1 dynamic programming topic 07 asst. prof. dr. bunyarit uyyanonvara it program, image and vision...
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1
Dynamic Programming
Topic 07
Asst. Prof. Dr. Bunyarit UyyanonvaraAsst. Prof. Dr. Bunyarit UyyanonvaraIT Program, Image and Vision Computing Lab.
School of Information and Computer Technology
Sirindhorn International Institute of Technology
Thammasat Universityhttp://www.siit.tu.ac.th/bunyarit
[email protected] 5013505 X 2005
ITS033 – Programming & Algorithms
2
ITS033Topic 01Topic 01 -- Problems & Algorithmic Problem SolvingProblems & Algorithmic Problem SolvingTopic 02Topic 02 – Algorithm Representation & Efficiency Analysis – Algorithm Representation & Efficiency AnalysisTopic 03Topic 03 - State Space of a problem - State Space of a problemTopic 04Topic 04 - Brute Force Algorithm - Brute Force AlgorithmTopic 05Topic 05 - Divide and Conquer - Divide and ConquerTopic 06Topic 06 -- Decrease and ConquerDecrease and ConquerTopic 07Topic 07 - Dynamics Programming - Dynamics ProgrammingTopic 08Topic 08 -- Transform and ConquerTransform and ConquerTopic 09Topic 09 - Graph Algorithms - Graph AlgorithmsTopic 10Topic 10 - Minimum Spanning Tree - Minimum Spanning TreeTopic 11Topic 11 - Shortest Path Problem - Shortest Path ProblemTopic 12Topic 12 - Coping with the Limitations of Algorithms Power - Coping with the Limitations of Algorithms Power
http://www.siit.tu.ac.th/bunyarit/its033.phphttp://www.siit.tu.ac.th/bunyarit/its033.phpand and http://www.vcharkarn.com/vlesson/showlesson.php?lessonid=7http://www.vcharkarn.com/vlesson/showlesson.php?lessonid=7
Midterm
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This Week Overview Dynamic Programming
Fibonacci Series
Knapsack Problem
Memory Function
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Dynamic Programming: Introduction
Topic 07.1
Asst. Prof. Dr. Bunyarit UyyanonvaraAsst. Prof. Dr. Bunyarit UyyanonvaraIT Program, Image and Vision Computing Lab.
School of Information and Computer Technology
Sirindhorn International Institute of Technology
Thammasat Universityhttp://www.siit.tu.ac.th/bunyarit
[email protected] 5013505 X 2005
ITS033 – Programming & Algorithms
5
Introduction
Dynamic programming is a technique for solving problems with overlapping sub-problems.
Typically, these sub-problems arise from a recurrence relating a solution to a given problem with solutions to its smaller sub-problems of the same type.
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Introduction Rather than solving overlapping sub-problems
again and again,
dynamic programming suggests solving each of the smaller sub-problems only once
and recording the results in a table from which we can then obtain a solution to the original problem.
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Introduction The Fibonacci numbers are the elements of the sequence 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, . . . ,
Algorithm fib(n) if n = 0 or n = 1 return 1 return fib(n − 1) + fib(n − 2)
The original problem F(n) is defined by F(n-1) and F(n-2)
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Introduction The Fibonacci numbers are the elements of the sequence 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, . . . ,
Algorithm fib(n) if n = 0 or n = 1 return 1 return fib(n − 1) + fib(n − 2)
If we try to use recurrence directly to compute the nth Fibonacci
number F(n) , we would have to recompute the same values of this function many times
in fact, we can avoid using an extra array to accomplish this task by recording the values of just the last two elements of the Fibonaccisequence
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Introduction
Notice that if we call, say, fib(5), we produce a call tree that calls the function on the same value many different times:
fib(5) fib(4) + fib(3) (fib(3) + fib(2)) + (fib(2) + fib(1)) ((fib(2) + fib(1)) + (fib(1) + fib(0))) + ((fib(1) + fib(0)) +
fib(1)) (((fib(1) + fib(0)) + fib(1)) + (fib(1) + fib(0))) + ((fib(1) +
fib(0)) + fib(1))
If we try to use recurrence directly to compute the nth Fibonacci number F(n) , we would have to recompute the same values of this function many times
10
Introduction Certain algorithms compute the nth Fibonacci number
without computing all the preceding elements of this sequence.
It is typical of an algorithm based on the classic bottom-up dynamic programming approach,
A top-down variation of it exploits so-called memory functions
The crucial step in designing such an algorithm remains the same => Deriving a recurrence relating a solution to the problem’s instance with solutions of its smaller (and overlapping) subinstances.
