How to understand the knapsack problem is NP
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.)
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").
Further References
In knapsack 0/1 problem, we need 2 inputs(1 array & 1 integer) to solve this problem:
Let's assume n=10 and W=8:
so the time complexity T(n) = O(nW) = O(10*8) = O(80)
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)
but as you double the size of W , it does not mean W=20, but the length is twice long:
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.
It all depends on which parameters you put inside O(...)
.
If target weight is limited by number W
, then problem has O(n*W)
complexity, as you mentioned.
But if weights are way too big and you need algorithm with complexity independent of W
, then problem is NP-complete. ( O(2^n*n)
in most naive implementation).
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