Concatenate algorithms

maths/binary_exp_mod.py

@@ -17,6 +17,74 @@ def bin_exp_mod(a, n, b):
     return (r * r) % b
 
 
+def binary_exponentiation_mod_multiplication(a, b, c):
+    """
+    * 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
+
+    * 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.
+
+    * @author chinmoy159
+    * @version 1.0 dated 10/08/2017
+    """
+
+    res = 0
+    while b > 0:
+        if b & 1:
+            res = ((res % c) + (a % c)) % c
+
+        a += a
+        b >>= 1
+
+    return res
+
+
+def binary_exponentiation_mod_powers(a, b, c):
+    """
+    * Binary Exponentiation for Powers
+    * 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 powers.
+    * 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) = 16 ^ 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 = 1
+    *
+    * 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
+    * @version 1.0 dated 10/08/2017
+    """
+
+    res = 1
+    while b > 0:
+        if b & 1:
+            res = ((res % c) * (a % c)) % c
+
+        a *= a
+        b >>= 1
+
+    return res
+
+
 if __name__ == "__main__":
     try:
         BASE = int(input("Enter Base : ").strip())

maths/binary_exponentiation.py

@@ -1,20 +1,78 @@
-"""Binary Exponentiation."""
+def binary_exponentiation_multiplication(a, b):
+    """
+    * 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 : Junth Basnet
-# Time Complexity : O(logn)
+    * 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
 
+    * @author chinmoy159
+    * @version 1.0 dated 10/08/2017
+    """
 
-def binary_exponentiation(a, n):
+    res = 0
+    while b > 0:
+        if b & 1:
+            res += a
 
-    if n == 0:
+        a += a
+        b >>= 1
+
+    return res
+
+
+def binary_exponentiation_powers(a, b):
+    """
+    * Binary Exponentiation for Powers
+    * 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 powers.
+    * 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) = 16 ^ 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 = 1
+
+    * @author chinmoy159
+    * @version 1.0 dated 10/08/2017
+    """
+
+    res = 1
+    while b > 0:
+        if b & 1:
+            res *= a
+
+        a *= a
+        b >>= 1
+
+    return res
+
+
+def binary_exponentiation_recursion(a, b):
+    """Binary Exponentiation with recursion.
+
+    * Time Complexity : O(logn)
+    * @author : Junth Basnet
+    """
+
+    if b == 0:
         return 1
 
-    elif n % 2 == 1:
-        return binary_exponentiation(a, n - 1) * a
+    elif b % 2 == 1:
+        return binary_exponentiation_recursion(a, b - 1) * a
 
     else:
-        b = binary_exponentiation(a, n / 2)
-        return b * b
+        return binary_exponentiation_recursion(a, b / 2) ** 2
 
 
 if __name__ == "__main__":
@@ -24,5 +82,5 @@ def binary_exponentiation(a, n):
     except ValueError:
         print("Invalid literal for integer")
 
-    RESULT = binary_exponentiation(BASE, POWER)
+    RESULT = binary_exponentiation_recursion(BASE, POWER)
     print(f"{BASE}^({POWER}) : {RESULT}")

maths/double_factorial_recursive.py

@@ -1,4 +1,33 @@
-def double_factorial(n: int) -> int:
+def double_factorial(num: int) -> int:
+    """
+    Compute double factorial using iterative method.
+
+    To learn about the theory behind this algorithm:
+    https://en.wikipedia.org/wiki/Double_factorial
+
+    >>> import math
+    >>> all(double_factorial(i) == math.prod(range(i, 0, -2)) for i in range(20))
+    True
+    >>> double_factorial(0.1)
+    Traceback (most recent call last):
+        ...
+    ValueError: double_factorial() only accepts integral values
+    >>> double_factorial(-1)
+    Traceback (most recent call last):
+        ...
+    ValueError: double_factorial() not defined for negative values
+    """
+    if not isinstance(num, int):
+        raise ValueError("double_factorial() only accepts integral values")
+    if num < 0:
+        raise ValueError("double_factorial() not defined for negative values")
+    value = 1
+    for i in range(num, 0, -2):
+        value *= i
+    return value
+
+
+def double_factorial_recursive(n: int) -> int:
     """
     Compute double factorial using recursive method.
     Recursion can be costly for large numbers.
@@ -22,7 +51,7 @@ def double_factorial(n: int) -> int:
         raise ValueError("double_factorial() only accepts integral values")
     if n < 0:
         raise ValueError("double_factorial() not defined for negative values")
-    return 1 if n <= 1 else n * double_factorial(n - 2)
+    return 1 if n <= 1 else n * double_factorial_recursive(n - 2)
 
 
 if __name__ == "__main__":