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Introduction Dynamic programming usually takes one of two
approaches:
Bottom-up approach: All subproblems that might be needed are solved in advance and then used to build up solutions to larger problems. This approach is slightly better in stack space and number of function calls, but it is sometimes not intuitive to figure out all the subproblems needed for solving the given problem.
Top-down approach: The problem is broken into subproblems, and these subproblems are solved and the solutions remembered, in case they need to be solved again. This is recursion and Memory Function combined together.
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Bottom Up In the bottom-up approach we calculate the smaller values
of Fibo first, then build larger values from them. This method also uses linear (O(n)) time since it contains a loop that repeats n − 1 times.
In both these examples, we only calculate fib(2) one time, and then use it to calculate both fib(4) and fib(3), instead of computing it every time either of them is evaluated.
Algorithm Fibo(n)
previousFib = 0, currentFib = 1 repeat n − 1 times
newFib = previousFib + currentFib previousFib = currentFib currentFib = newFib return currentFib
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Top-Down suppose we have a simple map object, m, which maps each
value of Fibo that has already been calculated to its result, and we modify our function to use it and update it. The resulting function requires only O(n) time instead of exponential time:
This technique of saving values that have already been calculated is called Memory Function; this is the top-down approach, since we first break the problem into subproblems and then calculate and store values
m [0] = 0m [1] = 1
Algorithm Fibo(n) if map m does not contain key n m[n] := Fibo(n − 1) + Fibo(n − 2) return m[n]
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Dynamic Programming: Knapsack Problem
Topic 07.2
Asst. Prof. Dr. Bunyarit UyyanonvaraAsst. Prof. Dr. Bunyarit UyyanonvaraIT Program, Image and Vision Computing Lab.
School of Information and Computer Technology
Sirindhorn International Institute of Technology
Thammasat Universityhttp://www.siit.tu.ac.th/bunyarit
[email protected] 5013505 X 2005
ITS033 – Programming & Algorithms
15
0-1 Knapsack problem Given a knapsack with maximum capacity W, and a set S consisting of n
items
Each item i has some weight wi and benefit value bi (all wi , bi and W are
integer values)
Problem: How to pack the knapsack to achieve maximum total value of packed items?
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0-1 Knapsack problem:
W = 20
wi bi
109
85
54
43
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Weight Benefit value
This is a knapsackMax weight: W = 20
Items
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0-1 Knapsack problem Problem, in other words, is to find
Ti
iTi
i Wwb subject to max
The problem is called a “0-1” problem, because each item must be entirely accepted or rejected.
Just another version of this problem is the “Fractional Knapsack Problem”, where we can take fractions of items. (see Greedy Algorithm)
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0-1 Knapsack problem: brute-force approach
Since there are n items, there are 2n possible combinations of items.
We go through all combinations and find the one with the most total value and with total weight less or equal to W
Running time will be O(2n)
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0-1 Knapsack problem
Can we do better?
Yes, with an algorithm based on dynamic programming
We need to carefully identify the subproblems
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The Knapsack Problem To design a dynamic programming
algorithm,
we need to derive a recurrence relation that expresses a solution to an instance of the knapsack problem
in terms of solutions to its smaller sub-instances.
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Recursive Formula for subproblems
It means, that the best subset of Sk that has total weight w is one of the two:
1) the best subset of Sk-1 that has total weight w, or
2) the best subset of Sk-1 that has total weight w-wk plus the item k
else }],1[],,1[max{
if ],1[],[
kk
k
bwwkBwkB
wwwkBwkB
Recursive formula for subproblems:
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Recursive Formula
The best subset of Sk that has the total weight w, either contains item k or not.
