

After almost 20 years, math problem falls - bchjam
http://web.mit.edu/newsoffice/2011/convexity-0715.html

======
AjayTripathy
From the abstract: "We show that unless P=NP, there exists no polynomial time
(or even pseudo-polynomial time) algorithm that can decide whether a
multivariate polynomial of degree four (or higher even degree) is globally
convex."

Can someone help me understand this? Isn't stating that there exists no
polynomial time algorithm asserting that P!=NP?

~~~
jeffcoat
We say an algorithm is in P if the running time grows as a polynomial function
(e.g., n^2, but not 2^n) of the input size (n, in my example; the number of
variables in the polynomial in this paper).

We say that an algorithm is in NP if checking a given solution can be done in
polynomial time. For example, if the problem is to find a path that visits all
the nodes of a graph (or cities on a salesman's route, or whatever) exactly
once that's less than some distance k, you might have to try every possible
route to find a solution, but given an answer, you can check it's distance and
compare it to k very quickly.

NP-hard means (even more informally than the above) that the problem is "at
least as hard as the hardest problem in NP". (Some problems in NP are easy;
e.g., if you have an algorithm in P to find a solution to a problem, then you
can obviously also check a given answer in polynomial time.)

It is not known whether or not all problems for which answers can be checked
in polynomial time (NP) can also be solved in polynomial time (P).

They're saying that if P=NP, there does exist an algorithm in P that can
decide the convexity of a degree 4 polynomial (even though they don't know
what that algorithm is right now), and if P!=NP, then there doesn't.

(*Edit: The last paragraph is wrong:

I can't find that they've proven that the determining convexity is in NP, just
that it's NP-hard and they didn't rule out it being in NP. (Recall that some
NP-hard problems are not in NP; halting problem is one.)

So it's possible that that both P=NP and there's still no algorithm in P that
can check convexity. But if P!=NP, that algorithm definitely doesn't exist.

Also, on closer reading, the relevant "size" of a problem that this paper is
talking about (n) is the number of bits needed to represent the coefficients
of the polynomial.)

~~~
AjayTripathy
"(Recall that some NP-hard problems are not in NP; halting problem is one.)"

Actually, this quite clears it up for me, I was confusing being in NP with
being NP hard.

