private bool Legalize(SweepContext tcx, Triangle t)
{
// To legalize a triangle we start by finding if any of the three edges
// violate the Delaunay condition
for (int i = 0; i < 3; i++)
{
if (t.DelaunayEdge[i])
{
continue;
}
Triangle ot = t.GetNeighbor(i);
if (ot != null)
{
TriPoint p = t.Points[i];
TriPoint op = ot.OppositePoint(t, p);
int oi = ot.Index(op);
// If this is a Constrained Edge or a Delaunay Edge(only during recursive legalization)
// then we should not try to legalize
if (ot.ConstrainedEdge[oi] || ot.DelaunayEdge[oi])
{
t.ConstrainedEdge[i] = ot.ConstrainedEdge[oi];
continue;
}
bool inside = Incircle(p, t.PointCCW(p), t.PointCW(p), op);
if (inside)
{
// Lets mark this shared edge as Delaunay
t.DelaunayEdge[i] = true;
ot.DelaunayEdge[oi] = true;
// Lets rotate shared edge one vertex CW to legalize it
RotateTrianglePair(t, p, ot, op);
// We now got one valid Delaunay Edge shared by two triangles
// This gives us 4 new edges to check for Delaunay
// Make sure that triangle to node mapping is done only one time for a specific triangle
bool not_legalized = !Legalize(tcx, t);
if (not_legalized)
{
tcx.MapTriangleToNodes(t);
}
not_legalized = !Legalize(tcx, ot);
if (not_legalized)
{
tcx.MapTriangleToNodes(ot);
}
// Reset the Delaunay edges, since they only are valid Delaunay edges
// until we add a new triangle or point.
// XXX: need to think about this. Can these edges be tried after we
// return to previous recursive level?
t.DelaunayEdge[i] = false;
ot.DelaunayEdge[oi] = false;
// If triangle have been legalized no need to check the other edges since
// the recursive legalization will handles those so we can end here.
return(true);
}
}
}
return(false);
}
