dynamic programming
DESCRIPTION
Dynamic Programming. 2012/11/20. Dynamic Programming (DP). Dynamic programming is typically applied to optimization problems . Problems that can be solved by dynamic programming satisfy the principle of optimality . Principle of optimality. - PowerPoint PPT PresentationTRANSCRIPT
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Dynamic Programming
2012/11/20
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P.2
Dynamic Programming (DP) Dynamic programming is
typically applied to optimization problems.
Problems that can be solved by dynamic programming satisfy the principle of optimality.
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Principle of optimality Suppose that in solving a problem, we
have to make a sequence of decisions D1, D2, …, Dn-1, Dn
If this sequence of decisions D1, D2, …, Dn-
1, Dn is optimal, then the last k, 1 k n, decisions must be optimal under the condition caused by the first n-k decisions.
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Dynamic method v.s.Greedy method
Comparison: In the greedy method, any decision is locally optimal.
These locally optimal solutions will finally add up to be a globally optimal solution.
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The Greedy Method E.g. Find a shortest path from v0 to v3.
The greedy method can solve this problem. The shortest path: 1 + 2 + 4 = 7.
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The Greedy Method E.g. Find a shortest path from v0 to v3 in the multi-
stage graph.
Greedy method: v0v1,2v2,1v3 = 23 Optimal: v0v1,1v2,2v3 = 7 The greedy method does not work for this
problem. This is because decisions at different stages
influence one another.
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Multistage graph A multistage graph G=(V,E) is a directed
graph in which the vertices are partitioned into k2 disjoint sets Vi, 1i k
In addition, if <u,v> is an edge in E then uVi and vVi+i for some i, 1i<k
The set V1 and Vk are such that V1 =Vk=1
The multistage graph problem is to find a minimum cost path from s in V1 to t in Vk
Each set Vi defines a stage in the graph
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Greedy Method vs. Multistage graph E.g.
The greedy method cannot be applied to this case: S A D T 1+4+18 = 23.
The shortest path is: S C F T 5+2+2 = 9.
S T132 B E
9
A D4
C F2
1
5
11
5
16
18
2
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Dynamic Programming Dynamic programming approach:
d(S, T) = min{1+d(A, T), 2+d(B, T), 5+d(C, T)}
S T2 B
A
C
1
5 d(C, T)
d(B, T)
d(A, T)
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Dynamic Programming
d(A, T) = min{4+d(D, T), 11+d(E, T)}
= min{4+18, 11+13} = 22.
A
T
4
E
D
11d(E, T)
d(D, T)S T132 B E
9
A D4
C F2
1
5
11
5
16
18
2
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Dynamic Programming d(B, T) = min{9+d(D, T), 5+d(E, T), 16+d(F,
T)} = min{9+18, 5+13, 16+2} = 18. d(C, T) = min{ 2+d(F, T) } = 2+2 = 4 d(S, T) = min{1+d(A, T), 2+d(B, T), 5+d(C,
T)} = min{1+22, 2+18, 5+4} = 9.
B T5 E
D
F
9
16 d(F, T)
d(E, T)
d(D, T)
S T132 B E
9
A D4
C F2
1
5
11
5
16
18
2
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Save computation For example, we never calculate (as a
whole) the length of the path
S B D T( namely, d(S,B)+d(B,D)+d(D,T) ) because we have found d(B, E)+d(E, T)<d(B,D)+d(D,T)
There are some more examples… Compare with the brute-force method…
S T132 B E
9
A D4
C F2
1
5
11
5
16
18
2
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The advantages of dynamic programming approach
To avoid exhaustively searching the entire solution space (to eliminate some impossible solutions and save computation).
To solve the problem stage by stage systematically.
To store intermediate solutions in a table (array) so that they can be retrieved from the table in later stages of computation.
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Comment If a problem can be described by a
multistage graph then it can be solved by dynamic programming.
