Comments on: Well, just take n=1...right?
http://www.metafilter.com/103143/Well-just-take-n1right/
Comments on MetaFilter post Well, just take n=1...right?Wed, 04 May 2011 09:23:20 -0800Wed, 04 May 2011 09:23:20 -0800en-ushttp://blogs.law.harvard.edu/tech/rss60Well, just take n=1...right?
http://www.metafilter.com/103143/Well-just-take-n1right
<a href="http://blog.computationalcomplexity.org/2011/05/forty-years-of-p-v-np.html">In the afternoon of May 4, 1971,</a> in the Stouffer's Somerset Inn in Shaker Heights, Ohio, Steve Cook presented his STOC paper proving that Satisfiability is <a href="http://en.wikipedia.org/wiki/NP-complete">NP-complete</a> and Tautology is <a href="http://en.wikipedia.org/wiki/NP-hard">NP-hard.</a> <br /><br />40 years ago today, Cook presented his paper on the problem we know today as the "<a href="http://en.wikipedia.org/wiki/P_versus_NP_problem">P v. NP</a>" problem, or the hypothesis "P = NP".
From the wikipedia link: "Suppose that solutions to a problem can be verified quickly. Then, can the solutions themselves also be computed quickly?" If you can answer this question, you'll get <a href="http://www.claymath.org/millennium/P_vs_NP/">a million dollars from the Clay Institute</a> and a ton of accolades.
Some interesting problems which are not known to be in P or NP-complete are the <a href="http://en.wikipedia.org/wiki/Integer_factorization">integer factorization problem</a>. Even still, if it turns out that P=NP, then we're going to have to retire <a href="http://www.ics.uci.edu/~eppstein/161/960312.html">RSA encryption</a>.
(<a href="http://www.metafilter.com/94553/Complex-matters-for-the-millenium">previously</a>, and the <a href="http://rjlipton.wordpress.com/2010/08/12/fatal-flaws-in-deolalikars-proof/">fatal</a> <a href="http://www.sciencenews.org/index/generic/activity/view/id/63252/title/Crowdsourcing_peer_review">flaws</a> in the proposed <a href="http://rjlipton.wordpress.com/2010/09/15/an-update-on-vinay-deolalikars-proof/">"proof"</a>)post:www.metafilter.com,2011:site.103143Wed, 04 May 2011 09:21:29 -0800King BeemathpnpbirthdayhardproblemsBy: madcaptenor
http://www.metafilter.com/103143/Well-just-take-n1right#3674460
<i>a million dollars from the Clay Institute </i>
I've been saying for a while that the prize should be $1,166,666.67. If Perelman's turning down the million for Poincare, shouldn't the other six winners (should they eventually exist) share his money?comment:www.metafilter.com,2011:site.103143-3674460Wed, 04 May 2011 09:23:20 -0800madcaptenorBy: oonh
http://www.metafilter.com/103143/Well-just-take-n1right#3674474
crivens, I grew up in Shaker. I know that building.comment:www.metafilter.com,2011:site.103143-3674474Wed, 04 May 2011 09:27:42 -0800oonhBy: jonp72
http://www.metafilter.com/103143/Well-just-take-n1right#3674482
If you can come up with an <a href="http://for.mat.bham.ac.uk/R.W.Kaye/minesw/ordmsw.htm">algorithm for solving Minesweeper</a>, you've also solved the P vs. NP problem.comment:www.metafilter.com,2011:site.103143-3674482Wed, 04 May 2011 09:31:19 -0800jonp72By: kmz
http://www.metafilter.com/103143/Well-just-take-n1right#3674497
I saw this on Slashdot earlier today and I was shocked at how many "what's P=NP and why should I care?" comments there were. I mean, I know it's Slashdot, but even from them I expected better. Sad.
<i>Even still, if it turns out that P=NP, then we're going to have to retire RSA encryption.</i>
Well, probably. But even then it could turn out that the actual solution is O(n^(2|||||||10)), where | is Knuth arrow notation. And the constant turns out to be Graham's number.
Realistically, I'd worry more about making sure you choose good factors for RSA, not to mention if/when quantum computing ever takes off.comment:www.metafilter.com,2011:site.103143-3674497Wed, 04 May 2011 09:36:48 -0800kmzBy: DU
http://www.metafilter.com/103143/Well-just-take-n1right#3674502
I thought factorization was already known to be NP-complete.
