How can I test for primality?

Question

I am writing a little library with some prime number related methods. As I've done the groundwork (aka working methods) and now I'm looking for some optimization. Ofcourse the internet is an excellent place to do so. I've, however, stumbled upon a rounding problem and I was wondering how to solve this.

In the loop I use to test a number for it's primality it's more efficient to search until sqrt(n) instead of n/2 or even n - 1. But due to rounding problems some number get skipped and thus some primes are skipped! For example, the 10000th prime should be: 104729, but the 'optimized' version ends up with: 103811.

Some code (it's open for more optimization, I know, but I can handle only one thing at a time):

/// <summary>
/// Method for testing the primality of a number e.g.: return IsPrime(29);
/// History:
/// 1. Initial version, most basic form of testing: m smaller then n -1
/// 2. Implemented m smaller then sqrt(n), optimization due to prime factoring
/// </summary>
/// <param name="test">Number to be tested on primality</param>
/// <returns>True if the number is prime, false otherwise</returns>
public static bool IsPrime(int test)
{
    // 0 and 1 are not prime numbers
    if (test == 0 || test == 1) return false;

    // 2 and 3 are prime numbers
    if (test == 2) return true;

    // all even numbers, save 2, are not prime
    if (test % 2 == 0) return false;

    double squared = Math.Sqrt(test);
    int flooredAndSquared = Convert.ToInt32(Math.Floor(squared));

    // start with 5, make increments of 2, even numbers do not need to be tested
    for (int idx = 3; idx < flooredAndSquared; idx++)
    {
        if (test % idx == 0)
        {
            return false;
        }
    }
    return true;
}

I know the squared part fails me (or I fail), tried Math.Ceiling as well, with about the same results.

Answer 1

Sadly, I haven't tried the algorithmic approaches before. But if you want to implement your approach efficiently, I'd suggest doing some caching. Create an array to store all prime numbers less than a defined threshold, fill this array, and search within/using it.

In the following example, finding whether a number is prime is O(1) in the best case (namely, when the number is less than or equal to maxPrime, which is 821,461 for a 64K buffer), and is somewhat optimized for other cases (by checking mod over only 64K numbers out of the first 820,000 -- about 8%).

(Note: Don't take this answer as the "optimal" approach. It's more of an example on how to optimize your implementation.)

public static class PrimeChecker
{
    private const int BufferSize = 64 * 1024; // 64K * sizeof(int) == 256 KB

    private static int[] primes;
    public static int MaxPrime { get; private set; }

    public static bool IsPrime(int value)
    {
        if (value <= MaxPrime)
        {
            return Array.BinarySearch(primes, value) >= 0;
        }
        else
        {
            return IsPrime(value, primes.Length) && IsLargerPrime(value);
        }
    }

    static PrimeChecker()
    {
        primes = new int[BufferSize];
        primes[0] = 2;
        for (int i = 1, x = 3; i < primes.Length; x += 2)
        {
            if (IsPrime(x, i))
                primes[i++] = x;
        }
        MaxPrime = primes[primes.Length - 1];
    }

    private static bool IsPrime(int value, int primesLength)
    {
        for (int i = 0; i < primesLength; ++i)
        {
            if (value % primes[i] == 0)
                return false;
        }
        return true;
    }

    private static bool IsLargerPrime(int value)
    {
        int max = (int)Math.Sqrt(value);
        for (int i = MaxPrime + 2; i <= max; i += 2)
        {
            if (value % i == 0)
                return false;
        }
        return true;
    }
}

Answer 2

I guess this is your problem:

for (int idx = 3; idx < flooredAndSquared; idx++)

This should be

for (int idx = 3; idx <= flooredAndSquared; idx++)

so you don't get square numbers as primes. Also, you can use "idx += 2" instead of "idx++" because you only have to test odd numbers (as you wrote in the comment directly above...).

Answer 3

I don't know if this is quite what you are looking for but if you are really concerned about speed then you should look into probablistic methods for testing primality rather than using a sieve. Rabin-Miller is a probabilistic primality test used by Mathematica.

Answer 4

I posted a class that uses the sieve or Eratosthenes to calculate prime numbers here:

Is the size of an array constrained by the upper limit of int (2147483647)?

Answer 5

As Mark said, the Miller-Rabin test is actually a very good way to go. An additional reference (with pseudo-code) is the Wikipedia article about it.

It should be noted that while it is probabilistic, by testing just a very small number of cases, you can determine whether a number is prime for numbers in the int (and nearly long) range. See this part of that Wikipedia article, or the Primality Proving reference for more details.

I would also recommend reading this article on modular exponentiation, as otherwise you're going to be dealing with very very large numbers when trying to do the Miller-Rabin test...

Answer 6

You might want to look into Fermat's little theorem.

Here is the pseudo code from the book Algorithms by S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani, where n is the number you are testing for primality.

Pick a positive integer a < n at random
if a^n-1 is equivalent to 1 (mod n)
   return yes
else
   return no

Implementing Fermat's theorem should be faster then the sieve solution. However, there are Carmichael numbers that pass Fermat's test and are NOT prime. There are workarounds for this. I recommend consulting Section 1.3 in the fore mentioned book. It is all about primality testing and might be helpful for you.

