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;
}
}
public void computeIntersection(Coordinate p, Coordinate p1, Coordinate p2) {
isProper = false;
// do between check first, since it is faster than the orientation test
if (Envelope.intersects(p1, p2, p)) {
if ((CGAlgorithms.orientationIndex(p1, p2, p) == 0)
&& (CGAlgorithms.orientationIndex(p2, p1, p) == 0)) {
isProper = true;
if (p.equals(p1) || p.equals(p2)) {
isProper = false;
}
result = POINT_INTERSECTION;
return;
}
}
result = NO_INTERSECTION;
}
/**
* Find the innermost enclosing shell EdgeRing containing the argument EdgeRing, if any. The
* innermost enclosing ring is the <i>smallest</i> enclosing ring. The algorithm used depends on
* the fact that: <br>
* ring A contains ring B iff envelope(ring A) contains envelope(ring B) <br>
* This routine is only safe to use if the chosen point of the hole is known to be properly
* contained in a shell (which is guaranteed to be the case if the hole does not touch its shell)
*
* @return containing EdgeRing, if there is one or null if no containing EdgeRing is found
*/
public static EdgeRing findEdgeRingContaining(EdgeRing testEr, List shellList) {
LinearRing testRing = testEr.getRing();
Envelope testEnv = testRing.getEnvelopeInternal();
Coordinate testPt = testRing.getCoordinateN(0);
EdgeRing minShell = null;
Envelope minShellEnv = null;
for (Iterator it = shellList.iterator(); it.hasNext(); ) {
EdgeRing tryShell = (EdgeRing) it.next();
LinearRing tryShellRing = tryShell.getRing();
Envelope tryShellEnv = tryShellRing.getEnvelopeInternal();
// the hole envelope cannot equal the shell envelope
// (also guards against testing rings against themselves)
if (tryShellEnv.equals(testEnv)) continue;
// hole must be contained in shell
if (!tryShellEnv.contains(testEnv)) continue;
testPt =
CoordinateArrays.ptNotInList(testRing.getCoordinates(), tryShellRing.getCoordinates());
boolean isContained = false;
if (CGAlgorithms.isPointInRing(testPt, tryShellRing.getCoordinates())) isContained = true;
// check if this new containing ring is smaller than the current minimum ring
if (isContained) {
if (minShell == null || minShellEnv.contains(tryShellEnv)) {
minShell = tryShell;
minShellEnv = minShell.getRing().getEnvelopeInternal();
}
}
}
return minShell;
}
/**
* Finds the endpoint of the segments P and Q which is closest to the other segment. This is a
* reasonable surrogate for the true intersection points in ill-conditioned cases (e.g. where two
* segments are nearly coincident, or where the endpoint of one segment lies almost on the other
* segment).
*
* <p>This replaces the older CentralEndpoint heuristic, which chose the wrong endpoint in some
* cases where the segments had very distinct slopes and one endpoint lay almost on the other
* segment.
*
* @param p1 an endpoint of segment P
* @param p2 an endpoint of segment P
* @param q1 an endpoint of segment Q
* @param q2 an endpoint of segment Q
* @return the nearest endpoint to the other segment
*/
private static Coordinate nearestEndpoint(
Coordinate p1, Coordinate p2, Coordinate q1, Coordinate q2) {
Coordinate nearestPt = p1;
double minDist = CGAlgorithms.distancePointLine(p1, q1, q2);
double dist = CGAlgorithms.distancePointLine(p2, q1, q2);
if (dist < minDist) {
minDist = dist;
nearestPt = p2;
}
dist = CGAlgorithms.distancePointLine(q1, p1, p2);
if (dist < minDist) {
minDist = dist;
nearestPt = q1;
}
dist = CGAlgorithms.distancePointLine(q2, p1, p2);
if (dist < minDist) {
minDist = dist;
nearestPt = q2;
}
return nearestPt;
}
protected int computeIntersect(Coordinate p1, Coordinate p2, Coordinate q1, Coordinate q2) {
isProper = false;
// first try a fast test to see if the envelopes of the lines intersect
if (!Envelope.intersects(p1, p2, q1, q2)) return NO_INTERSECTION;
// for each endpoint, compute which side of the other segment it lies
// if both endpoints lie on the same side of the other segment,
// the segments do not intersect
int Pq1 = CGAlgorithms.orientationIndex(p1, p2, q1);
int Pq2 = CGAlgorithms.orientationIndex(p1, p2, q2);
if ((Pq1 > 0 && Pq2 > 0) || (Pq1 < 0 && Pq2 < 0)) {
return NO_INTERSECTION;
}
int Qp1 = CGAlgorithms.orientationIndex(q1, q2, p1);
int Qp2 = CGAlgorithms.orientationIndex(q1, q2, p2);
if ((Qp1 > 0 && Qp2 > 0) || (Qp1 < 0 && Qp2 < 0)) {
return NO_INTERSECTION;
}
boolean collinear = Pq1 == 0 && Pq2 == 0 && Qp1 == 0 && Qp2 == 0;
if (collinear) {
return computeCollinearIntersection(p1, p2, q1, q2);
}
/**
* At this point we know that there is a single intersection point (since the lines are not
* collinear).
*/
/**
* Check if the intersection is an endpoint. If it is, copy the endpoint as the intersection
* point. Copying the point rather than computing it ensures the point has the exact value,
* which is important for robustness. It is sufficient to simply check for an endpoint which is
* on the other line, since at this point we know that the inputLines must intersect.
*/
if (Pq1 == 0 || Pq2 == 0 || Qp1 == 0 || Qp2 == 0) {
isProper = false;
/**
* Check for two equal endpoints. This is done explicitly rather than by the orientation tests
* below in order to improve robustness.
*
* <p>[An example where the orientation tests fail to be consistent is the following (where
* the true intersection is at the shared endpoint POINT (19.850257749638203
* 46.29709338043669)
*
* <p>LINESTRING ( 19.850257749638203 46.29709338043669, 20.31970698357233 46.76654261437082 )
* and LINESTRING ( -48.51001596420236 -22.063180333403878, 19.850257749638203
* 46.29709338043669 )
*
* <p>which used to produce the INCORRECT result: (20.31970698357233, 46.76654261437082, NaN)
*/
if (p1.equals2D(q1) || p1.equals2D(q2)) {
intPt[0] = p1;
} else if (p2.equals2D(q1) || p2.equals2D(q2)) {
intPt[0] = p2;
}
/** Now check to see if any endpoint lies on the interior of the other segment. */
else if (Pq1 == 0) {
intPt[0] = new Coordinate(q1);
} else if (Pq2 == 0) {
intPt[0] = new Coordinate(q2);
} else if (Qp1 == 0) {
intPt[0] = new Coordinate(p1);
} else if (Qp2 == 0) {
intPt[0] = new Coordinate(p2);
}
} else {
isProper = true;
intPt[0] = intersection(p1, p2, q1, q2);
}
return POINT_INTERSECTION;
}
/**
* Tests whether this ring is a hole. Due to the way the edges in the polyongization graph are
* linked, a ring is a hole if it is oriented counter-clockwise.
*
* @return <code>true</code> if this ring is a hole
*/
public boolean isHole() {
LinearRing ring = getRing();
return CGAlgorithms.isCCW(ring.getCoordinates());
}