Primality by trial division

Primality by trial division
You are encouraged to solve this task according to the task description, using any language you may know.

Write a boolean function that tells whether a given integer is prime. Remember that 1 and all non-positive numbers are not prime.

Use trial division. Even numbers may be eliminated right away. A loop from 3 to √(n) will suffice, but other loops are allowed.

Ada

function Is_Prime(Item : Positive) return Boolean is
   Result : Boolean := True;
   Test : Natural;
begin
   if Item /= 2 and Item mod 2 = 0 then
      Result := False;
   else
      Test := 3;
      while Test < Integer(Sqrt(Float(Item))) loop
         if Item mod Test = 0 then
            Result := False;
            exit;
         end if;
         Test := Test + 2;
      end loop;
  end if;
  return Result;
end Is_Prime;

ALGOL 68

main:(
  PROC is prime = ( INT n )BOOL:
  (
    IF n = 2 THEN 
      TRUE
    ELIF n <= 1 OR n MOD 2 = 0 THEN 
      FALSE
    ELSE
      BOOL prime := TRUE;
      FOR i FROM 3 BY 2 TO ENTIER sqrt(n) WHILE prime := n MOD i /= 0 DO
         SKIP
      OD;
      prime
    FI
  );
  
  INT upb=100;
  printf(($" The primes up to "g(-3)" are:"l$,upb));
  FOR i TO upb DO 
    IF is prime(i) THEN
      printf(($g(-4)$,i))
    FI
  OD;
  printf($l$)
)

Output:

The primes up to 100 are:
  2   3   5   7  11  13  17  19  23  29  31  37  41  43  47  53  59  61  67  71  73  79  83  89  97

BASIC

Compiler: QuickBasic 4.5

Going with the classic 1 for "true" and 0 for "false":

FUNCTION prime% (n!)
  IF n = 2 THEN prime = 1
  IF n <= 1 OR n MOD 2 = 0 THEN prime = 0
  FOR a = 3 TO INT(SQR(n)) STEP 2
    IF n MOD a = 0 THEN prime = 0
  NEXT a
  prime = 1
END FUNCTION

C

#include <math.h>
#define FALSE 0
#define TRUE  1
int isPrime( unsigned int n )
{
    unsigned int i;
    if ( n == 2 )
        return TRUE;
    if ( n <= 1 || ( n & 1 ) == 0 )
        return FALSE;
    for ( i = 3 ; i <= sqrt( n ) ; i += 2 )
        if ( n % i == 0 )
            return FALSE;
    return TRUE;
}

Common Lisp

(defun primep (a)
 (cond ((= a 2) T)
       ((or (<= a 1) (= (mod a 2) 0)) nil)
       ((loop for i from 3 to (sqrt a) by 2 do
               (if (= (mod a i) 0)
                   (return nil))) nil)
       (T T)))

D

import std.math: sqrt;
bool isPrime(int n) {
    if (n == 2)
        return true;
    if (n <= 1 || (n & 1) == 0)
        return false;

    for(int i = 3; i <= sqrt(cast(float)n); i += 2)
        if (n % i == 0)
            return false;
    return true;
}

Forth

: prime? ( n -- ? )
       dup 2 < if      drop false
  else dup 2 = if      drop true
  else dup 1 and 0= if drop false
  else 3
       begin 2dup dup * >=
       while 2dup mod 0=
             if       2drop false exit
             then 2 +
       repeat         2drop true
  then then then ;

Haskell

Without square roots:

divides k n = n `mod` k == 0

isPrime :: Integer -> Bool
isPrime n | n < 2 = False
isPrime n         = not $ any (`divides` n) $ takeWhile (\k -> k*k <= n) [2..]

J

Actually 1&p: would do, but the task calls for trial division, so:

isprime=: 3 : 'if. 3>:y do. 1<y else. 0 *./@:< y|~2+i.<.%:y end.'

Java

public static boolean prime(double a){
   if(a == 2){
      return true;
   }else if(a <= 1 || a % 2 == 0){
      return false;
   }
   for(double n= 3; n <= (long)Math.sqrt(a); n+= 2){
      if(a % n == 0){ return false; }
   }
   return true;
}

LSE64

over : 2 pick
2dup : over over
even? : 1 & 0 =

# trial n d yields "n d 0/1 false" or "n d+2 true"
trial : 2 +                 true
trial : 2dup % 0 =   then 0 false
trial : 2dup dup * < then 1 false
trial-loop : trial &repeat

# prime? n yields flag
prime? : 3 trial-loop >flag drop drop
prime? : dup even? then drop false
prime? : dup 2 =   then drop true
prime? : dup 2 <   then drop false

MAXScript

fn isPrime n =
(
    if n == 2 then
    (
        return true
    )
    else if (n <= 1) OR (mod n 2 == 0) then
    (
        return false
    )

    for i in 3 to (sqrt n) by 2 do
    (
        if mod n i == 0 then return false
    )
    true
)

Perl

sub prime {
  $a = shift;
  if ($a == 2) {
    return 1;
  }
  if ($a <= 1 || $a % 2 == 0) {
    return 0;
  }
  $d = 3;
  while ($d <= sqrt($a)) {
    if ($a % $d == 0) {
      return 0;
    }
    $d += 2;
  }
  return 1;
}

Python

The simplest primality test, using trial division:

Interpreter: Python 2.5

def prime(a):
    return not (a < 2 or any(a % x == 0 for x in range(2, int(a**0.5) + 1)))

Another test. Exclude even numbers first:

Interpreter: Python 2.5

def prime2(a):
    if a == 2: return True
    if a < 2 or a % 2 == 0: return False
    return not any(a % x == 0 for x in range(3, int(a**0.5) + 1, 2))

Yet another test. Exclude multiples of 2 and 3, see http://www.devx.com/vb2themax/Tip/19051:

Interpreter: Python 2.4

def prime3(a):
    if a < 2: return False
    if a == 2 or a == 3: return True # manually test 2 and 3   
    if a % 2 == 0 or a % 3 == 0: return False # exclude multiples of 2 and 3

    maxDivisor = a**0.5
    d, i = 5, 2
    while d <= maxDivisor:
        if a % d == 0: return False
        d += i 
        i = 6 - i # this modifies 2 into 4 and viceversa

    return True

Ruby

def prime(a)
  if a==2
   return true
  end

  if a<=1 || a%2==0
    return false
  end

  d=3
  while d <= Math.sqrt(a) do
    if a%d==0
      return false
    end
    d+=2
  end

return true
end

TI-83 BASIC

Calculator symbol translations:

"STO" arrow: →

Square root sign: √

Prompt A
If A=2:Then
Disp "PRIME"
Stop
End

If (fPart(A/2)=0 and A>0) or A<2:Then
Disp "NOT PRIME"
Stop
End

1→P
For(B,3,√(A))
If FPart(A/B)=0:Then
0→P
√(A)→B
End
B+1→B
End

If P=1:Then
Disp "PRIME"
Else
Disp "NOT PRIME"
End