Answer 7

Try this...

if (testVal == 2) return true;
if (testVal % 2 == 0) return false;

for (int i = 3; i <= Math.Ceiling(Math.Sqrt(testVal)); i += 2)
{
   if (testVal % i == 0)
       return false;
}

return true;

Ive used this quite a few times.. Not as fast as a sieve.. but it works.

Answer 8

Here is a halfway decent function I wrote to solve one of the Euler:

private static long IsPrime(long input)
{
    if ((input % 2) == 0)
    {
        return 2;
    }
    else if ((input == 1))
    {
        return 1;
    }
    else
    {
        long threshold = (Convert.ToInt64(Math.Sqrt(input)));
        long tryDivide = 3;
        while (tryDivide < threshold)
        {
            if ((input % tryDivide) == 0)
            {
                Console.WriteLine("Found a factor: " + tryDivide);
                return tryDivide;
            }
            tryDivide += 2;
        }
        Console.WriteLine("Found a factor: " + input);
        return -1;
    }
}

Answer 9

I have a fast algorithm for checking prime numbers. You can improve your algorithm if you know that primes are in the form 6k ± 1, with the exception of 2 and 3. So, first you can test if input is divisible by 2 or 3. Then, check all the numbers of form 6k ± 1 ≤ √input.

bool IsPrime(int input)
        {
            //2 and 3 are primes
            if (input == 2 || input == 3)
                return true; 
            else if (input % 2 == 0 || input % 3 == 0)
                return false;     //Is divisible by 2 or 3?
            else
            {
                for (int i = 5; i * i <= input; i += 6)
                {
                    if (input % i == 0 || input % (i + 2) == 0)
                        return false;
                }
                return true;
            }
        }

Answer 10

Try the sieve of eratosthenes -- that should take out getting the root and floating point issues.

As for the floor, you will be better served by taking the ceiling.

Answer 11

private static bool IsPrime(int number) {
    if (number <= 3)
        return true;
    if ((number & 1) == 0)
        return false;
    int x = (int)Math.Sqrt(number) + 1;
    for (int i = 3; i < x; i += 2) {
        if ((number % i) == 0)
            return false;
    }
    return true;
}

I can't get it any faster...

Answer 12

I thought Prime numbers and primality testing was useful and the AKS algorithm sounds interesting even if it isn't particularly practical compared to a probabily based tests.

Answer 13

This works very fast for testing primes (vb.net)

Dim rnd As New Random()
Const one = 1UL

    Function IsPrime(ByVal n As ULong) As Boolean
        If n Mod 3 = 0 OrElse n Mod 5 = 0 OrElse n Mod 7 = 0 OrElse n Mod 11 = 0 OrElse n Mod 13 = 0 OrElse n Mod 17 = 0 OrElse n Mod 19 = 0 OrElse n Mod 23 = 0 Then
           return false
        End If

        Dim s = n - one

        While s Mod 2 = 0
            s >>= one
        End While

        For i = 0 To 10 - 1
            Dim a = CULng(rnd.NextDouble * n + 1)
            Dim temp = s
            Dim m = Numerics.BigInteger.ModPow(a, s, n)

            While temp <> n - one AndAlso m <> one AndAlso m <> n - one
                m = (m * m) Mod n
                temp = temp * 2UL
            End While

            If m <> n - one AndAlso temp Mod 2 = 0 Then
                Return False
            End If
        Next i

        Return True
    End Function

Answer 14

In case anyone else is interested, here's my conversion of Mohammad's procedure above to VBA. I added a check to exclude 1, 0, and negative numbers as they are all defined as not prime.

I have only tested this in Excel VBA:

Function IsPrime(input_num As Long) As Boolean
    Dim i As Long
    If input_num < 2 Then '1, 0, and negative numbers are all defined as not prime.
        IsPrime = False: Exit Function
    ElseIf input_num = 2 Then
        IsPrime = True: Exit Function '2 is a prime
    ElseIf input_num = 3 Then
        IsPrime = True: Exit Function '3 is a prime.
    ElseIf input_num Mod 2 = 0 Then
        IsPrime = False: Exit Function 'divisible by 2, so not a prime.
    ElseIf input_num Mod 3 = 0 Then
        IsPrime = False: Exit Function 'divisible by 3, so not a prime.
    Else
        'from here on, we only need to check for factors where
        '6k ± 1 = square root of input_num:
        i = 5
        Do While i * i <= input_num
            If input_num Mod i = 0 Then
                IsPrime = False: Exit Function
            ElseIf input_num Mod (i + 2) = 0 Then
                IsPrime = False: Exit Function
            End If
            i = i + 6
        Loop
        IsPrime = True
    End If
End Function

Answer 15

Your for loop should look like this:

for (int idx = 3; idx * idx <= test; idx++) { ... }

That way, you avoid floating-point computation. Should run faster and it'll be more accurate. This is why the for statement conditional is simply a boolean expression: it makes things like this possible.