forked from enviPath/enviPy
393 lines
13 KiB
JavaScript
393 lines
13 KiB
JavaScript
'use strict';
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function Point(x, y) {
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this.x = x;
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this.y = y;
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}
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Point.prototype.add = function (point) {
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return new Point(this.x + point.x, this.y + point.y);
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};
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Point.prototype.sub = function (point) {
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return new Point(this.x - point.x, this.y - point.y);
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};
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Point.prototype.mul = function (value) {
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return new Point(this.x * value, this.y * value);
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};
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Point.prototype.div = function (value) {
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return new Point(this.x / value, this.y / value);
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};
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Point.prototype.dist = function () {
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return Math.sqrt(this.x*this.x + this.y*this.y);
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};
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Point.prototype.sqr = function () {
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return this.x*this.x + this.y*this.y;
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};
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Point.prototype.dot = function (point) {
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return this.x * point.x + this.y * point.y;
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};
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function calcPowerCoefficients(p1, c1, c2, p2) {
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// point(t) = p1*(1-t)^3 + c1*t*(1-t)^2 + c2*t^2*(1-t) + p2*t^3 = a*t^3 + b*t^2 + c*t + d
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// for each t value, so
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// a = (p2 - p1) + 3 * (c1 - c2)
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// b = 3 * (p1 + c2) - 6 * c1
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// c = 3 * (c1 - p1)
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// d = p1
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var a = p2.sub(p1).add(c1.sub(c2).mul(3));
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var b = p1.add(c2).mul(3).sub(c1.mul(6));
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var c = c1.sub(p1).mul(3);
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var d = p1;
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return [ a, b, c, d ];
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}
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function calcPoint(a, b, c, d, t) {
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// a*t^3 + b*t^2 + c*t + d = ((a*t + b)*t + c)*t + d
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return a.mul(t).add(b).mul(t).add(c).mul(t).add(d);
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}
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function calcPointQuad(a, b, c, t) {
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// a*t^2 + b*t + c = (a*t + b)*t + c
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return a.mul(t).add(b).mul(t).add(c);
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}
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function calcPointDerivative(a, b, c, d, t) {
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// d/dt[a*t^3 + b*t^2 + c*t + d] = 3*a*t^2 + 2*b*t + c = (3*a*t + 2*b)*t + c
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return a.mul(3*t).add(b.mul(2)).mul(t).add(c);
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}
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function quadSolve(a, b, c) {
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// a*x^2 + b*x + c = 0
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if (a === 0) {
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return (b === 0) ? [] : [ -c / b ];
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}
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var D = b*b - 4*a*c;
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if (D < 0) {
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return [];
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} else if (D === 0) {
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return [ -b/(2*a) ];
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}
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var DSqrt = Math.sqrt(D);
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return [ (-b - DSqrt) / (2*a), (-b + DSqrt) / (2*a) ];
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}
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function cubicRoot(x) {
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return (x < 0) ? -Math.pow(-x, 1/3) : Math.pow(x, 1/3);
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}
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function cubicSolve(a, b, c, d) {
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// a*x^3 + b*x^2 + c*x + d = 0
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if (a === 0) {
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return quadSolve(b, c, d);
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}
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// solve using Cardan's method, which is described in paper of R.W.D. Nickals
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// http://www.nickalls.org/dick/papers/maths/cubic1993.pdf (doi:10.2307/3619777)
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var xn = -b / (3*a); // point of symmetry x coordinate
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var yn = ((a * xn + b) * xn + c) * xn + d; // point of symmetry y coordinate
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var deltaSq = (b*b - 3*a*c) / (9*a*a); // delta^2
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var hSq = 4*a*a * Math.pow(deltaSq, 3); // h^2
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var D3 = yn*yn - hSq;
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if (D3 > 0) { // 1 real root
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var D3Sqrt = Math.sqrt(D3);
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return [ xn + cubicRoot((-yn + D3Sqrt)/(2*a)) + cubicRoot((-yn - D3Sqrt)/(2*a)) ];
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} else if (D3 === 0) { // 2 real roots
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var delta1 = cubicRoot(yn/(2*a));
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return [ xn - 2 * delta1, xn + delta1 ];
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}
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// 3 real roots
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var theta = Math.acos(-yn / Math.sqrt(hSq)) / 3;
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var delta = Math.sqrt(deltaSq);
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return [
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xn + 2 * delta * Math.cos(theta),
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xn + 2 * delta * Math.cos(theta + Math.PI * 2 / 3),
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xn + 2 * delta * Math.cos(theta + Math.PI * 4 / 3)
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];
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}
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function minDistanceToQuad(point, p1, c1, p2) {
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// f(t) = (1-t)^2 * p1 + 2*t*(1 - t) * c1 + t^2 * p2 = a*t^2 + b*t + c, t in [0, 1],
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// a = p1 + p2 - 2 * c1
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// b = 2 * (c1 - p1)
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// c = p1; a, b, c are vectors because p1, c1, p2 are vectors too
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// The distance between given point and quadratic curve is equal to
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// sqrt((f(t) - point)^2), so these expression has zero derivative by t at points where
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// (f'(t), (f(t) - point)) = 0.
