This was my code:

# include <stdio.h>
float absolute(float x){
  return x>0? x: (-1)*x;
}
float bisect(float a, float b){
  float k = (a b)*0.5;
   return k;
}
//For n>1
float find(float n, float precision){
  float l=1,m=n;
  float sqrt, k=bisect(l,m);
while(absolute(k*k-n)>precision){
  if(k*k-n>0){
    m=k;
    l=l;
     k=bisect(l,m);
  }
  else if(k*k-n<0)
  {l=k;
  m=m;
    k=bisect(k,m);
  }
 }
  return k;
}
int main(){
  float n, precision;
  scanf("%f %f", &n, &precision);
  printf("%f", find(n, precision));
  return 0;
}

Well, it gave me the square root alright, but not up to the precision I specified. For instance, the square root of 3 was given to be 1.734375 and not 1.732050, which is the correct value, in spite of setting the precision equal to 0.001. I spent nearly an hour trying to debug this and now I'm done. Please help me identify where the code went wrong. Thanks in advance!

CodePudding user response：

You use precision to compare k * k and n, while you want the precision for k and not for k * k.

Your while condition should be instead while (m - l > precision) { .... That is the correct way to make sure that the difference between k and the real value is less than precision

CodePudding user response：

You get 1.734375 if you ask for precision 0.01, not 0.001 as you said in your question. Note that your precision is the absolute error you accept in k*k-n, and not the absolute error you accept in sqrt(n) - k. Perhaps that's the source of your confusion?

If you want to bound the absolute error in the result, you can check that if k*k < n, then (k precision)*(k precision) >= n, or if k*k>n, then (k-precision)*(k-precision) <= n.

Or alternatively, you can check that abs(n - k^2 - p^2/4) < pk, where p is the precision. (This is the same as checking that (k-p/2)^2 < n < (k p/2)^2).