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Merge pull request #797 from iconexperience/New-getConvexHull

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Faster simpler and more reliable implementation of getConvexHull()
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Jürg Lehni 2015-10-06 16:03:45 +02:00
commit 6b421d5f72
1 changed files with 19 additions and 34 deletions
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src/path/Curve.js
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@ -1535,11 +1535,9 @@ new function() { // Scope for intersection using bezier fat-line clipping
p1 = [ 1 / 3, dq1 ],
p2 = [ 2 / 3, dq2 ],
p3 = [ 1, dq3 ],
// Find signed distance of p1 and p2 from line [ p0, p3 ]
getSignedDistance = Line.getSignedDistance,
dist1 = getSignedDistance(0, dq0, 1, dq3, 1 / 3, dq1),
dist2 = getSignedDistance(0, dq0, 1, dq3, 2 / 3, dq2),
flip = false,
// Find vertical signed distance of p1 and p2 from line [ p0, p3 ]
dist1 = dq1 - (dq0 + (dq3 - dq0) / 3),
dist2 = dq2 - (dq0 + 2 * (dq3 - dq0) / 3),
hull;
// Check if p1 and p2 are on the same side of the line [ p0, p3 ]
if (dist1 * dist2 < 0) {
@ -1549,40 +1547,27 @@ new function() { // Scope for intersection using bezier fat-line clipping
// The top part includes p1,
// we will reverse it later if that is not the case
hull = [[p0, p1, p3], [p0, p2, p3]];
flip = dist1 < 0;
} else {
// p1 and p2 lie on the same sides of [ p0, p3 ]. The hull can be
// a triangle or a quadrilateral and line [ p0, p3 ] is part of the
// hull. Check if the hull is a triangle or a quadrilateral.
// Also, if at least one of the distances for p1 or p2, from line
// [p0, p3] is zero then hull must at most have 3 vertices.
var pmax, cross = 0,
distZero = dist1 === 0 || dist2 === 0;
if (Math.abs(dist1) > Math.abs(dist2)) {
pmax = p1;
// apex is dq3 and the other apex point is dq0 vector dqapex ->
// dqapex2 or base vector which is already part of the hull.
cross = (dq3 - dq2 - (dq3 - dq0) / 3)
* (2 * (dq3 - dq2) - dq3 + dq1) / 3;
} else {
pmax = p2;
// apex is dq0 in this case, and the other apex point is dq3
// vector dqapex -> dqapex2 or base vector which is already part
// of the hull.
cross = (dq1 - dq0 + (dq0 - dq3) / 3)
* (-2 * (dq0 - dq1) + dq0 - dq2) / 3;
// hull. Check if the hull is a triangle or a quadrilateral. We have a
// triangle if the vertical distance of one of the middle points (p1, p2)
// is equal or less than half the vertical distance of the other middle point.
var distRatio = dist1 / dist2;
hull = [
// p2 is inside, the hull is a triangle.
distRatio >= 2 ? [p0, p1, p3]
// p1 is inside, the hull is a triangle.
: distRatio <= .5 ? [p0, p2, p3]
// Hull is a quadrilateral, we need all lines in correct order.
: [p0, p1, p2, p3],
// Line [p0, p3] is part of the hull.
[p0, p3]
];
}
// Compare cross products of these vectors to determine if the point
// is in the triangle [ p3, pmax, p0 ], or if it is a quadrilateral.
hull = cross < 0 || distZero
// p2 is inside the triangle, hull is a triangle.
? [[p0, pmax, p3], [p0, p3]]
// Convex hull is a quadrilateral and we need all lines in
// correct order where line [ p0, p3 ] is part of the hull.
: [[p0, p1, p2, p3], [p0, p3]];
flip = dist1 ? dist1 < 0 : dist2 < 0;
}
return flip ? hull.reverse() : hull;
// flip hull if dist1 is negative or if it is zero and dist2 is negative
return (dist1 || dist2) < 0 ? hull.reverse() : hull;
}
/**