Fast modular exponentiation

How can we calculate A^B mod C quickly if B is a power of 2 ?

Using modular multiplication rules:

i.e. A^2 mod C = (A * A) mod C = ((A mod C) * (A mod C)) mod C

We can use this to calculate 7^256 mod 13 quickly

7^1 mod 13 = 7
7^2 mod 13 = (7^1 *7^1) mod 13 = (7^1 mod 13 * 7^1 mod 13) mod 13

We can substitute our previous result for 7^1 mod 13 into this equation.

7^2 mod 13 = (7 *7) mod 13 = 49 mod 13 = 10
7^2 mod 13 = 10

7^4 mod 13 = (7^2 *7^2) mod 13 = (7^2 mod 13 * 7^2 mod 13) mod 13

We can substitute our previous result for 7^2 mod 13 into this equation.

7^4 mod 13 = (10 * 10) mod 13 = 100 mod 13 = 9
7^4 mod 13 = 9

7^8 mod 13 = (7^4 * 7^4) mod 13 = (7^4 mod 13 * 7^4 mod 13) mod 13

We can substitute our previous result for 7^4 mod 13 into this equation.

7^8 mod 13 = (9 * 9) mod 13 = 81 mod 13 = 3
7^8 mod 13 = 3

We continue in this manner, substituting previous results into our equations.

...after 5 iterations we hit:

7^256 mod 13 = (7^128 * 7^128) mod 13 = (7^128 mod 13 * 7^128 mod 13) mod 13
7^256 mod 13 = (3 * 3) mod 13 = 9 mod 13 = 9
7^256 mod 13 = 9

This has given us a method to calculate A^B mod C quickly provided that B is a power of 2.

However, we also need a method for fast modular exponentiation when B is not a power of 2.

How can we calculate A^B mod C quickly for any B ?


Step 1: Divide B into powers of 2 by writing it in binary


Start at the rightmost digit, let k=0 and for each digit:

If the digit is 1, we need a part for 2^k, otherwise we do not
Add 1 to k, and move left to the next digit

Step 2: Calculate mod C of the powers of two ≤ B

5^1 mod 19 = 5

5^2 mod 19 = (5^1 * 5^1) mod 19 = (5^1 mod 19 * 5^1 mod 19) mod 19
5^2 mod 19 = (5 * 5) mod 19 = 25 mod 19
5^2 mod 19 = 6

5^4 mod 19 = (5^2 * 5^2) mod 19 = (5^2 mod 19 * 5^2 mod 19) mod 19
5^4 mod 19 = (6 * 6) mod 19 = 36 mod 19
5^4 mod 19 = 17

5^8 mod 19 = (5^4 * 5^4) mod 19 = (5^4 mod 19 * 5^4 mod 19) mod 19
5^8 mod 19 = (17 * 17) mod 19 = 289 mod 19
5^8 mod 19 = 4

5^16 mod 19 = (5^8 * 5^8) mod 19 = (5^8 mod 19 * 5^8 mod 19) mod 19
5^16 mod 19 = (4 * 4) mod 19 = 16 mod 19
5^16 mod 19 = 16

5^32 mod 19 = (5^16 * 5^16) mod 19 = (5^16 mod 19 * 5^16 mod 19) mod 19
5^32 mod 19 = (16 * 16) mod 19 = 256 mod 19
5^32 mod 19 = 9

5^64 mod 19 = (5^32 * 5^32) mod 19 = (5^32 mod 19 * 5^32 mod 19) mod 19
5^64 mod 19 = (9 * 9) mod 19 = 81 mod 19
5^64 mod 19 = 5

Step 3: Use modular multiplication properties to combine the calculated mod C values

5^117 mod 19 = ( 5^1 * 5^4 * 5^16 * 5^32 * 5^64) mod 19
5^117 mod 19 = ( 5^1 mod 19 * 5^4 mod 19 * 5^16 mod 19 * 5^32 mod 19 * 5^64 mod 19) mod 19
5^117 mod 19 = ( 5 * 17 * 16 * 9 * 5 ) mod 19
5^117 mod 19 = 61200 mod 19 = 1
5^117 mod 19 = 1

https://www.khanacademy.org/computing/computer-science/cryptography/modarithmetic/a/fast-modular-exponentiation