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// Substituting quadratic curve as f(t) one could obtain a cubic equation
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// e3*t^3 + e2*t^2 + e1*t + e0 = 0 with following coefficients:
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// e3 = 2 * a^2
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// e2 = 3 * a*b
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// e1 = (b^2 + 2 * a*(c - point))
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// e0 = (c - point)*b
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// One of the roots of the equation from [0, 1], or t = 0 or t = 1 is a value of t
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// at which the distance between given point and quadratic Bezier curve has minimum.
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// So to find the minimal distance one have to just pick the minimum value of
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// the distance on set {t = 0 | t = 1 | t is root of the equation from [0, 1] }.
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var a = p1.add(p2).sub(c1.mul(2));
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var b = c1.sub(p1).mul(2);
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var c = p1;
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var e3 = 2 * a.sqr();
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var e2 = 3 * a.dot(b);
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var e1 = (b.sqr() + 2 * a.dot(c.sub(point)));
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var e0 = c.sub(point).dot(b);
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var candidates = cubicSolve(e3, e2, e1, e0).filter(function (t) { return t > 0 && t < 1; }).concat([ 0, 1 ]);
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var minDistance = 1e9;
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for (var i = 0; i < candidates.length; i++) {
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var distance = calcPointQuad(a, b, c, candidates[i]).sub(point).dist();
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if (distance < minDistance) {
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minDistance = distance;
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}
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}
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return minDistance;
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}
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function processSegment(a, b, c, d, t1, t2) {
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// Find a single control point for given segment of cubic Bezier curve
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// These control point is an interception of tangent lines to the boundary points
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// Let's denote that f(t) is a vector function of parameter t that defines the cubic Bezier curve,
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// f(t1) + f'(t1)*z1 is a parametric equation of tangent line to f(t1) with parameter z1
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// f(t2) + f'(t2)*z2 is the same for point f(t2) and the vector equation
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// f(t1) + f'(t1)*z1 = f(t2) + f'(t2)*z2 defines the values of parameters z1 and z2.
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// Defining fx(t) and fy(t) as the x and y components of vector function f(t) respectively
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// and solving the given system for z1 one could obtain that
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//
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// -(fx(t2) - fx(t1))*fy'(t2) + (fy(t2) - fy(t1))*fx'(t2)
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// z1 = ------------------------------------------------------.
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// -fx'(t1)*fy'(t2) + fx'(t2)*fy'(t1)
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//
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// Let's assign letter D to the denominator and note that if D = 0 it means that the curve actually
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// is a line. Substituting z1 to the equation of tangent line to the point f(t1), one could obtain that
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// cx = [fx'(t1)*(fy(t2)*fx'(t2) - fx(t2)*fy'(t2)) + fx'(t2)*(fx(t1)*fy'(t1) - fy(t1)*fx'(t1))]/D
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// cy = [fy'(t1)*(fy(t2)*fx'(t2) - fx(t2)*fy'(t2)) + fy'(t2)*(fx(t1)*fy'(t1) - fy(t1)*fx'(t1))]/D
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// where c = (cx, cy) is the control point of quadratic Bezier curve.
