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posted by takyon on Tuesday August 15 2017, @04:14AM   Printer-friendly
from the math-is-a-science dept.

Researchers claim A Solution of the P versus NP Problem ( https://arxiv.org/abs/1708.03486) :

Berg and Ulfberg and Amano and Maruoka have used CNF-DNF-approximators to prove exponential lower bounds for the monotone network complexity of the clique function and of Andreev's function. We show that these approximators can be used to prove the same lower bound for their non-monotone network complexity. This implies P not equal NP.

The full journal article is available as a pdf on arXiv ( https://arxiv.org/pdf/1708.03486 ).

I recall studying computational complexity in college, so am attuned to the concepts of P versus NP being an outstanding issue, but cannot talk to what these researchers have found. I'm hoping a fellow Soylentil could shed some light on this finding and its implications. I notice the claim of "a" solution rather than "the" solution. I suspect this may be significant, but am unsure.

Here is some background information from Wikipedia's coverage of the P versus NP problem:

The P versus NP problem is a major unsolved problem in computer science. Informally speaking, it asks whether every problem whose solution can be quickly verified by a computer can also be quickly solved by a computer.

The underlying issues were first discussed in the 1950s, in letters from John Forbes Nash Jr. to the National Security Agency, and from Kurt Gödel to John von Neumann. The precise statement of the P versus NP problem was introduced in 1971 by Stephen Cook in his seminal paper "The complexity of theorem proving procedures"[2] and is considered by many to be the most important open problem in the field.[3] It is one of the seven Millennium Prize Problems selected by the Clay Mathematics Institute to carry a US$1,000,000 prize for the first correct solution.

The informal term quickly, used above, means the existence of an algorithm solving the task that runs in polynomial time, such that the time to complete the task varies as a polynomial function on the size of the input to the algorithm (as opposed to, say, exponential time). The general class of questions for which some algorithm can provide an answer in polynomial time is called "class P" or just "P". For some questions, there is no known way to find an answer quickly, but if one is provided with information showing what the answer is, it is possible to verify the answer quickly. The class of questions for which an answer can be verified in polynomial time is called NP, which stands for "nondeterministic polynomial time".[Note 1]

Consider the subset sum problem, an example of a problem that is easy to verify, but whose answer may be difficult to compute. Given a set of integers, does some nonempty subset of them sum to 0? For instance, does a subset of the set {−2, −3, 15, 14, 7, −10} add up to 0? The answer "yes, because the subset {−2, −3, −10, 15} adds up to zero" can be quickly verified with three additions. There is no known algorithm to find such a subset in polynomial time (there is one, however, in exponential time, which consists of 2n-n-1 tries), but such an algorithm exists if P = NP; hence this problem is in NP (quickly checkable) but not necessarily in P (quickly solvable).

An answer to the P = NP question would determine whether problems that can be verified in polynomial time, like the subset-sum problem, can also be solved in polynomial time. If it turned out that P ≠ NP, it would mean that there are problems in NP (such as NP-complete problems) that are harder to compute than to verify: they could not be solved in polynomial time, but the answer could be verified in polynomial time.

Aside from being an important problem in computational theory, a proof either way would have profound implications for mathematics, cryptography, algorithm research, artificial intelligence, game theory, multimedia processing, philosophy, economics and many other fields.


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  • (Score: 3, Informative) by sea on Tuesday August 15 2017, @03:51PM (1 child)

    by sea (86) Subscriber Badge on Tuesday August 15 2017, @03:51PM (#554290) Homepage Journal

    It's only a polynomial if the number of terms is finite.

    Also, representing f(x) as an infinite series still doesn't put it in class P:

    Let f(x) = a + bx + cxx + dxxx + exxxx + ...

    Find the big-O complexity of f using a deterministic machine:

    Well, that's just O(a + bx + cxx + dxxx + exxxx + ...) = O(a) + O(bx) + O(cxxx) + ...

    Now suppose f(x) = O(x^m). Contradiction immediately, because what about all those terms x^(m+k) for k>0? Those are in the series, and thus, f(x) is bounded below by those, etc. and so it /can't/ be O(x^m). It must at least be O(x^(m+k)).

    but k was unspecified, only that k > 0. Thus, there's no finite exponent that works, and f(x) is not in P (by this infinite series method. It is still possible for a closed-form finite expression to reproduce f in polynomial time)

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  • (Score: 0) by Anonymous Coward on Tuesday August 15 2017, @04:08PM

    by Anonymous Coward on Tuesday August 15 2017, @04:08PM (#554293)

    Yup, this is accurate, infinite series don't work for efficient computation.

    Basically, you need to do the math to get the infinite series re-written as a finite function (sometimes this is even possible!) THEN you can evaluate its complexity.