dynamic programming complete by mumtaz ali (03154103173)

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Dynamic ProgrammingPrepared by :-

MumtazUsmanUsama

Contents: Meaning Defination What is dynamic programming used for Technique used in it Divide and Conquer Strategy General Divide and Conquer recurrence Common example Approaches of dynamic programming Elements of Dynamic Programming Dynamic Programming and Chain Matrix Multiplication Fibonacci Numbers Steps for problem solving Advantages and Disadvantages Good bye :p :p

What does Dynamic Mean ??

Characterized by continuous change activity

Characterized by much activity and vigor, especially in bringing about change energetic and forceful

The designing, scheduling, or planning of a program Ref http://www.thefreedictionary.com

So Combine Meaning Of Dynamic Programming is :

Change and Schedule a Solution for a problem

Defination :

Dynamic Programming refers to a very large class of algorithms. The idea is to break a large problem down (if possible) into incremental steps so that, at any given stage, optimal solutions are known to sub-problems.

Ref :- Monash University

DEFINATION :

Use of dynamic programming

Dynamic programming is used for problems requiring a sequence of interrelated decision. This means that to take another decision we have to depend on the previous decision or solution formed.

Technique used in dynamic programming

We Also Know this Strategy As

The Divide and Conquer Algorithm

• Divide_Conquer(problem P)• {• if Small(P) return S(P);• else {• divide P into smaller instances P1, P2, …, P k , k1;• Apply Divide_Conquer to each of these sub problems;

• return Combine(Divide_Conquer(P1), Divide_Conquer(P2),…, Divide Conquer(P k));

• }• }

14

Divide Conquer recurrence relation

The computing time of Divide Conquer is

T(n) is the time for Divide Conquer on any input size n. g(n) is the time to compute the answer directly (for small inputs) f(n) is the time for dividing P and combining the solutions.

)()(...)()()(

)(21 nfnTnTnT

ngnT

k

n small

otherwise

15

General Divide and Conquer recurrence

The Master Theorem

T(n) = aT(n/b) + f (n), where f (n) ∈ Θ(n k) 1. a < bk T(n) ∈ Θ(n k) 2. a = bk T(n) ∈ Θ(n k lg n ) 3. a > bk T(n) ∈ Θ(n log b a)

the time spent on solving a sub problem of size n/b.

the time spent on dividing the probleminto smaller ones and combining their solutions.

Difference between DP and Divide-and-Conquer

Using Divide-and-Conquer to solve these problems is inefficient because the same common sub problems have to be solved many times.

DP will solve each of them once and their answers are stored in a table for future use.

C o m m o n

i) Backward Recursionii) Forward Recursion

Two Approach of Dynamic Programming

It Contains Sequence of “n” decisions.

Each “n” corresponding to one of the decision.

Each stage of analysis is described by a set of elements decision, input state, output state and return.

Then symbolic representation of n stages of analysis using backward recursion so we can formalize the notation

Backward recursion

Cumulative return = Direct return + Cumulative returnthrough stage from stage through stage i-1

We use sb to denote the previous state Tb determines the state that came before s when the decision made to reach state s is d Db(s) is the set of decisions that can be used to enter state s

sb = Tb(s, d), //where d belongs to Db(s)

Backward recursion formula

What if we don’t have a final stage

Approach takes a problem Decomposed into a sequence of n stages Analyzes the problem starting with the first

stage in the sequence Working forward to the last stage it is also known as deterministic probability

approach

Forward recursion

Example

Elements of Dynamic Programming

i) Optimal substructureii) Overlapping sub problemiii) Memoization

Substructure

What IS

 a problem is said to have optimal substructure if an optimal solution can be constructed efficiently from optimal solutions of its sub problems. This property is used to determine the usefulness of dynamic programming

ref : wikipedia

Optimal Substructure

This is the necessary property , if this property is not present we cant use dynamic programming.

a problem p , with sub problems p1,p2.Solution of problem p is s , and s1 is the optimum solution of sub

problem p1,and s2 is the optimum solution of sub problem p2. Claim “s” is the optimal solution ,if both solution of the sub problems

are optimal only then the final solution is optimal.

