public void findEdge(List dirEdgeList) {
/**
* Check all forward DirectedEdges only. This is still general, because each edge has a forward
* DirectedEdge.
*/
for (Iterator i = dirEdgeList.iterator(); i.hasNext(); ) {
DirectedEdge de = (DirectedEdge) i.next();
if (!de.isForward()) continue;
checkForRightmostCoordinate(de);
}
/**
* If the rightmost point is a node, we need to identify which of the incident edges is
* rightmost.
*/
Assert.isTrue(
minIndex != 0 || minCoord.equals(minDe.getCoordinate()),
"inconsistency in rightmost processing");
if (minIndex == 0) {
findRightmostEdgeAtNode();
} else {
findRightmostEdgeAtVertex();
}
/** now check that the extreme side is the R side. If not, use the sym instead. */
orientedDe = minDe;
int rightmostSide = getRightmostSide(minDe, minIndex);
if (rightmostSide == Position.LEFT) {
orientedDe = minDe.getSym();
}
}
private void findRightmostEdgeAtVertex() {
/**
* The rightmost point is an interior vertex, so it has a segment on either side of it. If these
* segments are both above or below the rightmost point, we need to determine their relative
* orientation to decide which is rightmost.
*/
Coordinate[] pts = minDe.getEdge().getCoordinates();
Assert.isTrue(
minIndex > 0 && minIndex < pts.length,
"rightmost point expected to be interior vertex of edge");
Coordinate pPrev = pts[minIndex - 1];
Coordinate pNext = pts[minIndex + 1];
int orientation = CGAlgorithms.computeOrientation(minCoord, pNext, pPrev);
boolean usePrev = false;
// both segments are below min point
if (pPrev.y < minCoord.y
&& pNext.y < minCoord.y
&& orientation == CGAlgorithms.COUNTERCLOCKWISE) {
usePrev = true;
} else if (pPrev.y > minCoord.y
&& pNext.y > minCoord.y
&& orientation == CGAlgorithms.CLOCKWISE) {
usePrev = true;
}
// if both segments are on the same side, do nothing - either is safe
// to select as a rightmost segment
if (usePrev) {
minIndex = minIndex - 1;
}
}