How to understand the knapsack problem is NP-complete?

Question

We know that the knapsack problem can be solved in O(nW) complexity by dynamic programming. But we say this is a NP-complete problem. I feel it is hard to understand here.

(n is the number of items. W is the maximum volume.)

Answer 1

O(n*W) looks like a polynomial time, but it is not, it is pseudo-polynomial.

Time complexity measures the time that an algorithm takes as a function of the length in bits of its input. The dynamic programming solution is indeed linear in the value of W, but exponential in the length of W — and that's what matters!

More precisely, the time complexity of the dynamic solution for the knapsack problem is basically given by a nested loop:

// here goes other stuff we don't care about for (i = 1 to n)     for (j = 0 to W)         // here goes other stuff 

Thus, the time complexity is clearly O(n*W).

What does it mean to increase linearly the size of the input of the algorithm? It means using progressively longer item arrays (so n, n+1, n+2, ...) and progressively longer W (so, if W is x bits long, after one step we use x+1 bits, then x+2 bits, ...). But the value of W grows exponentially with x, thus the algorithm is not really polynomial, it's exponential (but it looks like it is polynomial, hence the name: "pseudo-polynomial").

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Further References

• http://www.cs.ship.edu/~tbriggs/dynamic/index.html
• http://websrv.cs.umt.edu/classes/cs531/index.php/Complexity_of_dynamic_programming_algorithm_for_the_0-1_knapsack_problem_3/27

Answer 2

In knapsack 0/1 problem, we need 2 inputs(1 array & 1 integer) to solve this problem:

1. a array of n items: [n1, n2, n3, ... ], each item with its value index and weight index.
2. integer W as maximum acceptable weight

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Let's assume n=10 and W=8:

1. n = [n1, n2, n3, ... , n10]
2. W = 1000 in binary term (4-bit long)

so the time complexity T(n) = O(nW) = O(10*8) = O(80)

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If you double the size of n:

n = [n1, n2, n3, ... , n10] -> n = [n1, n2, n3, ... , n20]

so time complexity T(n) = O(nW) = O(20*8) = O(160)

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but as you double the size of W, it does not mean W=16, but the length will be twice longer:

W = 1000 -> W = 10000000 in binary term (8-bit long)

so T(n) = O(nW) = O(10*128) = O(1280)

the time needed increases in exponential term, so it's a NPC problem.