Primality by trial division: Difference between revisions
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Write a boolean function that tells whether a given integer is prime. Remember that 1 and all non-positive numbers are not prime. |
Write a boolean function that tells whether a given integer is prime. Remember that 1 and all non-positive numbers are not prime. |
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Implement the simplest primality test, using trial division. |
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=={{header|Ada}}== |
=={{header|Ada}}== |
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prime = 1 |
prime = 1 |
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END FUNCTION |
END FUNCTION |
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=={{header|Haskell}}== |
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Without square roots: |
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divides k n = n `mod` k == 0 |
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isPrime :: Integer -> Bool |
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isPrime n | n < 2 = False |
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isPrime n = not $ any (`divides` n) $ takeWhile (\k -> k*k <= n) [2..] |
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=={{header|Java}}== |
=={{header|Java}}== |
Revision as of 12:32, 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.
Implement the simplest primality test, using trial division.
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
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..]
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