Hackerrank Solution: Little Panda Power

Original Problem

Little Panda has a thing for powers and modulus and he likes challenges. His friend Lucy, however, is impractical and challenges Panda to find both positive and negative powers of a number modulo a particular number. We all know that \(A^{-1}\bmod X\) refers to the modular inverse of \(A\) modulo \(X\).

Since Lucy is impractical, she says that \(A^{-n}\bmod X=(A^{-1}\bmod X)^n \bmod X\) for \(n>0\) .

Now she wants Panda to compute \(A^B\bmod X\).

She also thinks that this problem can be very difficult if the constraints aren't given properly. Little Panda is very confused and leaves the problem to the worthy programmers of the world. Help him in finding the solution.

Input Format
The first line contains \(T\), the number of test cases.
Then \(T\) lines follow, each line containing \(A\), \(B\) and \(X\).

Output Format
Output the value of \(A^B\bmod X\).

Constraints
\(1\leq T\leq 1000\)
\(1\leq A\leq 10^6\)
\(-10^6\leq B\leq 10^6\)
\(1\leq X\leq 10^6\)
\(A\) and \(X\) are coprime to each other

Sample Input

3
1 2 3
3 4 2
4 -1 5

Sample Output

1
1
4

Explanation
Case 1: \(1^2\bmod 3=1\bmod 3=1\)
Case 2: \(3^4\bmod 2=81\bmod 2=1\)
Case 3: \(4^{-1}\bmod 5=4\)

Solution

The solution is already implied by the statement. We need to check the sign of the exponent \(B\); if it's positive we can calculate the modular exponentation; if it's negative we need to calculate the modular inverse of \(A\) first. The modular inverse can be calculated quite easily using the extended euclidean algorithm, which finds the parameters \(s, t\) such that \(as+bt=\gcd(a,b)\). Implemented in Ruby, the egcd algorithm looks as follows:

def egcd(a, b)
 return 1, 0 if b == 0
 q, r = a.divmod b
 s, t = egcd(b, r)
 return t, s - q * t
end

Using the egcd, the modular inverse is given by:

def modinv(z, n)
 return mod(egcd(z, n)[0], n)
end

Using a mathematical modulo operator:

def mod(n, m)
 return (n % m + m) % m
end

The last piece to complete the task is an modular exponentation function:

def modpow(b, e, m)
 r = 1
 while e > 0 do
  if e % 2 == 1 then
   r = (r * b) % m
  end
  b = (b * b) % m
  e >>= 1
 end
 return r
end

With all these helper functions, we can finally formulate the solution like this:

gets.to_i.times{
 a, b, x = gets.split.map(&:to_i)
 if b >= 0 then
  puts modpow(a, b, x)
 else
  puts modpow(modinv(a, x), -b, x)
 end
}

Another way to calculate the modular inverse uses Euler's theorem. Given two coprime numbers \(a\) and \(m\), it states

\[a^{\phi(m)}\equiv 1\pmod{m}\]

Dividing both sides by \(a\) reveals

\[a^{\phi(m)-1}\equiv a^{-1}\pmod{m}\]

However, I did not use it, since it requires a prime factorization of \(m\) which limits a possible improvement over the egcd alot.