CODEWITHSUNDEEP
c++
Akash21
Akash21

Reputation: 39

My Square Root Function does not give accurate results for some numbers

I need to write a function that squares numbers for a course assignment. It calculates the square roots of numbers like 16 and 25 but does not accurately calculate the square root of 9.

Below is the code 

 double mysqrt (double x)
{
   double low, high, mid;

This if statement decides creates a range to determine the square root.

   if (x >= 0) {

      low = 0;
      high = x;

   }
   else {

     low = x;
     high = 1;

   }

This statement calculates the middle value

   mid = (high + low)/2.0;

The while loop is used to determine the square root. 

   while (abs(mid*mid - x) > 0.0001)
   {

      if (mid * mid > x)
          high = mid;

      else
          low = mid;

      mid = (high/2.0) + (low/2.0);
   }

   return mid;
 }

Upvotes: 0

Views: 387

Answers (1)

Christopher Oicles
Christopher Oicles

Reputation: 3107

Normally I don't just do someone's homework for them, but i wanted to challenge the idea that convergence should always be detected by comparing the error against a fixed epsilon value.

Keep in mind that the double data type, can only represent values taken from a finite range, so comparing for equivalence between two results can be used to detect convergence, where doing so with arbitrary precision could lead to infinite iteration

Also, a set of possible floating-point values is distributed so that the difference between one value and the next greater is smaller for small values than large values, so if you use an epsilon, it usually isn't appropriate to keep it fixed at some arbitrarily "small" value.

The code below performs the same dumb binary search, but the exit condition tests for a re-visiting of a previous iteration's result, which means that further iteration would just repeatedly cover the same search states forever and that the error is close to the minimum possible.

In this way, epsilon is automatically determined by the double data type itself, along with the input. If you replaced double with some kind of arbitrary-precision class, then you would indeed need to supply an epsilon, otherwise irrational roots might loop until some sort of failure, like an out-of-memory condition.

#include <iostream>
#include <iomanip>

double mysqrt(double x) {
    double low, high;
    if(x < 1) {
        if(x <= 0) return 0;
        low = x;
        high = 1;
    } else {
        low = 1;
        high = x;
    }
    for(;;) {
        const double mid = (low + high) / 2;
        if(high == mid || low == mid) return mid;
        if(mid * mid > x) {
            high = mid;
        } else {
            low = mid;
        }
    }
}

int main() {
    std::cout <<  std::setprecision(14) << mysqrt(3) << '\n';
}

Upvotes: 1

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