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'use strict'
// Precision used to check determinant in quad and cubic solvers,
// any number lower than this is considered to be zero.
// `8.67e-19` is an example of real error occurring in tests.
var epsilon = 1e-16
function Point (x, y) {
this.x = x
this.y = y
}
Point.prototype.add = function (point) {
return new Point(this.x + point.x, this.y + point.y)
}
Point.prototype.sub = function (point) {
return new Point(this.x - point.x, this.y - point.y)
}
Point.prototype.mul = function (value) {
return new Point(this.x * value, this.y * value)
}
Point.prototype.div = function (value) {
return new Point(this.x / value, this.y / value)
}
/*Point.prototype.dist = function () {
return Math.sqrt(this.x * this.x + this.y * this.y)
}*/
Point.prototype.sqr = function () {
return this.x * this.x + this.y * this.y
}
Point.prototype.dot = function (point) {
return this.x * point.x + this.y * point.y
}
function calcPowerCoefficients (p1, c1, c2, p2) {
// 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
// for each t value, so
// a = (p2 - p1) + 3 * (c1 - c2)
// b = 3 * (p1 + c2) - 6 * c1
// c = 3 * (c1 - p1)
// d = p1
var a = p2.sub(p1).add(c1.sub(c2).mul(3))
var b = p1.add(c2).mul(3).sub(c1.mul(6))
var c = c1.sub(p1).mul(3)
var d = p1
return [a, b, c, d]
}
function calcPowerCoefficientsQuad (p1, c1, p2) {
// point(t) = p1*(1-t)^2 + c1*t*(1-t) + p2*t^2 = a*t^2 + b*t + c
// for each t value, so
// a = p1 + p2 - 2 * c1
// b = 2 * (c1 - p1)
// c = p1
var a = c1.mul(-2).add(p1).add(p2)
var b = c1.sub(p1).mul(2)
var c = p1
return [a, b, c]
}
function calcPoint (a, b, c, d, t) {
// a*t^3 + b*t^2 + c*t + d = ((a*t + b)*t + c)*t + d
return a.mul(t).add(b).mul(t).add(c).mul(t).add(d)
}
function calcPointQuad (a, b, c, t) {
// a*t^2 + b*t + c = (a*t + b)*t + c
return a.mul(t).add(b).mul(t).add(c)
}
function calcPointDerivative (a, b, c, d, t) {
// 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
return a.mul(3 * t).add(b.mul(2)).mul(t).add(c)
}
function quadSolve (a, b, c) {
// a*x^2 + b*x + c = 0
if (a === 0) {
return (b === 0) ? [] : [-c / b]
}
var D = b * b - 4 * a * c
if (Math.abs(D) < epsilon) {
return [-b / (2 * a)]
} else if (D < 0) {
return []
}
var DSqrt = Math.sqrt(D)
return [(-b - DSqrt) / (2 * a), (-b + DSqrt) / (2 * a)]
}
/*function cubicRoot(x) {
return (x < 0) ? -Math.pow(-x, 1/3) : Math.pow(x, 1/3)
}
function cubicSolve(a, b, c, d) {
// a*x^3 + b*x^2 + c*x + d = 0
if (a === 0) {
return quadSolve(b, c, d)
}
// solve using Cardan's method, which is described in paper of R.W.D. Nickals
// http://www.nickalls.org/dick/papers/maths/cubic1993.pdf (doi:10.2307/3619777)
var xn = -b / (3*a) // point of symmetry x coordinate
var yn = ((a * xn + b) * xn + c) * xn + d // point of symmetry y coordinate
var deltaSq = (b*b - 3*a*c) / (9*a*a) // delta^2
var hSq = 4*a*a * Math.pow(deltaSq, 3) // h^2
var D3 = yn*yn - hSq
if (Math.abs(D3) < epsilon) { // 2 real roots
var delta1 = cubicRoot(yn/(2*a))
return [ xn - 2 * delta1, xn + delta1 ]
} else if (D3 > 0) { // 1 real root
var D3Sqrt = Math.sqrt(D3)
return [ xn + cubicRoot((-yn + D3Sqrt)/(2*a)) + cubicRoot((-yn - D3Sqrt)/(2*a)) ]
