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I was reading some of the notes (made easy and ACE)  and noticed that some teacher have taught that time complexity for Binary Knapsack O(2^(n/2)). which can not be reduced further. but What I have seen here that in worst case by using memoization we can reduce the time complexity upto O(n.W) where n is number of Object and W is capacity of knapsack.

What I am thinking that, Is it really Binary knapsack is NPC, Like Travelling Salesman problem because TSP tales O(n^2*2^n) times to calculate the tour by using dynamic method, which is efficient than o(n^n) of TSP brutforce techniques? Put some light, Am I missing something. Should I stop seeing those notes which people have posted over the internet. 

asked in Algorithms by Boss (35.5k points) | 1.1k views

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See AFAIK in DP we consider only the distinct function calls .So Using DP the complexity of Binary knapsack is O(nW).

But for the larger value of W the the complexity of Binary knapsack is O(2n). Because in 0/1 knapsack has very less repetition

So for the larger value W ,0/1 knapsack will generate n level complete binary tree So it will have  2n function calls and the time complexity

becomes O(2n). So it one of NP complete problem.

answered by Boss (38.8k points)
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@ManojK Can you explain more, What do you mean by larger value? and Why that O(nW) did not work in the larger value?
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 for n=5, W=100000  Here w has larger value.

Here n is polynomial in length but length of  w not polynomial it  can anything.

When comparing this with the  brute force algo take O(2n), but depending on W, either the dynamic programming algorithm is more efficient or the brute force algorithm could be more efficient. (For example, for n=5, W=100000, brute force is preferable, but for n=30 and W=1000, the dynamic programming solution is preferable.)

+1

@manojK sir, abhi samjha cheekylaugh

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