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
<lang 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;</lang>
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
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
<lang 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;
}</lang>
Clojure
The symbol # is a shortcut for creating lambda functions; the arguments in such a function are %1, %2, %3... (or simply % if there is only one argument). Thus, #(< (* % %) n) is equivalent to (fn [x] (< (* x x) n)) or more mathematically f(x) = x * x < n.
(defn divides [k n] (= (rem n k) 0))
(defn prime? [n]
(if (< n 2)
false
(nil? (filter #(divides % n) (take-while #(< (* % %) n) (range 2 n))))))
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
<lang 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;
}</lang>
Version with excluded multiplies of 2 and 3
<lang d>/**
* to compile: * $ dmd -run prime_trial.d * to optimize: * $ dmd -O -inline -release prime_trial.d */
module prime_trial;
import std.conv : to; import std.stdio : writefln;
/// Adapted from: http://www.devx.com/vb2themax/Tip/19051 bool isprime(Integer)(in Integer number) {
/* manually test 1, 2, 3 and multiples of 2 and 3 */
if (number == 2 || number == 3)
return true;
else if (number < 2 || number % 2 == 0 || number % 3 == 0)
return false;
/* we can now avoid to consider multiples
* of 2 and 3. This can be done really simply
* by starting at 5 and incrementing by 2 and 4
* alternatively, that is:
* 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35, 37, ...
* we don't need to go higher than the square root of the number */
for (Integer divisor = 5, increment = 2; divisor*divisor <= number;
divisor += increment, increment = 6 - increment)
if (number % divisor == 0)
return false;
return true; // if we get here, the number is prime
}
/// print all prime numbers less then a given limit void main(char[][] args) {
const limit = (args.length == 2) ? to!(uint)(args[1]) : 100;
for (uint i = 0; i < limit; ++i)
if (isprime(i))
writefln(i);
}</lang>
E
<lang e> def isPrime(n :int) {
if (n == 2) {
return true
} else if (n <= 1 || n %% 2 == 0) {
return false
} else {
def limit := (n :float64).sqrt().ceil()
var divisor := 1
while ((divisor += 2) <= limit) {
if (n %% divisor == 0) {
return false
}
}
return true
}
}</lang>
Factor
<lang factor> USING: combinators kernel math math.functions math.ranges sequences ;
- prime? ( n -- ? )
{
{ [ dup 2 < ] [ drop f ] }
{ [ dup even? ] [ 2 = ] }
[ 3 over sqrt 2 <range> [ mod 0 > ] with all? ]
} cond ;
</lang>
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 ;
Fortran
<lang fortran> FUNCTION isPrime(number)
LOGICAL :: isPrime
INTEGER, INTENT(IN) :: number
INTEGER :: i
IF(number==2) THEN
isPrime = .TRUE.
ELSE IF(number < 2 .OR. MOD(number,2) == 0) THEN
isPRIME = .FALSE.
ELSE
isPrime = .TRUE.
DO i = 3, INT(SQRT(REAL(number))), 2
IF(MOD(number,i) == 0) THEN
isPrime = .FALSE.
EXIT
END IF
END DO
END IF
END FUNCTION</lang>
Haskell
Without square roots:
divides k n = n `mod` k == 0
isPrime :: Integer -> Bool
isPrime n | n < 2 = False
| otherwise = not $ any (`divides` n) $ takeWhile (\k -> k*k <= n) (2:[3,5..])
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.'
Joy
From http://www.latrobe.edu.au/philosophy/phimvt/joy/jp-imper.html
prime ==
2
[ [dup * >] nullary [rem 0 >] dip and ]
[ succ ]
while
dup * < ;
Java
<lang java>public static boolean prime(double a){
if(a == 2){
return true;
}else if(a <= 1 || a % 2 == 0){
return false;
}
for(long n= 3; n <= (long)Math.sqrt(a); n+= 2){
if(a % n == 0){ return false; }
}
return true;
}</lang>
Logo
to prime? :n if :n < 2 [output "false] if :n = 2 [output "true] if equal? 0 modulo :n 2 [output "false] for [i 3 [sqrt :n] 2] [if equal? 0 modulo :n :i [output "false]] output "true end
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
)
OCaml
<lang ocaml>let is_prime n =
if n = 2 then true
else if n < 2 || n mod 2 = 0 then false
else
let rec loop k =
if k * k > n then true
else if n mod k = 0 then false
else loop (k+2)
in loop 3</lang>
Octave
This function works on vectors and matrix.
<lang octave>function b = isthisprime(n)
for r = 1:rows(n)
for c = 1:columns(n)
b(r,c) = false;
if ( n(r,c) == 2 )
b(r,c) = true;
elseif ( (n(r,c) < 2) || (mod(n(r,c),2) == 0) )
b(r,c) = false;
else
b(r,c) = true; for i = 3:2:sqrt(n(r,c)) if ( mod(n(r,c), i) == 0 ) b(r,c) = false; break; endif endfor
endif endfor endfor
endfunction
% as test, print prime numbers from 1 to 100 p = [1:100]; pv = isthisprime(p); disp(p( pv ));</lang>
Perl
<lang 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;
}
$d += 2;
}
return 1;
}</lang>
Python
The simplest primality test, using trial division:
<lang python>def prime(a):
return not (a < 2 or any(a % x == 0 for x in xrange(2, int(a**0.5) + 1)))</lang>
Another test. Exclude even numbers first:
<lang python>def prime2(a):
if a == 2: return True if a < 2 or a % 2 == 0: return False return not any(a % x == 0 for x in xrange(3, int(a**0.5) + 1, 2))</lang>
Yet another test. Exclude multiples of 2 and 3, see http://www.devx.com/vb2themax/Tip/19051:
<lang python>def prime3(a):
if a < 2: return False if a == 2 or a == 3: return True # manually test 2 and 3 if a % 2 == 0 or a % 3 == 0: return False # exclude multiples of 2 and 3
maxDivisor = a**0.5
d, i = 5, 2
while d <= maxDivisor:
if a % d == 0: return False
d += i
i = 6 - i # this modifies 2 into 4 and viceversa
return True</lang>
Ruby
<lang ruby>def prime(a)
if a == 2
true
elsif a <= 1 || a % 2 == 0
false
else
divisors = Enumerable::Enumerator.new(3..Math.sqrt(a), :step, 2)
!divisors.any? { |d| a % d == 0 }
end
end</lang>
Scheme
<lang scheme>(define (is-prime? x)
(cond ((< x 2) #f)
((= x 2) #t)
((zero? (remainder x 2)) #f)
(else
(let loop ((c 3))
(cond ((> (* c c) x) #t)
((zero? (remainder x c)) #f)
(else (loop (+ c 2))))))))</lang>
Standard ML
<lang sml>fun is_prime n =
if n = 2 then true
else if n < 2 orelse n mod 2 = 0 then false
else let
fun loop k =
if k * k > n then true
else if n mod k = 0 then false
else loop (k+2)
in loop 3
end</lang>
Tcl
<lang tcl>proc is_prime n {
if {$n <= 1} {return false}
if {$n == 2} {return true}
if {$n % 2 == 0} {return false}
for {set i 3} {$i <= sqrt($n)} {incr i 2} {
if {$n % $i == 0} {return false}
}
return true
}</lang>
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+2→B End If P=1:Then Disp "PRIME" Else Disp "NOT PRIME" End
V
like joy
[prime?
2
[[dup * >] [true] [false] ifte [% 0 >] dip and]
[succ]
while
dup * <].
Using it
|11 prime? =true |4 prime? =false