Ch 5 Dynamic Programming

1
Ch 5 Dynamic Programming
2
Making change

• How do we use the minimum number of coins to make
  change for 8 dollars with coins 1, 4, and 6
  dollars?
• Using greedy method, we can make change for 8
  dollars with 1 6-dollar and 2 1-dollars, which
  needs 3 coins.
• However, we can make change for 8 dollars with 2
  4-dollars.
• Therefore, greedy method can not guarantee to
  derive the minimum number of coins for this
  problem.

3
Brute-and-Force Method
4
Using Dynamic Programming for Making Change

• ci,j the minimum number of coins required to
  pay an amount of j dollars using only coins d1,
  d2, , di.
• ci,j min( ci-1,j, 1ci,j-di )
• Amount 0 1 2 3 4 5 6 7
  8
• d1 1 0 1 2 3 4 5 6 7
  8
• d2 4 0 1 2 3 1 2 3 4
  2
• d3 6 0 1 2 3 1 2 1 2
  2

1
1
c2,6 min c1,6, 1c2,2 min 6, 12
3 c3,8 min c2,8, 1c3,2 min 2,
12 2
5
Using Dynamic Programming for Making Change
Function coins(N) N dollars for making change
d1..n the dollars of coin i c1..n,0..m
the minimum number of coins for i ? 1 to n do
ci,0 ? 0 for i ? 1 to n do for j ? 1
to m do ci,j ? if i 1 and j lt
di then 8 else if i 1 then 1
c1, j d1 else if j lt di then
ci-1,j else min( ci - 1, j, 1
ci,j - di )
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Multistage Graphs P254, P257

• Principle of optimality (p254)
• Exhaustive search can guarantee to find an
  optimal solution.
• However, dynamic programming finds optimal
  solutions for all scales of sub-problems and
  finally find an optimal solution.
• That is to say, the global optimum comes from the
  optimums of all sub-problems.
• A multistage is a directed graph in which the
  vertices are partitioned into k ? 2 disjoint
  sets.
• The multistage graph problem is to find a
  minimum-cost path from s to t.

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Multistage Graphs P259, Figure 5.2
Let cost(i,j) be the cost of a minimum-cost
path from vertex j in Vi to destination vertex t.
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stage i
stage i1
.
l
j
t
7 min 47, 25, 17
7 min 46, 25
4
9
16
5
2
18
7
5
15
9
P259 P261
stage i
stage i1
.
l
j
t
10
P262, Algorithm 5.1
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Multistage Graphs

• The time for the for loop of line 7 is T(V
  E), and the time for the for loop of line 16 is
  T(k), Hence, the total time is T(V E).
• The backward trace from vertex 1 to n also works.
• The algorithm also works for the edges crossing
  more than 1 stage.

12
Efficient allocation of chain-like tasks on
chain-like network computers

• Jang-Ping SHEU
• Zen-Fu CHIANG
• Information Processing Letters 36 (1990)241-245

13
Introduction

• The purpose of task allocation on distributed
  computing systems is to reduce the task
  turnaround time by increasing the system
  throughput.
• Assumption
• cost for computation is known
• communication cost is known
• linear precedence relationship
• Let be the task
  turnaround time of allocation A, then

tasks or modules
system or processors
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Notation

• m the number of modules
• n the number of processors
• ei the computation cost for module i
• cj,j1 the communication cost between modules j
  and j1 if modules j and j 1 are allocated to
  different processors
• m, n, ei, and cj,j1 are known

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Allocation of a nine-module chain-like task on a
four-processor chain-like network computer

• m 9, n 4
• The turnaround time of processor 2 for task
  allocation as shown in the above figure is equal
  to

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Bokharis task allocation graph

• A node labeled lth,igt is connected to all nodes
  lti1,jgt in the layer below it.
• If a path contains the node lth,igt of layer k,
  this represents the assignment of modules h
  through i to processor k.
• This method expands all possible solutions in the
  allocation graph and there are a lot of paths and
  redundant nodes in the graph.

• A path s ? lt1,1gt ? lt2,3gt ? lt4,6gt
• lt7,9gt ? d represents the possible
• allocation depicted in the previous page

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The improved task allocation graph

• m-k1 nodes
• The edge that connects node i in layer k and node
  j in layer k 1 represents the assignment of
  modules i 1 through j to processor k.
• The weight of edge joining node i in layer k and
  j in layer k 1 equals the sum of all
  computation costs of modules i 1 through j and
  communication costs between modules i and i 1,
  and between modules j and j 1.

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Alogrithm(1/5)

• Example m 4, n 4
• Step 1
• 1. There is one distinguished node called the
  source node, denoted s.
• 2. There are min(m,n) layers in the task
  allocation graph.

