

Is P=NP an Ill Posed Problem?  - michael_nielsen
http://rjlipton.wordpress.com/2009/07/03/is-pnp-an-ill-posed-problem/

======
dmg_83
According to the Time Hierarchy theorem (
<http://en.wikipedia.org/wiki/Time_hierarchy_theorem> ), for each number k >
1, there exists a problem in P that can be decided (solved) in O(n^k), and not
decided (solved) in O(n^j) for each j in [1, k).

This was the first time I heard of this theorem (was googling for something I
thought would weaken the author's point), and it really strengthens the
authors argument - I thought his example of O(n^10) algorithms was sort of a
fallacy but it appears not. If I'm properly understanding the Time Hierarchy
Theorem, it is a very strong argument for what the author is suggesting (and
somewhat disappointing for me, because I wish he were wrong).

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tome
If David Hilbert woke up in 1000 years he wouldn't ask "Does P=NP?" but "what
progress have we made on P=NP?". Once the Riemann Hypothesis is done with,
it's done with, but P=NP or P!=NP opens up whole new lines of discussion.

That said, P=NP could be independent of ZF[1], and if that was shown we'd be
scratching our heads asking what P=NP actually means.

[1] <http://www.scottaaronson.com/papers/pnp.pdf>

------
Confusion
Doesn't argue that "P=?NP" is an ill posed problem, but rather that solving it
may not be very important, as the details matter at lot for the implications
the result may have. There are a lot of solutions that would settle the
problem, without having any influence on the practical state of affairs.

Introduces the term 'SAT' at some point, without explaining what it means.
Wikipedia isn't helpful.

~~~
alextp
SAT is the boolean satisfiability problem, a simple NP-complete problem often
used to prove the other problems are NP-complete. An instance of SAT is a
boolean formula, and the answer to it is yes if and only if the formula can be
true for some assignment of variables. 3SAT is the same problem, only with the
formula in a conjunctive normal form and each clause having exactly three
variables.

Also, it seems the poster is really just trying to get attention. Complexity
theorists do know all that and have spent a lot of time trying to see which
postulated consequences of P?=NP are necessary and which are not.

(If I were to argue it is an ill-posed question, I'd go more in the direction
of CO-NP (which is the set of problems for which "NO" answers have short
certificates), and why this really strong emphasys on "yes" certificates for
NP problems)

Doron Zeilberger's take on that (
<http://www.math.rutgers.edu/~zeilberg/Opinion98.html> ) is a lot more
interesting.

And, also, the "easy" equals "polynomial time" is (a) a good heuristic, if you
assume computational power increases exponentially (hence, a problem will be
solvable in the foreseeable future if all it needs is polynomially more
processing power for larger instances); (b) it is an easy to study class,
since it is composable; (c) not the only thing complexity theorists argue
about (ie, P=NP means practical bounds for some problems and collapses a lot
of classes that are not trivially seen to be equal, like AM and MA)

~~~
anatoly
I think your evaluation of the post may have been somewhat rushed.

\- The poster is a notable complexity theorist and is not "just trying to get
attention".

\- Doron Zeilberger's "a lot more interesting" take is completely subsumed in
Richard Lipton's blogpost, in that it mentions other possibilities beyond the
basic "what if it's polynomial with a really huge exponent" scenario.

\- Assuming computational power to increase exponentially indefinitely is
completely ridiculous, and without the "indefinitely" part this assumption no
longer works to justify "easy = P".

\- It is trivial to say that if P=NP, then AM=MA, because both AM and MA are
in the polynomial hierarchy, and if P=NP, the hierarchy collapses completely.

