On the wikipedia page for the knapsack problem it says that the runtime is $\mathcal{O} (nW)$ and goes on to say that this doesn't violate its classification as NP because the input size is related to $\log W$, where, I believe, $W$ is the size of the knapsack. Why is the size of the input related logarithmically to $W$?
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Polynomial time means that the running time is bounded by a polynomial in the length of the input. The running time here is bounded by $nW$. $n$, the number of items, is surely less than the length of the input, so that part is fine. But $W$, the target weight, is a number that appears in the input, in binary. In $\ell$ bits, you can write a number up to $2^\ell$ so $W$ is potentially exponential in the length of the input, not polynomial.
David Richerby
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Because you normally would not encode the sizes unary. Note that the number "100" would need 100 bit to encode unary (as opposed to the normal 7 bit), so the size of your Knapsack problem would be gigantic, and relative to the then-gigantic size the runtime would not be that bad.
john_leo
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Knapsack is a weakly-NP Complete problem because when one encodes it in unary, it runs in polynomial time. However, this does not imply that P=NP it just means that you have artificially increased the input length and so it gives the illusion of running in polynomial time. Also, keep in mind that on small inputs the dynamic programming algorithm appears to be polynomial.
AKASH VETRIVEL
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Think about it this way. Say you want to run the algorithm on an instance with $W=1,000,000$. When the program will ask you: ``Please input the value of $W$'', are you going to type 7 keys on your keyboard, or 1 million keys?
Vincenzo
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