The document discusses the 0/1 knapsack problem and dynamic programming algorithm to solve it. The 0/1 knapsack problem involves selecting a subset of items to pack in a knapsack that maximizes the total value without exceeding the knapsack's weight capacity. The dynamic programming algorithm solves this by building up a table where each entry represents the maximum value for a given weight. It iterates through items, checking if including each item increases the maximum value for that weight.

2. Dynamic Programming
It is a multistage optimizing decision
making problem closely related to “divide
& conquer” technique.
It solves the sub-problem only once &
stores the result in a table instead of
solving it recursively.

3. 0/1 Knapsack Problem
 Given a set of n items and a knapsack having
capacity w, each item has weight wi and
value vi.
 The problem is to pack the knapsack in such
a way so that maximum total value is
achieved and the total weight should not be
more than the fixed weight.
 The problem is called 0/1 knapsack because
the items are either accepted or rejected.

4. Algorithm
 Loop, for w←0 to W(total capacity)
Set Kp[0,w] ←0.
 Loop, for i ←1 to n(set of items)
Set Kp[i,0] ←0.
 Check whether K is a part of solution or not-
if (wi<=w)(can be part of the solution)
if (vi+kp[i-1,w-wi])>Kp(i-1,w)
Set kp[i,w] ←vi
+kp[i-1,w-wi]
Else
Set Kp[i,w] ←Kp[i-1,w]
 Exit

5. Consider a knapsack with weight capacity
W=3 and total items 3 such that
S=3
wi={1,2,3}
vi={2,3,4}
On applying knapsack algorithm,
Item wi
vi
1 1 2
2 2 3
3 3 4

6. i/w 0 1 2 3
0
1
2
3
for w ←0 to W, for i←1 to n
Set Kp[0,w] ←0, Set Kp[i,0]←0
i/w 0 1 2 3
0 0 0 0 0
1 0
2 0
3 0

7. i=1,vi=2,wi=1,w=1,w-wi=0
if (wi<=w), set kp[i,w] ←vi
+kp[i-1,w-wi]
Kp[1,1]←2+Kp[0,0]=2
i/w 0 1 2 3
0 0 0 0 0
1 0
2 0
3 0
i/w 0 1 2 3
0 0 0 0 0
1 0 2
2 0
3 0

8. i=1,vi=2,wi=1,w=2,w-wi=1
if (wi<=w), set kp[i,w] ←vi
+kp[i-1,w-wi]
Kp[1,1]←2+Kp[1,1]=2
i/w 0 1 2 3
0 0 0 0
↓
0
1 0 2 2
2 0
3 0

9. i=1,vi=2,wi=1,w=3,w-wi=2
(wi<=w), Kp[1,3]←2+Kp[0,2]=2
i=2,vi=3,wi=2,w=1,w-wi=-1
if (wi>w), Set Kp[i,w] ←Kp[i-1,w]
kp[2,1]←Kp[1,1]=2
i/w 0 1 2 3
0 0 0 0 0
1 0 2 2 2
2 0 2
3 0

10. i=2,vi=3,wi=2,w=2,w-wi=0
(wi<=w), Kp[2,2]←3+Kp[1,0]=3
i=2,vi=3,wi=2,w=3,w-wi=1
(wi<=w), Kp[2,3]←3+Kp[1,1]=3+2=5
i/w 0 1 2 3
0 0 0 0 0
1 0 2 2 2
2 0 2 3 5
3 0

11. i=3,vi=4,wi=3,w=1,w-wi=-2
if (wi>w), Set Kp[i,w] ←Kp[i-1,w]=2
i=3,vi=4,wi=3,w=2,w-wi=-1
if (wi>w), Set Kp[i,w] ←Kp[i-1,w]=3
i/w 0 1 2 3
0 0 0 0 0
1 0 2 2 2
2 0 2 3 5
3 0 2 3

12. i=3,vi=4,wi=3,w=3,w-wi=0 (wi<=w),
if (vi+kp[i-1,w-wi])>Kp(i-1,w)=4<5
Kp[3,3]←5
i/w 0 1 2 3
0 0 0 0 0
1 0 2 2 2
2 0 2 3 5
3 0 2 3 5

13. Now we have computed the maximum
possible values that can be taken in a
knapsack i.e Kp[n,W]
In order to select items that can take part
in making this maximum value let
i=n,k=W.

14. Algorithm
 Set i←n
set k←W
 Loop to check K is a part of knapsack
while (i>0 and k>0)
if Kp[i,k]≠Kp[i-1,k] then
mark the ith item in knapsack
set i ←i-1
set k ←k-wi
Else
set i ←i-1
 Exit

15. i=3,k=3,vi=4,wi=3,Kp[i,k]=5,Kp[i-1,k]=5
i←i-1
i/w 0 1 2 3
0 0 0 0 0
1 0 2 2 2
2 0 2 3 5↕
3 0 2 3 5↕

16. i=2,k=3,vi=3,wi=2,Kp[i,k]=5,Kp[i-1,k]=2
Set k←k-wi=1
Set i←i-1
i/w 0 1 2 3
0 0 0 0 0
1 0 2 2 2↕
2 0 2 3 5↕
3 0 2 3 5

17. i=1,k=1,vi=2,wi=1,Kp[i,k]=2,Kp[i-1,k]=0
Set k←k-wi=0
Set i←i-1
i=0
K=0
i/w 0 1 2 3
0 0 0↕ 0 0
1 0 ↖2↕ 2 2↕
2 0 2 3 ↖5↕
3 0 2 3 5

18. The optimal knapsack contains
S= {1,2}
wi= {1,2}={3} <= W
vi= {2,3}= {5}» Optimum Value

19. THANKS!!!!