Question

Programming

implement and test 3 different algorithms for computing modular exponentiation:

1. Implement a function powerDirect(a, b, m) that takes three integers a, b, and m > 1. The function can compute the modular exponentiation ab modulo m.

2. Implement a function powerMemoryEfficient(a, b, m) that takes three integers a, b, and m > 1. The function can compute the modular exponentiation ab modulo m. This method requires more operations than the first method, because the required memory is substantially less, however, operations take less time than before. The end result is that the algorithm is faster.

3. Implement a function powerRightLeftBinary(a, b, m) that takes three integers a, b, and m > 1. The function can compute the modular exponentiation ab modulo m. This method drastically reduces the number of operations to perform modular exponentiation, while keeping the same memory footprint as in method 2. It is a combination of the method 2 and a more general principle called exponentiation by squaring (also known as binary exponentiation).

Required Information The following Wikipedia page includes required information about “Direct”, “Memoryefficient” and “Right-to-Left binary” algorithms:

https://en.wikipedia.org/wiki/Modular_exponentiation

Answer 1

1.powerDirect(a,b,m){

res=1

if(m>1){

while(y>0){

a=a%m

if (y%2==1)

res=(res*x)%m

b=b-1

x=(x*x)%p

}

}

return res

}

2.powerMemoryEfficient(a,b,m)

{

    if (a== 0)

        return 0;

    if (b == 0)

        return 1;

  

    // If B is even

    long y;

    if (a % 2 == 0) {

        y = powerMemoryEfficient(a, b / 2, c);

        y = (y * y) % c;

    }

  

    // If B is odd

    else {

        y = A % C;

        y = (y * exponentMod(a, b - 1, c) % c) % c;

    }

  

    return (int)((y + c) % c);

}

3.powerRightLeftBinary(a, b, m){

    if m = 1 then return 0
    Assert :: (m - 1) * (m - 1) 
    result = 1
    a =a %m
    while b > 0
        if (e%2 == 1):
           result := (result * a) %m
        b = b>> 1
        a= (a * a)%m
    return result

}