1. MCS 312: NP Completeness and Approximation algorthms
Instructor
Neelima Gupta

2. Table of Contents
FPTAS for the Knapsack Problem
Reference: Vazirani.

3. Definition
Given a set {a1, a an} of n objects with profit p(ai) and size s(ai). And, a knapsack with capacity B. Find a collection of objects fitting into the knapsack that maximizes the profit.

4. Pseudo Polynomial Algorithms

5. Pseudo Polynomial for KS
Let P = max { p(ai) }
Let S(i, p) denotes a subset of {a1, a ai} with profit exactly equal to p that minimizes the size.
Let A(i,p) denotes the size of S(i, p). A(i,p) = infinity if no such set exists. A(1,p) is easy to compute (in fact known directly) for every p in {1 … nP}
A(i+1,p) = min{ A(i, p), size (ai+1, p – p(ai+1) } if p(ai) < p
= A(i, p) otherwise

6. Example n = 4 B = 5 Elements (size, profit):
(2,3), (3,4), (4,5), (5,6)
Thanks to : Neha Katyal (17, MCS '11), Shikha Walia (27, MCS'11)

7. (2,3), (3,4), (4,5), (5,6) W i 1 2 3 4
Thanks to : Neha Katyal (17, MCS '11), Shikha Walia (27, MCS'11)

8. As w<w1 ; m[1,1] = m[1-1,1] = m[0,1]
(2,3), (3,4), (4,5), (5,6)
As w<w1 ; m[1,1] = m[1-1,1] = m[0,1] W i 1 2 3 4
inf 2
Thanks to : Neha Katyal (17, MCS '11), Shikha Walia (27, MCS'11)

9. As w<w2 ; m[2,1] = m[2-1,1] = m[1,1]
(2,3), (3,4), (4,5), (5,6)
As w<w2 ; m[2,1] = m[2-1,1] = m[1,1] W i 1 2 3 4
Thanks to : Neha Katyal (17, MCS '11), Shikha Walia (27, MCS'11)

10. As w<w3 ; m[3,1] = m[3-1,1] = m[2,1]
(2,3), (3,4), (4,5), (5,6)
As w<w3 ; m[3,1] = m[3-1,1] = m[2,1] W i 1 2 3 4
Thanks to : Neha Katyal (17, MCS '11), Shikha Walia (27, MCS'11)

11. As w<w4 ; m[4,1] = m[4-1,1] = m[3,1]
(2,3), (3,4), (4,5), (5,6)
As w<w4 ; m[4,1] = m[4-1,1] = m[3,1] W i 1 2 3 4
Thanks to : Neha Katyal (17, MCS '11), Shikha Walia (27, MCS'11)

12. As w>=w1 ; m[1,2] = max{m[1-1,2] , m[1-1,2-2]+3} =max{ 0,0+3}
(2,3), (3,4), (4,5), (5,6)
As w>=w1 ; m[1,2] = max{m[1-1,2] , m[1-1,2-2]+3}
=max{ 0,0+3} W i 1 2 3 4 3
Thanks to : Neha Katyal (17, MCS '11), Shikha Walia (27, MCS'11)

13. As w<w2 ; m[2,2] = m[2-1,2] = m[1,2]
(2,3), (3,4), (4,5), (5,6)
As w<w2 ; m[2,2] = m[2-1,2] = m[1,2] W i 1 2 3 4 3
Thanks to : Neha Katyal (17, MCS '11), Shikha Walia (27, MCS'11)

14. As w<w3 ; m[3,2] = m[3-1,2] = m[2,2]
(2,3), (3,4), (4,5), (5,6)
As w<w3 ; m[3,2] = m[3-1,2] = m[2,2] W i 1 2 3 4 3
Thanks to : Neha Katyal (17, MCS '11), Shikha Walia (27, MCS'11)

15. As w<w4 ; m[4,2] = m[4-1,2] = m[3,2]
(2,3), (3,4), (4,5), (5,6)
As w<w4 ; m[4,2] = m[4-1,2] = m[3,2] W i 1 2 3 4 3
Thanks to : Neha Katyal (17, MCS '11), Shikha Walia (27, MCS'11)

16. As w>=w1 ; m[1,3] = max{m[1-1,3] , m[1-1,3-2]+3} =max{ 0,0+3}
(2,3), (3,4), (4,5), (5,6)
As w>=w1 ; m[1,3] = max{m[1-1,3] , m[1-1,3-2]+3}
=max{ 0,0+3} W i 1 2 3 4 3
Thanks to : Neha Katyal (17, MCS '11), Shikha Walia (27, MCS'11)

17. As w>=w2 ; m[2,3] = max{m[2-1,3] , m[2-1,3-3]+4} =max{ 3,0+4}
(2,3), (3,4), (4,5), (5,6)
As w>=w2 ; m[2,3] = max{m[2-1,3] , m[2-1,3-3]+4}
=max{ 3,0+4} W i 1 2 3 4 3 4
Thanks to : Neha Katyal (17, MCS '11), Shikha Walia (27, MCS'11)

18. As w<w3 ; m[3,3] = m[3-1,3] = m[2,3]
(2,3), (3,4), (4,5), (5,6)
As w<w3 ; m[3,3] = m[3-1,3] = m[2,3] W i 1 2 3 4 3 4
Thanks to : Neha Katyal (17, MCS '11), Shikha Walia (27, MCS'11)

