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### Welcome to our presentation

### What is dynamic programming?

Presentation topic : Dynamic programming

Presented by: 1. Shahin Pervin

(072-20-105)

2. Mostakima Yesmin

(092-15-793)

3. Jyotirmoyee Saha

(092-15-804)

Dynamic programming is a technique for solving problem and come up an algorithm. Dynamic programming divide the problem into subparts and then solve the subparts and use the solutions of the subparts to come to a solution. The main difference between dynamic programming and divide and conquer design technique is that the partial solutions are stored in dynamic programming but are not stored and used in divide and conquer technique.

It’s history

- The term dynamic programming was originally used in the 1940s by Richard Bellman to describe the process of solving problems where one needs to find the best decisions one after another.

Why does it apply?

- Dynamic programming is typically applied to optimization problems. In such problems there can be many possible solution. Each solution has a value, and we wish to find a solution with the optimal (minimum or maximum) value.

It’s step

The development of dynamic programming algorithm can be broken into a sequence of four steps-

- Characterize the structure of an optimal solution.
- Recursively define the value of an optimal solution.
- Compute the value of an optimal solution in a bottom-up fashion.
- Construct an optimal solution from computed information.

Example

- Matrix-chain multiplication----

If the chain of matrices is (A1,A2,A3,A4), the product A1A2A3A4 can be fully parenthesized in five distinct ways:

(A1(A2(A3A4)))

(A1((A2A3)A4))

((A1A2)(A3A4))

((A1(A2A3))A4)

(((A1A2)A3)A4)

Matrix-chain Multiplication Problem

- If columns A equal the numbers of rows of B

For A=p*q and B=q*r. the resulting matrix C is a p*r.

The matrix-chain multiplication problem-

it’s not actually multiplying matrices.It’s goal is only to determine an order for multiplying matrices that has the lowest or highest cost.

Matrix-chain Multiplication Apply

- We have many options because matrix multiplication is associative. In other words, no matter how we parenthesize the product, the result will be the same. For example, if we had four matrices A, B andC we would have:
- (A(BC))= ((AB)C) = ((AC)B) = ....
- However, the order in which we parenthesize the product affects the number of simple arithmetic operations needed to compute the product, or the efficiency. For example, suppose A is a (10*100) matrix, B is a (100*5) matrix, and C is a (5 × 50) matrix. Then,
- ((AB)C)= (10*5*100) + (10*50*5) = 5000 + 2500 = 7500 operations
- (A(BC)) = (100*50*5) + (50*10*100) = 25000 + 50000 = 75000 operations.

Structure of an Optimal parenthesization

- The structure of an optimal parenthesization
- Notation: Ai..j = result from evaluating AiAi+1…Aj (i j)
- Any parenthesization of AiAi+1…Aj must split the product between Akand Ak+1 for some integer k in the range i k < j
- Cost = cost of computing Ai..k + cost of computing Ak+1..j + cost of multiplying Ai..k and Ak+1..j together.

A recursive solution

- m[i, j ] = m[i, k] + m[k+1, j ] + pi-1pk pj for i ≤ k < j
- m[i, i ] = 0 for i=1,2,…,n

A recursive solution

But… optimal parenthesization occurs at one value of k among all possible i ≤ k < j

Check all these and select the best one

0 if i=j

m[i, j ] =

min {m[i, k] + m[k+1, j ] + pi-1pk pj}if i<j

i≤ k< j

11-11

Algorithm to Compute Optimal Cost

First computes costs for chains of length l=1

Then for chains of length l=2,3, … and so on

Computes the optimal cost bottom-up

Input: Array p[0…n] containing matrix dimensions and n

Result: Minimum-cost table m and split table s

MATRIX-CHAIN-ORDER(p[ ], n)

for i ← 1 ton

m[i, i]← 0

for l ← 2 ton

for i ← 1 ton-l+1

j← i+l-1

m[i, j]←

for k ← itoj-1

q ← m[i, k] + m[k+1, j] + p[i-1] p[k] p[j]

ifq < m[i, j]

m[i, j]← q

s[i, j]← k

returnm and s

Takes O(n3) time

Requires O(n2) space

11-12

l = 2

35*15*5=2625

10*20*25=5000

m[3,4]+m[5,5] + 15*10*20 =750 + 0 + 3000 = 3750

m[3,5] = min

m[3,3]+m[4,5] + 15*5*20=0 + 1000 + 1500 = 2500

Constructing an optimal solution

Each entry s[i, j] records the value of k such that the optimal parenthesization of AiAi+1…Aj splits the product between Ak and Ak+1

A1..n A1..s[1..n] As[1..n]+1..n

A1..s[1..n] A1..s[1, s[1..n]] As[1, s[1..n]]+1..s[1..n]

Recursive…

Optimal parenthesization:

((A1(A2A3))((A4 A5)A6))

The end

- Thank You

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