] by using dynamic
programming
„
„
„
1. Establish a recursive property
„
„
D(0) =W
D(n) =D
compute D(k) from D(k-1)
2. Solve in a bottom-up fashion
„
D(0), D(1), D(2), ..., D(n)
W
D
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3.2 Floyd’s Algorithm for Shortest Paths
„
Step 1: two cases:
„
(1) one shortest path from vi to vj not passing vk
„
„
„
„
D(k)[i][j] = D(k-1)[i][j]
e.g., D(5)[1][3] = D(4)[1][3]=3
→ [v1,v4, v3] is still shortest even including v5
(2) all shortest paths from vi to vj passing vk
„
„
Fig. 3.4 pp. 98 (a sub-path of a shortest path is still
shortest)
D(k)[i][j] = D(k-1)[i][k]+ D(k-1)[k][j]
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3.2 Floyd’s Algorithm for Shortest Paths
„
Case 1
D(k)[i][j] = min(D(k-1)[i][j],
D(k-1)[i][k]+ D(k-1)[k][j])
Case 2
„
Step2:
„
„
create the sequence of arrays for
computing D(n) from D(0)
e.g., Example 3.3 pp. 101
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3.2 Floyd’s Algorithm for Shortest Paths
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3.2 Floyd’s Algorithm for Shortest Paths
„
„
Every-case time complexity of Algorithm
3.3, pp.103
Algorithm 3.4 (Floyd‘s algorithm for
shortest paths 2) for output a shortest
path
„
„
e.g., Fig. 3.5 pp.104
Algorithm 3.5 (Print Shortest Path)
„
„
e.g., in Fig. 3.5 if q=5, r=3
path [v5, v1, v4, v3]
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3.2 Floyd’s Algorithm for Shortest Paths
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3.2 Floyd’s Algorithm for Shortest Paths
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3.2 Floyd’s Algorithm for Shortest Paths
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3.3 Dynamic Programming
and Optimization Problems
3.3 Dynamic Programming and
Optimization Problems
„ 1. Establish a recursive property that
gives the optimal solution to an
instance of the problem.
„
„
2. Compute the value of an optimal
solution in bottom-up fashion.
3. Construct an optimal solution in a
bottom-up fashion.
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3.3 Dynamic Programming and
Optimization Problems
„ Principle of Optimality:
„
„
„
an optimal solution to an instance always
contains optimal solutions to all
subinstances.
e.g., pp. 106, 4th paragraph
e.g., pp. 106 Example 3.4
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3.4 Chained Matrix
Multiplication
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3.4 Chained Matrix Multiplication
„
„
2×3×4 multiplications (standard method)
(i×j) ＊ (j×k) matrix
„
Æ i×j ×k multiplications
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3.4 Chained Matrix Multiplication
×
A
20×2
×
×
B
C
2×30
30×12
D
12×8
- A(B(CD) = 30×12×8＋2×30×8＋20×2×8 =3680
- (AB)(CD)=
...........
=8800
- A((BC)D)=
...........
=1232
- ((AB)C)D=
...........
=10320
- (A(BC)D)=
...........
=3120
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3.4 Chained Matrix Multiplication
„
how to determine the optimal order in
A1 × A2 × ... × An ?
„
„
consider all possible order and take the
minimum (exponential-time)
tn: the number of different order in A1
× A2 × ... × An
„
„
„
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A1×(A2×A3×...×An) → tn-1
(A1×A2×A3×...)×An → tn-1
tn ≥ tn-1＋tn-1＝2tn-1 ; t2 = 1
tn ≥2n-2
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3.4 Chained Matrix Multiplication
„
„
principle of optimality in A1×A2×...×An
If A1((((A2A3)A4)A5)A6) is optimal
order,
„
„
(A2A3)A4 must be optimal too
using dynamic programming to solve
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3.4 Chained Matrix Multiplication
„
„
Let d0 = # of rows in A0
dk = # of columns in Ak, 1<=k<=n
„
e.g., Figure 3.7
A1 × A2 × A3 × A4
d0xd1 d1xd2 d2xd3 d4xd5
„
Example 3.5 pp. 109
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3.4 Chained Matrix Multiplication
„
„
„
M[i][j]= min # of multiplications
needed in Ai×...×Aj
M[i][i]= 0
Example 3.5 pp. 109
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3.4 Chained Matrix Multiplication
„
„
„
„
„
consider multiplying six matrices (i.e.,
computing M[1][6])
1. A1×(A2×A3×A4×A5×A6)
2. (A1×A2)(A3×A4×A5×A6)
..............................
If (A1×A2)(A3×A4×A5×A6) is optimal,
„
„
„
both (A1×A2) and (A3×A4×A5×A6) are
optimal
M[1][6] = M[1][2]+M[3][6] + d0d2d6
M[1][6] = M[1][k]+M[k+1][6] +
d0dkd6
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3.4 Chained Matrix Multiplication
(
)
M [i ][ j ] = min M [i ][k ] + M [k + 1][ j ] + d i −1d k d j , if i < j
i ≤ k ≤ j −1
M [i ][i ] = 0
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3.4 Chained Matrix Multiplication
„
„
„
„
„
how to compute array M[i][j] in steps ?
M[i][j] is computed all entries on row i
but left it and all entries in column j but
beneath it.
compute entries in the diagonal 0, 1,...,j
See Fig. 3.8 pp. 111
See Example 3.6 pp. 111
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3.4 Chained Matrix Multiplication
„
Every-Case Time Complexity of
Algorithm 3.6, pp. 113
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3.4 Chained Matrix Multiplication
„
See Fig. 3.9, pp. 114
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3.6 The Traveling Salesperson
Problem
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3.6 Traveling Salesperson Problem
„
„
„
„
finding a shortest simple cycle which
passes all vertices
tour (Hamiltonian circuit)
a path from a vertex to itself passing all
vertices exactly once
optimal tour
„
„
a tour with minimum length
See Fig. 3.16 pp. 126
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3.6 Traveling Salesperson Problem
„
Length of an optimal tour (V1→→→V1)
[ ][
]
min (W [1][ j ] + D v j V − {v1 , v j } )
2≤ j ≤ n
where
(
[ ][ { }])
D[vi ][A] = min W [i ][ j ] + D v j A − v j , if A ≠ φ
j :v j ∈ A
D[vi ][φ ] = W [i ][1]
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3.6 Traveling Salesperson Problem
„
„
„
See Example 3.11, pp. 128
Algorithm 3.11
Every-Case Time and Space Complexity
of Algorithm 3.11, pp. 130
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The End
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