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 findRightmostEdgeAtNode() {
Node node = minDe.getNode();
DirectedEdgeStar star = (DirectedEdgeStar) node.getEdges();
minDe = star.getRightmostEdge();
// the DirectedEdge returned by the previous call is not
// necessarily in the forward direction. Use the sym edge if it isn't.
if (!minDe.isForward()) {
minDe = minDe.getSym();
minIndex = minDe.getEdge().getCoordinates().length - 1;
}
}
/**
* Computes the list of coordinates which are contained in this ring. The coordinatea are computed
* once only and cached.
*
* @return an array of the {@link Coordinate}s in this ring
*/
private Coordinate[] getCoordinates() {
if (ringPts == null) {
CoordinateList coordList = new CoordinateList();
for (Iterator i = deList.iterator(); i.hasNext(); ) {
DirectedEdge de = (DirectedEdge) i.next();
PolygonizeEdge edge = (PolygonizeEdge) de.getEdge();
addEdge(edge.getLine().getCoordinates(), de.getEdgeDirection(), coordList);
}
ringPts = coordList.toCoordinateArray();
}
return ringPts;
}
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;
}
}
private int getRightmostSideOfSegment(DirectedEdge de, int i) {
Edge e = de.getEdge();
Coordinate coord[] = e.getCoordinates();
if (i < 0 || i + 1 >= coord.length) return -1;
if (coord[i].y == coord[i + 1].y) return -1; // indicates edge is parallel to x-axis
int pos = Position.LEFT;
if (coord[i].y < coord[i + 1].y) pos = Position.RIGHT;
return pos;
}
private void checkForRightmostCoordinate(DirectedEdge de) {
Coordinate[] coord = de.getEdge().getCoordinates();
for (int i = 0; i < coord.length - 1; i++) {
// only check vertices which are the start or end point of a non-horizontal segment
// <FIX> MD 19 Sep 03 - NO! we can test all vertices, since the rightmost must have a
// non-horiz segment adjacent to it
if (minCoord == null || coord[i].x > minCoord.x) {
minDe = de;
minIndex = i;
minCoord = coord[i];
}
// }
}
}