5
2
2
3
19
Alogrithm(2/5)

• 3. The kth layer consists of m-k1 nodes and
  sequence numbers k, k1, , m is labeled at each
  node from left to right.
• 4. A node Labeled i in layer k is connected to
  all nodes, in layer k 1, labeled number greater
  than i. All nodes in the first layer are
  connected to node s.

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Alogrithm(3/5)
5
2
2
3

• 5. Each edge joining source node s to node j, for
  1?j?m, has weight
• 6. Each edge joining node i in layer k to node j
  in layer k 1 has weight

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Alogrithm(4/5)

• Step 2
• Initially all nodes in the allocation graph are
  given infinite labels except the source node,
  where it is labeled zero.

5
2
2
3
L(s) 0
L(1) 8
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Alogrithm(5/5)

• Step 3.
• For k 1 to min(m,n) do
• L(j)minL(j), max(W(i,j), L(i))
• where L(j) is the label of node j and W(i,j) is
  the weight of the edge connecting from i to j.

5
2
2
3
8
11
11
12
9
10
9
10 min8,max8,10 10 min10,max11,9 10
min10,max11,6
Task turnaround time table
9
9
9
The method avoids a lot of computations of
redundant paths and nodes
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P268, Example 5.15 Figure 5.6
We requires that G has no cycles with negative
length
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All Pairs Shortest PathsP267, Algorithm 5.3
O(n3)
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Single-Source Shortest Paths General
WeightsP271 P272, Figure 5.10
Let distku be the length of a shortest path
from the source vertex v to vertex u containing
at most k edges.
distk-1(u)
..
i
v
...
u
cost(i,u)
distk-1(i)
We requires that G has no cycles with negative
length
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P273, Algorithm 5.4
O(n3) for adjacency matrix and O(ne) for
adjacency list
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String EditingP284
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Let cost(i,j) be the minimum cost of edit
sequence for transforming string x1, x2, , xi
into y1, y2, , yj.
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cost(0,1) D(x1) x1 a ? D(x1a) ? F ?
cost(0,1) I(y1b) ? b y1 cost(1,0) I(y1)
x1 a ? I(y1b) ? ab ? cost(1,0) D(x1a) ? b
y1
cost(0,2) D(x1) x1 a ? D(x1 a) ? F ?
cost(0,2) I(y1y2 ba) ? ba y1 y2 cost(0,1)
C(x1a,y2 a) x1 a ? C(x1a,y2 a)0 ? a ?
cost(0,1) I(y1a) ? ba y1 y2
Y
b
a
b
b
X
C(xi,y2j) 2 or 0
2
1
D(xj) 1
a
1
a
I(yi) 1
b
The time complexity is O(mn) where m and n denote
the length of strings X and Y respectively.
a
b
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0-1 Knapsack P287
0-1 knapsack where xi 0 or 1
For example m 11, using greedy method we will
choose objects 5, 2, and 1 with total profit
35. i 1 2 3 4 5 p 1 6
18 22 28 w 1 2 5 6 7
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0-1 Knapsack

• Let Pi,j be the maximum profit using object 1,
  2, , i, with knapsacks capacity j.
• Pi,j max( Pi 1,j, Pi 1, j wi pi )
• Let the knapsacks capacity m 11.
• Weight limit 0 1 2 3 4 5 6
  7 8 9 10 11
• w11, p11 0 1 1 1 1 1 1 1
  1 1 1 1
• w22, p26 0 1 6 7 7 7 7 7
  7 7 7 7
• w35, p218 0 1 6 7 7 18 19 24
  25 25 25 25
• w46, p422 0 1 6 7 7 18 22 24
  28 29 29 40
• w57, p528 0 1 6 7 7 18 22 28
  29 34 35 40
• P5,11 max( P4,11, P4,428 ) max( 40,
  728) 40
• P4,11 max( P3,11, P3,522 ) max( 25,
  1822) 40
• P3,5 max( P2,5, P2,06 ) max( 7, 018)
  18

Time complexity is O(mn)
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Traveling Salesperson ProblemP298

• A salesperson would like to travel from one city
  to the other (n 1) cities just once then back
  to the original city, what is the minimum
  distance for this travel?
• The brute-and-force method is trying all possible
  (n 1)! permutations of (n 1) cities and
  picking the path with the minimum distance.
• There are a lot of redundant computations for the
  brute-and-force method such as the permutations
  of 6 cities are 1234561, , 1243561, ., 1324561,
  , 1342561, , 1423561, , 1432561,

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Traveling Salesperson ProblemP299

• The method computes from vertex 1 backward to the
  other vertices then return to vertex 1.
• Let g(i,S) be the length of a shortest path
  starting at vertex i, going through all vertices
  in S, and terminating at vertex 1.

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Compute from vertex 1 backward to the other
vertices then return to vertex 1