What does the following algorithm approximate?
(Assume m > 1, $$ \in > 0$$)

x = m;
y = 1;
while(x - y > ε){
 x = (x + y) / 2;
 y = m/x;
}
print(x);

The time complexity of the following C function is (assume n > 0)

int recursive(int n){
 if(n == 1){
   return (1);
 }
 return (recursive(n - 1) + recursive(n - 1));
}

The cube root of a natural number n is defined as the largest natural number m such that $${m^3} \le n$$. The complexity of computing the cube root of n (n is represented in binary notation) is

Consider the following algorithm for searching for a given number x in an unsorted array A[1..n] having n distinct values:
1. Choose an i uniformly at random fro 1..n;
2. If A[i]=x then stop else Goto 1;
Assuming that x is present A, what is the expected number of comparisons made by the algorithm before it terminates?