}
// 3 real roots
var theta = Math.acos(-yn / Math.sqrt(hSq)) / 3
var delta = Math.sqrt(deltaSq)
return [
xn + 2 * delta * Math.cos(theta),
xn + 2 * delta * Math.cos(theta + Math.PI * 2 / 3),
xn + 2 * delta * Math.cos(theta + Math.PI * 4 / 3)
]
}*/
/*
* Calculate a distance between a `point` and a line segment `p1, p2`
* (result is squared for performance reasons), see details here:
* https://stackoverflow.com/questions/849211/shortest-distance-between-a-point-and-a-line-segment
*/
function minDistanceToLineSq (point, p1, p2) {
var p1p2 = p2.sub(p1)
var dot = point.sub(p1).dot(p1p2)
var lenSq = p1p2.sqr()
var param = 0
var diff
if (lenSq !== 0) param = dot / lenSq
if (param <= 0) {
diff = point.sub(p1)
} else if (param >= 1) {
diff = point.sub(p2)
} else {
diff = point.sub(p1.add(p1p2.mul(param)))
}
return diff.sqr()
}
/*function minDistanceToQuad(point, p1, c1, p2) {
// f(t) = (1-t)^2 * p1 + 2*t*(1 - t) * c1 + t^2 * p2 = a*t^2 + b*t + c, t in [0, 1],
// a = p1 + p2 - 2 * c1
// b = 2 * (c1 - p1)
// c = p1; a, b, c are vectors because p1, c1, p2 are vectors too
// The distance between given point and quadratic curve is equal to
// sqrt((f(t) - point)^2), so these expression has zero derivative by t at points where
// (f'(t), (f(t) - point)) = 0.
// Substituting quadratic curve as f(t) one could obtain a cubic equation
// e3*t^3 + e2*t^2 + e1*t + e0 = 0 with following coefficients:
// e3 = 2 * a^2
// e2 = 3 * a*b
// e1 = (b^2 + 2 * a*(c - point))
// e0 = (c - point)*b
// One of the roots of the equation from [0, 1], or t = 0 or t = 1 is a value of t
// at which the distance between given point and quadratic Bezier curve has minimum.
// So to find the minimal distance one have to just pick the minimum value of
// the distance on set {t = 0 | t = 1 | t is root of the equation from [0, 1] }.
var a = p1.add(p2).sub(c1.mul(2))
var b = c1.sub(p1).mul(2)
var c = p1
var e3 = 2 * a.sqr()
var e2 = 3 * a.dot(b)
var e1 = (b.sqr() + 2 * a.dot(c.sub(point)))
var e0 = c.sub(point).dot(b)
var candidates = cubicSolve(e3, e2, e1, e0).filter(function (t) { return t > 0 && t < 1 }).concat([ 0, 1 ])
var minDistance = 1e9
for (var i = 0; i < candidates.length; i++) {
var distance = calcPointQuad(a, b, c, candidates[i]).sub(point).dist()
if (distance < minDistance) {
minDistance = distance
}
}
return minDistance
}*/
function processSegment (a, b, c, d, t1, t2) {
// Find a single control point for given segment of cubic Bezier curve
// These control point is an interception of tangent lines to the boundary points
// Let's denote that f(t) is a vector function of parameter t that defines the cubic Bezier curve,
// f(t1) + f'(t1)*z1 is a parametric equation of tangent line to f(t1) with parameter z1
// f(t2) + f'(t2)*z2 is the same for point f(t2) and the vector equation
// f(t1) + f'(t1)*z1 = f(t2) + f'(t2)*z2 defines the values of parameters z1 and z2.
// Defining fx(t) and fy(t) as the x and y components of vector function f(t) respectively
// and solving the given system for z1 one could obtain that
//
// -(fx(t2) - fx(t1))*fy'(t2) + (fy(t2) - fy(t1))*fx'(t2)
// z1 = ------------------------------------------------------.
