# Testing primality

### From HaskellWiki

(Difference between revisions)

(→Primality Test and Integer Factorization: mention one-off invocation, on odds.) |
(→Primality Test and Integer Factorization: add one-liner 'factors') |
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== Primality Test and Integer Factorization == |
== Primality Test and Integer Factorization == |
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− | Given an infinite list of prime numbers, we can implement primality test and integer factorization by trial division: |
+ | Simplest primality test and integer factorization is by trial division: |

+ | <haskell> |
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+ | import Data.List (unfoldr) |
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+ | import Data.Maybe (listToMaybe) |
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+ | |||

+ | factors n = unfoldr (\(d, n) -> listToMaybe [(x, (x, div n x)) |
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+ | | n > 1, x <- [d..isqrt n] ++ [n], rem n x == 0]) (2,n) |
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+ | |||

+ | isPrime n = factors n == [n] |
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+ | isqrt n = floor . sqrt . fromIntegral $ n |
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+ | </haskell> |
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+ | |||

+ | Of course there's no need to try any even numbers above 2. Given an infinite list of primes we can avoid any composites: |
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<haskell> |
<haskell> |
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− | -- isPrime n = head (primeFactors n) == n |
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isPrime n = n > 1 && |
isPrime n = n > 1 && |
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foldr (\p r -> p*p > n || ((n `rem` p) /= 0 && r)) |
foldr (\p r -> p*p > n || ((n `rem` p) /= 0 && r)) |

## Revision as of 09:22, 3 June 2014

# 1 Testing Primality

(for a context to this see Prime numbers).

## 1.1 Primality Test and Integer Factorization

Simplest primality test and integer factorization is by trial division:

import Data.List (unfoldr) import Data.Maybe (listToMaybe) factors n = unfoldr (\(d, n) -> listToMaybe [(x, (x, div n x)) | n > 1, x <- [d..isqrt n] ++ [n], rem n x == 0]) (2,n) isPrime n = factors n == [n] isqrt n = floor . sqrt . fromIntegral $ n

Of course there's no need to try any even numbers above 2. Given an infinite list of primes we can avoid any composites:

isPrime n = n > 1 && foldr (\p r -> p*p > n || ((n `rem` p) /= 0 && r)) True primes primeFactors n | n > 1 = go n primes -- or go n (2:[3,5..]) where -- for one-off invocation go n ps@(p:t) | p*p > n = [n] | r == 0 = p : go q ps | otherwise = go n t where (q,r) = quotRem n p

When trying to factorize only one number or two, it will be faster to just use `(2:[3,5..])`

as a source of possible divisors instead of first finding prime numbers and then using them. For more than a few factorizations, when no other primes source is available, just use

primes = 2 : filter isPrime [3,5..]

More at Prime numbers#Optimal trial division.

## 1.2 Miller-Rabin Primality Test

-- (eq. to) find2km (2^k * n) = (k,n) find2km :: Integral a => a -> (a,a) find2km n = f 0 n where f k m | r == 1 = (k,m) | otherwise = f (k+1) q where (q,r) = quotRem m 2 -- n is the number to test; a is the (presumably randomly chosen) witness millerRabinPrimality :: Integer -> Integer -> Bool millerRabinPrimality n a | a <= 1 || a >= n-1 = error $ "millerRabinPrimality: a out of range (" ++ show a ++ " for "++ show n ++ ")" | n < 2 = False | even n = False | b0 == 1 || b0 == n' = True | otherwise = iter (tail b) where n' = n-1 (k,m) = find2km n' b0 = powMod n a m b = take (fromIntegral k) $ iterate (squareMod n) b0 iter [] = False iter (x:xs) | x == 1 = False | x == n' = True | otherwise = iter xs -- (eq. to) pow' (*) (^2) n k = n^k pow' :: (Num a, Integral b) => (a->a->a) -> (a->a) -> a -> b -> a pow' _ _ _ 0 = 1 pow' mul sq x' n' = f x' n' 1 where f x n y | n == 1 = x `mul` y | r == 0 = f x2 q y | otherwise = f x2 q (x `mul` y) where (q,r) = quotRem n 2 x2 = sq x mulMod :: Integral a => a -> a -> a -> a mulMod a b c = (b * c) `mod` a squareMod :: Integral a => a -> a -> a squareMod a b = (b * b) `rem` a -- (eq. to) powMod m n k = n^k `mod` m powMod :: Integral a => a -> a -> a -> a powMod m = pow' (mulMod m) (squareMod m)

Example:

-- check if '1212121' is prime with several witnesses > map (millerRabinPrimality 1212121) [5432,1265,87532,8765,26] [True,True,True,True,True]