CODEWITHSUNDEEP
c++algorithmfloating-pointgeometry
Alessandro Jacopson
Alessandro Jacopson

Reputation: 18675

How to convert a 2D line equation in General Form to a Slope-intercept Form, in C++

I have the equation of a 2D line in the General Form a x + b y + c = 0 and I need to convert it to the proper Slope-intercept Form; with proper I mean I can choose between y = m x + q and x = m y + q.

My idea is to check if the line appears "more" horizontal or vertical and consequently choose one of the two Slope-intercept Form.

This is a sample code:

#include <iostream>
#include <cmath>

void abc2mq( double a, double b, double c, double& m, double& q, bool& x2y )
{
   if ( fabs(b) >= fabs(a) ) {
      x2y = true;
      m = -a/b;
      q = -c/b;
   } else {
      x2y = false;
      m = -b/a;
      q = -c/a;
   }
}

void test(double a, double b, double c)
{
   double m,q;
   bool x2y;
   abc2mq( a, b, c, m, q, x2y );
   std::cout << a << " x + " << b << " y + " << c << " = 0\t";
   if ( x2y ) {
      std::cout << "y = " << m << " x + " << q << "\n";
   } else {
      std::cout << "x = " << m << " y + " << q << "\n";
   }
}

int main(int argc, char* argv[])
{
   test(0,0,0);
   test(0,0,1);
   test(0,1,0);
   test(0,1,1);
   test(1,0,0);
   test(1,0,1);
   test(1,1,0);
   test(1,1,1);

   return 0;
}

And this is the output

0 x + 0 y + 0 = 0       y = -1.#IND x + -1.#IND
0 x + 0 y + 1 = 0       y = -1.#IND x + -1.#INF
0 x + 1 y + 0 = 0       y = -0 x + -0
0 x + 1 y + 1 = 0       y = -0 x + -1
1 x + 0 y + 0 = 0       x = -0 y + -0
1 x + 0 y + 1 = 0       x = -0 y + -1
1 x + 1 y + 0 = 0       y = -1 x + -0
1 x + 1 y + 1 = 0       y = -1 x + -1

Any different or better idea? In particular, how can I handle the first two "degenerate" lines?

Upvotes: 1

Views: 879

Answers (3)

salva
salva

Reputation: 10242

The equations corresponding to the two degenerated cases do not represent lines but the full plane (ℝ2) and the empty set (∅) respectively. The right thing to do is probably to discard them or to throw an error.

For the non degenerate cases, you are already handling them properly.

Upvotes: 1

Viktor Latypov
Viktor Latypov

Reputation: 14467

You're almost done, just handle the degenerate case. Add the check for a and b to be non-zero.

if(fabs(a) > DBL_EPSILON && fabs(b) > DBL_EPSILON)
{
    ... non-degenerate line handling
} else
{
    // both a and b are machine zeros
    degenerate_line = true;
}

Then add the parameter 'degenerate_line':

void abc2mq( double a, double b, double c, double& m, double& q, bool& x2y, bool& degenerate_line)
{
    if(fabs(a) > DBL_EPSILON && fabs(b) > DBL_EPSILON)
    {
       if ( fabs(b) >= fabs(a) ) {
           x2y = true;
           m = -a/b;
           q = -c/b;
        } else {
           x2y = false;
           m = -b/a;
           q = -c/a;
        }

        degenerate_line = false;
    } else
    {
        degenerate_line = true;
    }
}

And then check for the line to be empty set:

void test(double a, double b, double c)
{
    double m,q;
    bool x2y, degenerate;
    abc2mq( a, b, c, m, q, x2y, degenerate );
    std::cout << a << " x + " << b << " y + " << c << " = 0\t";
    if(!degenerate)
    {
       if ( x2y ) {
           std::cout << "y = " << m << " x + " << q << std::endl;
       } else {
           std::cout << "x = " << m << " y + " << q << std::endl;
       }
    } else
    {
       if(fabs(c) > DBL_EPSILON)
       {
          std::cout << "empty set" << std::endl
       } else
       {
          std::cout << "entire plane" << std::endl
       }
    }
}

If all you need is to draw the line, just use Thorsten's advice - use the rasterization algorithm instead.

Upvotes: 2

Thorsten Karrer
Thorsten Karrer

Reputation: 1365

If you are looking for a good way to draw these lines, I would recommend using Bresenham's algorithm instead of sampling the result of the slope-intercept form of your line equation. Apologies if this is not what you are trying to do.

Upvotes: 4

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