tag:blogger.com,1999:blog-3722233.post5602882563612927913..comments2024-09-16T11:29:06.673-05:00Comments on Computational Complexity: Factorization in coNP- in other domains?Lance Fortnowhttp://www.blogger.com/profile/06752030912874378610noreply@blogger.comBlogger9125tag:blogger.com,1999:blog-3722233.post-66094245446309559712014-04-16T11:51:26.546-05:002014-04-16T11:51:26.546-05:00Malcin is correct, and Andy is correct, so I used ...Malcin is correct, and Andy is correct, so I used Andy's correction.<br /><br />GASARCHhttps://www.blogger.com/profile/06134382469361359081noreply@blogger.comtag:blogger.com,1999:blog-3722233.post-62872447811153588492014-04-16T10:39:07.423-05:002014-04-16T10:39:07.423-05:00What about adding MathJax to the site? (one line o...What about adding MathJax to the site? (one line of code :-)Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-3722233.post-64402389415549209602014-04-16T09:22:33.240-05:002014-04-16T09:22:33.240-05:00Here http://npcomplete-001-site1.myasp.net/ resolv...Here http://npcomplete-001-site1.myasp.net/ resolved Exact cover ploblem, on line solver aviable novakohttps://www.blogger.com/profile/05306915643108888136noreply@blogger.comtag:blogger.com,1999:blog-3722233.post-5822312009285915112014-04-16T00:33:12.319-05:002014-04-16T00:33:12.319-05:00maybe someone should answer malcin's concern ?...maybe someone should answer malcin's concern ? Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-3722233.post-28898436323536135662014-04-15T22:54:39.677-05:002014-04-15T22:54:39.677-05:00It looks like you're right.
Couldn't we ad...It looks like you're right.<br />Couldn't we add in the condition that each p_i is prime? PRIMES is clearly in co-NP: for all q, r, q*r=p -> q=1 or r=1.Andy Parrishhttps://www.blogger.com/profile/12252029594014518238noreply@blogger.comtag:blogger.com,1999:blog-3722233.post-57832094708645875612014-04-15T01:08:43.178-05:002014-04-15T01:08:43.178-05:00Maybe I'm confused, but is Jesse's proof r...Maybe I'm confused, but is Jesse's proof really correct? A pair (n=64, m=6) belongs to FACT (2 is a factor). But in this definition, for L=2, p1=p2=8 we have that it does not belong.malcinhttps://www.blogger.com/profile/13527592176910207210noreply@blogger.comtag:blogger.com,1999:blog-3722233.post-44971884010906669972014-04-13T22:55:46.404-05:002014-04-13T22:55:46.404-05:00YES, I garbled things and also my editor garbled t...YES, I garbled things and also my editor garbled things. But its fixed now. I hope.GASARCHhttps://www.blogger.com/profile/06134382469361359081noreply@blogger.comtag:blogger.com,1999:blog-3722233.post-91717431341961176972014-04-13T22:43:19.695-05:002014-04-13T22:43:19.695-05:00Has Jesse's proof been garbled? If I am under...Has Jesse's proof been garbled? If I am understanding it right, it says FACT is the set of (n,m) such that every factorization of n contains a factor between 2 and m, which is not correct.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-3722233.post-20352557216198474932014-04-13T22:05:08.955-05:002014-04-13T22:05:08.955-05:00The various definitions of FACT and FACT-bar seem ...The various definitions of FACT and FACT-bar seem have the form { (n,m) | }. I'm pretty good at figuring out what you mean, but I can't tell what Jesse means (Are there more typos in there? Did you really mean it when you said "n \ne p_1 p_2 ... p_L"?)<br /><br />It would also be helpful to get an inline LaTeX parser for the blog!Andy Parrishhttps://www.blogger.com/profile/12252029594014518238noreply@blogger.com