Can the Subset Sum Problem Be Solved in Polynomial Time?

[RagingHadron]

So I really know very little about the subject but from the little I could gather online...
Consider the subset problem on wikipedia. Does a subset of {−2, −3, 15, 14, 7, −10} equal zero? It shows the work for you and then says that no algorithm to find it in polynomial time is known, only in exponential (with (2^n)-1 tries) It says that an algorithm can only exist in polynomial time if P=NP. So now, can we not set (2^n)-1=n^x so that the algorithm in polynomial time is n^((log((2^n)-1)+2i∏c)/(log(n)) where c∈Z, Z being the set of integers. Does that make any sense?

 

[Mark44]

So I really know very little about the subject but from the little I could gather online...
Consider the subset problem on wikipedia.

http://en.wikipedia.org/wiki/Subset-sum_problem

Does a subset of {−2, −3, 15, 14, 7, −10} equal zero?

As you have written this, it doesn't make sense. Each subset of some other set is itself a set, and a set is not equal to a number. The actual description is "is there a non-empty subset whose sum is zero?"

It shows the work for you and then says that no algorithm to find it in polynomial time is known, only in exponential (with (2^n)-1 tries) It says that an algorithm can only exist in polynomial time if P=NP. So now, can we not set (2^n)-1=n^x so that the algorithm in polynomial time is n^((log((2^n)-1)+2i∏c)/(log(n)) where c∈Z, Z being the set of integers. Does that make any sense?

 

[RagingHadron]

The actual description is "is there a non-empty subset whose sum is zero?"

Yeah that's what I meant. But what was wrong with the rest of it?

 

[Mark44]

Which wikipedia article were you reading? I provided a link to the one I thought you were referring to, but I don't see in that one some of what you're talking about.

 

[RagingHadron]

It was in the p versus np problem page specifically, http://en.m.wikipedia.org/wiki/P_versus_NP_problem here. It's in the third paragraph. But was the work that I did correct/incorrect? I'm sure that there's a flaw in my approach to the problem somewhere seeing as it's so simple...

 

[RagingHadron]

Never mind, I saw what my flaw was.