Find the square of a number using recursion with one parameter function [closed]

Answer 1

In order to implement the square operation as a recursive function, you need first to express the operation in terms of itself:

(n-1)² = n² - 2n + 1 --> n² = (n-1)² + 2n - 1

Then, in order to avoid the operator *:

2n = n + n

Therefore, n² = (n-1)² + n + n - 1

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With that in mind, you can easily implement square() as a recursive function that does not use the operator *:

unsigned int square(unsigned int n) {
   if (n == 0)
      return 0; // base case

   return square(n-1) + n + n - 1; // recursive case
}

Or just with a single statement using the ternary operator:

unsigned int square(unsigned int n) {
   return n? square(n-1) + n + n - 1: 0;
}

n equal to zero is the base case (i.e., when the recursion stops). It returns zero for this case, since 0² is zero.