Talk:Prime numbers
Here's an interesting question: will the program go faster if we replace all those (n >)
expressions with (\x -> floor (sqrt n) > x)
?
On one hand, a composite integer cannot possess a factor greater than its square root.
On the other hand, since the list we're looking through contains all possible prime numbers, we are guaranteed to find a factor or an exact match eventually, so do we need the takeWhile
at all?
Throwing this over to somebody with a bigger brain than me...
MathematicalOrchid 16:41, 5 February 2007 (UTC)
a composite can indeed have factors greater than its square root, and indeed most do. what you mean is that a composite will definitely have at least one factor smaller-equal than its square root.
why not use (\x -> n > x*x)
--Johannes Ahlmann 21:18, 5 February 2007 (UTC)