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#1 |
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Jul 2018
3·13 Posts |
https://rosettacode.org/wiki/Miller–...primality_test
very bad algorithm! so very slow, if R is very big number! in chinese, about 4000 years ago: if R prime than: (2^R-2) mod R = 0 and: (2^R-2) mod R>0 then R is non-prime! but: (2^R-2) mod R=0 then, R is may be a prime and not? but there was more! and some arabian mathematicians translated it, arabic lang. thousand year ago. and a few hundred years ago, some mathematicians translated it, european lang. and very bad, wrong manuplation! (2^R-1) mod R =? 1 if chinese method think: MM127=2^(2^127-1)-1 is a prime, only half page prof, and very simple! any R prime test by chinese method a few minutes, if R countable! please think it! |
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#2 | |
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"Forget I exist"
Jul 2009
Dumbassville
26·131 Posts |
Quote:
Last fiddled with by science_man_88 on 2018-07-28 at 19:30 |
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#3 | |
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Aug 2006
3×1,993 Posts |
Quote:
See, e.g., http://mathworld.wolfram.com/ChineseHypothesis.html for the origin of this misunderstanding. |
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#4 |
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Feb 2017
Nowhere
4,643 Posts |
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#5 |
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Aug 2006
3·1,993 Posts |
On the mathematical content, Fermat's test is no faster than the Miller-Rabin test, but provides less certainty and (unlike M-R) has pseudoprimes.
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#6 |
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1976 Toyota Corona years forever!
"Wayne"
Nov 2006
Saskatchewan, Canada
10010010001002 Posts |
When I read the title I first thought of this:
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#7 |
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"Rashid Naimi"
Oct 2015
Remote to Here/There
80716 Posts |
I think it is interesting that all known primality tests other than sieving (not exactly a test) and trial-by-division and the LL test are based on and have as a bottleneck the powermod operation (corrections/counterexamples would be appreciated).
It is also interesting that at least in theory the powermod operation can be made to run in parallel, but there does not seem to be any such publicised/published effort (corrections are appreciated).
Last fiddled with by a1call on 2018-08-02 at 02:35 |
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#8 |
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Jul 2003
wear a mask
2×829 Posts |
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