Daa:Dynamic Programing

General Strategy

 Used for optimization problems: often minimizing or maximizing.
 Solves problems by combining solutions to sub-problems.
 Sub-problems are not independent.
 Reduces computation by
◦ Solving sub-problems in a bottom-up fashion.
◦ Storing solution to a sub-problem the first time it is solved.
◦ Looking up the solution when subproblem is encountered again.
 Key: determine structure of optimal solutions

Steps in Dynamic Programming

 Characterize structure of an optimal solution.
 Define value of optimal solution recursively.
 Compute optimal solution values caching in bottom-
up manner from table.
 Construct an optimal solution from computed values.

Principle of optimality
 Principle of optimality states that “ In an
optimal sequence of decisions or choices each
sub-sequences must be optimal.”
e.g. D

A

C B

 Suppose that in solving a problem, we have to make
a sequence of decisions D1, D2,…, Dn. Then
according to Principle of Optimality, if this
sequence is optimal, then the last k decisions,
1< k< n must be optimal.

Knapsack Problem
 A thief robbing a store and can carry a maximal weight of W into
their knapsack. There are n items and ith item weight wi and is
worth vi dollars. What items should thief take?

 Fractional Knapsack
◦ The items are allowed to be broken into smaller pieces so that
knapsack contains fraction of xi of item i, where 0 ≤ xi ≤ 1

 0/1 Knapsack
◦ Either to include an item in knapsack or to leave it (binary
choice), but may not take a fraction of an item i.e. 0 or 1.

0/1 Knapsack Problem
Let i=1,2,3,…,n are items available.
wi - weight of ith item
vi – value of ith item
m – capacity of knapsack
 Now the problem is to maximize the value or profit with
capacity constraints
Let {x1,x2,…,xn} is the optimal solution
Where xi ∈ {0,1}

=0 if i=0 or j=0
=-∞ if j<0
C[i,j] =c[i-1,j] if wi ≥ 0

=max{vi+c[i-1,j-wi],c[i-1,wi]} if i>0 & j>wi

Consider the E.g. : w={1,2,5,6,7}
v={1,6,18,22,28}
m=11

How to calculate the cost table

C[1,1]=max{C[1-1,1-1]+1,C[1-1,1]}
=max{1,0}=1
C[1,2]=max{C[1-1,2-1]+1,C[1-1,1]}
=max{1,0}=1
C[3,3]=max{C[3-1,3-5]+18,C[3-1,3]}
=max{-∞,C[2,3]}
=C[2,3]

 Thus Optimal Solution Vector is, {0,0,1,1,0}
 And Optimal Profit is, 0+18+22+0=40

Dknap(i,j,m)
{
w :=[w1,w2,…,wn] ; // array of weights
v := [v1,v2,…,vn]; // array of values
C:= [1,2,…,n,1,2,…,m]; // Knapsack array(2-D)
// 1,2,…i,…,n are objects
// 1,2,…,j,…,m are capacities of knapsack

for i :=1 to n do
C[i,0]:=0; //Knapsack with capacity 0
for i:=1 to n do
{
for j:=1 to m do
{
if (i=1 and j<w[i]) then
C[i,j]=0;
else
if (i=1) then
C[i,j]=v[i];
else
if(j<w[i]) then
C[i,j]=C[i-1,j];
else
if(C[i-1,j] > C[i-1,j-w[i]]+v[i]) then
C[i,j]= C[i-1,j];
else
C[i,j]= C[i-1,j-w[i]]+v[i];
}
}
Traceback(C)
}

Traceback(C)
{ //opt[] array is used to store optimal solution vector
i:=n; j:=m;
while(i≥1)
{ k:=i; //k is the index of opt[] array
while(j≥0)
{
if(C[i,j]=C[i-1,j]) then
{
i:=i-1;
opt[k]:=0;
else
{
opt[k]=1;
i:=i-1;
j:=j-w[i];
}
}

Analysis of 0/1Knapsack
Time Complexity is Θ(nm) as there are n m
entries in table.

Using Dynamic Programming determine the
optimal profit and solution vector
w={2,3,4} v ={1,2,5} m=6

Optimal solution vector is {1,0,1} and Profit is 6

Daa:Dynamic Programing

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