/**
* Returns true if the edge (v0, v1) intersects the given longitude
* interval, and then saves 'v1' to be used as the next 'v0'.
*/
public bool Intersects(S2Point v1)
{
var lng1 = S2LatLng.Longitude(v1).Radians;
var result = interval.Intersects(S1Interval.FromPointPair(lng0, lng1));
lng0 = lng1;
return(result);
}
/**
* Return true if the edge AB intersects the given edge of constant latitude.
*/
private static bool IntersectsLatEdge(S2Point a, S2Point b, double lat,
S1Interval lng)
{
// Return true if the segment AB intersects the given edge of constant
// latitude. Unfortunately, lines of constant latitude are curves on
// the sphere. They can intersect a straight edge in 0, 1, or 2 points.
// assert (S2.isUnitLength(a) && S2.isUnitLength(b));
// First, compute the normal to the plane AB that points vaguely north.
var z = S2Point.Normalize(S2.RobustCrossProd(a, b));
if (z.Z < 0)
{
z = -z;
}
// Extend this to an orthonormal frame (x,y,z) where x is the direction
// where the great circle through AB achieves its maximium latitude.
var y = S2Point.Normalize(S2.RobustCrossProd(z, new S2Point(0, 0, 1)));
var x = S2Point.CrossProd(y, z);
// assert (S2.isUnitLength(x) && x.z >= 0);
// Compute the angle "theta" from the x-axis (in the x-y plane defined
// above) where the great circle intersects the given line of latitude.
var sinLat = Math.Sin(lat);
if (Math.Abs(sinLat) >= x.Z)
{
return(false); // The great circle does not reach the given latitude.
}
// assert (x.z > 0);
var cosTheta = sinLat / x.Z;
var sinTheta = Math.Sqrt(1 - cosTheta * cosTheta);
var theta = Math.Atan2(sinTheta, cosTheta);
// The candidate intersection points are located +/- theta in the x-y
// plane. For an intersection to be valid, we need to check that the
// intersection point is contained in the interior of the edge AB and
// also that it is contained within the given longitude interval "lng".
// Compute the range of theta values spanned by the edge AB.
var abTheta = S1Interval.FromPointPair(Math.Atan2(
a.DotProd(y), a.DotProd(x)), Math.Atan2(b.DotProd(y), b.DotProd(x)));
if (abTheta.Contains(theta))
{
// Check if the intersection point is also in the given "lng" interval.
var isect = (x * cosTheta) + (y * sinTheta);
if (lng.Contains(Math.Atan2(isect.Y, isect.X)))
{
return(true);
}
}
if (abTheta.Contains(-theta))
{
// Check if the intersection point is also in the given "lng" interval.
var intersection = (x * cosTheta) - (y * sinTheta);
if (lng.Contains(Math.Atan2(intersection.Y, intersection.X)))
{
return(true);
}
}
return(false);
}
/**
* Return the minimum distance (measured along the surface of the sphere) to
* the given S2LatLngRect. Both S2LatLngRects must be non-empty.
*/
public S1Angle GetDistance(S2LatLngRect other)
{
var a = this;
var b = other;
Preconditions.CheckState(!a.IsEmpty);
Preconditions.CheckArgument(!b.IsEmpty);
// First, handle the trivial cases where the longitude intervals overlap.
if (a.Lng.Intersects(b.Lng))
{
if (a.Lat.Intersects(b.Lat))
{
return(S1Angle.FromRadians(0)); // Intersection between a and b.
}
// We found an overlap in the longitude interval, but not in the latitude
// interval. This means the shortest path travels along some line of
// longitude connecting the high-latitude of the lower rect with the
// low-latitude of the higher rect.
S1Angle lo, hi;
if (a.Lat.Lo > b.Lat.Hi)
{
lo = b.LatHi;
hi = a.LatLo;
}
else
{
lo = a.LatHi;
hi = b.LatLo;
}
return(S1Angle.FromRadians(hi.Radians - lo.Radians));
}
// The longitude intervals don't overlap. In this case, the closest points
// occur somewhere on the pair of longitudinal edges which are nearest in
// longitude-space.
