Modular Exponentiation in Python

Given three numbers x, y and p, compute (x^y) % p

Examples:

Input:  x = 2, y = 3, p = 5
Output: 3
Explanation: 2^3 % 5 = 8 % 5 = 3.

Input:  x = 2, y = 5, p = 13
Output: 6
Explanation: 2^5 % 13 = 32 % 13 = 6.

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

The problem with above solutions is, overflow may occur for large value of n or x. Therefore, power is generally evaluated under modulo of a large number.

Naive multiplication is O(n) with a very low constant factor with %m.
Pow function calculates in O(log n) time in python but it takes a lot of time when numbers are large enough if you first calculate the value of x^y and then mod it with p to get (x^y) % p evaluated.

# Simple python code that first calls pow()  
# then applies % operator. 
a = 2
b = 100
p = (int)(1e9+7) 
  
# pow function used with % 
d = pow(a, b) % p 
print d 

Output:

976371285

While computing with large numbers modulo, the (%) operator takes a lot of time, so a Fast Modular Exponentiation is used. Python has pow(x, e, m) to get the modulo calculated which takes a lot less time. [Please refer Python Docs for details]

# Fast python code that first calls pow()  
# then applies % operator 
a = 2
b = 100
p = (int)(1e9+7) 
  
# Using direct fast method to compute  
# (a ^ b) % p. 
d = pow(a, b, p) 
print d 

Output:

976371285

The fast modular exponentiation algorithm has been explained more briefly in link

My Personal Notes

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