scratchfoundation/paper.js
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Remove tolerance parameter from solveCubic() / solveQuadratic() since we should always use EPSILON.

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This commit is contained in:
Jürg Lehni 2013-04-20 19:58:42 -07:00
parent 14aa8e5dea
commit c45ad26b67
2 changed files with 15 additions and 14 deletions
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6
src/path/Curve.js
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@ -533,8 +533,7 @@ statics: {
c = 3 * (c1 - p1),
b = 3 * (c2 - c1) - c,
a = p2 - p1 - c - b;
return Numerical.solveCubic(a, b, c, p1 - val, roots,
/*#=*/ Numerical.EPSILON);
return Numerical.solveCubic(a, b, c, p1 - val, roots);
},
getParameterOf: function(v, x, y) {
@ -636,8 +635,7 @@ statics: {
var a = 3 * (v1 - v2) - v0 + v3,
b = 2 * (v0 + v2) - 4 * v1,
c = v1 - v0,
count = Numerical.solveQuadratic(a, b, c, roots,
/*#=*/ Numerical.TOLERANCE),
count = Numerical.solveQuadratic(a, b, c, roots),
// Add some tolerance for good roots, as t = 0 / 1 are added
// seperately anyhow, and we don't want joins to be added with
// radiuses in getStrokeBounds()

23
src/util/Numerical.js
View file

@ -132,16 +132,17 @@ var Numerical = this.Numerical = new function() {
*
* a*x^2 + b*x + c = 0
*/
solveQuadratic: function(a, b, c, roots, tolerance) {
solveQuadratic: function(a, b, c, roots) {
// Code ported over and adapted from Uintah library (MIT license).
// If problem is actually linear, return 0 or 1 easy roots
if (abs(a) < tolerance) {
if (abs(b) >= tolerance) {
var epsilon = this.EPSILON;
// If a is 0, equation is actually linear, return 0 or 1 easy roots.
if (abs(a) < epsilon) {
if (abs(b) >= epsilon) {
roots[0] = -c / b;
return 1;
}
// If all the coefficients are 0, we have infinite solutions!
return abs(c) < tolerance ? -1 : 0; // Infinite or 0 solutions
return abs(c) < epsilon ? -1 : 0; // Infinite or 0 solutions
}
var q = b * b - 4 * a * c;
if (q < 0)
@ -161,10 +162,12 @@ var Numerical = this.Numerical = new function() {
*
* a*x^3 + b*x^2 + c*x + d = 0
*/
solveCubic: function(a, b, c, d, roots, tolerance) {
solveCubic: function(a, b, c, d, roots) {
// Code ported over and adapted from Uintah library (MIT license).
if (abs(a) < tolerance)
return Numerical.solveQuadratic(b, c, d, roots, tolerance);
var epsilon = this.EPSILON;
// If a is 0, equation is actually quadratic.
if (abs(a) < epsilon)
return Numerical.solveQuadratic(b, c, d, roots);
// Normalize to form: x^3 + b x^2 + c x + d = 0:
b /= a;
c /= a;
@ -178,8 +181,8 @@ var Numerical = this.Numerical = new function() {
D = q * q - ppp;
// Substitute x = y - b/3 to eliminate quadric term: x^3 +px + q = 0
b /= 3;
if (abs(D) < tolerance) {
if (abs(q) < tolerance) { // One triple solution.
if (abs(D) < epsilon) {
if (abs(q) < epsilon) { // One triple solution.
roots[0] = - b;
return 1;
} else { // One single and one double solution.