////////////////////////////////////////////////////////////////////////////////////////////////////
/// <summary> Sort out zero length edge issues. </summary>
/// <remarks>
/// Zero length edges are a pain and have to be dealt with specially since they don't sort
/// properly using normal geometrical position nor can "sidedness" be determined solely from
/// their geometry (a zero length line has no "sides"). Instead, we have to look at the non-zero
/// length edges around them and determine this information by extrapolating from those edges
/// topological connection to this edge.
/// </remarks>
////////////////////////////////////////////////////////////////////////////////////////////////////
internal void HandleZeroLengthEdges()
{
// If no zero length edges
if (!FZeroLengthEdge)
{
// We're outta here
return;
}
// For every edge in the polygon
for (var i = 0; i < VertexCount; i++)
{
// Retrieve the edge
var edgeCheck = FortuneEdges[i];
// If it's zero length
if (edgeCheck.FZeroLength())
{
// Remove the edge from both this polygon and the polygon "across" the zero length edge
DetachEdge(edgeCheck);
FortuneEdges.Remove(edgeCheck);
edgeCheck.OtherPoly(this).FortuneEdges.Remove(edgeCheck);
// We have to back up one because we deleted edge i
i--;
}
}
}
////////////////////////////////////////////////////////////////////////////////////////////////////
/// <summary> Sort the edges in Clockwise order. </summary>
/// <remarks>
/// We do this partially by knowing that all the polygons in a Voronoi diagram are convex. That
/// means we can sort edges by measuring their angle around the generator for the polygon. We
/// have to pick the point to measure this angle carefully which is what
/// WEEdge.PolyOrderingTestPoint() does. We also have to make a special case for the rare doubly
/// infinite lines (such as that created with only two generators).
/// Darrellp, 2/18/2011.
/// </remarks>
////////////////////////////////////////////////////////////////////////////////////////////////////
internal void SortEdges()
{
// If there are only two edges
if (FortuneEdges.Count == 2)
{
// If they are split
if (FortuneEdges[0].FSplit)
{
// If they need reordering
//
// I don't think this is really necessary. It makes sense to order more than 2
// edges because they really have an "order" which will be maintained no matter which
// polygon they're ordered by (though the starting edge may change). With two oppositely
// directed rays as in this case, there is no such "order" independent of the polygon that
// does the ordering. I'm leaving it in but I don't think it's necessary.
// TODO: Check on this!
if (ICcw(
FortuneEdges[0].PolyOrderingTestPoint,
FortuneEdges[1].PolyOrderingTestPoint,
VoronoiPoint) < 0)
{
// Reorder them
var edgeT = FortuneEdges[0];
FortuneEdges[0] = FortuneEdges[1];
FortuneEdges[1] = edgeT;
}
}
else
{
// I think this represents an infinite polygon with only two edges.
// If not ordered around the single base point properly
if (ICcw(FortuneEdges[0].VtxStart.Pt,
FortuneEdges[0].PolyOrderingTestPoint,
FortuneEdges[1].PolyOrderingTestPoint) > 0)
{
// Swap the edges
var edgeT = FortuneEdges[0];
FortuneEdges[0] = FortuneEdges[1];
FortuneEdges[1] = edgeT;
}
}
}
else if (FortuneEdges.Count == 4 && FortuneEdges[0].FSplit)
{
// I think the nontransitivity of the "order" here will cause the CLR Sort()
// to screw up our ordering in this case so we handle it specially...
if (ICcw(FortuneEdges[0].VtxStart.Pt,
FortuneEdges[0].PolyOrderingTestPoint,
FortuneEdges[1].PolyOrderingTestPoint) > 0)
{
var edgeT = FortuneEdges[0];
FortuneEdges[0] = FortuneEdges[1];
FortuneEdges[1] = edgeT;
}
if (ICcw(FortuneEdges[2].VtxStart.Pt,
FortuneEdges[2].PolyOrderingTestPoint,
FortuneEdges[3].PolyOrderingTestPoint) > 0)
{
var edgeT = FortuneEdges[2];
FortuneEdges[2] = FortuneEdges[3];
FortuneEdges[3] = edgeT;
}
}
else
{
// More than 3 vertices just get a standard CLR sort
// TODO: Is nontransitivity screwing up here too and we're just not noticing?
FortuneEdges.Sort();
}
}
