"""
Binary Multiplication
This is a method to find a*b in a time complexity of O(log b)
This is one of the most commonly used methods of finding result of multiplication.
Also useful in cases where solution to (a*b)%c is required,
where a,b,c can be numbers over the computers calculation limits.
Done using iteration, can also be done using recursion

Let's say you need to calculate a * b
RULE 1 : a * b = (a+a) * (b/2) ---- example : 4 * 4 = (4+4) * (4/2) = 8 * 2
RULE 2 : IF b is odd, then ---- a * b = a + (a * (b - 1)), where (b - 1) is even.
Once b is even, repeat the process to get a * b
Repeat the process until b = 1 or b = 0, because a*1 = a and a*0 = 0

As far as the modulo is concerned,
the fact : (a+b) % c = ((a%c) + (b%c)) % c
Now apply RULE 1 or 2, whichever is required.

@author chinmoy159
"""


def binary_multiply(a: int, b: int) -> int:
 """
 Multiply 'a' and 'b' using bitwise multiplication.

 Parameters:
 a (int): The first number.
 b (int): The second number.

 Returns:
 int: a * b

 Examples:
 >>> binary_multiply(2, 3)
 6
 >>> binary_multiply(5, 0)
 0
 >>> binary_multiply(3, 4)
 12
 >>> binary_multiply(10, 5)
 50
 >>> binary_multiply(0, 5)
 0
 >>> binary_multiply(2, 1)
 2
 >>> binary_multiply(1, 10)
 10
 """
 res = 0
 while b > 0:
 if b & 1:
 res += a

 a += a
 b >>= 1

 return res


def binary_mod_multiply(a: int, b: int, modulus: int) -> int:
 """
 Calculate (a * b) % c using binary multiplication and modular arithmetic.

 Parameters:
 a (int): The first number.
 b (int): The second number.
 modulus (int): The modulus.

 Returns:
 int: (a * b) % modulus.

 Examples:
 >>> binary_mod_multiply(2, 3, 5)
 1
 >>> binary_mod_multiply(5, 0, 7)
 0
 >>> binary_mod_multiply(3, 4, 6)
 0
 >>> binary_mod_multiply(10, 5, 13)
 11
 >>> binary_mod_multiply(2, 1, 5)
 2
 >>> binary_mod_multiply(1, 10, 3)
 1
 """
 res = 0
 while b > 0:
 if b & 1:
 res = ((res % modulus) + (a % modulus)) % modulus

 a += a
 b >>= 1

 return res


if __name__ == "__main__":
 import doctest

 doctest.testmod()