Power modulo golf.

Question

Write a function f(a,b,c) that calculates a^b (mod c) in less than 10 seconds.

• a, b < 10^100
• c < 10000.

examples:

f(10^50, 10^50, 54123) = 46555
f(5^99, 10^99, 777) = 1

Answer 1

dc 1 or 5 chars

In GNU dc, there is the operator |, doing exactly this. Quoted from the manual:

| Pops three values and computes a modular exponentiation. The first value popped is used as the reduction modulus; this value must be a non-zero number, and should be an integer. The second popped is used as the exponent; this value must be a non-negative number, and any fractional part of this exponent will be ignored. The third value popped is the base which gets exponentiated, which should be an integer. For small integers this is like the sequence Sm^Lm%, but, unlike ^, this command will work with arbitrarily large exponents.

You may assign it to "function" as following:

[|]sf

(also 5 chars...) This assigns | to f. You may call it like lfx.

Answer 2

Python - 5 chars

f=pow

Test

>>> f(10**50, 10**50, 54123)
46555L
>>> f(5**99, 10**99, 777)
1L

Answer 3

Haskell, 65

Takes the liberty of taking the modulus only on set bits of binary b, but given the range constraints that won't be a problem.

f a b c|b==0=1|odd b=mod(a*f a(b-1)c)c|0<1=f(mod(a^2)c)(div b 2)c

Answer 4

Java 56 chars

My humble attempt with Java.

Object a(BigInteger...a){return a[0].modPow(a[1],a[2]);}

How to use :

a(new BigInteger("95"), new BigInteger("56"), new BigInteger("67"));

Answer 5

Haskell (54)

f a=g a a
g d a b c|b>0=g(mod(d*a)c)a(b-1)c|True=mod d c

This is my algorithm:

1. let d = a
2. while b > 0:
   • d <- (d*a)%c
   • b --
3. Return d%b

Answer 6

Python 80 Characters

def f(a,b,c):
 x=1;y=a
 while b:
  if b%2:x=(x*y)%c
  y=(y*y)%c;b/=2
 return x%c