Matakuliah
Tahun
: K0414 / Riset Operasi Bisnis dan Industri
: 2008 / 2009
Penerapan Dinamik Programming (DP)
Pertemuan 16 (GSLC)
Learning Outcomes
• Mahasiswa dapat mendemonstrasikan penyelesaian
masalah dinamik programming, dengan menggunakan
bantuan program komputer.
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Outline Materi:
• Penerapan Dinamik Programming (DP)
• Penyelesaian masalah DP dengan program komputer.
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Dynamic Programming
• An extraordinarily powerful approach for some
classes of problems.
• The trick is: You must be able to express the answer
for your current problem in terms of smaller problems
for which you already know the answer.
• Sometimes the possibility of using dynamic
programming presents itself when we arrange our
data in matrix form.
• Let’s look at the “rock game” proposed by the
authors.
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The “Rock Game”
• Rules:
– Start with two piles of rocks, n in one pile and m in
the other.
– Players take turns; I start. On each turn, a player
can either take a single rock (taken from either
pile), or two rocks (one from each pile).
– Player who takes the last rock wins!
• Goal: We want to find an optimal strategy, so that
given an initial collection of n and m rocks, we are
sure to win (provided that’s possible, of course!)
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Encoding the Rocks game
• We can use a matrix to represent the current state of
the game; being in row i and column j represents
having i rocks in the first pile and j in the second.
• I put a W in each matrix cell if there exists a winning
strategy for me, given the corresponding configuration
of rocks, and an L if a competent opponent is sure to
beat me.
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Moves in the Rock game
• Each possible action that I can take “reduces” the
size of the game, and corresponds to a move from
my current cell to
– the cell immediately above;
– the cell immediately to the left;
– the cell on the diagonal, one space to the left and
one up.
• Suppose that I move, and enter a cell that already
has a W; that symbol means that there exists a
winning strategy for me, provided that I move first.
Since it is now my opponent’s move, I have just lost
the game!
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Trying out the Rock Game example
• Let’s fill in the matrix, following the lead of our authors.
• Note how we easily build on the known solutions for
smaller problems to progressively fill in the matrix.
• Let’s go a step further, and plot the possible moves that
correspond to a winning strategy. Note that a particular
game will be represented by a particular path through
our matrix.
• Can we write a rule for expressing the form of a winning
game, expressed as a path?
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Revisiting USChange as a DP Problem
• Recall: Given a set of coin denominations
{c1,c2,…,cd}, and a monetary value M to return as
change, return the smallest number of coins possible.
• Here’s how to recast the problem by reference to
“smaller” problems. We define the “best number of
coins” for M in terms of the “best number of coins” for
M - ci, for each of the denominations.
bestNumCoinsM c  1
1


