I have developed a theorem that proposes a method to build algorithms. All the algorithms produced by this method are in P ... they never go up to more than $6(n^{12})$ operations.

Following that, I have defined an algorithm to resolve the TSP-metric. I agree that $6(n^{12})$ is a lot, but it is $P$ and it is the limit of n trending to infinite. For $n < 45$ the performance is $5(n^9)$.

First question: Assuming the theorem is correct, Is this enough to proof $P = NP$ ?

Before publishing the results and the method in a paper, I want to show on-line that the algorithm is in $P$. I can produce an exact result of 45 cities in less than a half of a day since the Held-Karp algorithm will need more than 16 days.

I will certify with a cloud computing provider the host hardware used, but source code will be black-boxed. Inputs will be provided by anyone and she/he/they may be the only one(s) that know(s) the solution.

Second question: Is this "show" good enough/idea to prove that I actually have solid evidence of it is not a fake?

Important to note that its purpose is only to claim attention showing evidence not substituting the proof of the theorem, I´ll make it publish (public-free) as soon as it will be demanded, I won't keep it in secret. The reason to do this "show" is that being no one if I publish the "theorem" no one gives it a chance. I have checked that way.

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    $\begingroup$ Can you prove that your algorithm runs in polynomial time and always produces an optimal solution? $\endgroup$ Feb 23, 2020 at 1:49
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    $\begingroup$ By the deterministic time hierarchy theorem there are problems in $P(f(n)^2)$ not in $P(f(n))$. Note that $P(n^{24}) \subset NP$, thus there are problems in NP not solvable in $O(n^12)$. Thus there can exist no method of producing algorithms with complexity $O(n^{12})$ that works for all problems in NP (though this doesn't exclude the possibility of a $O(n^{12})$ algorithm to just TSP). $\endgroup$ Feb 23, 2020 at 2:30
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    $\begingroup$ It's not clear what you mean by "resolve the TSP-metric". What do you mean by "show on-line"? $\endgroup$
    – D.W.
    Feb 23, 2020 at 3:27
  • $\begingroup$ @AnttiRöyskö you are right the method (of the theorem) may need for some problems up to $n^{24}$ operations to produce an algorithm, and "n" is a number depending of the kernel of the problem. But when it "optimize" a boolean function it produces an algorithm that throws an exact solution with less than $6(n^{12})$ operations. $\endgroup$
    – Ixer
    Feb 23, 2020 at 12:17
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    $\begingroup$ @D.W. I think with ‚on-line’ he means putting a solver on a homepage, such that anybody could use it. On-line as in ‚putting on the internet‘. (He is not referring to an online-algorithm in the ‚usual‘ sense of the class of algorithms that aren‘t fed with the complete onput from the beginning). $\endgroup$ Feb 23, 2020 at 17:18

1 Answer 1

  1. Metric TSP is NP-complete - hence yes, assuming you can solve the metric TSP in $6(n^{12})$, this is good enough to prove $P=NP$.

  2. Giving a blackbox that can do that, would be a quite strong evidence (if it actually works). But there are some considerations one should take into account: it actually could be somewhat dangerous, as someone else could figure out a reduction of crypto-algos to metric-TSP. It could break ‚secure communication‘, blockchain (bitcoin), etc... All these technologies are based on the assumption computer scientists and mathematicians have had for decades now, namely that that $P \neq NP$

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    $\begingroup$ thx for your feedback. You are right about its implications, this theorem changes a lot of things. However, to resolve a hash of 512 bits it will need $6(512^10)$ (note that each bit is an independent decision) so even with a lot of supercomputers you will need more than a couple of centuries. Using quantum computers with 512 qbits you can break a hash instantaneously, with a 32 qbit one it will be broken in a few hours. $\endgroup$
    – Ixer
    Feb 23, 2020 at 7:41
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    $\begingroup$ @Ixer that is exactly my point: many of the cryptographic tools rely on P != NP. If you manage to solve an NP-Complete problem in poly-time, then this solver can also be used to solve other NP-problems in polytime (like for instance breaking crypto-tools). To do this, you only need to figure out a poly-time reduction of whatever problem you are trying to solve to metric TSP, which might not be trivial but isn‘t that hard either. One basically can ‚translate‘ the crypto-problem into metric TSP, and the än use your blackbox to solve it. See en.m.wikipedia.org/wiki/Turing_reduction $\endgroup$ Feb 23, 2020 at 17:01

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