// -fx'(t1)*fy'(t2) + fx'(t2)*fy'(t1)
//
// Let's assign letter D to the denominator and note that if D = 0 it means that the curve actually
// is a line. Substituting z1 to the equation of tangent line to the point f(t1), one could obtain that
// cx = [fx'(t1)*(fy(t2)*fx'(t2) - fx(t2)*fy'(t2)) + fx'(t2)*(fx(t1)*fy'(t1) - fy(t1)*fx'(t1))]/D
// cy = [fy'(t1)*(fy(t2)*fx'(t2) - fx(t2)*fy'(t2)) + fy'(t2)*(fx(t1)*fy'(t1) - fy(t1)*fx'(t1))]/D
// where c = (cx, cy) is the control point of quadratic Bezier curve.
var f1 = calcPoint(a, b, c, d, t1)
var f2 = calcPoint(a, b, c, d, t2)
var f1_ = calcPointDerivative(a, b, c, d, t1)
var f2_ = calcPointDerivative(a, b, c, d, t2)
var D = -f1_.x * f2_.y + f2_.x * f1_.y
if (Math.abs(D) < 1e-8) {
return [f1, f1.add(f2).div(2), f2] // straight line segment
}
var cx = (f1_.x * (f2.y * f2_.x - f2.x * f2_.y) + f2_.x * (f1.x * f1_.y - f1.y * f1_.x)) / D
var cy = (f1_.y * (f2.y * f2_.x - f2.x * f2_.y) + f2_.y * (f1.x * f1_.y - f1.y * f1_.x)) / D
return [f1, new Point(cx, cy), f2]
}
/*function isSegmentApproximationClose(a, b, c, d, tmin, tmax, p1, c1, p2, errorBound) {
// a,b,c,d define cubic curve
// tmin, tmax are boundary points on cubic curve
// p1, c1, p2 define quadratic curve
// errorBound is maximum allowed distance
// Try to find maximum distance between one of N points segment of given cubic
// and corresponding quadratic curve that estimates the cubic one, assuming
// that the boundary points of cubic and quadratic points are equal.
//
// The distance calculation method comes from Hausdorff distance defenition
// (https://en.wikipedia.org/wiki/Hausdorff_distance), but with following simplifications
// * it looks for maximum distance only for finite number of points of cubic curve
// * it doesn't perform reverse check that means selecting set of fixed points on
// the quadratic curve and looking for the closest points on the cubic curve
// But this method allows easy estimation of approximation error, so it is enough
// for practical purposes.
var n = 10 // number of points + 1
var dt = (tmax - tmin) / n
for (var t = tmin + dt; t < tmax - dt; t += dt) { // don't check distance on boundary points
// because they should be the same
var point = calcPoint(a, b, c, d, t)
if (minDistanceToQuad(point, p1, c1, p2) > errorBound) {
return false
}
}
return true
}*/
/*
* Divide cubic and quadratic curves into 10 points and 9 line segments.
* Calculate distances between each point on cubic and nearest line segment
* on quadratic (and vice versa), and make sure all distances are less
* than `errorBound`.
*
* We need to calculate BOTH distance from all points on quadratic to any cubic,
* and all points on cubic to any quadratic.
