Realz Slaw
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 Mar4 awarded Popular Question Feb24 accepted Can we easily compute the percentile of a ratio, using the percentiles of the values in the ratio? Feb24 comment Can we easily compute the percentile of a ratio, using the percentiles of the values in the ratio? Would restricting the values to $x \ge 1, y \ge 1$ for everyone in the group have any effect? Feb24 asked Can we easily compute the percentile of a ratio, using the percentiles of the values in the ratio? Dec15 awarded Caucus Sep24 awarded Autobiographer Sep5 awarded Yearling Jul2 awarded Curious Dec10 revised Convert Circuit SAT to 3-SAT Fixed broken imgur images :( Dec3 awarded Cleanup Dec3 revised Plane intersection by a mapping and different cases rolled back to a previous revision Dec3 answered NP-complete: One proof to rule them all Dec2 comment How to prove that $P \neq NP$ @rewritten ah ok. I assume you mean that it is "NP-complete", in that a proof of it would be easy to recognize, thus a NTM could find it in polynomial time of the length of the proof. Dec2 revised How to prove that $P \neq NP$ added 141 characters in body Dec2 comment How to prove that $P \neq NP$ @user43400 what did I say that gave you that idea? Dec2 comment How to prove that $P \neq NP$ @user43400 where do I say so? Dec2 revised How to prove that $P \neq NP$ added 117 characters in body Dec2 answered How to prove that $P \neq NP$ Dec2 comment How to prove that $P \neq NP$ @rewritten what does that have to do with Kevin's answer? Nov29 comment relative size of most factors of semiprimes, close? @Amzoti from poncho's answer: "BTW: the recommendation $\Delta > 2^{k/2-100}$ doesn't mean 'p and q should differ by a number which is at least 100 bits long', it means closer to 'p and q should differ somewhere in their 100 most-significant bits'." i.e. it isn't talking about the difference in bit-length, as vzn is. Rather it is talking about having at least one different bit in one of upper significant bits.