
Ask HN: If you prove that P=NP, dare you announce it? - waitingkuo
If you solve the P=NP? problem that find that P=NP. Dare you announce it publicly?
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jepler
Purported proofs of P=NP are published multiple times a year. They just always
turn out to have a flaw, just like the P≠NP proofs.
[http://www.win.tue.nl/~gwoegi/P-versus-
NP.htm](http://www.win.tue.nl/~gwoegi/P-versus-NP.htm) collects them, though
it hasn't been updated since 2 May 2015, probably when the refutation of the
most recent purported P=NP proof was added.

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33a
Not unless you were very certainly sure it was true. The probability that you
got it right and didn't mess up some detail is so vanishingly small that it
might as well be 0.

If you go around shouting about yet another half-baked "solution", people will
think you're a crank (and they might even be right). Exercise some restraint,
check it and wait.

One thing about a positive P=NP proof is that it would necessarily be
constructive, so you could try to actually implement it and maybe do some SAT
solving competition. If you start stomping the competition, you would get
noticed and be in a much stronger position to announce something like this.

~~~
Someone
_" One thing about a positive P=NP proof is that it would necessarily be
constructive"_

I don't see why that would necessarily be true. For example, let's say that I
show that, if P != NP, the set of all NP-complete problems can be split into N
equivalence classes and then produce N+1 NP-complete problems that are
pairwise in different equivalence classes.

Alternatively (and, I'm guessing, slightly more realistic), one might define a
Turing-equivalent machine and show that, if there is any room between P and
NP, for that machine, any program that solves a NP-complete problem can be
reduced to run for one less time step.

Can you explain?

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balazsdavid987
A proof that P=NP would go against our everyday experience and it seems so
unlikely that it would take you to the Gödelian level of fame and prestige.
Would guys in black suits trying to kill you? No. Would you get 100s of job
offers? Yes. Go for it, it would be a huge advancement for humanity.

Personally, I have a strong feeling that no one will ever prove that P=NP.
There's that story that out of 100 math professors 10 or so say that P equals
NP, but many of them admitted that they just wanted to be controversial. My
suspicion is that the existence of P and NP as different complexity groups is
a direct consequence of the way Boolean algebra is built up and the way
operations are defined, but I far from being an expert on this.

------
PeterWhittaker
If I thought I had such a proof (and I expect the proof would be complex,
relying on many leading-edge pieces of wildly different branches of
mathematics, and therefore comprehensible in the short to medium term to but a
few), I would first see if I could develop a practical implementation of the
proof for a well-known NP problem, test that out, and see if it worked.

If so, I would have that reviewed by trusted colleagues. ("Hey, buddy, I have
a P space solution to travelling salesman!" "Get outta town!", "No
seriously..."). That would at least demonstrate that the proof works.

Next step? Well, there's a bit of the responsible disclosure argument at play:
If P=NP and you have a practical implementation of a P-time algorithm for an
NP-complete problem, translating that to another will be less work, I should
think, than the original proof or the original implementation...

...meaning much crypto would break soon after publication. Give the proof and
the implementation to 100 well-known and trustworthy mathematicians from
around the world, have them agree to your disclosure strategy, then announce
what you've got, with their backing, and tell the world that you won't publish
for 6 months. Or 12. Whatever.

The Fields Medal will wait.

That will give the world time to adjust to its new reality.

------
rumcajz
Knuth suspects that P=NP. However, theoretical feasibility doesn't necessarily
mean that you'll know how to solve NP problems in P time, he says.

