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Co-authored-by: pre-commit-ci[bot] <66853113+pre-commit-ci[bot]@users.noreply.github.com>
Co-authored-by: Tianyi Zheng <tianyizheng02@gmail.com>

maths/binary_exponentiation_2.py

@@ -1,50 +0,0 @@
-"""
-* Binary Exponentiation with 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
-
-* @author chinmoy159
-* @version 1.0 dated 10/08/2017
-"""
-
-
-def b_expo(a: int, b: int) -> int:
-    res = 0
-    while b > 0:
-        if b & 1:
-            res += a
-
-        a += a
-        b >>= 1
-
-    return res
-
-
-def b_expo_mod(a: int, b: int, c: int) -> int:
-    res = 0
-    while b > 0:
-        if b & 1:
-            res = ((res % c) + (a % c)) % c
-
-        a += a
-        b >>= 1
-
-    return res
-
-
-"""
-* Wondering how this method works !
-* It's pretty simple.
-* 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 till 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.
-"""

maths/binary_multiplication.py

@@ -0,0 +1,101 @@
+"""
+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()