First case: wk>w. Item k can’t be part of the solution, since if it was, the total weight would
be > w, which is unacceptable
Second case: wk <=w. Then the item k can be in the solution, and we choose the case with
greater value
else }],1[],,1[max{
if ],1[],[
kk
k
bwwkBwkB
wwwkBwkB
23
0-1 Knapsack Algorithm
for w = 0 to W
B[0,w] = 0
for i = 0 to n
B[i,0] = 0
for w = 0 to W
if wi <= w // item i can be part of the solution
if bi + B[i-1,w-wi] > B[i-1,w]
B[i,w] = bi + B[i-1,w- wi]
else
B[i,w] = B[i-1,w]
else B[i,w] = B[i-1,w] // wi > w
24
Example
Let’s run our algorithm on the following data:
n = 4 (# of elements)W = 5 (max weight)Elements (weight, benefit):(2,3), (3,4), (4,5), (5,6)
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Example (2)
for w = 0 to WB[0,w] = 0
0
0
0
0
0
0
W0
1
2
3
4
5
i 0 1 2 3
4
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Example (3)
for i = 0 to nB[i,0] = 0
0
0
0
0
0
0
W0
1
2
3
4
5
i 0 1 2 3
0 0 0 0
4
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Example (4)
if wi <= w // item i can be part of the solution if bi + B[i-1,w-wi] > B[i-1,w] B[i,w] = bi + B[i-1,w- wi] else B[i,w] = B[i-1,w] else B[i,w] = B[i-1,w] // wi > w
0
0
0
0
0
0
W0
1
2
3
4
5
i 0 1 2 3
0 0 0 0i=1bi=3
wi=2
w=1w-wi =-1
Items:1: (2,3)2: (3,4)3: (4,5) 4: (5,6)
4
0
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Example (5)
if wi <= w // item i can be part of the solution if bi + B[i-1,w-wi] > B[i-1,w] B[i,w] = bi + B[i-1,w- wi] else B[i,w] = B[i-1,w] else B[i,w] = B[i-1,w] // wi > w
0
0
0
0
0
0
W0
1
2
3
4
5
i 0 1 2 3
0 0 0 0i=1bi=3
wi=2
w=2w-wi =0
Items:1: (2,3)2: (3,4)3: (4,5) 4: (5,6)
4
0
3
29
Example (6)
if wi <= w // item i can be part of the solution if bi + B[i-1,w-wi] > B[i-1,w] B[i,w] = bi + B[i-1,w- wi] else B[i,w] = B[i-1,w] else B[i,w] = B[i-1,w] // wi > w
0
0
0
0
0
0
W0
1
2
3
4
5
i 0 1 2 3
0 0 0 0i=1bi=3
wi=2
w=3w-wi=1
Items:1: (2,3)2: (3,4)3: (4,5) 4: (5,6)
4
0
3
3
30
Example (7)
if wi <= w // item i can be part of the solution if bi + B[i-1,w-wi] > B[i-1,w] B[i,w] = bi + B[i-1,w- wi] else B[i,w] = B[i-1,w] else B[i,w] = B[i-1,w] // wi > w
0
0
0
0
0
0
W0
1
2
3
4
5
i 0 1 2 3
0 0 0 0i=1bi=3
wi=2
w=4w-wi=2
Items:1: (2,3)2: (3,4)3: (4,5) 4: (5,6)
4
0
3
3
3
31
Example (8)
if wi <= w // item i can be part of the solution if bi + B[i-1,w-wi] > B[i-1,w] B[i,w] = bi + B[i-1,w- wi] else B[i,w] = B[i-1,w] else B[i,w] = B[i-1,w] // wi > w
0
0
0
0
0
0
W0
1
2
3
4
5
i 0 1 2 3
0 0 0 0i=1bi=3
wi=2
w=5w-wi=2
Items:1: (2,3)2: (3,4)3: (4,5) 4: (5,6)
4
0
3
3
3
3
32
Example (9)
if wi <= w // item i can be part of the solution if bi + B[i-1,w-wi] > B[i-1,w] B[i,w] = bi + B[i-1,w- wi] else B[i,w] = B[i-1,w] else B[i,w] = B[i-1,w] // wi > w
0
0
0
0
0
0
W0
1
2
3
4
5
i 0 1 2 3
0 0 0 0i=2bi=4
wi=3
w=1w-wi=-2
Items:1: (2,3)2: (3,4)3: (4,5) 4: (5,6)
4
0
3
3
3
3
0
33
Example (10)
if wi <= w // item i can be part of the solution if bi + B[i-1,w-wi] > B[i-1,w] B[i,w] = bi + B[i-1,w- wi] else B[i,w] = B[i-1,w] else B[i,w] = B[i-1,w] // wi > w
0
0
0
0
0
0
W0
1
2
3
4
