How to compute this modulus when there is an integer overflow

Question

(10^{17}-1)*(10^{17}-1) mod 10^{18}

I am solving a programming problem and I hold my integers in 64 bit long long integers. Above is a particular case I am unable to solve. (ab)mod m = (a mod m)(b mod m) mod m, doesn't hold here as (a mod m)(b mod m) would still overflow a 64 bit integer. How do I solve this? I took 17th power only as an example. The problem holds even for all the integers in the range (10^{10}, 10^{18}-1).

Edit: I am using C++ for solving this problem. This problem can be solved without using a library for handling big integers.

Looks like you might need to use a library: stackoverflow.com/questions/117429/handling-large-numbers-in-c unless you want the experience of writing one yourself. — sircodesalot
– sircodesalot, Commented Jan 4, 2014 at 5:06
Possible duplicate of stackoverflow.com/questions/14857702/… (which has an accepted answer). — halfbit
– halfbit, Commented Jan 4, 2014 at 5:26
You said 17 in the exponent is just an example. Do you need a solution for this particular structure of the expression or in general for large integer mod operations? — user3072164
– user3072164, Commented Jan 4, 2014 at 5:29

B-Con · Accepted Answer · 2014-01-04 08:01:57Z

You can use the identity you quoted, you just need another similar identity: (a+b) mod m = (a mod m) + (b mod m).

The goal is to multiply x*y mod m without any intermediate values exceeding the overflow limit (in this case 2^64), where x is starting less than m (if it isn't, reduce it mod m), y is possibly larger than m, and x*y can potentially overflow. We can do this if m is less than half of the overflow limit.

A solution is simple: Just perform basic multiplication bit-by-bit for x*y and do every step modulo m.

Start with x and y less than m (if either isn't, reduce it first). Write y in the form a_0 * 2^0 + a_1 * 2^1 + a_2 * 2^2 + ... , where a_n is either 0 or 1 (indicating the term is present or not). (Aka, write y in binary form.) Now we have:

x * (a_0 * 2^0 + a_1 * 2^1 + a_2 * 2^2 + ...) mod m

Distribute x over each of the terms of y:

(x * a_0 * 2^0) + (x * a_1 * 2^1) + (x * a_2 * 2^2) + ... mod m

Then use the original multiplication identity: For each term above, multiply x by 2 mod m until you reach the desired power of 2 for that term. (Since x < m and 2 * m < 2^64, then 2 * x < 2^64, so we can multiply by 2 without overflowing.) When you are done, add the result for each term mod m (you can keep a running sum as you go).

None of those operations will exceed 2^64 and thus will not overflow. This will work for any value of m less than 2^64 / 2 = 2^63 and any integers x and y less than m.

This is not necessarily the fastest way to do it, feel free to find something more efficient. For starters, the smaller m is compared to the overflow limit, the bigger the radix for the terms we can rewrite y as.

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