Determining if a given number is a prime in haskell

Question

So I have devised the following function for seeing if a given number is a prime in Haskell (it assumes the first prime is 2):

isPrime k = length [ x | x <- [2..k], k `mod` x == 0] == 1

it has the obvious pitfall of continuing the evaluation even if it is divisible by several numbers :(. Is there any sane way of "cutting" the evaluation when it finds more than one solution, using list comprehensions?

Also, which other implementations would you you try on? I'm not looking for performance here, I'm just trying to see if there are other more "haskellish" ways of doing the same thing.

Tango

possible duplicate of Lazy List of Prime Numbers – Mileage

Answer 1

A quick change to your code that will 'short circuit' the evaluation, and relies on the laziness of Haskell Lists, is:

isPrime k = if k > 1 then null [ x | x <- [2..k - 1], k `mod` x == 0] else False

The very first divisor of k will cause the list to be non-empty, and the Haskell implementation of null will only look at the first element of the list.

You should only need to check up to sqrt(k) however:

isPrime k = if k > 1 then null [ x | x <- [2..isqrt k], k `mod` x == 0] else False

Of course, if you are looking to do high-performance primality testing, a library is preferred.

Hydroponics

@Can your edit had broken this post. please don't do that. – Marven

@JamesKPolk thank you for rejecting a nonsensical edit on this post. :) – Marven

@WillNess I re-checked my edit. Yet again I don't see how it deviates from the original answer. All of the information from the original author is preserved. As my edit note stated, it was just a minor formatting edit that fixed some wordings and styling. – Center

@WillNess It could be nonsensical to you but it wasn't for me, and certainly not for others as well since it also got approved by two others. All of the time, I edit posts on SE to make them more readable and approachable; and they get approved. These changes I've made could seem pointless to you but they're important indeed. Learning is crucial and information should be presented as best as it could. – Center

@Center "yet again" (? it's our first communication that I know of.) two things you broke: 1. you changed isqrt to sqrt which doesn't work with Integral types, so the code would cause an error. 2. you changed "relies" to "rely" which broke the meaning of the sentence (in a bit subtle, but still real, sense). – Marven

@WillNess Anyway, I don't care if you've reverted my edit. However, as a last statement, I would like to say that calling something that others have also supported nonsensical is both weird and more importantly rude. Make it a habit to first ask why. – Center

@Center it is an unfortunate fact that your edit broke the post in two ways which I showed to you above, after you've asked. It is good to engage in a dialog, to clear things up. You shouldn't take personally the critique of an artifact, even if produced by you. See dreamsongs.com/10ideas.html for more about that. – Marven

@WillNess rest assured, I'm definitely not taking it personally but just stating what I think. True, changing isqrt to sqrt(hey, learned something new, thank you) was a mistake and I missed the word rely when I was typing fast but those weren't the only changes. Anyway, I really don't want to take this further. Thanks. – Center

@Center happy trails! – Marven

See https://mcmap.net/q/433488/-get-sqrt-from-int-in-haskell for a definition of isqrt, which is missing in the answer. – Boat

isqrt :: Integer -> Integer isqrt n = floor (sqrt (fromIntegral n)) – Santanasantayana

Answer 2

Here is the best resource for prime numbers in haskell in haskell.org

and here prime.hs github project

Rolfston

Answer 3

I like this approach:

First make function to get all factors of n:

factors n = [x | x <- [1..n], mod n x == 0]

Then check if factors are only the given number and 1, if so, the number is prime:

prime n = factors n == [1,n]

Paean

Answer 4

It's perhaps not directly relevant, but on the topic of finding primes in functional languages I found Melissa E. O'Neill's The Genuine Sieve of Eratosthenes very interesting.

Schnauzer

Answer 5

Ignoring the primes issue, and focusing on the narrow point of a more efficient method of length xs == n:

hasLength :: Integral count => [a] -> count -> Bool
_        `hasLength` n | n < 0 = False
[]       `hasLength` n         = n == 0
(_ : xs) `hasLength` n         = xs `hasLength` (pred n)

isPrime k = [ x | x <- [2..k], k `mod` x == 0)] `hasLength` 1

Confrere

Answer 6

This may be silly and inefficient (I'm a complete Haskell newby), but the function isMyNumberPrime (in ghci) seems to tell you if a number is prime or not.

factors n = [x | x <- [2..(n`div` 2)], mod n x == 0]
factormap n = fmap factors $ factors n
isMyNumberPrime n = case factormap n of [] -> True; _ -> False

Sogdiana

For a complete Haskell newbie not so bad. I tested it for some small numbers, and it worked. However, how does this work, anyway? Which factors does factors provide? It doesn't seem to be prime factors, nor all the factors. For instance the prime factors of 8 are [2,2,2] and the factors are [1, 2, 4, 8] - but your function provides something different. The next question is why map - factors of factors? Please, if you don't mind than I would edit your post or please explain better. Thank you. – Moorfowl

Factors can be evaluated by factors n = [ n' | n' <- [1..n], n `mod` n' == 0] – Moorfowl

Primes are numbers that just have 1 and the prime as factors and hence isPrime n = factors n == [1,n]. – Moorfowl

The code factors n = [x | x <- [2..(n`div` 2)], mod n x == 0], and factormap n = fmap factors $ factors n, isMyNumberPrime n = case factormap n of [] -> True; _ -> False does not work for 1. One is not a prime number, and isMyNumberPrime 1 evaluates True. – Moorfowl

Thank you for your comment, Jorg. Unfortunately, it was so long ago that I posted the function that I've totally forgotten how it works (and much of Haskell, for that matter) – Sogdiana