5
i 0 1 2 3
0 0 0 0i=2bi=4
wi=3
w=2w-wi=-1
Items:1: (2,3)2: (3,4)3: (4,5) 4: (5,6)
4
0
3
3
3
3
0
3
34
Example (11)
if wi <= w // item i can be part of the solution if bi + B[i-1,w-wi] > B[i-1,w] B[i,w] = bi + B[i-1,w- wi] else B[i,w] = B[i-1,w] else B[i,w] = B[i-1,w] // wi > w
0
0
0
0
0
0
W0
1
2
3
4
5
i 0 1 2 3
0 0 0 0i=2bi=4
wi=3
w=3w-wi=0
Items:1: (2,3)2: (3,4)3: (4,5) 4: (5,6)
4
0
3
3
3
3
0
3
4
35
Example (12)
if wi <= w // item i can be part of the solution if bi + B[i-1,w-wi] > B[i-1,w] B[i,w] = bi + B[i-1,w- wi] else B[i,w] = B[i-1,w] else B[i,w] = B[i-1,w] // wi > w
0
0
0
0
0
0
W0
1
2
3
4
5
i 0 1 2 3
0 0 0 0i=2bi=4
wi=3
w=4w-wi=1
Items:1: (2,3)2: (3,4)3: (4,5) 4: (5,6)
4
0
3
3
3
3
0
3
4
4
36
Example (13)
if wi <= w // item i can be part of the solution if bi + B[i-1,w-wi] > B[i-1,w] B[i,w] = bi + B[i-1,w- wi] else B[i,w] = B[i-1,w] else B[i,w] = B[i-1,w] // wi > w
0
0
0
0
0
0
W0
1
2
3
4
5
i 0 1 2 3
0 0 0 0i=2bi=4
wi=3
w=5w-wi=2
Items:1: (2,3)2: (3,4)3: (4,5) 4: (5,6)
4
0
3
3
3
3
0
3
4
4
7
37
Example (14)
if wi <= w // item i can be part of the solution if bi + B[i-1,w-wi] > B[i-1,w] B[i,w] = bi + B[i-1,w- wi] else B[i,w] = B[i-1,w] else B[i,w] = B[i-1,w] // wi > w
0
0
0
0
0
0
W0
1
2
3
4
5
i 0 1 2 3
0 0 0 0i=3bi=5
wi=4
w=1..3
Items:1: (2,3)2: (3,4)3: (4,5) 4: (5,6)
4
0
3
3
3
3
00
3
4
4
7
0
3
4
38
Example (15)
if wi <= w // item i can be part of the solution if bi + B[i-1,w-wi] > B[i-1,w] B[i,w] = bi + B[i-1,w- wi] else B[i,w] = B[i-1,w] else B[i,w] = B[i-1,w] // wi > w
0
0
0
0
0
0
W0
1
2
3
4
5
i 0 1 2 3
0 0 0 0i=3bi=5
wi=4
w=4w- wi=0
Items:1: (2,3)2: (3,4)3: (4,5) 4: (5,6)
4
0 00
3
4
4
7
0
3
4
5
3
3
3
3
39
Example (15)
if wi <= w // item i can be part of the solution if bi + B[i-1,w-wi] > B[i-1,w] B[i,w] = bi + B[i-1,w- wi] else B[i,w] = B[i-1,w] else B[i,w] = B[i-1,w] // wi > w
0
0
0
0
0
0
W0
1
2
3
4
5
i 0 1 2 3
0 0 0 0i=3bi=5
wi=4
w=5w- wi=1
Items:1: (2,3)2: (3,4)3: (4,5) 4: (5,6)
4
0 00
3
4
4
7
0
3
4
5
7
3
3
3
3
40
Example (16)
if wi <= w // item i can be part of the solution if bi + B[i-1,w-wi] > B[i-1,w] B[i,w] = bi + B[i-1,w- wi] else B[i,w] = B[i-1,w] else B[i,w] = B[i-1,w] // wi > w
0
0
0
0
0
0
W0
1
2
3
4
5
i 0 1 2 3
0 0 0 0i=3bi=5
wi=4
w=1..4
Items:1: (2,3)2: (3,4)3: (4,5) 4: (5,6)
4
0 00
3
4
4
7
0
3
4
5
7
0
3
4
5
3
3
3
3
41
Example (17)
if wi <= w // item i can be part of the solution if bi + B[i-1,w-wi] > B[i-1,w] B[i,w] = bi + B[i-1,w- wi] else B[i,w] = B[i-1,w] else B[i,w] = B[i-1,w] // wi > w
0
0
0
0
0
0
W0
1
2
3
4
5
i 0 1 2 3
0 0 0 0i=3bi=5
wi=4
w=5
Items:1: (2,3)2: (3,4)3: (4,5) 4: (5,6)
4
0 00
3
4
4
7
0
3
4
5
7
0
3
4
5
7
3
3
3
3
42
Comments
This algorithm only finds the max possible value that can be carried in the knapsack
To know the items that make this maximum value, an addition to this algorithm is necessary
We can trace back to see how to extract this data from the table we built
43
Running timefor w = 0 to W
B[0,w] = 0
for i = 0 to n
B[i,0] = 0
for w = 0 to W
< the rest of the code >
What is the running time of this algorithm?