<i>It is suspected to be outside of all three of the complexity classes P, NP-complete, and co-NP-complete. If it could be proved that it is in either NP-Complete or co-NP-Complete, that would imply NP = co-NP. That would be a very surprising result, and therefore integer factorization is widely suspected to be outside both of those classes</i>
Harder than NP-complete?! Jeepers.comment:www.metafilter.com,2011:site.103143-3674502Wed, 04 May 2011 09:37:44 -0800DUBy: kmz
http://www.metafilter.com/103143/Well-just-take-n1right#3674504
<i>If you can come up with an algorithm for solving Minesweeper, you've also solved the P vs. NP problem.</i>
Holy shit, <a href="http://www.activewin.com/winxp/tips/misc/3.shtml">P=NP!</a>comment:www.metafilter.com,2011:site.103143-3674504Wed, 04 May 2011 09:38:25 -0800kmzBy: King Bee
http://www.metafilter.com/103143/Well-just-take-n1right#3674519
<i>But even then it could turn out that the actual solution is O(n^(2|||||||10)), where | is Knuth arrow notation. And the constant turns out to be Graham's number.</i>
Imagine me snorting water through my nose as I read this. Good one. =)comment:www.metafilter.com,2011:site.103143-3674519Wed, 04 May 2011 09:42:14 -0800King BeeBy: Wolfdog
http://www.metafilter.com/103143/Well-just-take-n1right#3674596
<i>But even then it could turn out that the actual solution is O(n^(2|||||||10)), where | is Knuth arrow notation. And the constant turns out to be Graham's number.</i>
Your exponent might be correct, but I don't think the case <i>n=1</i> is going to be quite that difficult.comment:www.metafilter.com,2011:site.103143-3674596Wed, 04 May 2011 10:06:46 -0800WolfdogBy: DU
http://www.metafilter.com/103143/Well-just-take-n1right#3674602
Factoring numbers of 1 digit isn't that hard anyway.comment:www.metafilter.com,2011:site.103143-3674602Wed, 04 May 2011 10:08:03 -0800DUBy: DU
http://www.metafilter.com/103143/Well-just-take-n1right#3674603
Oh wait...that's your joke. D'oh.comment:www.metafilter.com,2011:site.103143-3674603Wed, 04 May 2011 10:08:30 -0800DUBy: Wolfdog
http://www.metafilter.com/103143/Well-just-take-n1right#3674611
Humor is P-Space.comment:www.metafilter.com,2011:site.103143-3674611Wed, 04 May 2011 10:10:18 -0800WolfdogBy: madcaptenor
http://www.metafilter.com/103143/Well-just-take-n1right#3674622
Well, wouldn't O(n^bignumber) be slower than trial division for any number you can write down? (If not, make bignumber bigger.)comment:www.metafilter.com,2011:site.103143-3674622Wed, 04 May 2011 10:13:22 -0800madcaptenorBy: hellojed
http://www.metafilter.com/103143/Well-just-take-n1right#3674644
In prepping me for a possible interview at Amazon, one programmer told me that they'd send in one of their more senior engineers to ask me increasingly difficult questions "until I broke". And that during this juncture they might throw me a P=NP problem. I said that if it car to that I would say I had solved it, just to piss the guy off.comment:www.metafilter.com,2011:site.103143-3674644Wed, 04 May 2011 10:21:11 -0800hellojedBy: Serf
http://www.metafilter.com/103143/Well-just-take-n1right#3674660
<i>Harder than NP-complete?! Jeepers.</i>
Factoring is definitely in NP, because it's easy to verify an answer. So it's definitely not harder than NP-complete.comment:www.metafilter.com,2011:site.103143-3674660Wed, 04 May 2011 10:25:27 -0800SerfBy: spamguy
http://www.metafilter.com/103143/Well-just-take-n1right#3674788
I went to Case Western Reserve University, just down the road, and had an entire half a semester dedicated to P/NP. I had no idea it was born mere miles away though. Very cool.comment:www.metafilter.com,2011:site.103143-3674788Wed, 04 May 2011 11:01:58 -0800spamguyBy: delmoi
http://www.metafilter.com/103143/Well-just-take-n1right#3674928
<i>Even still, if it turns out that P=NP, then we're going to have to retire RSA encryption.</i>
No. First of all, no one has ever proven that prime factorization is outside of P to begin with. Second of all there are methods you can use to factor a prime number within a given probability bound that are in P time.
And third of all P time doesn't mean "fast". Algorithms can be in P time but still grow quickly enough to be useless for large inputs.comment:www.metafilter.com,2011:site.103143-3674928Wed, 04 May 2011 11:39:44 -0800delmoiBy: Serf
http://www.metafilter.com/103143/Well-just-take-n1right#3675011
This is probably a good place to mention Impagliazzo's <a href="http://cseweb.ucsd.edu/~russell/average.ps">A Personal View of Average-Case Complexity</a> (1995, linked file is PostScript). The paper goes one step further than just contemplating the consequences of P=NP or P!=NP by also considering how hard <i>typical</i> problems are. It lays out five possible scenarios, all the way from Algorithmica (P=NP), to Cryptomania (public key cryptography is possible, believed to be closest to reality).
It's an engaging read if you like this sort of thing.comment:www.metafilter.com,2011:site.103143-3675011Wed, 04 May 2011 12:01:01 -0800SerfBy: Joe in Australia
http://www.metafilter.com/103143/Well-just-take-n1right#3676061
<em>If you can come up with an algorithm for solving Minesweeper, you've also solved the P vs. NP problem.</em>
<ol><li>Click really fast in a bunch of random places.</li><li>The map is probably now easy to solve.</li><li>If not, click some more.</li><li>If you hit a mine, no biggie! Just restart the game.</li><li>Now the game is easy to solve and you can click the remaining squares, no problem.</li><li>Epic win!</li></ol>Where's my million bucks?comment:www.metafilter.com,2011:site.103143-3676061Wed, 04 May 2011 18:27:33 -0800Joe in AustraliaBy: Crabby Appleton
http://www.metafilter.com/103143/Well-just-take-n1right#3676261
<i>Second of all there are methods you can use to factor a prime number within a given probability bound that are in P time. </i>
Sure. Let <i>P</i> be a prime number. Then the factors of <i>P</i> are <i>P</i> and 1.comment:www.metafilter.com,2011:site.103143-3676261Wed, 04 May 2011 20:18:34 -0800Crabby Appleton