Dynamic Programming

Chain Matrix

Multiplication

Dynamic Programming and Chain Matrix Multiplication

In mathematics or computer science, Dynamic Programming is a method for solving complex problems by breaking them down into simpler sub-problems. So, Matrix chain multiplication is an ideal example that demonstrates utility of dynamic programming.

Engineering applications often have to multiply a large chain of matrices with large dimensions, for example: 100 matrices of dimension 100×100. We can multiply this chain of matrices in many different ways, but we will go for that way which takes lower computations.

Dynamic Programming of Chain Matrix MultiplicationFor example, we are going to multiply 4 matrices:

M1 = 2 x 3M2 = 3 x 4M3 = 4 x 5M4 = 5 x 7

And we have conditions for multiplying matrices:• We can multiply only two matrices at a time.• When we go to multiply 2 matrices, the number of columns of 1st

matrix should be same as the number of rows of 2nd matrix.

M1 = 2 x 3M2 = 3 x 4M3 = 4 x 5M4 = 5 x 7

( M1 M2 )( M3 M4 )= 220(( M1 M2 ) M3 ) M4= 134M1 ( M2 ( M3 M4 ) = 266( M1 ( M2 M3 ) M4 = 160M1 (( M2 M3 ) M4 ) = 207

We can multiply the chain of matrices by following those conditions in these ways:

Numbers of the rightmost side is number of total scalar multiplication. So we have realized that we can reduce the number of total multiplication and this reduced time is a fact for a large chain of matrices.

Algorithm and Mechanism

Renaming matrices as Mi and dimensions as Pi - 1 x Pi , we have got:

M1 = P0 x P1

M2 = P1 x P2

M3 = P2 x P3

M4 = P3 x P4

| | |Mi = Pi – 1 x Pi

We will use a formula:

Where C i, j means Mi to Mj .i.e.: C 1, 4 means M1 to M4 .

And we will use a variable 'k' as follows:M1 |k=1 M2 M3 M4

M1 M2 |k=2 M3 M4

M1 M2 M3 |k=3 M4

The thing we’re going to do is to apply above formula for every 'k' in the range 'i' to 'j' and pick the lowest value every step.

C 1 , 4 = min ( C1 , 1 + C2 , 4 + P0 * P1 * P4 , C1 , 2 + C3 , 4 + P0 * P2 * P4 , C1 , 3 + C4 , 4 + P0 * P3 * P4 ) = min ( 207, 220, 134 ) = 134C 2, 4 = min ( C2 , 2 + C3 , 4 + P1 * P2 * P4 , C2 , 3 + C4 , 4 + P1 * P3 * P4 ) = min ( 224, 165 ) = 165C 1, 3 = min ( C1 , 1 + C2 , 3 + P0 * P1 * P3 , C1 , 2 + C3 , 3 + P0 * P2 * P3 ) = min ( 90, 64 ) = 64C 1, 2 = P0 * P1 * P2 = 24C 2, 3 = P1 * P2 * P3 = 60C 3, 4 = P2 * P3 * P4 = 140

Pseudocode

1. int Chain( int i, int j )

2. {

3. int min = 10000, value, k;

4. if( i == j ){

5. return 0;

6. }

7. else{

8. for( k = i; k < j; k++ ){

9. value = (Chain(i, k) + Chain(k + 1, j) + (dimensions[i-1] * dimensions[k] * dimensions[j]));

10. if( min > value ){

11. min = value;

12. mat[i][j] = k;

13. }

14. }

15. }

16. return min;

17. }

1. int main(void)

2. {

3. int result, i;

4. printf("Enter number of matrices: ");

5. scanf("%d", &n);

6. printf("Enter dimensions : ");

7. for( i = 0; i <= n; i++ ){

8. scanf("%d", &dimensions[i]);

9. }

10. result = Chain(1, n);

11. printf("\nTotal number of multiplications: %d and\n", result);

12. printf("Multiplication order is: ");

13. PrintOrder( 1, n );

14. printf("\n");

15. }

Input and Output in Console App

Fibonacci Numbers

Fibonacci Numbers Fn= Fn-1+ Fn-2 n ≥ 2 F0 =0, F1 =1 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, …

Straightforward recursive procedure is slow!

Let’s draw the recursion tree

Fibonacci Numbers

Fibonacci Numbers

How many summations are there? Using Golden Ratio

As you go farther and farther to the right in this sequence, the ratio of a term to the one before it will get closer and closer to the Golden Ratio.

