tag:blogger.com,1999:blog-2293611840857369720.post7079292272814369704..comments2021-07-23T14:59:07.938+03:00Comments on Computer Blindness: Theorem 8.10. P ≠ NP.Anonymoushttp://www.blogger.com/profile/01422787855469629259noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-2293611840857369720.post-43488184484665457622010-08-21T15:38:06.546+04:002010-08-21T15:38:06.546+04:00@hr0nix First of all, one should not be afraid of ...@hr0nix First of all, one should not be afraid of a 100 pages paper, because there is a lot of background information. The idea of the proof is concentrated in the introduction. Second, most of background info is known to us (graphical models, 3-SAT etc). Specifically, no additional knowledge in complexity theory is needed. The problem is more with logics (FMT) and statistical mechanics (though it is used mostly as analogy).<br /><br />After all, I didn't understand all the details, but only the general framework. One should spend few more days for that..Anonymoushttps://www.blogger.com/profile/01422787855469629259noreply@blogger.comtag:blogger.com,1999:blog-2293611840857369720.post-78291869626250666062010-08-20T01:42:33.861+04:002010-08-20T01:42:33.861+04:00Thanks to you, we don't need "broken P \n...Thanks to you, we don't need "broken P \neq NP proof understanding party" anymore =)<br /><br />It wasn't that hard to understand I thought it would be, was it?hr0nixhttp://sexdrugsandappliedscience.comnoreply@blogger.comtag:blogger.com,1999:blog-2293611840857369720.post-7404816121219256232010-08-20T01:27:40.974+04:002010-08-20T01:27:40.974+04:00If it was the case (P=NP can be neither proved nor...If it was the case (P=NP can be neither proved nor disproved) then using the assumption P not equal NP (as most of theorems in this domain do) we could get a contradiction. Up to now no such contradictions have been reported.Dmitry Vetrovhttp://vetrovd.narod.runoreply@blogger.com