0/1 Knapsack Problem in Dynamic Programming

1. 0/1 KNAPSACK
PROBLEM
-RACKSAVI.R,
I M.Sc Information Technology,
V.V.Vanniaperumal college for women,
Virudhunagar.

2. OBJECTIVES
 Dynamic Programming
 0/1 Knapsack Problem
 Algorithm
 Example Problem

3. DYNAMIC PROGRAMMING
Dynamic Programming
is used where we have
problems, which can be
divided into similar sub-
problems.

4. 0/1 KNAPSACK PROBLEM
 The 0/1 Knapsack Problem is a
classical dynamic programming
problem.
 The goal is to maximize the value
of a Knapsack that can hold at most W
units.

5. Contd…
The objective is to
maximize the profit:-
• The capacity of the
bag is C.
• We have n items.
• The ith item has Value
vi and Weight wi.

6. ALGORITHM
for w = 0 to w
V[0,w] = 0 O(W)
for i = 1 to n
V[i ,0] = 0 O(n)
for i = 1 to n // Repeat n times
for w = 1 to w
if wi <= w O(W)
//item i can be part of
the solution

7. If bi + V[i-1,w-wi] > V[i-1,w]
V[i,w] = bi + V[i-1,w-wi]
Else
V[i,w] = V[i-1,w]
Else
V[i,w] = V[i-1,w] //wi > w

8. EXAMPLE PROBLEM
GIVEN:-
Weight = {3,4,6,5}
Profits = {2,3,1,4}
W = 8
N = 4

9. p
iw 0 1 2 3 4 5 6 7 8
0 0 0 0 0 0 0 0 0 0
1 0
2 0
3 0
4 0
Pi
____
2
3
4
1
Wi
_____
3
4
5
6

10. iw 0 1 2 3 4 5 6 7 8
Pi Wi 0 0 0 0 0 0 0 0 0 0
2 3 1 0 0 0 2
3 4 2 0
4 5 3 0
1 6 4 0
The bag of capacity
W = 4
Value of wi = 3
i.e., Solution is 2

11. iW 0 1 2 3 4 5 6 7 8
Pi Wi 0 0 0 0 0 0 0 0 0 0
2 3 1 0 0 0 2 2 2 2 2 2
3 4 2 0 0 0 2 3 3 3 5 5
4 5 3 0
1 6 4 0
= Max( 3+0,2)
= Max(3,2)
= 3
= Max (3+2,2)
=Max (5,2)
= 5

12. iw 0 1 2 3 4 5 6 7 8
Pi Wi 0 0 0 0 0 0 0 0 0 0
2 3 1 0 0 0 2 2 2 2 2 2
3 4 2 0 0 0 2 3 3 3 5 5
4 5 3 0 0 0 2 3 4 4 5 6
1 6 4 0
= Max(4+0,3) = Max(4+0,5)
= Max(4+0,3) = Max(4+2,5)

13. iw 0 1 2 3 4 5 6 7 8
Pi Wi 0 0 0 0 0 0 0 0 0 0
2 3 1 0 0 0 2 2 2 2 2 2
3 4 2 0 0 0 2 3 3 3 5 5
4 5 3 0 0 0 2 3 4 4 5 6
1 6 4 0 0 0 2 3 4 4 5 6
= Max(1+0,4)
= Max(1+0,5)
= Max(1+0,6)

14. The Maximum profit is 6.
The formula for filling a
particular value of the cell:
“m[i,w] = max( m[i-1,w], m[i-
1,w-w[i] ] + p[i] ) “
Eg:
m(4,7) = max( m[3,7] , m[3,
7-6]+ 1)
m(4,7) = max(5,0+1)
m(4,7) = max(5,1)
=> 5