Knapsack Problem (DP & GREEDY)

COURSE TITLE: ALGORITHM
TOPIC: KNAPSACK PROBLEM
(DP & GREEDY)

CONTENTS
Introduction
Explanation & Example
Algorithm & Complexity
Application

Knapsack problem states that: Given a set of items,
each with a weight and a profit, determine the
number of each item to include in a collection so that
the total weight is less than or equal to a given limit
and the total profit is as large as possible.
Definition:

Version of knapsack:
There are two versions of knapsack problem:
1. 0/1 Knapsack Problem: Items are indivisible;
you either take them or not. And it is solved
using Dynamic Programming(DP).
2. Fractional Knapsack Problem: Items are
divisible; you can take any fraction of an item.
And it is solved using Greedy Algorithm.

Explanation:
Necklace
Weight: 1kg
Value: 3000tk
Camera
Weight: 2kg
Value: 1000tk
Laptop
Weight: 3kg
Value: 20000tk
Vase
Weight: 4kg
Value: 4000tk
Knapsack
Weight: 7kg
????

example of 0/1 knapsack:
DP approach to solve 0/1 knapsack problem:
 Identify the smaller sub-problems.
 Recursively define the value of an optimal solution in terms
of solutions to smaller problems.
Initial condition:
B[0,w]=0 for w>=0
B[i,0]=0 for i>=0
Recursive condition:
B[k,w]= B[k-1,w]; if w<wk
max{B[k-1,w], B[k-1, w-wk]+bk}; else
Example: Number of n=5, capacity m=6 & the weights and profits ar
given below:Weight(wk) 1 4 3 2 5
Profit(bk) 3 6 4 4 6
Now find the chosen item with maximum profit.

Solution:
Step 1: Identify the smaller sub-problems.
Step 2: Use the initial and recursive condition to find the maximum prof
Step 3: Use recursion to find the chosen item.
0 0 0 0 0 0 0
0 3 3 3 3 3 3
0 3 4 7 7 7 7
0 3 4 7 7 8 11
0 3 4 7 7 9 11
0 3 4 7 7 9 11
W 0 1 2 3 4 5 6i
0
1
2
3
4
5
(1,3)
(2,4)
(3,4)
(4,6)
(5,6)
Maximum
Profit
Ans : Maximum Profit=11. Item Taken :
Weight 1 2 3
Profit 3 4 4

Example of fractional knapsack:
Example: Number of items n=3, Capacity m=45
&
the weights and profits are given below:Weight (wi) 20 20 20
Profit (pi) 40 60 50
Now find out the fraction of chosen items with maximum p
Greedy approach to solve
Fractional knapsack problem:
Find the unit ui using the formula ui=pi/wi.
Find the fraction of the items xi that will be taken
in order to get maximum profit.

SOLUTION:
Step 1: Find the unit ui.
Step 2: Sort the item in descending order of ui.
Step 3: Find the maximum profit & the fraction xi of the items.
wi 20 20 20
pi 60 50 40
ui=pi/xi 3 2.5 2
xi 1 1 1/4
Therefore, maximum profit=(60*1)+(50*1)+40*1/4)=120.
Ans: Maximum Profit =120.
Taken fraction of the item:
Weight 20 20 20
Profit 60 50 40
Fraction 1 1 1/4

Algorithm& Complexityof 0/1 knapsack (DP):
Algorithm:
Knapsack_DP(n,W)
1. for w = 1 to m
2. do B[0,w] 0
3. for i = 1 to n
4. do B[i,0] 0
5. for w = 1 to m
6. do if (Wi < = w && B[i-1,w-Wi] +bk > B[i-1,w]
7. then B[i,w] B[i-1,w-Wi]+bk
8. else B[i,w] B[i-1,w]
Complexity:
Clearly the dynamic programming algorithm for the knapsack
problem has a time complexity of O(n.w) .
Where , n= the number of items &
w= the capacity of the knapsack.

Algorithm& Complexity of FRACTIONAL
knapsack(GREEDY):
Algorithm:
1. for i=1 to n
2. do x[i]=0.0
3. for i=1 to n
4. {
5. if(w[i]>m) then break
6. x[i]=1.0; m=m-w[i]
7. }
8. if(i<=n) then x[i]= m/w[i]
Complexity:
First sort according to profit to weight ratio time required : n log n
Then we need one for loop to find out maximum profit and for that
time needed is: n .
Overall Time complexity : n log n + n = O(n log n)

Application of knapsack:
Knapsack problems appear in real
world decision making processes in a
wide variety of fields, such as finding
the least wasteful way to cut raw
materials , selection of
investment and selection of assets.

ANY QUESTIONS ?
THANK YOU
EVERYONE

Knapsack Problem (DP & GREEDY)

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