# 99 questions/Solutions/31

(**) Determine whether a given integer number is prime.

Well, a natural number *k* is a prime number if it is larger than **1** and no natural number *n >= 2* with *n^2 <= k* is a divisor of *k*. However, we don't actually need to check all natural numbers *n <= sqrt k*. We need only check the * primes p <= sqrt k*:

```
isPrime :: Integral a => a -> Bool
isPrime k = k > 1 &&
foldr (\p r -> p*p > k || k `rem` p /= 0 && r)
True primesTME
```

This uses

```
{-# OPTIONS_GHC -O2 -fno-cse #-}
-- tree-merging Eratosthenes sieve
-- producing infinite list of all prime numbers
primesTME = 2 : gaps 3 (join [[p*p,p*p+2*p..] | p <- primes'])
where
primes' = 3 : gaps 5 (join [[p*p,p*p+2*p..] | p <- primes'])
join ((x:xs):t) = x : union xs (join (pairs t))
pairs ((x:xs):ys:t) = (x : union xs ys) : pairs t
gaps k xs@(x:t) | k==x = gaps (k+2) t
| True = k : gaps (k+2) xs
```

The tree-merging Eratosthenes sieve here seems to strike a good balance between efficiency and brevity. More at Prime numbers haskellwiki page. The semi-standard `union`

function is readily available from `Data.List.Ordered`

package, put here just for reference:

```
-- duplicates-removing union of two ordered increasing lists
union (x:xs) (y:ys) = case (compare x y) of
LT -> x : union xs (y:ys)
EQ -> x : union xs ys
GT -> y : union (x:xs) ys
```

Here is another solution, intended to be extremely short while still being reasonably fast.

```
isPrime :: (Integral a) => a -> Bool
isPrime n | n < 4 = n > 1
isPrime n = all ((/=0).mod n) $ 2:3:[x + i | x <- [6,12..s], i <- [-1,1]]
where s = floor $ sqrt $ fromIntegral n
```

This one does not go as far as the previous, but it does observe the fact that you only need to check numbers of the form 6k +/- 1 up to the square root. And according to some quick tests (nothing extensive) this version can run a bit faster in some cases, but slower in others; depending on optimization settings and the size of the input.

There is a subtle bug in the version above. The above version will fail on 25, because the bound of s is incorrect. It is x+i that is bounded by the sqrt of the argument, not x. This version will work correctly:

```
isPrime n | n < 4 = n /= 1
isPrime n = all ((/=0) . mod n) $ takeWhile (<= m) candidates
where candidates = (2:3:[x + i | x <- [6,12..], i <- [-1,1]])
m = floor . sqrt $ fromIntegral n
```