P = Np Conjecture (反)

The flaw of this paper is exactly what the paper emphasis, "the problem has a special mathematical structure". The proof assumes every pair of subsets has to be compared and used sorting to optimize number of comparisons. However, not every pair of subsets has to be compared. A large set of subsets can be removed when one comparison fails. [One can prune by odd/even, >0 or <0, or even better, not using that comparison method at all]
In particular, one cannot assert the lower bound of the problem (i.e. the least amount of time needed to solve the problem) by unnecessary constrainting the way to compute it.

2008-03-02 11:29:43 補充：
It is known that every pair of subsets must be compared.
For example, take S = {1,2,4,8,16,... 2^n}
Then each subsets will give you a different sum. So looking into all subsets is necessary.

2008-03-02 11:29:54 補充：
That is wrong. Different sums doesn't mean I have to compare.
The subsets that can be pruned does not mean they have the same sum, the subsets that can be pruned are those known to fail.

2008-03-02 11:29:59 補充：
For example, if there is a subset [of all +ve numbers] has a sum larger than b, then all the supersets of that subsets can be pruned. if there is no such subset (i.e. the sum of all numbers are smaller than B), the algorithm can return false.

2008-03-11 20:21:32 補充：
Let me be more precise and point to the paper. "Explanation: As we discussed earlier, in order to search a set of size N in a more efficient way than a brute-force search by a single computer processor, it is necessary for many processors to search the set in parallel, ... ".

2008-03-11 20:21:40 補充：
There is no need to search that set of all subsets, these subsets has some relationship within themselves that can help solve the problem.