~~~
brudgers
The citation:
[http://www.informit.com/articles/article.aspx?p=2213858](http://www.informit.com/articles/article.aspx?p=2213858)

 _I 've come to believe that P = NP, namely that there does exist an integer M
and an algorithm that will solve every n-bit problem belonging to the class NP
in nM elementary steps._

 _Some of my reasoning is admittedly naïve: It 's hard to believe that P ≠ N P
and that so many brilliant people have failed to discover why. On the other
hand if you imagine a number M that's finite but incredibly large—like say the
number_ [really big number] _discussed in my paper on "coping with
finiteness"—then there's a humongous number of possible algorithms that do nM
bitwise or addition or shift operations on n given bits, and it's really hard
to believe that all of those algorithms fail._

 _My main point, however, is that I don 't believe that the equality P = NP
will turn out to be helpful even if it is proved, because such a proof will
almost surely be nonconstructive. Although I think M probably exists, I also
think human beings will never know such a value. I even suspect that nobody
will even know an upper bound on M._

 _Mathematics is full of examples where something is proved to exist, yet the
proof tells us nothing about how to find it. Knowledge of the mere existence
of an algorithm is completely different from the knowledge of an actual
algorithm._

 _For example, RSA cryptography relies on the fact that one party knows the
factors of a number, but the other party knows only that factors exist.
Another example is that the game of N × N Hex has a winning strategy for the
first player, for all N. John Nash found a beautiful and extremely simple
proof of this theorem in 1952. But Wikipedia tells me that such a strategy is
still unknown when N = 9, despite many attempts. I can 't believe anyone will
ever know it when N is 100._

 _More to the point, Robertson and Seymour have proved a famous theorem in
graph theory: Any class of graphs that is closed under taking minors has a
finite number of minor-minimal graphs. (A minor of a graph is any graph
obtainable by deleting vertices, deleting edges, or shrinking edges to a
point. A minor-minimal graph H for is a graph whose smaller minors all belong
to although H itself doesn 't.) Therefore there exists a polynomial-time
algorithm to decide whether or not a given graph belongs to : The algorithm
checks that G doesn't contain any of 's minor-minimal graphs as a minor._

 _But we don 't know what that algorithm is, except for a few special classes
, because the set of minor-minimal graphs is often unknown. The algorithm
exists, but it's not known to be discoverable in finite time._

 _This consequence of Robertson and Seymour 's theorem definitely surprised
me, when I learned about it while reading a paper by Lovász. And it tipped the
balance, in my mind, toward the hypothesis that P = NP._

 _The moral is that people should distinguish between known (or knowable)
polynomial-time algorithms and arbitrary polynomial-time algorithms. People
might never be able to implement a polynomial-time-worst-case algorithm for
satisfiability, even though P happens to equal NP._

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ddingus
Yes.

Either you have a solution, and that's a great thing, or you are close to a
solution and one will be found more quickly and that too is a great thing, or
you don't have a solution. That's not such a great thing, but it's a
contribution to the set of cases not known to be solutions, potentially
hinting at where the solution may lie.

This assumes you have done the thought work and want to know. Don't you?

~~~
waitingkuo
It's definitely a huge contribution. But P=NP potentially means that RSA is
solvable in Polynomial time. Then countless servers will be attackable. Not
sure whether it's good to announce "publicly".

~~~
srean
Polynomial time does not necessarily make it easy, the degree of the
polynomial could be plenty high making large problems still sufficiently
expensive.

~~~
Someone
It need not even be of high degree. With a large enough constant, even a
constant-time algorithm could be way out of reach.

Suppose that somebody shows that, once you are past a googol^googol (not a big
number, as numbers in mathematics go), factoring doesn't get harder at all,
that would be merely a curiosity in practice (It also would be a hugely
surprising result that would inspire mathematicians to start looking for ways
to bring that limit down)

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mcnamaratw
If just knowing P=NP were enough, for applications people could go ahead and
just make the assumption now. The problem is finding an algorithm.

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seiji
No. [http://www.antipope.org/charlie/blog-
static/fiction/toast/to...](http://www.antipope.org/charlie/blog-
static/fiction/toast/toast.html#antibodies)

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pvaldes
if N=1, yes

Can someone explain better the problem for people like me? What is P and what
is N here?

~~~
dagw
[https://en.wikipedia.org/wiki/P_versus_NP_problem](https://en.wikipedia.org/wiki/P_versus_NP_problem)

