Recursive power function: approach

Question

I\'m programming for a while now(beginner), and recursive functions are a somewhat abstract concept for me. I would not say I\'m stuck, program works fine, I\'m just wonderi

Answer 1

You want to avoid using pow(), right? So you used power() instead, leading to a recursive call within the argument list. This led to a segmentation fault.

First of all, let us understand the cause of the problem. I did a pen and paper run of the algorithm and the result is pretty interesting. It turns out that for any values of x and n, after a certain number of recursions one always ends up getting power(1, 2). This also means that power(1, 2) also leads to power (1, 2) after a certain number of recursions. Thus, this power() within a power() leads to an infinite recursion and thus the stack overflow.

Now, to your question. Yes, this can be done without using pow() because pow(a, 2) can simply be written as a*a. So, here is a slight modification to your code:

int power(int x, int n)
{
    if (n == 0) return 1;
    if (n % 2 == 0) return power(x, n / 2) * power(x, n / 2);
    else return x * power(x, n - 1);
}

But then, why do it this way? A more efficient way would be as follows.

int power(int x, int n)
{
    if (n == 0) return 1;
    if (n % 2 == 0) return power(x * x, n / 2);
    else return x * power(x * x, n / 2);
}

This reduces the number of recursions needed, making the code more time and space efficient. Hope this helps!

Answer 2

double result = 1;
int count = 1;

public double power(double baseval, double exponent) {
  if (count <= Math.Abs(exponent)){
    count++;
    result *= exponent<0 ?1/baseval:baseval;
    power(baseval, exponent);
  }
  return result;
}

This works with positive, negative, and 0 value

Answer 3

int pow(int a, int n) {
    if(n == 0) return 1;
    if(n == 1) return a;
    int x = pow(a, n/2);
    if(n%2 == 0) {
        return x*x;
    }
    else {
        return a*x*x;
    }
}