19. As w<w4 ; m[4,3] = m[4-1,3] = m[3,3]
(2,3), (3,4), (4,5), (5,6)
As w<w4 ; m[4,3] = m[4-1,3] = m[3,3] W i 1 2 3 4 3 4
Thanks to : Neha Katyal (17, MCS '11), Shikha Walia (27, MCS'11)

20. As w>=w1 ; m[1,4] = max{m[1-1,4] , m[1-1,4-2]+3} =max{ 0,0+3}
(2,3), (3,4), (4,5), (5,6)
As w>=w1 ; m[1,4] = max{m[1-1,4] , m[1-1,4-2]+3}
=max{ 0,0+3} W i 1 2 3 4 3 4
Thanks to : Neha Katyal (17, MCS '11), Shikha Walia (27, MCS'11)

21. As w>=w2 ; m[2,4] = max{m[2-1,4] , m[2-1,4-3]+4} =max{ 3,0+4}
(2,3), (3,4), (4,5), (5,6)
As w>=w2 ; m[2,4] = max{m[2-1,4] , m[2-1,4-3]+4}
=max{ 3,0+4} W i 1 2 3 4 3 4
Thanks to : Neha Katyal (17, MCS '11), Shikha Walia (27, MCS'11)

22. As w>=w3 ; m[3,4] = max{m[3-1,4] , m[3-1,4-4]+5} =max{ 4,0+5}
(2,3), (3,4), (4,5), (5,6)
As w>=w3 ; m[3,4] = max{m[3-1,4] , m[3-1,4-4]+5}
=max{ 4,0+5} W i 1 2 3 4 3 4 5
Thanks to : Neha Katyal (17, MCS '11), Shikha Walia (27, MCS'11)

23. As w<w4 ; m[4,4] = m[4-1,4] = m[3,4]
(2,3), (3,4), (4,5), (5,6)
As w<w4 ; m[4,4] = m[4-1,4] = m[3,4] W i 1 2 3 4 3 4 5
Thanks to : Neha Katyal (17, MCS '11), Shikha Walia (27, MCS'11)

24. As w>=w1 ; m[1,5] = max{m[1-1,5] , m[1-1,5-2]+3} =max{ 0,0+3}
(2,3), (3,4), (4,5), (5,6)
As w>=w1 ; m[1,5] = max{m[1-1,5] , m[1-1,5-2]+3}
=max{ 0,0+3} W i 1 2 3 4 3 4 5
Thanks to : Neha Katyal (17, MCS '11), Shikha Walia (27, MCS'11)

25. As w>=w2 ; m[2,5] = max{m[2-1,5] , m[2-1,5-3]+4} =max{ 3,3+4}
(2,3), (3,4), (4,5), (5,6)
As w>=w2 ; m[2,5] = max{m[2-1,5] , m[2-1,5-3]+4}
=max{ 3,3+4} W i 1 2 3 4 3 4 7 5
Thanks to : Neha Katyal (17, MCS '11), Shikha Walia (27, MCS'11)

26. As w>=w3 ; m[3,5] = max{m[3-1,5] , m[3-1,5-4]+5} =max{ 7,0+5}
(2,3), (3,4), (4,5), (5,6)
As w>=w3 ; m[3,5] = max{m[3-1,5] , m[3-1,5-4]+5}
=max{ 7,0+5} W i 1 2 3 4 3 4 7 5
Thanks to : Neha Katyal (17, MCS '11), Shikha Walia (27, MCS'11)

27. As w>=w4 ; m[4,5] = max{m[4-1,5] , m[4-1,5-5]+6} =max{ 7,0+6}
(2,3), (3,4), (4,5), (5,6)
As w>=w4 ; m[4,5] = max{m[4-1,5] , m[4-1,5-5]+6}
=max{ 7,0+6} W i 1 2 3 4 3 4 7 5
Thanks to : Neha Katyal (17, MCS '11), Shikha Walia (27, MCS'11)

28. (2,3), (3,4), (4,5), (5,6) W i 1 2 3 4 3 4 7 5
Thanks to : Neha Katyal (17, MCS '11), Shikha Walia (27, MCS'11)

29. FPTAS Given \epsilon, let k = \epsilon P /n Let p’(a) = floor (p(a)/k)
Find an optimal solution, say S’, using DP with these profits.
Output S’
Claim: p(S’) >= (1-\epsilon) OPT

30. kp’(a) <= p(a) p’(a) = floor (p(a)/k) >= p(a)/k - 1.
Thus, kp’(a) >= p(a) - k
i.e. p(a) - kp’(a) <= k
i.e. k p’(a) >= p(a) - k
Let O be an optimal solution. The O is a solution for the problem with the new profits as well. Thus,
p’(S’) >= p’(O)
Thus,
p(S) >= k p’(S’) >= kp’(O) >= p(O) – k|O|
>= p(O) – kn
= OPT - \epsilon P
>= (1 - \epsilon) OPT (as OPT <= P)

31. Running Time O(n^2 P’) = O(n^2 floor(P/k)) = O(n^2 floor(n/\epsilon))
Which is polynomial in n and \epsilon.

32. Thank You!
Consolidated by: Sumedha Upadhyaya( 42), Prachi Nagpal(41) (MCS '09)