Welcome to our presentation
Presentation topic : Dynamic programming
Presented by: 1. Shahin Pervin
(072-20-105)
2. Mostakima Yesmin
(092-15-793)
3. Jyotirmoyee Saha
(092-15-804)
What is dynamic programming?
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 problemit’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 and C 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 Ak and 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
m[i, j ] =
0
if 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

Takes O(n3) time
MATRIX-CHAIN-ORDER(p[ ], n)
Requires O(n2)
for i ← 1 to n
m[i, i] ← 0
for l ← 2 to n
for i ← 1 to n-l+1
j ← i+l-1
m[i, j] ← 
for k ← i to j-1
q ← m[i, k] + m[k+1, j] + p[i-1] p[k] p[j]
if q < m[i, j]
m[i, j] ← q
s[i, j] ← k
return m and s
space
11-12
l =3
l=2
35*15*5=
2625
10*20*25
=5000
m[3,5] = min
m[3,4]+m[5,5] + 15*10*20
=750 + 0 + 3000 = 3750
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