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Talk:Primality by trial division: Difference between revisions
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→"simplest"?: corrected an indention as per convention. -- ~~~~
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:: Even that isn't particularly precise. Should I do trial-division by every integer 2 through n-1? Or am I allowed to factor only up to (sqrt(n)) which is obviously sufficient but introduces the added complexity of invoking a square root. So is it still the "simplest" solution? Is simplicity preferred? Actually it would be entirely sufficient to trial-factor only by all <i>prime</i> numbers up to sqrt(n) which immediately leads to a very elegant recursive implementation. Which I'm pretty sure isn't what the creator of the task had in mind ;-) but which is entirely allowed when you just say "use trial division"... [[User:Sgeier|Sgeier]] 19:21, 4 February 2008 (MST)
::: It isn't necessary to invoke (use) a SQRT function to find the square root to limit the factors. See the REXX example. [[User:Gerard Schildberger|Gerard Schildberger]] 23:06, 18 December 2010 (UTC)
::: Also, trial division by three's can be skipped. See the REXX example and others. -- [[User:Gerard Schildberger|Gerard Schildberger]] 17:12, 15 May 2012 (UTC)
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