Primality by trial division: Difference between revisions
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=={{header|Ada}}== |
=={{header|Ada}}== |
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function Is_Prime(Item : Positive) return Boolean is |
function Is_Prime(Item : Positive) return Boolean is |
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Result : Boolean := True; |
Result : Boolean := True; |
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prime = 1 |
prime = 1 |
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END FUNCTION |
END FUNCTION |
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=={{header|Ruby}}== |
=={{header|Ruby}}== |
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return true |
return true |
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end |
end |
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Revision as of 07:51, 21 November 2007
Primality by trial division
You are encouraged to solve this task according to the task description, using any language you may know.
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.
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;
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
Java
public static boolean prime(long 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; }
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