Prime Number Formula

Question

I am trying to write a prime number function in C# and I am wondering if the follow code will work. It \"appears\" to work with the first 50 numbers or so. I just want to ma

Answer 1

Here we must have to consider the square root factor. A prime number can be verified if it is not divisible by any number less than the value of square root of any near number.

static bool isPrime(long number) 
{
    if (number == 1) return false;
    if (number == 2) return true;
    if (number % 2 == 0) return false; //Even number     
    long nn= (long) Math.Abs(Math.Sqrt(number));
    for (long i = 3; i < nn; i += 2)  {
       if (number % i == 0) return false;
    }
    return true;
}

Answer 2

this is a simple one

only odd numbers are prime....so

static bool IsPrime(int number)
{
int i;
if(number==2)
    return true;                    //if number is 2 then it will return prime
for(i=3,i<number/2;i=i+2)           //i<number/2 since a number cannot be 
  {                                     //divided by more then its half
    if(number%i==0)                 //if number is divisible by i, then its not a prime
          return false;
  }
return true;                        //the code will only reach here if control
}                                       //is not returned false in the for loop

Answer 3

    static List<long> PrimeNumbers = new List<long>();

    static void Main(string[] args)
    {
        PrimeNumbers.Add(2);
        PrimeNumbers.Add(3);
        PrimeNumbers.Add(5);
        PrimeNumbers.Add(7);
        for (long i = 11; i < 10000000; i += 2)
        {
            if (i % 5 != 0)
                if (IsPrime(i))
                    PrimeNumbers.Add(i);
        }
    }

    static bool IsPrime(long number)
    {
        foreach (long i in PrimeNumbers)
        {
            if (i <= Math.Sqrt(number))
            {
                if (number % i == 0)
                    return false;
            }
            else
                break;
        }
        return true;
    }

Answer 4

This approach definitely won't work, unless your if statement explicitly enumerates all the prime numbers between 0 and sqrt(INT_MAX) (or the C# equivalent).

To properly check for primality, you basically need to attempt to divide your number by every prime number less than its square root. The Sieve of Eratosthenes algorithm is your best bet.

Answer 5

No it won't work! Try 121 = 11 * 11 for example which obviously isn't a prime.

For any number given to your function, that is a product of the prime numbers X1, X2, ..., Xn(where n >= 2) with all of them being greater or equal to 11, your function will return true. (And also, as already said, 9 isn't a prime).

From wikipedia you can see that:

In mathematics, a prime number (or a prime) is a natural number that has exactly two distinct natural number divisors: 1 and itself.

so a very simple and naive algorithm on checking whether a number is prime could be:

public bool CalcIsPrime(int number) {

    if (number == 1) return false;
    if (number == 2) return true;

    if (number % 2 == 0) return false; // Even number     

    for (int i = 2; i < number; i++) { // Advance from two to include correct calculation for '4'
       if (number % i == 0) return false;
    }

    return true;

}

For better algorithms check here: Primality Test

If you want to check your code, do inlcude a test, here's a test case written in xunit.

        [Theory]
        [MemberData(nameof(PrimeNumberTestData))]
        public void CalcIsPrimeTest(int number, bool expected) {
            Assert.Equal(expected, CalcIsPrime(number));
        }

        public static IEnumerable<object[]> PrimeNumberTestData() {
            yield return new object[] { 0, false };
            yield return new object[] { 1, false };
            yield return new object[] { 2, true };
            yield return new object[] { 3, true };
            yield return new object[] { 4, false };
            yield return new object[] { 5, true };
            yield return new object[] { 6, false };
            yield return new object[] { 7, true };
            yield return new object[] { 8, false };
            yield return new object[] { 9, false };
            yield return new object[] { 10, false };
            yield return new object[] { 11, true };
            yield return new object[] { 23, true };
            yield return new object[] { 31, true };
            yield return new object[] { 571, true };
            yield return new object[] { 853, true };
            yield return new object[] { 854, false };
            yield return new object[] { 997, true };
            yield return new object[] { 999, false };
        }

Answer 6

There are some basic rules you can follow to check if a number is prime

1. Even numbers are out. If x % 2 = 0, then it is not prime
2. All non-prime numbers have prime factors. Therefore, you only need test a number against primes to see if it factors
3. The highest possible factor any number has is it's square root. You only need to check if values <= sqrt(number_to_check) are even divisible.

Using that set of logic, the following formula calculates 1,000,000 Primes Generated in: 134.4164416 secs in C# in a single thread.

    public IEnumerable<long> GetPrimes(int numberPrimes)
    {
      List<long> primes = new List<long> { 1, 2, 3 };
      long startTest = 3;

      while (primes.Count() < numberPrimes)
      {
        startTest += 2;
        bool prime = true;
        for (int pos = 2; pos < primes.Count() && primes[pos] <= Math.Sqrt(startTest); pos++)
        {
          if (startTest % primes[pos] == 0)
          {
            prime = false;
          }
        }
        if (prime)
          primes.Add(startTest);
      }
      return primes;
    }

Bear in mind, there is lots of room for optimization in the algorithm. For example, the algorithm could be parallelized. If you have a prime number (let's say 51), you can test all the numbers up to it's square (2601) for primeness in seperate threads as all it's possible prime factors are stored in the list.