bestNumCoinsM  min bestNumCoinsM c2  1
bestNumCoins


1
M c d

10
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Recursive Implementation
RecursiveChange(M,c,d)
if M = 0
return M ;
bestNumCoins <- Infinity
for i <- 1 to d
{
if M >= ci
{
numCoins <RecursiveChange(M-ci,c,d)
if numCoins + 1 < numBestCoins
numBestCoins <- numCoins + 1
}
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}
return bestNumCoins
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Drawbacks of Recursion
• This recursive algorithm suffers the same problem of
the recursive Fibonacci program; it computes the
same values multiple times.
– For example, if M=10, we will generate a call for M
- 5 = 5; we will also generate a call for
M - 1 = 9, which will generate (M - 1) - 1, etc until
eventually there is another call with argument 5.
• We are better off to start from M=0, and build up the
best coin counts for increasing M, always referring
back to the optimal values already recorded. This is
the essence of dynamic programming.
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Example…
• Suppose we have already determined the optimal number of coins to
return for the following values of M:
M=1
1 (1 cent)
M=2
2 (2 X 1 cent)
M=3
3 ( 3 X 1 cent)
M=4
4 (4 X 1 cent)
M=5
1 (1 nickel)
For M=6, we check
Optimum(6 - 1 cent) + 1 = Optimum(5) + 1
= 2 (1 nickel + 1 cent)
Optimum(6 - 1 nickel) + 1 = Optimum(1) + 1
= 2 (1 cent + 1 nickel)
So, the optimum number of coins in change for 6 cents is 2
(1 nickel + 1 cent)
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DP Implementation
DPChange(M,c,d)
bestNumCoins0 <- 0
for m <- 1 to M
{
bestNumCoinsm <- Infinity
for i <- 1 to d
{
if m >= ci
{
if (bestNumCoinsm-ci + 1
< bestNumCoinsm)
bestNumCoinsm <bestNumCoinsm-ci + 1
}
}
}
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return bestNumCoinsM
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The Manhattan Tourist Problem
Source
3
1
3
4
2
0
2
6
0
4
7
3
4
0
4
3
3
1
2
8
1
2
5
2
3
4
0
3
2
2
5
6
1
2
5
4
5
4
3
2
Edge
weight
Sink
Node/Vertex
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A graph
Edge
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The Greedy Solution
3
0
1
3
4
2
3
0
6
0
4
4
9
4
13
3
3
1
4
15
19
2
0
1
2
?
8
3
2
3
2
5
6
1
2
0
5
4
5
5
7
3
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2
4
20
5
2
?
3
2
23
We see immediately that the greedy solution is
not optimal!
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Approach to Dynamic Programming
• Start with the more general problem - find the best
path from vertex i to vertex j. Call the length of the
best path (the sum of the edge weights along it) sij.
• With analogy to the DPChange algorithm, we
discover that we can efficiently find the solution by
finding the answer for all the sub-problems (all
possible sink vertices), and then picking out the
solution we want.
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Getting started - paths along the edges
are completely constrained!
Column 0
Row 0
3
0
1
1
3
4
0
4
14
4
4
3
3
2
3
4
0
1
2
5
2
9
1
2
8
3
0
9
2
5
6
1
2
5
4
5
4
5
7
3
9
2
6
0
5
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2
3
3
2
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Now we can find s11…
Can arrive at (1,1) either from above (N) or the left (W). We know
the best scores for each of those possible sources, and the
weights of the edges that connect (1,1) to them…
3
0
1
1
3
4
0
4
4
14
4
4
3
3
2
3
4
0
1
2
5
2
9
1
2
8
3
0
9
2
5
6
1
2
5
4
5
4
5
7
3
9
2
6
0
5
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2
3
3
2
19
…and the rest of column 1
3
0
1
1
3
2
3
0
4
4
6
0
5
4
3
9
10
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4
5
4
4
3
3
2
3
4
0
1
2
5
2
9
1
2
8
3
0
9
2
5
6
1
2
7
14
5
14
2
4
5
3
2
20
…and the next column
3
0
1
1
3
2
3
0
4
4
6
0
5
4
3
9
10
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4
20
7
4
17
3
22
30
4
3
2
3
4
0
1
2
5
2
9
1
2
8
3
0
9
2
5
6
1
2
5
7
14
5
14
2
4
5
3
2
21
…and the next
3
0
1
1
3
2
3
0
4
4
6
0
5
4
3
9
10
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4
20
7
4
4
13
17
3
22
3
22
30
1
2
5
2
32
3
1
2
0
9
4
20
8
3
2
2
5
6
1
2
0
9
5
7
14
5
14
2
4
5
3
2
22
…and the last
3
0
1
1
3
2
3
0
4
4
6
0
5
4
3
9
10
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4
20
7
4
4
13
17
3
22
3
30
3
15
1
4
20
2
0
22
8
3
2
9
2
5
6
1
2
0
9
5
7
14
5
14
2
4
5
2
24
25
5
2
32
1
3
2
34
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Recurrence relation
si1, j  weight((i 1, j)  (i, j)
si, j  max 

si, j1  weight((i, j 1)  (i, j)
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Manhattan DP Algorithm
ManhattanTouristDP(vert, horiz, n, m)
s(0,0) <- 0
for i <- 1 to n
s(i,0) <- s(i-1,0) + vert(i,0)
for j <- 1 to m
s(0,j) <- s(0,j-1) + horiz(0,j)
for i <- 1 to n
{
for j <- 1 to m
{
s(i,j) <- max{ s(i-1,j) + vert(i,j),
s(i,j-1) + horiz(i,j) }
}
}
return s(n,m)
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What if the grid is not regular?
• Model the street system as a directed graph. The
intersections are vertices, and all the streets are oneway.
– We assume that there are no cycles in the graph; it is a
directed acyclic graph (DAG)
• Represent the graph as a set of vertices and a set of
weighted edges: G = (V,E)
• Number vertices from 1 to |V| with a single integer;
we drop the matrix formalism.
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Longest Path DAG Problem:
• Input: A weighted DAG G with source and sink
vertices.
• Output: the longest (optimal) path between source
and sink.
• Notation:
– Write a directed edge as a pair of vertices, (u,v).
– The indegree of a vertex is the number of
incoming edges; the outdegree the number that
exit.
– Vertex u is a predecessor to v if (u,v) is in E; if v
has indegree k that means it has k predecessors.
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
Modified recurrence relation:
sv 
max
uPredecessors(v)
su  weight(u  v)
A problem: In what order to we visit the vertices?? By
the time vertex v is visited, all of its predecessors must
have already been visited!
An ordering of vertices v1,v2,…,vn is topological if for
every edge (vi,vj) in the DAG, i < j. If the ordering is
topological, all is well because whenever we visit a
vertex, we are guaranteed to have already visited its
predecessors.
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