Question

I need to work out a very large power modulo (2^32), i.e. I want the result of:

y = (p^n) mod (2^32)

p is a prime number
n is a large integer

Is there a trick to doing this efficiently in Java?

Or am I stuck with doing it in a loop with n iterations?

Answer 1

The simple way to mod 2^32 is to use & 0xFFFFFFFFL. Also, there happens to be a type which naturally keeps the lowest 32-bit called int 😉 If you use that you don’t even need to perform the & until you have the result (so the answer is unsigned) For this reason you only need to keep the last 32 bit of the answer. To speed up the ^n you can calculate the square, it’s square and it’s square etc, e.g if n is 0b11111 then you need to multiply p^16 * p^8 * p^4 * p^2 * p.

In short, you can use plain int as you only need 32-bit of accuracy and values with a cost of O(ln n) where n is the power.

int prime = 2106945901;
for (int i = 0; i < 10; i++) {
    long start = System.nanoTime();
    long answer1 = BigInteger.valueOf(prime)
                             .modPow(
                                 BigInteger.valueOf(prime), 
                                 BigInteger.valueOf(2).pow(32)).longValue();

    long mid = System.nanoTime();
    int answer2 = 1;
    int p = prime;
    for (int n = prime; n > 0; n >>>= 1) {
        if ((n & 1) != 0)
            answer2 *= p;
        p *= p;
    }
    long end = System.nanoTime();
    System.out.printf("True answer %,d took %.3f ms, quick answer %,d took %.3f ms%n",
            answer1, (mid - start) / 1e6, answer2 & 0xFFFFFFFFL, (end - mid) / 1e6);
}

prints finally

True answer 4,169,684,317 took 0.233 ms, quick answer 4,169,684,317 took 0.002 ms