Our recursion tree has only 0s and 1s as leaves, thus we have 1.6n summations

Running time is exponential!

Fibonacci Numbers We can calculate Fn in linear time by remembering solutions to the

solved subproblems – dynamic programming

Compute solution in a bottom-up fashion

In this case, only two values need to be remembered at any time

problem solving

Steps for problem solving

There are four steps of problem solving Optimal Solution Structure Recursive Solution Optimal Solution Value Optimal Solution

Problem Definition

Problem: Given all these costs, what stations should be chosen from line 1 and from line 2 for minimizing the total time for car assembly.

“Brute force” is to try all possibilities. requires to examine Omega(2n) possibilities Trying all 2n subsets is infeasible when n is large.

Simple example : 2 station (2n) possibilities =4

start end

Step 1: Optimal Solution Structure

optimal substructure : choosing the best path to Sij.

The structure of the fastest way through the factory (from the starting point)

The fastest possible way to get through Si,1 (i = 1, 2) Only one way: from entry starting point to Si,1 take time is entry time (ei)

Step 1: Optimal Solution Structure The fastest possible way to get through Si,j (i = 1, 2) (j = 2, 3, ...,

n). Two choices:

Stay in the same line: Si,j-1 Si,j Time is Ti,j-1 + ai,j

If the fastest way through Si,j is through Si,j-1, it must have taken a fastest way through Si,j-1

Transfer to other line: S3-i,j-1 Si,j Time is T3-i,j-1 + t3-i,j-1 + ai,j

Same as above

Step 1: Optimal Solution Structure An optimal solution to a problem

finding the fastest way to get through Si,j contains within it an optimal solution to sub-problems

finding the fastest way to get through either Si,j-1 or S3-i,j-1 Fastest way from starting point to Si,j is either:

The fastest way from starting point to Si,j-1 and then directly from Si,j-1 to Si,jor The fastest way from starting point to S3-i,j-1 then a transfer from line 3-i to

line i and finally to Si,j

Optimal Substructure.

Example

Step 2: Recursive Solution

Define the value of an optimal solution recursively in terms of the optimal solution to sub-problems

Sub-problem here finding the fastest way through station j on both lines (i=1,2) Let fi [j] be the fastest possible time to go from starting point through Si,j

The fastest time to go all the way through the factory: f*

x1 and x2 are the exit times from lines 1 and 2, respectively

Step 2: Recursive Solution

The fastest time to go through Si,j e1 and e2 are the entry times for lines 1 and 2

Example

Example

Step 2: Recursive Solution

To help us keep track of how to construct an optimal solution, let us define li[j ]: line # whose station j-1 is used in a fastest way through Si,j (i = 1, 2,

and j = 2, 3,..., n)

we avoid defining li[1] because no station precedes station 1 on either lines.

We also define l*: the line whose station n is used in a fastest way through the entire

factory

Step 2: Recursive Solution

Using the values of l* and li[j] shown in Figure (b) in next slide, we would trace a fastest way through the factory shown in part (a) as follows

The fastest total time comes from choosing stations Line 1: 1, 3, & 6 Line 2: 2, 4, & 5

Step 3: Optimal Solution Value

Step 3: Optimal Solution Value

Step 3: Optimal Solution Value

Step 3: Optimal Solution Value

Step 3: Optimal Solution Value

Step 3: Optimal Solution Value

Step 3: Optimal Solution Value

Step 3: Optimal Solution Value

Step 4: Optimal Solution

Constructing the fastest way through the factory

Dynamic Programming

1)`the process of breaking down a complex problem into a series of interrelated sub problems often provides insight into the nature of problem

2) Because dynamic programming is an approach to optimization rather than a technique it has flexibility that allows application to other types of mathematical programming problems

3) The computational procedure in dynamic programming allows for a built in form of sensitivity analysis based on state variables and on variables represented by stages

4)Dynamic programming achieves computational savings over complete enumeration.

Advantages

1.)more expertise is required in solving dynamic programming problem then using other methods

2.)lack of general algorithm like the simplex method. It restricts computer codes necessary for inexpensive and widespread use

3.)the biggest problem is dimensionality. This problems occurs when a particular application is characterized by multiple states. It creates lot of problem for computers capabilities & is time consuming

Advantages

Dis

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