\- The most important point to be taken away from the blogpost is not just
that "easy = P" is an assumption that's been held on too unreflectively and
may not turn out to be true. It's that studying what this assumption means,
how it could break down, and why it hasn't, by and large, broken down so far,
may get us closer to understanding P ?= NP.

~~~
alextp
You have a point. My closeted complexity theorist persona thinks it would be a
lot more interesting to study the special cases of P (not just LOG, but seeing
how O(n) and O(n²) or "NO(n)" problems behave) instead of the most general
cases.

------
jganetsk
I once heard Manuel Blum say that it could be shown that the result can be
neither proven nor disproven given our framework of mathematics, similar to
the Continuum Hypothesis. In this case, I would call P=NP an ill-posed
problem.

I think the OP makes great points in his article, but the title of the blog
post is wrong. He really speaks of how the scope of P=NP is misconceived by
most people.

~~~
jerf
The problem is that if it can be proved that P=NP can't be proved or
disproved, that constitutes a proof that P!=NP, since it proves that you won't
ever come up with an algorithm, which is what the question is really about in
the first place.

This being math, one can imagine ever more finely sliced arguments being made,
but a "simple" proof that P=NP is forever unsolvable is itself a proof ("the
probability that we will find such an algorith is 1 in 10^100"), for all
practical and most theoretical (though not all) purposes.

~~~
frig
I wouldn't be so sure that P=NP not provable <=> de facto P!=NP.

It's at least not _obviously inconceivable_ that one might stumble upon an
algorithm for SAT or traveling salesman whose running time on all known inputs
_appears to be_ polynomial but for which a proof of its asymptotic running
time might prove extremely _elusive_ (in much the same way that so far as we
know the riemann hypothesis _appears_ to hold but proof remains elusive).

~~~
jerf
And that would simply "not be a proof". That doesn't disprove my point at all.
You have to try harder than that to get one of the funky edge cases.

~~~
frig
You're really moving the goalposts here: if it's already proven that P==NP is
undecidable (your supposition) then of course you couldn't prove that the
algorithm, say, solved arbitrary instances of SAT in time polynomial in their
size; such a proof would prove P==NP => contradiction of undecidability of
P==NP.

I was specifically critiquing this assertion:

\- since it (P=?NP shown unprovable) proves that you won't ever come up with
an algorithm

...which isn't exactly right (you're drawing too strong a conclusion from your
supposition): in a "P==NP is undecidable" universe there's nothing
(apparently) stopping there being polynomial-time algorithms for NP-complete
problems, just a barrier preventing _proof_ that a given algorithm's
asymptotic performance puts it in P.

~~~
anatoly
or - another fun possibility - you could have a provably polynomial-time
algorithm that seems to solve SAT, but you can't prove or refute its
correctness.

~~~
frig
Actually: I think you meant what I'm saying in the earlier version of this
comment. But yes: if "P==NP is undecidable" holds then for _any_ provably-
polynomial sat-"solver" the following possibilities hold:

\- you'd be able to find a proof it was an invalid "solver" (eg: the algorithm
that assigns 'YES' to each of the N variables is provably polynomial and
provably incorrect)

\- you _wouldn't_ be able to find a proof it was incorrect (in a strong sense:
this isn't 'because you aren't clever enough' but b/c you've stumbled upon one
of the 'true' polynomial-time algorithms, ergo there can't be a valid proof
it's an invalid solver)

I'd guess you couldn't prove that a given polynomial-time algorithm couldn't
be shown to be "provably known to be incapable of invalidation"; such a proof
=> doesn't exist a counterexample (in this case a SAT instance it doesn't get
the right answer for), b/c such a counterexample _is_ a refutation => proof
the algorithm is correct => contradiction of "P==NP undecidable", but I won't
remove the 'I'd guess' from that unless I had a better understanding of the
field, as this kind of reasoning is the kind of thing where the technicalities
_are_ the important things, and I'm not up to speed on them.

Thus in the "P==NP undecidable" universe if you found a _creepy_ algorithm --
in your example provably-polynomial but not known to be correct or incorrect,
but "empirically" correct over a very extended period of time -- you'd know
not to waste time trying to prove it correct (b/c such a proof would be
obviously impossible) but you'd see every attempt at proving it invalid also
fail, like you said.

Edited for clarity. Edited again for correction.

