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Check my proof of P=NP for errors: Let X be NP. Then by *link removed* (np-complete-5) exists X’ in P such that X’(Z)=Y. But X’ in NP. Therefore applying it again, exists X’’ in P such that X’’(Y)=Z. So X’’ produces the same result as X in polynomial time. So, P=NP. #complexity

(Here "generalized NP" is assumed: not yes/no, but arbitrary polynomial-sized data for my NP-complete problem.)

Posted

This proof in details: http://arweave.net/S-o7FdXfiHu9waP0T4SAJ1vpNAQiBQzpSdQkTjpXyOA

45 minutes ago, Ghideon said:

You are assuming P=NP . It is equally valid* to assume that PNP . How does the proof handle that?

Oh, soory, I mean I not assume P=NP but use my own modified definition of NP in the proof (that does not change the end result).

 *) Or even better, according to a majority of researchers in a poll:

image.png.fc60a63a88c83e3e2a9e6cf1027cff35.png

https://mags.acm.org/communications/201205?pg=12#pg12

 

Oh, sorryL

Posted
50 minutes ago, porton said:

Oh, soory, I mean I not assume P=NP but use my own modified definition of NP in the proof (that does not change the end result).

Please, @porton, do not embed your own words in a quote by other user. It's very confusing.

Posted (edited)

 

47 minutes ago, porton said:

My proof contained a big error. Nevermind.

Ok. Feel free to post an updated version for further discussion.

 

9 hours ago, porton said:

That's a bug of the forum software,

You can report the issue, edit the post (within 60 minutes) or post a followup with clarifications.

FTFY:

The quote of Ghideon is unfortunately incorrect. Red strikethrough does not belong there:

cut.thumb.png.28a5200c4ee5d64216afa927568c37f6.png

 

9 hours ago, joigus said:

do not embed your own words in a quote by other user.

Thanks for observing that!

 

Edited by Ghideon

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