Binary Exponentiation

other/binary_exponentiation.py

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+"""
+* 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
+
+* @author chinmoy159
+* @version 1.0 dated 10/08/2017
+"""
+
+
+def b_expo(a, b):
+    res = 1
+    while b > 0:
+        if b&1:
+            res *= a
+
+        a *= a
+        b >>= 1
+
+    return res
+
+
+def b_expo_mod(a, b, c):
+    res = 1
+    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) = 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.
+"""

other/binary_exponentiation_2.py

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+"""
+* 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, b):
+    res = 0
+    while b > 0:
+        if b&1:
+            res += a
+
+        a += a
+        b >>= 1
+
+    return res
+
+
+def b_expo_mod(a, b, c):
+    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.
+"""