Remainder of euclidean division algorithm

Question

I use JavaScript with my own BigInteger library, I have a problem with the complexity of the mod function.

// r = big1 - (big2 * (big1/big2))
function mod(big1, big2){
return subs(big1, multiply(big2, divide(big1,big2))); 
}

Answer 1

//r = b1 % b2

a different approach is to operate modulus by parts (say your limit for a number is 9 digits per number and say b2 < 100):

1. Starting from the leftmost digit of b1, construct a number using the first 9 digits and call it N.
2. Calculate N mod b2, giving a result in the range 00 to 99 (remember the example is b2 < 100).
3. Construct a new 9-digit N by concatenating above result (step 2) with the next 7 digits of b1. If there are fewer than 7 digits remaining in b1 but at least one, then construct a new N, which will have less than 9 digits, from the above result (step 2) followed by the remaining digits of b1
4. Repeat steps 2–3 until all the digits of b1 have been processed

final outcome is b1%b2. If also b2 is very large this will not work, but if only b1 is very large you can treat it is a string for the purpose of the algorithm.

coding it should be easy with iteration since you can know the number of iterations beforehand, don't even need recursion.