*
* If we do it only one way, it may lead to an error if the entire original curve
* falls within errorBound (then **any** quad will erroneously treated as good):
* https://github.com/fontello/svg2ttf/issues/105#issuecomment-842558027
*
* - a,b,c,d define cubic curve (power coefficients)
* - tmin, tmax are boundary points on cubic curve (in 0-1 range)
* - p1, c1, p2 define quadratic curve (control points)
* - errorBound is maximum allowed distance
*/
function isSegmentApproximationClose (a, b, c, d, tmin, tmax, p1, c1, p2, errorBound) {
var n = 10 // number of points
var t, dt
var p = calcPowerCoefficientsQuad(p1, c1, p2)
var qa = p[0], qb = p[1], qc = p[2]
var i, j, distSq
var errorBoundSq = errorBound * errorBound
var cubicPoints = []
var quadPoints = []
var minDistSq
dt = (tmax - tmin) / n
for (i = 0, t = tmin; i <= n; i++, t += dt) {
cubicPoints.push(calcPoint(a, b, c, d, t))
}
dt = 1 / n
for (i = 0, t = 0; i <= n; i++, t += dt) {
quadPoints.push(calcPointQuad(qa, qb, qc, t))
}
for (i = 1; i < cubicPoints.length - 1; i++) {
minDistSq = Infinity
for (j = 0; j < quadPoints.length - 1; j++) {
distSq = minDistanceToLineSq(cubicPoints[i], quadPoints[j], quadPoints[j + 1])
minDistSq = Math.min(minDistSq, distSq)
}
if (minDistSq > errorBoundSq) return false
}
for (i = 1; i < quadPoints.length - 1; i++) {
minDistSq = Infinity
for (j = 0; j < cubicPoints.length - 1; j++) {
distSq = minDistanceToLineSq(quadPoints[i], cubicPoints[j], cubicPoints[j + 1])
minDistSq = Math.min(minDistSq, distSq)
}
if (minDistSq > errorBoundSq) return false
}
return true
}
function _isApproximationClose (a, b, c, d, quadCurves, errorBound) {
var dt = 1 / quadCurves.length
for (var i = 0; i < quadCurves.length; i++) {
var p1 = quadCurves[i][0]
var c1 = quadCurves[i][1]
var p2 = quadCurves[i][2]
if (!isSegmentApproximationClose(a, b, c, d, i * dt, (i + 1) * dt, p1, c1, p2, errorBound)) {
return false
}
}
return true
}
function fromFlatArray (points) {
var result = []
var segmentsNumber = (points.length - 2) / 4
for (var i = 0; i < segmentsNumber; i++) {
result.push([
new Point(points[4 * i], points[4 * i + 1]),
new Point(points[4 * i + 2], points[4 * i + 3]),
new Point(points[4 * i + 4], points[4 * i + 5])
])
}
return result
}
function toFlatArray (quadsList) {
var result = []
result.push(quadsList[0][0].x)
result.push(quadsList[0][0].y)
for (var i = 0; i < quadsList.length; i++) {
result.push(quadsList[i][1].x)
result.push(quadsList[i][1].y)
result.push(quadsList[i][2].x)
result.push(quadsList[i][2].y)
}
return result
}
function isApproximationClose (p1x, p1y, c1x, c1y, c2x, c2y, p2x, p2y, quads, errorBound) {
// TODO: rewrite it in C-style and remove _isApproximationClose
var pc = calcPowerCoefficients(
new Point(p1x, p1y),
new Point(c1x, c1y),
new Point(c2x, c2y),
new Point(p2x, p2y)
)
return _isApproximationClose(pc[0], pc[1], pc[2], pc[3], fromFlatArray(quads), errorBound)
}
/*
* Split cubic bézier curve into two cubic curves, see details here:
* https://math.stackexchange.com/questions/877725
*/
function subdivideCubic (x1, y1, x2, y2, x3, y3, x4, y4, t) {
var u = 1 - t
var v = t
var bx = x1 * u + x2 * v
var sx = x2 * u + x3 * v
var fx = x3 * u + x4 * v
var cx = bx * u + sx * v
var ex = sx * u + fx * v
var dx = cx * u + ex * v
var by = y1 * u + y2 * v
var sy = y2 * u + y3 * v
var fy = y3 * u + y4 * v
var cy = by * u + sy * v
var ey = sy * u + fy * v
var dy = cy * u + ey * v
return [
[x1, y1, bx, by, cx, cy, dx, dy],
[dx, dy, ex, ey, fx, fy, x4, y4]
]
}
function byNumber (x, y) { return x - y }
/*
* Find inflection points on a cubic curve, algorithm is similar to this one:
* http://www.caffeineowl.com/graphics/2d/vectorial/cubic-inflexion.html
*/
function solveInflections (x1, y1, x2, y2, x3, y3, x4, y4) {
var p = -(x4 * (y1 - 2 * y2 + y3)) + x3 * (2 * y1 - 3 * y2 + y4) +
x1 * (y2 - 2 * y3 + y4) - x2 * (y1 - 3 * y3 + 2 * y4)
var q = x4 * (y1 - y2) + 3 * x3 * (-y1 + y2) + x2 * (2 * y1 - 3 * y3 + y4) - x1 * (2 * y2 - 3 * y3 + y4)
var r = x3 * (y1 - y2) + x1 * (y2 - y3) + x2 * (-y1 + y3)
return quadSolve(p, q, r).filter(function (t) { return t > 1e-8 && t < 1 - 1e-8 }).sort(byNumber)
}
/*
* Approximate cubic Bezier curve defined with base points p1, p2 and control points c1, c2 with
* with a few quadratic Bezier curves.