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var f1 = calcPoint(a, b, c, d, t1);
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var f2 = calcPoint(a, b, c, d, t2);
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var f1_ = calcPointDerivative(a, b, c, d, t1);
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var f2_ = calcPointDerivative(a, b, c, d, t2);
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var D = -f1_.x * f2_.y + f2_.x * f1_.y;
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if (Math.abs(D) < 1e-8) {
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return [ f1, f1.add(f2).div(2), f2 ]; // straight line segment
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}
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var cx = (f1_.x*(f2.y*f2_.x - f2.x*f2_.y) + f2_.x*(f1.x*f1_.y - f1.y*f1_.x)) / D;
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var cy = (f1_.y*(f2.y*f2_.x - f2.x*f2_.y) + f2_.y*(f1.x*f1_.y - f1.y*f1_.x)) / D;
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return [ f1, new Point(cx, cy), f2 ];
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}
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function isSegmentApproximationClose(a, b, c, d, tmin, tmax, p1, c1, p2, errorBound) {
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// a,b,c,d define cubic curve
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// tmin, tmax are boundary points on cubic curve
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// p1, c1, p2 define quadratic curve
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// errorBound is maximum allowed distance
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// Try to find maximum distance between one of N points segment of given cubic
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// and corresponding quadratic curve that estimates the cubic one, assuming
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// that the boundary points of cubic and quadratic points are equal.
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//
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// The distance calculation method comes from Hausdorff distance defenition
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// (https://en.wikipedia.org/wiki/Hausdorff_distance), but with following simplifications
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// * it looks for maximum distance only for finite number of points of cubic curve
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// * it doesn't perform reverse check that means selecting set of fixed points on
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// the quadratic curve and looking for the closest points on the cubic curve
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// But this method allows easy estimation of approximation error, so it is enough
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// for practical purposes.
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var n = 10; // number of points + 1
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var dt = (tmax - tmin) / n;
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for (var t = tmin + dt; t < tmax - dt; t += dt) { // don't check distance on boundary points
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// because they should be the same
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var point = calcPoint(a, b, c, d, t);
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if (minDistanceToQuad(point, p1, c1, p2) > errorBound) {
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return false;
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}
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}
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return true;
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}
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function _isApproximationClose(a, b, c, d, quadCurves, errorBound) {
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var dt = 1/quadCurves.length;
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for (var i = 0; i < quadCurves.length; i++) {
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var p1 = quadCurves[i][0];
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var c1 = quadCurves[i][1];
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var p2 = quadCurves[i][2];
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if (!isSegmentApproximationClose(a, b, c, d, i * dt, (i + 1) * dt, p1, c1, p2, errorBound)) {
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return false;
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}
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}
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return true;
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}
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function fromFlatArray(points) {
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var result = [];
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var segmentsNumber = (points.length - 2) / 4;
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for (var i = 0; i < segmentsNumber; i++) {
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result.push([
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new Point(points[4 * i], points[4 * i + 1]),
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new Point(points[4 * i + 2], points[4 * i + 3]),
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new Point(points[4 * i + 4], points[4 * i + 5])
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]);
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}
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return result;
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}
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function toFlatArray(quadsList) {
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var result = [];
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result.push(quadsList[0][0].x);
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result.push(quadsList[0][0].y);
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for (var i = 0; i < quadsList.length; i++) {
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result.push(quadsList[i][1].x);
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result.push(quadsList[i][1].y);
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result.push(quadsList[i][2].x);
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result.push(quadsList[i][2].y);
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}
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return result;
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}
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function isApproximationClose(p1x, p1y, c1x, c1y, c2x, c2y, p2x, p2y, quads, errorBound) {
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// TODO: rewrite it in C-style and remove _isApproximationClose
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var pc = calcPowerCoefficients(
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new Point(p1x, p1y),
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new Point(c1x, c1y),
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new Point(c2x, c2y),
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new Point(p2x, p2y)
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);
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return _isApproximationClose(pc[0], pc[1], pc[2], pc[3], fromFlatArray(quads), errorBound);
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}
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/*
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* Split cubic bézier curve into two cubic curves, see details here:
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* https://math.stackexchange.com/questions/877725
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*/
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function subdivideCubic(x1, y1, x2, y2, x3, y3, x4, y4, t) {
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var u = 1-t, v = t;
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var bx = x1*u + x2*v;
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var sx = x2*u + x3*v;
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var fx = x3*u + x4*v;
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var cx = bx*u + sx*v;
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var ex = sx*u + fx*v;
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var dx = cx*u + ex*v;
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var by = y1*u + y2*v;
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var sy = y2*u + y3*v;
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var fy = y3*u + y4*v;
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var cy = by*u + sy*v;
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var ey = sy*u + fy*v;
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var dy = cy*u + ey*v;
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return [
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[ x1, y1, bx, by, cx, cy, dx, dy ],
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[ dx, dy, ex, ey, fx, fy, x4, y4 ]
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];
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}
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function byNumber(x, y) { return x - y; }
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/*
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* Find inflection points on a cubic curve, algorithm is similar to this one:
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* http://www.caffeineowl.com/graphics/2d/vectorial/cubic-inflexion.html
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*/
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function solveInflections(x1, y1, x2, y2, x3, y3, x4, y4) {
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var p = -(x4 * (y1 - 2 * y2 + y3)) + x3 * (2 * y1 - 3 * y2 + y4)
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+ x1 * (y2 - 2 * y3 + y4) - x2 * (y1 - 3 * y3 + 2 * y4);
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var q = x4 * (y1 - y2) + 3 * x3 * (-y1 + y2) + x2 * (2 * y1 - 3 * y3 + y4) - x1 * (2 * y2 - 3 * y3 + y4);
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var r = x3 * (y1 - y2) + x1 * (y2 - y3) + x2 * (-y1 + y3);
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return quadSolve(p, q, r).filter(function (t) { return t > 1e-8 && t < 1 - 1e-8; }).sort(byNumber);
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}
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/*
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* Approximate cubic Bezier curve defined with base points p1, p2 and control points c1, c2 with
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* with a few quadratic Bezier curves.
