Back when I started messing around with code, I was told to try out Project Euler to improve my coding skills (something I suggest everyone do). I was able to progress fairly well along but I always struggled with the questions involving prime numbers because I could never get my code to run efficiently on them.

I looked for a solution and repeatedly only found mention to the Sieve of Eratosthenes, which is fairly intuitive and appeared to be the go-to primality test. However, in my early days of programming, I still struggled to implement the algorithm so I ultimately pushed those prime-based questions off to the side.

Years later, while reading Brian Christian’s and Tom Griffiths’ Algorithms to Live By, I happened across the Miller-Rabin Primality Test, an alternative to the Sieve of Eratosthenes. The algorithm works as follows: say you wanted to test if \(n\) (an odd integer) is a prime number. To do so, we can first find \(r\) and \(s\) such that \(n = 2^r s + 1\). Next, we can pick a random integer \(a\) such that \(1 \leq a \leq n - 1\). With \(a, r, s\) all defined, we next test if \(a^s = 1 \textrm{ mod } n\) or \(a^{2js} = -1 \textrm{ mod } n\) for some \(0 \leq j \leq r - 1\). If either of these equalities are true, then we can say that \(n\) passes the test and any prime number will pass the test. It is fairly trivial to test this for pretty much any \(n\). However, the trick with this algorithm is that non-prime numbers will also pass this test (return a false-positive) \(\frac{1}{4}\) of the time.

To correct for this, we can try the test again with a new randomly-selected \(a\). If \(n\) again passes the test, we know that there is a \(\frac{1}{4} \cdot \frac{1}{4} = \frac{1}{16}\) chance that \(n\) is not actually a prime. Therefore, if we repeated the test \(N\) times successfully, we can say that there is a \(\frac{1}{4^N}\) chance that \(n\) is not actually a prime number. Using this method, we can determine if a particular number is prime with fairly high certainty after only a few iterations. For example, if we repeat the test \(N = 5\) times, we can declare \(n\) to be prime with 99.9% certainty. If \(n\) passes after \(N = 10\) iterations, we know with 99.9999% certainty.

Using this, I took another crack at those pesky prime-based Project Euler problems and what do you know, it worked like a charm! While there is a very, very, very slim chance that this still returns a false-positive (I’m using \(N = 8\) so a 0.00153% chance), given the non-critical context of the application, I’ll definitely take the speed increase over the non-100% certainty.

For anyone interested, I’ve included a sample implementation in Python of the algorithm below.

# takes in a number n and a number of iterations iters
# to test if n is prime with (1 - (1 / 4^iters)) certainty
def miller_rabin(n, iters):
    # find r and s
    # such that: n = (2^r)s + 1
    s = n - 1
    r = 0
    while s % 2 == 0:
        s = s / 2
        r += 1

    # test n multiple times
    for i in range(iters):
        a = random.randint(1, n - 1)
        for j in range(r):
            x = a**(s * 2**j) % n
            if ((j == 0 and x == 1) or (x == n - 1)):
            elif (j == s - 1):
                return False
    return True