Simple Stretching: Mapping a circular disc to a square

We can map the circular disc to a square by just linearly stretching each point radially so that the rim of the circle matches the rim of the square.

The equations for the mapping are provided below

where sgn(x) is the signum function.

(u,v) are circular coordinates in the unit disc.

(x,y) are coordinates in the square [-1,1] x [-1,1]

[C++ implementation source code below]

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// sample code accompanying the paper:
// "Analytical Methods for Squaring the Disc"
// http://arxiv.org/abs/1509.06344
#include <stdio.h>
#include <math.h>// Simple Stretching map
// mapping a circular disc to a square region
// input: (u,v) coordinates in the circle
// output: (x,y) coordinates in the square
void stretchDiscToSquare(float u, float v, float& x, float& y)
{
    if ( (fabs(u) < epsilon) || (fabs(v) < epsilon))  {
        x = u;
        y = v;
        return;
    } 

    float u2 = u * u;
    float v2 = v * v;
    float r = sqrt(u2 + v2);

    // a trick based on Dave Cline's idea
    // link Peter Shirley's blog
    if (u2 >= v2) {
        float sgnu = sgn(u);
        x = sgnu * r;
        y = sgnu * r * v / u;
    } else {
        float sgnv = sgn(v);
        x = sgnv * r * u / v;
        y = sgnv * r;
    }
    
}


// Simple Stretching map
// mapping a square region to a circular disc
// input: (x,y) coordinates in the square
// output: (u,v) coordinates in the circle
void stretchSquareToDisc(float x, float y, float& u, float& v)
{
    if ( (fabs(x) < epsilon) || (fabs(y) < epsilon))  {
        u = x;
        v = y;
        return;
    }
    
    float x2 = x*x;
    float y2 = y*y;
    float hypothenusSquared = x2 + y2;

    // code can use fast reciprocal sqrt floating point trick
    // https://en.wikipedia.org/wiki/Fast_inverse_square_root