The document describes the 0-1 knapsack problem and provides an example of solving it using dynamic programming. Specifically, it defines the 0-1 knapsack problem, provides the formula for solving it dynamically using a 2D array C, walks through populating the C array and backtracking to find the optimal solution for a sample problem instance, and analyzes the time complexity of the dynamic programming algorithm.

1. KNAPSACK PROBLEM
USING DYNAMIC
PROGRAMMING
BY: KHUSHBOO JETHWA
ENROLLMENT NO. : 140950107028
DEPARTMENT : CSE-A
BATCH: A1

2. 0-1 KNAPSACK PROBLEM
• Given weights and value of n items, put these items in a knapsack of
capacity W to get the maximum total value in the knapsack.
• In other words,
•
• Given: value[0….n-1],weight[0….n-1] and W.
• Find: maximum value subset of value[0….n-1] such that sum of weights of
this subset is smaller than or equal to W.
• 0-1 Property: You cannot break an item, either pick it or don’t.

3. Basic example of 0-1 Knapsack Problem

4. Simple Solution

5. Formula
• Let i be the highest-numbered item in an optimal
solution S for W pounds. Then S` = S - {i} is an optimal solution for W -
wi pounds and the value to the solution S is Vi plus the value of the
subproblem.
• We can express this fact in the following formula: define c[i, w] to be
the solution for items 1,2, . . . , i and maximum weight w. Then
0 if i = 0 or w = 0
c[i,w] = c[i-1, w] if wi ≥ 0
max [vi + c[i-1, w-wi], c[i-1, w]] if i>0 and w ≥ wi

6. Explanation
• This says that the value of the solution to i items either include ith
item, in
which case it is vi plus a subproblem solution for (i - 1) items and the weight
excluding wi, or does not include ith
item, in which case it is a subproblem's
solution for (i - 1) items and the same weight.
• The algorithm takes as input the maximum weight W, the number of items
n, and the two sequences v = <v1, v2, . . . , vn> and w = <w1, w2, . . . , wn>. It
stores the c[i, j]values in the table, that is, a two dimensional
array, c[0 . . n, 0 . . w] whose entries are computed in a row-major order.
That is, the first row of c is filled in from left to right, then the second row,
and so on. At the end of the computation, c[n, w] contains the maximum
value that can be picked into the knapsack.
•

7. Dynamic 0-1 knapsack(v,w,n,W)
FOR w = 0 TO W
DO c[0, w] = 0
FOR i=1 to n
DO c[i, 0] = 0
FOR w=1 TO W
DO IFf wi ≤ w
THEN IF vi + c[i-1, w-wi]
THEN c[i, w] = vi + c[i-1, w-wi]
ELSE c[i, w] = c[i-1, w]
ELSE
c[i, w] = c[i-1, w]
•The set of items to take can be deduced from the table, starting at c[n. w] and
tracing backwards where the optimal values came from. If c[i, w] = c[i-1, w] item i is
not part of the solution, and we are continue tracing with c[i-1, w]. Otherwise
item i is part of the solution, and we continue tracing with c[i-1, w-W].
IF c[i,k] != c[i-1,k]
{
then mark the ith article as 1
k=k-wi
i=i-1
}
Else
{
mark the ith article as 0
i=i-1
}

8. Analysis
• This dynamic-0-1-kanpsack algorithm takes θ(nw) times, broken up as
follows: θ(nw) times to fill the c-table, which has (n +1).(w +1) entries,
each requiring θ(1) time to compute. O(n) time to trace the solution,
because the tracing process starts in row n of the table and moves up
1 row at each step.

9. Example
• Find optimal solution for 0-1 Knapsack problem where n=3,
(w1,w2,w3)={2,3,3}, (p1,p2,p3)-{1,2,4}, m=6
Article Value Weight
1 1 2
2 2 3
3 4 3

10. • c[1,1] w=1 wi=2 (wi>w)
c[0,1] = 0
• c[1,2] w=2 wi=2 (w=wi)
max{1+ c[0,0], c[0,2]}
max{1,0}= 1
• c[1,3] w=3 wi=2 (w>wi)
max{1+ c[0,1], c[0,3]}
max{1,0}= 1
• c[1,4] w=4 wi=2 (w>wi)
max{1+ c[0,2], c[0,4]}
max{1,0}= 1
• c[1,5] w=5 wi=2 (w>wi)
max{1+ c[0,3], c[0,5]}
max{1,0}= 1
• c[1,6] w=6 wi=2 (w>wi)
max{1+ c[0,4], c[0,6]}
max{1,0}= 1

11. • c[2,1] w=1 wi=3 (wi>w)
c[1,1] = 0
• c[2,2] w=2 wi=3 (wi>w)
c[1,2] = 1
• c[2,3] w=3 wi=3 (w=wi)
max{2+ c[1,0], c[1,3]}
max{2,2}= 2
• c[2,4] w=4 wi=3 (w>wi)
max{2+ c[1,1], c[1,4]}
max{2,1}= 2
• c[2,5] w=5 wi=3 (w>wi)
max{2+ c[1,2], c[1,5]}
max{3,1}= 3
• c[2,6] w=6 wi=3 (w>wi)
max{2+ c[1,3], c[1,6]}
max{3,1}= 3

12. • c[3,1] w=1 wi=3 (wi>w)
c[2,1] = 0
• c[3,2] w=2 wi=3 (wi>w)
c[2,2] = 1
• c[3,3] w=3 wi=3 (w=wi)
max{4+ c[2,0], c[2,3]}
max{4,2}= 4
• c[3,4] w=4 wi=3 (w>wi)
max{4+ c[2,1], c[2,4]}
max{4,2}= 4
• c[3,5] w=5 wi=3 (w>wi)
max{4+ c[2,2], c[2,5]}
max{5,3}= 5
• c[3,6] w=6 wi=3 (w>wi)
max{4+ c[2,3], c[2,6]}
max{6,3}= 6

13. 0 1 2 3 4 5 6
0 0 0 0 0 0 0 0
1 0 0 1 1 1 1 1
2 0 0 1 2 2 3 3
3 0 0 1 4 4 5 6
W(weight)
i(article)

14. • Maximum value = c[3,6]=6
so compare it with c[2,6]=3
both are not equal so include article 3
FOR i=3
k=6-3=3
i=3-1=2
c[2,3]=2
Compare it with c[1,3]=1
Both are not equal so include article 2
FOR i=2
k=3-3=0
i=2-1=1
c[1,0]=0
So we can’t include article 1
ANSWER: A1 A2 A3
0 1 1

15. THANK YOU