Thursday, September 3, 2015

Mapping a Circle to a Square

Philip Nowell (
devised equations for mapping a square to a circle. I will provide the inverse to the mapping below.

For this blog entry, I use a slightly different notation from his equations; i.e.
u = x'
v = y'

(u,v) are circular coordinates in the domain {(u,v) | u² + v² ≤ 1}
(x,y) are square coordinates in the range [-1,1] x [-1,1]

Here are the inverse mapping equations:

x = ½ √( 2 + 2u√2 + u² - v² ) - ½ √( 2 - 2u√2 + u² - v² )
y = ½ √( 2 + 2v√2 - u² + v² ) - ½ √( 2 - 2v√2 - u² + v² )

or more legibly

The forward equations for the mapping are:

u = x √( 1 - ½ y² )
v = y √( 1 - ½ x² )

Here are some examples of the forward & inverse mappings at work.

For a proof/derivation of the inverse equations, see my paper in
arxiv: "Analytical Methods for Squaring the Disc"

In this blog, I shall refer to this mapping as the elliptical grid mapping.  I chose this name because the mapping has a predilection to map a rectilinear grid into a grid of elliptical arcs in the circle.  See Nowell's blog for a discussion and derivation of the elliptical arcs.

Here are more examples of the mapping:

Starbucks logo converted to a square
Andy Warhol's pop art "Marilyn" as a circle

[ C++ implementation source code below]

// sample code accompanying the paper:
// "Analytical Methods for Squaring the Disc"

#include <stdio.h>
#include <math.h>

// Elliptical Grid mapping
// mapping a circular disc to a square region
// input: (u,v) coordinates in the circle
// output: (x,y) coordinates in the square
void ellipticalDiscToSquare(double u, double v, double& x, double& y)
    double u2 = u * u;
    double v2 = v * v;
    double twosqrt2 = 2.0 * sqrt(2.0);
    double subtermx = 2.0 + u2 - v2;
    double subtermy = 2.0 - u2 + v2;
    double termx1 = subtermx + u * twosqrt2;
    double termx2 = subtermx - u * twosqrt2;
    double termy1 = subtermy + v * twosqrt2;
    double termy2 = subtermy - v * twosqrt2;
    x = 0.5 * sqrt(termx1) - 0.5 * sqrt(termx2);
    y = 0.5 * sqrt(termy1) - 0.5 * sqrt(termy2);

// Elliptical Grid mapping
// mapping a square region to a circular disc
// input: (x,y) coordinates in the square
// output: (u,v) coordinates in the circle
void ellipticalSquareToDisc(double x, double y, double& u, double& v)
    u = x * sqrt(1.0 - y*y/2.0);
    v = y * sqrt(1.0 - x*x/2.0);    

int main()
    double x,y;
    double u,v;

    printf("%f %f\n",u,v);
    printf("%f %f\n",x,y);

    ellipticalDiscToSquare(0.31415, -0.926535,x,y);
    printf("%f %f\n",x,y);
    printf("%f %f\n",u,v);

    return 0;


  1. If u2 == v2, I find that termx2 and termy2 can be negative and result in errors.

    1.  I'm assuming you mean when u = v = sqrt(2) on the circle. The error you are getting with regards to a negative argument to a square root is because of numerical imprecision.  If you observe the value of termx2 and termy2 in the source code for the case u=v=sqrt(2), they are very small and close to zero.  You can insert this code before the square root calls in order to avoid numerical imprecision of floating point numbers.

      double epsilon = 0.0000001;
      if (fabs(termx2)< epsilon) {
         termx2 = 0.0;
      if (fabs(termy2)< epsilon) {
         termy2 = 0.0;

      In a perfect world of floating point numbers without numerical imprecision, termx2 and termy2 should be equal to zero for this case

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  3. thanks for the write up.
    Made a javascript/Canvas2D based version on codepen

    1. cool demo! Thanks for sharing your work!! It's nice to see an online implementation in javascript. I couldn't see the difference after changing the resampling factor but maybe it's just my poor eyes.

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  5. Are there any good ways of interpolating values? So you get a slightly rounded square for instance?
    For the regular version I've already found a simple solution that works quite well.
    where n is your interpolator.
    But I can't find a very elegant way of doing the same for the inverse.

    1. Actually, I found a similar solution for the inverse.

      x = ( √( 2 + 2un√2 + (un)² - (vn)² ) - √( 2 - 2un√2 + (un)² - (vn)² ) )/2n

      Where n is the interpolator.
      Also, by replacing each instance of n with n^0.48 in this function, you get nearly the original image back when using both formula's with the same value for n.

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