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The longest common subsequence (LCS or LCSS) problem
A sequence of symbols A = b a c a d A subsequence of A: deleting 0 or more
symbols (not necessarily consecutive) from A.
E.g., ad, ac, bac, acad, bacad, bcd. Common subsequences of A = b a c a d and B = a c c b a d c b : ad, ac, bac, acad. The longest common subsequence of A and
B: a c a d.
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P.16
DNA Matching DNA = {A|C|G|T}*S1=ACCGGTCGAGTGCGGCCGAAGCCGGC
CGAAS2=GTCGTTCGGAATGCCGTTGCTGTAAA
Are S1 and S2 similar DNAs?
The question can be answered by figuring out the longest common subsequence.
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Networked virtual environments (NVEs)
virtual worlds full of numerous virtual objects to simulate a variety of real world scenes
allowing multiple geographically distributed users to assume avatars to concurrently interact with each other via network connections.
E.G., MMOGs: World of Warcraft (WoW), Second Life (SL)
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Avatar Path Clustering Because of similar personalities, interests, or
habits, users may possess similar behavior patterns, which in turn lead to similar avatar paths within the virtual world.
We would like to group similar avatar paths as a cluster and find a representative path (RP) for them.
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How similar are two paths in Freebies island of Second Life?
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SeqA:C60.C61.C62.C63.C55.C47.C39.C31.C32
LCSS-DC - path transfers sequence
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SeqA :C60.C61.C62.C63.C55.C47.C39.C31.C32SeqB :C60.C61.C62.C54.C62.C63.C64LCSSAB :C60.C61.C62. C63
LCSS-DC - similar path thresholds
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P.22
Longest-common-subsequence problem: We are given two sequences X =
<x1,x2,...,xm> and Y = <y1,y2,...,yn> and wish to find a maximum length common subsequence of X and Y.
We define Xi = < x1,x2,...,xi > and Yj = <y1,y2,...,yj>.
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Brute Force Solution
m * 2n = O(2n ) or n * 2m = O(2m)
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P.24
A recursive solution to subproblem Define c [i, j] is the length of the
LCS of Xi and Yj .
ïî
ïí
ì
¹--+--=
ji
ji
yxi,j>jicjic=yx i,j>jic
j=i=
jic and 0 if]},1[],1,[max{
and 0if1]1,1[0or 0 if0
],[
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P.25
Computing the length of an LCSLCS_LENGTH(X,Y)1 m length[X]2 n length[Y]3 for i 1 to m4 do c[i, 0] 05 for j 1 to n6 do c[0, j] 0
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P.26
7 for i 1 to m8 for j 1 to n9 if xi = yj
10 then c[i, j] c[i-1, j-1]+111 b[i, j] “”12 else if c[i–1, j] c[i, j-1]13 then c[i, j] c[i-1, j]14 b[i, j] “”15 else c[i, j] c[i, j-1]16 b[i, j] “”17 return c and b
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P.27Complexity: O(mn) rather than O(2m) or O(2n) of Brute force method
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P.28
PRINT_LCS PRINT_LCS(b, X, i, j )1 if i = 0 or j = 02 then return3 if b[i, j] = “”4 then PRINT_LCS(b, X, i-1, j-1)5 print xi
6 else if b[i, j] = “”7 then PRINT_LCS(b, X, i-1, j)8 else PRINT_LCS(b, X, i, j-1)
Complexity: O(m+n)
By calling PRINT_LCS(b, X, length[X], length[Y]) to print LCS
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Chapter 15P.29
Matrix-chain multiplication How to compute where
is a matrix for every i. Example:
A A An1 2 ... Ai
A A A A1 2 3 4
( ( ( ))) ( (( ) ))(( )( )) (( ( )) )((( ) ) )
A A A A A A A AA A A A A A A AA A A A
1 2 3 4 1 2 3 4
1 2 3 4 1 2 3 4
1 2 3 4
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Chapter 15P.30
MATRIX MULTIPLY MATRIX MULTIPLY(A,B)1 if columns[A] rows[B]2 then error “incompatible dimensions”3 else for to rows[A]4 for to columns[B]5 6 for to columns[A]7 8 return C
¹
i 1j 1
0],[ jiCk 1
],[],[],[],[ jkBkiAjiCjiC +
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Chapter 15P.31
Complexity:
Let A be a matrix, and B be
a
matrix. Then the complexity
of
A xB is .
p q
q r
p q r
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Chapter 15P.32
Example: is a matrix, is a
matrix, and is a matrix. Then
takes time. However takes
time.
