scratchfoundation/paper.js
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Further simplify getConvexHull() by using p0 … p3 point objects.

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Jürg Lehni 2013-05-26 17:39:13 -07:00
parent 487c4c4aef
commit 90f0679700
1 changed files with 18 additions and 29 deletions
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47
src/path/Curve.js
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@ -1301,26 +1301,24 @@ new function() { // Scope for methods that require numerical integration
function getConvexHull(dq0, dq1, dq2, dq3) {
var getSignedDistance = Line.getSignedDistance,
distq1 = getSignedDistance(0, dq0, 1, dq3, 1 / 3, dq1),
distq2 = getSignedDistance(0, dq0, 1, dq3, 2 / 3, dq2);
// Check if [1/3, dq1] and [2/3, dq2] are on the same side of line
// [0,dq0, 1,dq3]
distq2 = getSignedDistance(0, dq0, 1, dq3, 2 / 3, dq2),
p0 = [ 0, dq0 ],
p1 = [ 1 / 3, dq1 ],
p2 = [ 2 / 3, dq2 ],
p3 = [ 1, dq3 ];
// Check if p1 and p2 are on the same side of the line [ p0, p3 ]
if (distq1 * distq2 < 0) {
// dq1 and dq2 lie on different sides on [0, q0, 1, q3]. The hull is
// dq1 and dq2 lie on different sides on [ p0, p3 ]. The hull is
// a quadrilateral and line [0, q0, 1, q3] is NOT part of the hull
// so we are pretty much done here.
return [
[ 2 / 3, dq2 ],
[ 0, dq0 ],
[ 1 / 3, dq1 ],
[ 1, dq3 ]
];
return [ p2, p0, p1, p3 ];
}
// dq1 and dq2 lie on the same sides on [0, q0, 1, q3]. The hull can be
// a triangle or a quadrilateral and line [0, q0, 1, q3] is part of the
// dq1 and dq2 lie on the same sides on [ 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.
var dqmax, cross;
var pmax, cross;
if (Math.abs(distq1) > Math.abs(distq2)) {
dqmax = [ 1 / 3, dq1 ];
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 = (vqa1a2X * vqa1MinY - vqa1a2Y * vqa1MinX)
@ -1328,29 +1326,20 @@ new function() { // Scope for methods that require numerical integration
cross = (dq3 - dq2 - (dq3 - dq0) / 3)
* (2 * (dq3 - dq2) - dq3 + dq1) / 3;
} else {
dqmax = [ 2 / 3, dq2 ];
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;
}
// Compare cross products of these vectors to determine, if
// point is in triangles [ dq3, dqMax, dq0 ] or [ dq0, dqMax, dq3 ]
// point is in triangles [ p3, pmax, p0 ] or [ p0, pmax, p3 ]
return cross < 0
// Point [2/3, dq2] is inside the triangle, hull is a triangle.
? [
[ 0, dq0 ],
dqmax,
[ 1, dq3 ]
]
// p2 is inside the triangle, hull is a triangle.
? [ p0, pmax, p3 ]
// Convexhull is a quadrilateral and we need all lines in the
// correct order where line [0, q0, 1, q3] is part of the hull.
: [
[ 0, dq0 ],
[ 1 / 3, dq1 ],
[ 2 / 3, dq2 ],
[ 1, dq3 ]
];
// correct order where line [ p1, p3 ] is part of the hull.
: [ p0, p1, p2, p3 ];
}
/*#*/ } // options.fatline