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102 lines
2.2 KiB
Python
102 lines
2.2 KiB
Python
"""
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Binary Multiplication
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This is a method to find a*b in a time complexity of O(log b)
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This is one of the most commonly used methods of finding result of multiplication.
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Also useful in cases where solution to (a*b)%c is required,
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where a,b,c can be numbers over the computers calculation limits.
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Done using iteration, can also be done using recursion
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Let's say you need to calculate a * b
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RULE 1 : a * b = (a+a) * (b/2) ---- example : 4 * 4 = (4+4) * (4/2) = 8 * 2
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RULE 2 : IF b is odd, then ---- a * b = a + (a * (b - 1)), where (b - 1) is even.
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Once b is even, repeat the process to get a * b
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Repeat the process until b = 1 or b = 0, because a*1 = a and a*0 = 0
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As far as the modulo is concerned,
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the fact : (a+b) % c = ((a%c) + (b%c)) % c
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Now apply RULE 1 or 2, whichever is required.
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@author chinmoy159
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"""
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def binary_multiply(a: int, b: int) -> int:
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"""
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Multiply 'a' and 'b' using bitwise multiplication.
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Parameters:
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a (int): The first number.
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b (int): The second number.
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Returns:
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int: a * b
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Examples:
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>>> binary_multiply(2, 3)
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6
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>>> binary_multiply(5, 0)
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0
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>>> binary_multiply(3, 4)
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12
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>>> binary_multiply(10, 5)
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50
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>>> binary_multiply(0, 5)
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0
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>>> binary_multiply(2, 1)
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2
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>>> binary_multiply(1, 10)
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10
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"""
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res = 0
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while b > 0:
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if b & 1:
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res += a
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a += a
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b >>= 1
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return res
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def binary_mod_multiply(a: int, b: int, modulus: int) -> int:
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"""
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Calculate (a * b) % c using binary multiplication and modular arithmetic.
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Parameters:
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a (int): The first number.
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b (int): The second number.
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modulus (int): The modulus.
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Returns:
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int: (a * b) % modulus.
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Examples:
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>>> binary_mod_multiply(2, 3, 5)
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1
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>>> binary_mod_multiply(5, 0, 7)
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0
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>>> binary_mod_multiply(3, 4, 6)
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0
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>>> binary_mod_multiply(10, 5, 13)
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11
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>>> binary_mod_multiply(2, 1, 5)
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2
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>>> binary_mod_multiply(1, 10, 3)
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1
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"""
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res = 0
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while b > 0:
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if b & 1:
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res = ((res % modulus) + (a % modulus)) % modulus
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a += a
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b >>= 1
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return res
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if __name__ == "__main__":
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import doctest
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doctest.testmod()
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