* The function uses tangent method to find quadratic approximation of cubic curve segment and
* simplified Hausdorff distance to determine number of segments that is enough to make error small.
* In general the method is the same as described here: https://fontforge.github.io/bezier.html.
*/
function _cubicToQuad (p1x, p1y, c1x, c1y, c2x, c2y, p2x, p2y, errorBound) {
var p1 = new Point(p1x, p1y)
var c1 = new Point(c1x, c1y)
var c2 = new Point(c2x, c2y)
var p2 = new Point(p2x, p2y)
var pc = calcPowerCoefficients(p1, c1, c2, p2)
var a = pc[0], b = pc[1], c = pc[2], d = pc[3]
var approximation
for (var segmentsCount = 1; segmentsCount <= 8; segmentsCount++) {
approximation = []
for (var t = 0; t < 1; t += (1 / segmentsCount)) {
approximation.push(processSegment(a, b, c, d, t, t + (1 / segmentsCount)))
}
if (segmentsCount === 1 &&
(approximation[0][1].sub(p1).dot(c1.sub(p1)) < 0 ||
approximation[0][1].sub(p2).dot(c2.sub(p2)) < 0)) {
// approximation concave, while the curve is convex (or vice versa)
continue
}
if (_isApproximationClose(a, b, c, d, approximation, errorBound)) {
break
}
}
return toFlatArray(approximation)
}
/*
* If this curve has any inflection points, split the curve and call
* _cubicToQuad function on each resulting curve.
*/
function cubicToQuad (p1x, p1y, c1x, c1y, c2x, c2y, p2x, p2y, errorBound) {
var inflections = solveInflections(p1x, p1y, c1x, c1y, c2x, c2y, p2x, p2y)
if (!inflections.length) {
return _cubicToQuad(p1x, p1y, c1x, c1y, c2x, c2y, p2x, p2y, errorBound)
}
var result = []
var curve = [p1x, p1y, c1x, c1y, c2x, c2y, p2x, p2y]
var prevPoint = 0
var quad, split
for (var inflectionIdx = 0; inflectionIdx < inflections.length; inflectionIdx++) {
split = subdivideCubic(
curve[0], curve[1], curve[2], curve[3],
curve[4], curve[5], curve[6], curve[7],
// we make a new curve, so adjust inflection point accordingly
1 - (1 - inflections[inflectionIdx]) / (1 - prevPoint)
)
quad = _cubicToQuad(
split[0][0], split[0][1], split[0][2], split[0][3],
split[0][4], split[0][5], split[0][6], split[0][7],
errorBound
)
result = result.concat(quad.slice(0, -2))
curve = split[1]
prevPoint = inflections[inflectionIdx]
}
quad = _cubicToQuad(
curve[0], curve[1], curve[2], curve[3],
curve[4], curve[5], curve[6], curve[7],
errorBound
)
return result.concat(quad)
}
module.exports = cubicToQuad
// following exports are for testing purposes
module.exports.isApproximationClose = isApproximationClose
//module.exports.cubicSolve = cubicSolve
module.exports.quadSolve = quadSolve