A1 10 100 A2 100 5A3 5 50
(( ) )A A A1 2 3 10 100 5 10 5 50 7500 + =( ( ))A A A1 2 3
100 5 50 10 100 50 75000 + =
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Chapter 15P.33
The matrix-chain multiplication problem: Given a chain of n
matrices, where for i=0,1,…,n, matrix Ai has dimension pi-1pi, fully parenthesize the product in a way that minimizes the number of scalar multiplications.
A product of matrices is fully parenthesized if it is either a single matrix, or a product of two fully parenthesized matrix product, surrounded by parentheses.
nAAA ,...,, 21
A A An1 2 ...
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Chapter 15P.34
Counting the number of parenthesizations:
[Catalan number]
P nif n
P k p n k if nk
n( ) ( ) ( )==
-
ìíï
îï =
-1 1
21
1
P n C n( ) ( )= - 1
=+
=112 4
3 2nn
n n
n( )/
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Chapter 15P.35
Step 1: The structure of an optimal parenthesization
(( ... )( ... ))A A A A A Ak k k n1 2 1 2+ +
Optimal
Combine
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Chapter 15P.36
Step 2: A recursive solution Define m[i, j]= minimum number
of scalar multiplications needed to compute the matrix
goal m[1, n]
A A A Ai j i i j.. ..= +1
m i j[ , ] =
îíì
¹+++=
- jipppjkmkimji
jkijki }],1[],[{min0
1
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Chapter 15P.37
Step 3: Computing the optimal costs Instead of computing the solution
to the recurrence recursively, we compute the optimal cost by using a tabular, bottom-up approach.
The procedure uses an auxiliary table m[1..n, 1..n] for storing the m[i, j] costs and an auxiliary table s[1..n, 1..n] that records which index of k achieved the optimal cost in computing m[i, j].
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Chapter 15P.38
MATRIX_CHAIN_ORDER MATRIX_CHAIN_ORDER(p)1 n length[p] –1 2 for i 1 to n3 do m[i, i] 04 for l 2 to n5 do for i 1 to n – l + 16 do j i + l – 1 7 m[i, j] 8 for k i to j – 1 9 do q m[i, k] + m[k+1, j]+ pi-1pkpj
10 if q < m[i, j]11 then m[i, j] q12 s[i, j] k13 return m and s
Complexity: O n( )3
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Chapter 15P.39
Example:
656
545
434
323
212
101
25202010
10551515353530
ppAppAppAppAppAppA
======
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Chapter 15P.40
the m and s table computed by MATRIX-CHAIN-ORDER for n=6
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Chapter 15P.41
m[2,5]=min{m[2,2]+m[3,5]+p1p2p5=0+2500+351520=13
000,m[2,3]+m[4,5]+p1p3p5=2625+1000+35520=
7125,m[2,4]+m[5,5]+p1p4p5=4375+0+351020=11
374}=7125
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Chapter 15P.42
MATRIX_CHAIN_MULTIPLYPRINT_OPTIMAL_PARENS(s, i, j)1 if i=j2 then print “A”i3 else print “(“4 PRINT_OPTIMAL_PARENS(s, i, s[i,j])5 PRINT_OPTIMAL_PARENS(s, s[i,j]+1,
j)6 print “)” Example:
(( ( ))(( ) ))A A A A A A1 2 3 4 5 6
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Q&A43