From 88f70ac2146f0c113dae3d375fc19da75f136ec3 Mon Sep 17 00:00:00 2001 From: Karl Wright Date: Thu, 5 May 2016 16:16:55 -0400 Subject: [PATCH] LUCENE-7241: More performance improvements --- .../apache/lucene/spatial3d/geom/Plane.java | 151 +++++++++++++++++- 1 file changed, 150 insertions(+), 1 deletion(-) diff --git a/lucene/spatial3d/src/java/org/apache/lucene/spatial3d/geom/Plane.java b/lucene/spatial3d/src/java/org/apache/lucene/spatial3d/geom/Plane.java index 759f88c93d7..670a2df96c3 100755 --- a/lucene/spatial3d/src/java/org/apache/lucene/spatial3d/geom/Plane.java +++ b/lucene/spatial3d/src/java/org/apache/lucene/spatial3d/geom/Plane.java @@ -1746,7 +1746,156 @@ public class Plane extends Vector { //System.err.println(" no notable points inside found; no intersection"); return false; } - return findIntersections(planetModel, q, bounds, moreBounds).length > 0; + + // Save on allocations; do inline instead of calling findIntersections + //System.err.println("Looking for intersection between plane "+this+" and plane "+q+" within bounds"); + // Unnormalized, unchecked... + final Vector lineVector = new Vector(y * q.z - z * q.y, z * q.x - x * q.z, x * q.y - y * q.x); + if (Math.abs(lineVector.x) < MINIMUM_RESOLUTION && Math.abs(lineVector.y) < MINIMUM_RESOLUTION && Math.abs(lineVector.z) < MINIMUM_RESOLUTION) { + // Degenerate case: parallel planes + //System.err.println(" planes are parallel - no intersection"); + return false; + } + + // The line will have the equation: A t + A0 = x, B t + B0 = y, C t + C0 = z. + // We have A, B, and C. In order to come up with A0, B0, and C0, we need to find a point that is on both planes. + // To do this, we find the largest vector value (either x, y, or z), and look for a point that solves both plane equations + // simultaneous. For example, let's say that the vector is (0.5,0.5,1), and the two plane equations are: + // 0.7 x + 0.3 y + 0.1 z + 0.0 = 0 + // and + // 0.9 x - 0.1 y + 0.2 z + 4.0 = 0 + // Then we'd pick z = 0, so the equations to solve for x and y would be: + // 0.7 x + 0.3y = 0.0 + // 0.9 x - 0.1y = -4.0 + // ... which can readily be solved using standard linear algebra. Generally: + // Q0 x + R0 y = S0 + // Q1 x + R1 y = S1 + // ... can be solved by Cramer's rule: + // x = det(S0 R0 / S1 R1) / det(Q0 R0 / Q1 R1) + // y = det(Q0 S0 / Q1 S1) / det(Q0 R0 / Q1 R1) + // ... where det( a b / c d ) = ad - bc, so: + // x = (S0 * R1 - R0 * S1) / (Q0 * R1 - R0 * Q1) + // y = (Q0 * S1 - S0 * Q1) / (Q0 * R1 - R0 * Q1) + double x0; + double y0; + double z0; + // We try to maximize the determinant in the denominator + final double denomYZ = this.y * q.z - this.z * q.y; + final double denomXZ = this.x * q.z - this.z * q.x; + final double denomXY = this.x * q.y - this.y * q.x; + if (Math.abs(denomYZ) >= Math.abs(denomXZ) && Math.abs(denomYZ) >= Math.abs(denomXY)) { + // X is the biggest, so our point will have x0 = 0.0 + if (Math.abs(denomYZ) < MINIMUM_RESOLUTION_SQUARED) { + //System.err.println(" Denominator is zero: no intersection"); + return false; + } + final double denom = 1.0 / denomYZ; + x0 = 0.0; + y0 = (-this.D * q.z - this.z * -q.D) * denom; + z0 = (this.y * -q.D + this.D * q.y) * denom; + } else if (Math.abs(denomXZ) >= Math.abs(denomXY) && Math.abs(denomXZ) >= Math.abs(denomYZ)) { + // Y is the biggest, so y0 = 0.0 + if (Math.abs(denomXZ) < MINIMUM_RESOLUTION_SQUARED) { + //System.err.println(" Denominator is zero: no