Need programming logic

Mayur Gujrathi

15y1.8k.NET

I need to do program for calculating square root of number programatically not using any function

4 Replies

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Vulpes15y ago
Accepted answer
Well, it's not very efficient but here's a way of doing it without using predefiined functions:

using System;

class SquareRoot
{
   static void Main()
   {
   Console.WriteLine("Enter a negative number to quit");

   while (true)
   {
   Console.Write("\nInput a number : ");
   double n;
   double.TryParse(Console.ReadLine(), out n);
   if (n < 0) return;
   double sqrRoot = SqrRoot(n);
   Console.WriteLine("The square root is {0}", sqrRoot);
   }
   }

   static double SqrRoot(double n)
   {
   double r = 0.0;
   double diff = 1.0;
   while (true)
   {
   while ((r * r) < n) r += diff;
   if ( (r * r - n) < 1.0e-10) return r;
   r -= diff;
   diff /= 2.0;
   }
   }
}
Vulpes15y ago
I don't know whether you noticed but the previous version had problems converging on the square root of some numbers.

Here's a much improved version (still using simple arithmetic) in which, hopefully, I've eliminated those problems. The output of Math.Sqrt is also shown for comparison:

using System;

class SquareRoot
{
   static void Main()
   {
   Console.WriteLine("Enter a negative number to quit");

   while (true)
   {
   Console.Write("\nInput a number : ");
   double n;
   double.TryParse(Console.ReadLine(), out n);
   if (n < 0) return;
   double sqrRoot = SqrRoot(n);
   Console.WriteLine("The square root is {0}", sqrRoot);
   Console.WriteLine("Math.Sqrt produces {0}", Math.Sqrt(n));
   }
   }

   static double SqrRoot(double n)
   {
   double r = 0.0;
   double p = 0.0;
   double pow2 = 1.0;

   if (n >= 2) // obtain a better starting increment
   {
   while (pow2 * pow2 < n) pow2 *= 2.0;
   }
   else if (n <= 0.5)
   {
   while (pow2 * pow2 > n) pow2 /= 2.0;
   pow2 *= 2;
   }

   if (pow2 * pow2 == n) return pow2;

   double diff = pow2;
   double tol = pow2 * 1.0e-15;

   while (true)
   {
   while (r * r < n) r += diff;
   if (r - p < tol) return r;
   p = r - diff;
   diff /= 2.0;
   r = p + diff;
   }
   }
}
Mayur Gujrathi15y ago
i said not using any predefined function
Frogleg15y ago
double i = 25;
double j = Math.Sqrt(i);