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)

LOL! That is indeed what I meant.

It turns out my comment above is correct - the takeWhile filtering in factors is in fact unecessary. The function works just fine without it. (Notice I have made some edits to correct the multiple bugs in the primes function. Oops!)

Now the only use of takeWhile is in the is_prime function, which could be changed to 'give up' the search a lot faster and hence confirm large primes with much less CPU time and RAM usage. Maybe I'll wrap my brain around that later.

MathematicalOrchid 10:17, 6 February 2007 (UTC)

The section Simple Prime Sieve II is not a sieve in the same sense that the first one is. It really implements a primality test as a filter.

A more "sieve-like" version of the simple sieve which exploits the fact that we need not check for primes larger than the square root would be

primes :: [Integer] primes = sieve [2..] where sieve (p:xs) = p : sieve [x | x<-xs, (x< p*p) || (x `mod` p /= 0)]

However, this runs even slower than the original!

Kapil Hari Paranjape 06:51, 4 February 2009 (UTC)