dipityPix app

Tuesday, January 18, 2011

Improved code for concentric map

If you need to warp points on a square to a disk, many people use Ken Chiu's and my code from jgt:

/* seedx, seedy is point on [0,1]^2. x, y is point on radius 1 disk */
void to_unit_disk( double seedx, double seedy, double *x, double *y )

double phi, r;

double a = 2*seedx - 1; >/* (a,b) is now on [-1,1]^2 */
double b = 2*seedy - 1;

if (a > -b) { /* region 1 or 2 */
if (a > b) { /* region 1, also |a| > |b| */
r = a;
phi = (M_PI/4 ) * (b/a);
else { /* region 2, also |b| > |a| */
phi = (M_PI/4) * (2 - (a/b));
else { /* region 3 or 4 */
if (a < b) /* region 3, also |a| >= |b|, a != 0
r = -a;
phi = (M_PI/4) * (4 + (b/a));
else { /* region 4, |b| >= |a|, but a==0 and b==0 could occur.
r = -b;
if (b != 0)
phi = (M_PI/4) * (6 - (a/b));
phi = 0;

*x = r * cos(phi);
*y = r * sin(phi);


Dave Cline recently sent me a neat trick that uses negative radii and I think is correct. Let me know if you try it. (cut and pasted from his mail)

Vector2 ToUnitDisk(Vector2 O) {
float phi,r;
float a = 2*O.x - 1;
float b = 2*O.y - 1;
if (a*a> b*b) { // use squares instead of absolute values
r = a;
phi = (PI/4)*(b/a);
} else {
r = b;
phi = (PI/4)*(a/b) + (PI/2);
return Vector2( r*cos(phi), r*sin(phi) );


franz said...


As is, the improved formulation leads to clumping and alignment issues when fed samples coming from a Hammersley sequence. It's easy to fix: in the second branch of the if, replace

phi = (PI/4)*(a/b) + (PI/2);


phi = (PI/2) - (PI/4)*(a/b);

With this modification, the new function will return the exact same results as the original jgt one.


Greg Ward said...

Nice! Looks like we missed the case where (a==0&&b==0), so we'll have a divide-by-zero in the else section unless we add a test.

BTW, I'm using this heavily in my new BSDF sampling scheme. Very handy!

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jui fardin said...

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C. Fong said...
This comment has been removed by the author.
C. Fong said...

That's a nice trick by Dave Cline! I just updated my code to use his trick. The C++ code and the resulting equations are much simpler! Here are some images showing the results:


Manuel said...

This is a neat implementation and produces the same results as the jgt function (with the fix given in the first comment) with much less branching. I am just wondering why the test (a*a > b*b) is considered better than (abs(a) > abs(b)). The latter avoids two floating point multiplies and simply involves zeroing the sign bits in the binary representations for a and b.

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