Determining if a number is prime

Question

I have perused a lot of code on this topic, but most of them produce the numbers that are prime all the way up to the input number. However, I need code which only checks w

Answer 1

bool check_prime(int num) {
    for (int i = num - 1; i > 1; i--) {
        if ((num % i) == 0)
            return false;
    }
    return true;
}

checks for any number if its a prime number

Answer 2

This code only checks if the number is divisible by two. For a number to be prime, it must not be evenly divisible by all integers less than itself. This can be naively implemented by checking if it is divisible by all integers less than floor(sqrt(n)) in a loop. If you are interested, there are a number of much faster algorithms in existence.

Answer 3

//simple function to determine if a number is a prime number
//to state if it is a prime number


#include <iostream>

using namespace std;


int isPrime(int x);  //functioned defined after int main()


int main()
{
 int y;
    cout<<"enter value"<<endl;

    cin>>y;

    isPrime(y);    

  return 0;

 } //end of main function


//-------------function

  int isPrime(int x)
 {
   int counter =0;
     cout<<"factors of "<<x<<" are "<<"\n\n";    //print factors of the number

     for (int i =0; i<=x; i++)

     {
       for (int j =0; j<=x; j++)

         {
           if (i * j == x)      //check if the number has multiples;
                {

                  cout<<i<<" ,  ";  //output provided for the reader to see the
                                    // muliples
                  ++counter;        //counts the number of factors

                 }


          }


    }
  cout<<"\n\n";

  if(counter>2) 
     { 
      cout<<"value is not a prime number"<<"\n\n";
     }

  if(counter<=2)
     {
       cout<<"value is a prime number"<<endl;
     }
 }

Answer 4

I would guess taking sqrt and running foreach frpm 2 to sqrt+1 if(input% number!=0) return false; once you reach sqrt+1 you can be sure its prime.

Answer 5

My own IsPrime() function, written and based on the deterministic variant of the famous Rabin-Miller algorithm, combined with optimized step brute forcing, giving you one of the fastest prime testing functions out there.

__int64 power(int a, int n, int mod)
{
 __int64 power=a,result=1;

 while(n)
 {
  if(n&1) 
   result=(result*power)%mod;
  power=(power*power)%mod;
  n>>=1;
 }
 return result;
}

bool witness(int a, int n)
{
 int t,u,i;
 __int64 prev,curr;

 u=n/2;
 t=1;
 while(!(u&1))
 {
  u/=2;
  ++t;
 }

 prev=power(a,u,n);
 for(i=1;i<=t;++i)
 {
  curr=(prev*prev)%n;
  if((curr==1)&&(prev!=1)&&(prev!=n-1)) 
   return true;
  prev=curr;
 }
 if(curr!=1) 
  return true;
 return false;
}

inline bool IsPrime( int number )
{
 if ( ( (!(number & 1)) && number != 2 ) || (number < 2) || (number % 3 == 0 && number != 3) )
  return (false);

 if(number<1373653)
 {
  for( int k = 1; 36*k*k-12*k < number;++k)
  if ( (number % (6*k+1) == 0) || (number % (6*k-1) == 0) )
   return (false);

  return true;
 }

 if(number < 9080191)
 {
  if(witness(31,number)) return false;
  if(witness(73,number)) return false;
  return true;
 }


 if(witness(2,number)) return false;
 if(witness(7,number)) return false;
 if(witness(61,number)) return false;
 return true;

 /*WARNING: Algorithm deterministic only for numbers < 4,759,123,141 (unsigned int's max is 4294967296)
   if n < 1,373,653, it is enough to test a = 2 and 3.
   if n < 9,080,191, it is enough to test a = 31 and 73.
   if n < 4,759,123,141, it is enough to test a = 2, 7, and 61.
   if n < 2,152,302,898,747, it is enough to test a = 2, 3, 5, 7, and 11.
   if n < 3,474,749,660,383, it is enough to test a = 2, 3, 5, 7, 11, and 13.
   if n < 341,550,071,728,321, it is enough to test a = 2, 3, 5, 7, 11, 13, and 17.*/
}

To use, copy and paste the code into the top of your program. Call it, and it returns a BOOL value, either true or false.

if(IsPrime(number))
{
    cout << "It's prime";
}

else
{
    cout<<"It's composite";
}

If you get a problem compiling with "__int64", replace that with "long". It compiles fine under VS2008 and VS2010.

How it works: There are three parts to the function. Part checks to see if it is one of the rare exceptions (negative numbers, 1), and intercepts the running of the program.

Part two starts if the number is smaller than 1373653, which is the theoretically number where the Rabin Miller algorithm will beat my optimized brute force function. Then comes two levels of Rabin Miller, designed to minimize the number of witnesses needed. As most numbers that you'll be testing are under 4 billion, the probabilistic Rabin-Miller algorithm can be made deterministic by checking witnesses 2, 7, and 61. If you need to go over the 4 billion cap, you will need a large number library, and apply a modulus or bit shift modification to the power() function.

If you insist on a brute force method, here is just my optimized brute force IsPrime() function:

inline bool IsPrime( int number )
{
 if ( ( (!(number & 1)) && number != 2 ) || (number < 2) || (number % 3 == 0 && number != 3) )
  return (false);

 for( int k = 1; 36*k*k-12*k < number;++k)
  if ( (number % (6*k+1) == 0) || (number % (6*k-1) == 0) )
   return (false);
  return true;
 }
}

How this brute force piece works: All prime numbers (except 2 and 3) can be expressed in the form 6k+1 or 6k-1, where k is a positive whole number. This code uses this fact, and tests all numbers in the form of 6k+1 or 6k-1 less than the square root of the number in question. This piece is integrated into my larger IsPrime() function (the function shown first).

If you need to find all the prime numbers below a number, find all the prime numbers below 1000, look into the Sieve of Eratosthenes. Another favorite of mine.

As an additional note, I would love to see anyone implement the Eliptical Curve Method algorithm, been wanting to see that implemented in C++ for a while now, I lost my implementation of it. Theoretically, it's even faster than the deterministic Rabin Miller algorithm I implemented, although I'm not sure if that's true for numbers under 4 billion.

Answer 6

if(number%2!=0)
      cout<<"Number is prime:"<<endl;

The code is incredibly false. 33 divided by 2 is 16 with reminder of 1 but it's not a prime number...