Question

The algorithm I’m using at the moment runs into extremely high numbers very quickly. A step in the algorithm I’m to raises x to the result of the totient function applied to y. The result is that you can run into very large numbers.

Eg. When calculating the multiplicative order of 10 modulo 53:

10^totient(53) == 10^52 == 1 * 10^52 

The following algorithm fares a bit better either in terms of avoiding large numbers, but it still fails where 10^mOrder is greater than the capacity of the data type:

  mOrder = 1   while 10^mOrder % 53 != 1       if mOrder >= i           mOrder = 0;           break       else           mOrder = mOrder + 1 

Answer 1

Using Modular exponentiation, it is possible to calculate (10 ^ mOrder % 53) or in general, any (a ^ b mod c) without getting values much bigger than c. See Wikipedia for details, there’s this sample code, too:

Bignum modpow(Bignum base, Bignum exponent, Bignum modulus) {      Bignum result = 1;      while (exponent > 0) {         if ((exponent & 1) == 1) {             // multiply in this bit's contribution while using modulus to keep result small             result = (result * base) % modulus;         }         // move to the next bit of the exponent, square (and mod) the base accordingly         exponent >>= 1;         base = (base * base) % modulus;     }      return result; }