S1Angle aLng, bLng;
var loHi = S1Interval.FromPointPair(a.Lng.Lo, b.Lng.Hi);
var hiLo = S1Interval.FromPointPair(a.Lng.Hi, b.Lng.Lo);
if (loHi.Length < hiLo.Length)
{
aLng = a.LngLo;
bLng = b.LngHi;
}
else
{
aLng = a.LngHi;
bLng = b.LngLo;
}
// The shortest distance between the two longitudinal segments will include
// at least one segment endpoint. We could probably narrow this down further
// to a single point-edge distance by comparing the relative latitudes of the
// endpoints, but for the sake of clarity, we'll do all four point-edge
// distance tests.
var aLo = new S2LatLng(a.LatLo, aLng).ToPoint();
var aHi = new S2LatLng(a.LatHi, aLng).ToPoint();
var aLoCrossHi =
S2LatLng.FromRadians(0, aLng.Radians - S2.PiOver2).Normalized.ToPoint();
var bLo = new S2LatLng(b.LatLo, bLng).ToPoint();
var bHi = new S2LatLng(b.LatHi, bLng).ToPoint();
var bLoCrossHi =
S2LatLng.FromRadians(0, bLng.Radians - S2.PiOver2).Normalized.ToPoint();
return(S1Angle.Min(S2EdgeUtil.GetDistance(aLo, bLo, bHi, bLoCrossHi),
S1Angle.Min(S2EdgeUtil.GetDistance(aHi, bLo, bHi, bLoCrossHi),
S1Angle.Min(S2EdgeUtil.GetDistance(bLo, aLo, aHi, aLoCrossHi),
S2EdgeUtil.GetDistance(bHi, aLo, aHi, aLoCrossHi)))));
}
/**
* Returns true if this rectangle intersects the given cell. (This is an exact
* test and may be fairly expensive, see also MayIntersect below.)
*/
public bool Intersects(S2Cell cell)
{
// First we eliminate the cases where one region completely contains the
// other. Once these are disposed of, then the regions will intersect
// if and only if their boundaries intersect.
if (IsEmpty)
{
return(false);
}
if (Contains(cell.Center))
{
return(true);
}
if (cell.Contains(Center.ToPoint()))
{
return(true);
}
// Quick rejection test (not required for correctness).
if (!Intersects(cell.RectBound))
{
return(false);
}
// Now check whether the boundaries intersect. Unfortunately, a
// latitude-longitude rectangle does not have straight edges -- two edges
// are curved, and at least one of them is concave.
// Precompute the cell vertices as points and latitude-longitudes.
var cellV = new S2Point[4];
var cellLl = new S2LatLng[4];
for (var i = 0; i < 4; ++i)
{
cellV[i] = cell.GetVertex(i); // Must be normalized.
cellLl[i] = new S2LatLng(cellV[i]);
if (Contains(cellLl[i]))
{
return(true); // Quick acceptance test.
}
}
for (var i = 0; i < 4; ++i)
{
var edgeLng = S1Interval.FromPointPair(
cellLl[i].Lng.Radians, cellLl[(i + 1) & 3].Lng.Radians);
if (!_lng.Intersects(edgeLng))
{
continue;
}
var a = cellV[i];
var b = cellV[(i + 1) & 3];
if (edgeLng.Contains(_lng.Lo))
{
if (IntersectsLngEdge(a, b, _lat, _lng.Lo))
{
return(true);
}
}
if (edgeLng.Contains(_lng.Hi))
{
if (IntersectsLngEdge(a, b, _lat, _lng.Hi))
{
return(true);
}
}
if (IntersectsLatEdge(a, b, _lat.Lo, _lng))
{
return(true);
}
if (IntersectsLatEdge(a, b, _lat.Hi, _lng))
{
return(true);
}
}
return(false);
}
/**
* Convenience method to construct the minimal bounding rectangle containing
* the two given points. This is equivalent to starting with an empty
* rectangle and calling AddPoint() twice. Note that it is different than the
* S2LatLngRect(lo, hi) constructor, where the first point is always used as
* the lower-left corner of the resulting rectangle.
*/
public static S2LatLngRect FromPointPair(S2LatLng p1, S2LatLng p2)
{
// assert (p1.isValid() && p2.isValid());
return(new S2LatLngRect(R1Interval.FromPointPair(p1.Lat.Radians, p2.Lat.Radians), S1Interval.FromPointPair(p1.Lng.Radians, p2.Lng.Radians)));
}