O(W)
O(W)
Repeat n times
O(n*W)
Remember that the brute-force algorithm takes O(2n)
44
Example
capacity W = 5
Build a Dynamic Programming Table for this Knapsack Problem
45
Example – Dynamic Programming Table
capacity W = 5
46
Example
Thus, the maximal value is V [4, 5]= $37. We can find the composition of an optimal subset by tracing back the computations of this entry in the table.
Since V [4, 5] is not equal to V [3, 5], item 4 was included in an optimal solution along with an optimal subset for filling 5 - 2 = 3 remaining units of the knapsack capacity.
capacity W = 5
47
Example
The remaining is V[3,3] Here V[3,3] = V[2,3] so item 3 is not included V[2,3] V[1,3] so item 2 is included
capacity W = 5
48
Example
The remaining is V[1,2] V[1,2] V[0,2] so item 1 is included
The solution is {item 1, item 2, item 4} Total weight is 5 Total value is 37
capacity W = 5
49
The Knapsack Problem The time efficiency and space efficiency of this algorithm
are both in θ(nW).
The time needed to find the composition of an optimal solution is in O(n + W).
50
Dynamic Programming: Memory Function
Lecture 07.3
Asst. Prof. Dr. Bunyarit UyyanonvaraAsst. Prof. Dr. Bunyarit UyyanonvaraIT Program, Image and Vision Computing Lab.
School of Information and Computer Technology
Sirindhorn International Institute of Technology
Thammasat Universityhttp://www.siit.tu.ac.th/bunyarit
[email protected] 5013505 X 2005
ITS033 – Programming & Algorithms
51
Memory Function The classic dynamic programming approach, fills
a table with solutions to all smaller subproblems but each of them is solved only once.
An unsatisfying aspect of this approach is that solutions to some of these smaller subproblems are often not necessary for getting a solution to the problem given.
52
Memory Function Since this drawback is not present in the top-
down approach, it is natural to try to combine the strengths of the top-down and bottom-up approaches.
The goal is to get a method that solves only subproblems that are necessary and does it only once. Such a method exists; it is based on using memory functions
53
Memory Function Initially, all the table’s entries are initialized with a special “null”
symbol to indicate that they have not yet been calculated.
Thereafter, whenever a new value needs to be calculated, the method checks the corresponding entry in the table first: if this entry is not “null,” it is simply retrieved from the table;
otherwise, it is computed by the recursive call whose result is then recorded in the table.
54
Memory Function for solving Knapsack Problem
55
Memory Function for solving Knapsack Problem
56
Memory Function In general, we cannot expect more than a constant-factor
gain in using the memory function method for the knapsack problem because its time efficiency class is the same as that of the bottom-up algorithm
A memory function method may be less space-efficient than a space efficient version of a bottom-up algorithm.
57
Conclusion Dynamic programming is a useful technique of solving certain kind of
problems
When the solution can be recursively described in terms of partial solutions, we can store these partial solutions and re-use them as necessary
Running time (Dynamic Programming algorithm vs. naïve algorithm): 0-1 Knapsack problem: O(W*n) vs. O(2n)
58
ITS033Topic 01Topic 01 -- Problems & Algorithmic Problem SolvingProblems & Algorithmic Problem SolvingTopic 02Topic 02 – Algorithm Representation & Efficiency Analysis – Algorithm Representation & Efficiency AnalysisTopic 03Topic 03 - State Space of a problem - State Space of a problemTopic 04Topic 04 - Brute Force Algorithm - Brute Force AlgorithmTopic 05Topic 05 - Divide and Conquer - Divide and ConquerTopic 06Topic 06 -- Decrease and ConquerDecrease and ConquerTopic 07Topic 07 - Dynamics Programming - Dynamics ProgrammingTopic 08Topic 08 -- Transform and ConquerTransform and ConquerTopic 09Topic 09 - Graph Algorithms - Graph AlgorithmsTopic 10Topic 10 - Minimum Spanning Tree - Minimum Spanning TreeTopic 11Topic 11 - Shortest Path Problem - Shortest Path ProblemTopic 12Topic 12 - Coping with the Limitations of Algorithms Power - Coping with the Limitations of Algorithms Power
http://www.siit.tu.ac.th/bunyarit/its033.phphttp://www.siit.tu.ac.th/bunyarit/its033.phpand and http://www.vcharkarn.com/vlesson/showlesson.php?lessonid=7http://www.vcharkarn.com/vlesson/showlesson.php?lessonid=7
Midterm
59
End of Chapter 7
Thank you!