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Talk:Prime numbers

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(An interesting thought)

Revision as of 21:18, 5 February 2007

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
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)