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* The function uses tangent method to find quadratic approximation of cubic curve segment and
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* simplified Hausdorff distance to determine number of segments that is enough to make error small.
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* In general the method is the same as described here: https://fontforge.github.io/bezier.html.
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*/
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function _cubicToQuad(p1x, p1y, c1x, c1y, c2x, c2y, p2x, p2y, errorBound) {
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var p1 = new Point(p1x, p1y);
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var c1 = new Point(c1x, c1y);
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var c2 = new Point(c2x, c2y);
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var p2 = new Point(p2x, p2y);
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var pc = calcPowerCoefficients(p1, c1, c2, p2);
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var a = pc[0], b = pc[1], c = pc[2], d = pc[3];
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var approximation;
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for (var segmentsCount = 1; segmentsCount <= 8; segmentsCount++) {
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approximation = [];
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for (var t = 0; t < 1; t += 1/segmentsCount) {
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approximation.push(processSegment(a, b, c, d, t, t + 1/segmentsCount));
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}
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if (segmentsCount === 1 && (
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approximation[0][1].sub(p1).dot(c1.sub(p1)) < 0 ||
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approximation[0][1].sub(p2).dot(c2.sub(p2)) < 0)) {
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// approximation concave, while the curve is convex (or vice versa)
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continue;
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}
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if (_isApproximationClose(a, b, c, d, approximation, errorBound)) {
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break;
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}
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}
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return toFlatArray(approximation);
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}
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/*
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* If this curve has any inflection points, split the curve and call
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* _cubicToQuad function on each resulting curve.
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*/
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function cubicToQuad(p1x, p1y, c1x, c1y, c2x, c2y, p2x, p2y, errorBound) {
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var inflections = solveInflections(p1x, p1y, c1x, c1y, c2x, c2y, p2x, p2y);
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if (!inflections.length) {
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return _cubicToQuad(p1x, p1y, c1x, c1y, c2x, c2y, p2x, p2y, errorBound);
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}
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var result = [];
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var curve = [ p1x, p1y, c1x, c1y, c2x, c2y, p2x, p2y ];
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var prevPoint = 0;
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var quad, split;
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for (var inflectionIdx = 0; inflectionIdx < inflections.length; inflectionIdx++) {
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split = subdivideCubic(
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curve[0], curve[1], curve[2], curve[3],
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curve[4], curve[5], curve[6], curve[7],
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// we make a new curve, so adjust inflection point accordingly
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1 - (1 - inflections[inflectionIdx]) / (1 - prevPoint)
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);
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quad = _cubicToQuad(
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split[0][0], split[0][1], split[0][2], split[0][3],
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split[0][4], split[0][5], split[0][6], split[0][7],
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errorBound
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);
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result = result.concat(quad.slice(0, -2));
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curve = split[1];
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prevPoint = inflections[inflectionIdx];
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}
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quad = _cubicToQuad(
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curve[0], curve[1], curve[2], curve[3],
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curve[4], curve[5], curve[6], curve[7],
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errorBound
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);
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return result.concat(quad);
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}
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module.exports = cubicToQuad;
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// following exports are for testing purposes
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module.exports.isApproximationClose = isApproximationClose;
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module.exports.cubicSolve = cubicSolve;
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