Thread: please help me with logic

I was trying to solve the following problem for a very long time but not getting the logic of how to solve the problem.

Code:

int binsearch(int x, int v[], int n)
{
int low, high, mid;
low = 0;
high = n - 1;
while (low <= high) 
{
mid = (low+high)/2;
if (x < v[mid])
high = mid + 1;
else if (x > v[mid])
low = mid + 1;
else /* found match */
return mid;
}
return -1; /* no match */
}

The binary search makes two tests inside the loop, when one would suffice (at the price of more tests outside.) Write a version with only one test inside the loop and measure the difference in run-time? Please suggest me the logic.

To my best understanding, you could drop the second comparison inside the loop, and test for equality against v[mid] outside the loop.
This should take you from 3log(n) comparisons to 2log(n)+1.