Question

I have a homework that include a new operation (a [n] b), given:

• a [1] b = a^b
• a [n] 1 = a
• 2 [2] 3 = 2 [2-1] (2 [2-1] 2) 2 repeated 3 times = 2 [1] (2 [1] 2) = 2²² = 16
• 2 [2] 2 = 2 [2-1] 2 2 repeated 2 times = 2 [1] 2 = (2²) = 4
  
• 4 [3] 3 = 4 [3-1] (4 [3-1] 4) = 4 [2] (4 [2] 4) 4 repeated 3 times = 4 [2] (4 [1] (4 [1] ( 4 [1] 4))) =
  = 4 [2] 4 ⁴⁴⁴

I don’t need a solution, I just need advice so that I can solve it myself.

Answer 1

What is being said here can be rephrased as follows.

For any positive integers a, b and n define

    a [n] b = a [n-1] ( a [n-1] ( ... a ) ) taken b times

In a C-like language

int myoperator (a, n, b) {
    int x, i;
    x = a;
    if (n == 1){
       x = pow(a,b);
    } else {
       for(i = 1; i < b; i++){
           x = myoperator (a, [n-1], x);
       }
    return x;
}

Note that the values will grow quickly and get out of the range of machine integers very soon.

Note also that a[n]b can be defined as

 a [n] b = a [n-1] ( a [n-1] (b-1) ).

using this definition the for loop above can be eliminated.

int myoperator (a, n, b) 
int a,n,b;
{
    int x;
    x = a;
    if (n == 1) 
        x = pow(a,b);
    else if (b == 1) 
                x = a;
         else {
              assert(b>1 && n>1);
              x = myoperator (a, n-1, myoperator(a,n-1,b-1));
         }
    return x;
}