The document discusses the knapsack problem and algorithms to solve it. It describes the 0-1 knapsack problem, which does not allow breaking items, and the fractional knapsack problem, which does. It provides an example comparing the two. The document then explains the greedy algorithm approach to solve the fractional knapsack problem by calculating value to weight ratios and filling the knapsack with the highest ratio items first. Pseudocode for the greedy fractional knapsack algorithm is provided along with analysis of its time complexity.

1. KNAPSACK
PROBLEM ALGORIT
HM
Department of Masters Of Computer Science(MCA)
Central university of Karnataka
Presented by: Prajjaval Rao Chintal
2018MCA09

2. WHAT ?
• GIVEN WEIGHTS AND VALUES OF N ITEMS, WE NEED TO PUT THESE ITEMS IN A
KNAPSACK OF CAPACITY W TO GET THE MAXIMUM TOTAL VALUE IN THE
KNAPSACK.

3. TYPES
• 0-1 KNAPSACK PROBLEM
In the 0-1 knapsack problem, we are not allowed to break items. We either
take the whole item or don’t take it.
• FRACTIONAL KNAPSACK
In fractional knapsack, we can break items for maximizing the total value
of knapsack. This problem in which we can break an item is also called the
fractional knapsack problem.
THE OPTIMAL KNAPSACK ALGORITHM

4. EXAMPLE
0-1 KNAPSACK
Take B and C
Total weight =20+30=50
Total value
=100+120=220
FRACTIONAL KNAPSACK
Take A,B and 2/3rd of C
Total weight =10+20+(30* 2/3)=50
Total value =60+100+(120* 2/3)=240

5. GREEDY APPROACH
• The basic idea of the greedy approach is to calculate the ratio value/weight for
each item.
• Sort the item on basis of this ratio.
• Then take the item with the highest ratio and add them until we can’t add the
next item as a whole.
• At the end add the next item as much (fraction) as we can.

6. GREEDY APPROACH SOLUTION
Ratio=
6
Ratio=
5
Ratio=
4
1. calculate the ratio
value/weight.
2. Sort (descending) item on basis
of ratio.
3. Take item with highest ratio
and add to knapsack until we
cant add the next item as
whole.
4. At the end add the next item
Capacity left =40
Value = 60
Capacity left =20
Value = 160
Take 2/3rd of C
Weight = 2/3 *30 =20
Value = 2/3 *120 =80
Capacity left =0
Value = 240

7. THE OPTIMAL KNAPSACK ALGORITHM
• INPUT: AN INTEGER N
• Positive values wi and vi such that 1 <= i <= n
• Positive value W.
• OUTPUT:
• N values of xi such that 0 <= xi <=1
• Total profit

8. GREEDY KNAPSACK ALGORITHM
Algorithm: Greedy-Fractional-Knapsack (w[1..n], p[1..n], W)
for i = 1 to n
do x[i] = 0 // X is ratio of items
weight = 0 //remaining weight of knapsack
for i = 1 to n
if weight + w[i] ≤ W //W is total capacity of bag
then x[i] = 1
weight = weight + w[i] //w[i] is the weight of item
else
x[i] = (W - weight) / w[i]
weight = W
break
return x
Time complexity
• If the provided items are already sorted into a decreasing order of then the
whileloop takes a time in O(n).
• As main time taking step is sorting, the whole problem can be solved in
O(n log n) only.
• It can be O(n^2),O(n^3) anything greater than O(nlogn).
• BUT we usually consider the tightest upper bound that is O(nlogn).

9. Thank you