intersection"); + return false; + } + final double denom = 1.0 / denomXZ; + x0 = (-this.D * q.z - this.z * -q.D) * denom; + y0 = 0.0; + z0 = (this.x * -q.D + this.D * q.x) * denom; + } else { + // Z is the biggest, so Z0 = 0.0 + if (Math.abs(denomXY) < MINIMUM_RESOLUTION_SQUARED) { + //System.err.println(" Denominator is zero: no intersection"); + return false; + } + final double denom = 1.0 / denomXY; + x0 = (-this.D * q.y - this.y * -q.D) * denom; + y0 = (this.x * -q.D + this.D * q.x) * denom; + z0 = 0.0; + } + + // Once an intersecting line is determined, the next step is to intersect that line with the ellipsoid, which + // will yield zero, one, or two points. + // The ellipsoid equation: 1,0 = x^2/a^2 + y^2/b^2 + z^2/c^2 + // 1.0 = (At+A0)^2/a^2 + (Bt+B0)^2/b^2 + (Ct+C0)^2/c^2 + // A^2 t^2 / a^2 + 2AA0t / a^2 + A0^2 / a^2 + B^2 t^2 / b^2 + 2BB0t / b^2 + B0^2 / b^2 + C^2 t^2 / c^2 + 2CC0t / c^2 + C0^2 / c^2 - 1,0 = 0.0 + // [A^2 / a^2 + B^2 / b^2 + C^2 / c^2] t^2 + [2AA0 / a^2 + 2BB0 / b^2 + 2CC0 / c^2] t + [A0^2 / a^2 + B0^2 / b^2 + C0^2 / c^2 - 1,0] = 0.0 + // Use the quadratic formula to determine t values and candidate point(s) + final double A = lineVector.x * lineVector.x * planetModel.inverseAbSquared + + lineVector.y * lineVector.y * planetModel.inverseAbSquared + + lineVector.z * lineVector.z * planetModel.inverseCSquared; + final double B = 2.0 * (lineVector.x * x0 * planetModel.inverseAbSquared + lineVector.y * y0 * planetModel.inverseAbSquared + lineVector.z * z0 * planetModel.inverseCSquared); + final double C = x0 * x0 * planetModel.inverseAbSquared + y0 * y0 * planetModel.inverseAbSquared + z0 * z0 * planetModel.inverseCSquared - 1.0; + + final double BsquaredMinus = B * B - 4.0 * A * C; + if (Math.abs(BsquaredMinus) < MINIMUM_RESOLUTION_SQUARED) { + //System.err.println(" One point of intersection"); + final double inverse2A = 1.0 / (2.0 * A); + // One solution only + final double t = -B * inverse2A; + // Maybe we can save ourselves the cost of construction of a point? + final double pointX = lineVector.x * t + x0; + final double pointY = lineVector.y * t + y0; + final double pointZ = lineVector.z * t + z0; + for (final Membership bound : bounds) { + if (!bound.isWithin(pointX, pointY, pointZ)) { + return false; + } + } + for (final Membership bound : moreBounds) { + if (!bound.isWithin(pointX, pointY, pointZ)) { + return false; + } + } + return true; + } else if (BsquaredMinus > 0.0) { + //System.err.println(" Two points of intersection"); + final double inverse2A = 1.0 / (2.0 * A); + // Two solutions + final double sqrtTerm = Math.sqrt(BsquaredMinus); + final double t1 = (-B + sqrtTerm) * inverse2A; + final double t2 = (-B - sqrtTerm) * inverse2A; + // Up to two points being returned. Do what we can to save on object creation though. + final double point1X = lineVector.x * t1 + x0; + final double point1Y = lineVector.y * t1 + y0; + final double point1Z = lineVector.z * t1 + z0; + boolean point1Valid = true; + for (final Membership bound : bounds) { + if (!bound.isWithin(point1X, point1Y, point1Z)) { + point1Valid = false; + break; + } + } + if (point1Valid) { + for (final Membership bound : moreBounds) { + if (!bound.isWithin(point1X, point1Y, point1Z)) { + point1Valid = false; + break; + } + } + } + if (point1Valid) { + return true; + } + final double point2X = lineVector.x * t2 + x0; + final double point2Y = lineVector.y * t2 + y0; + final double point2Z = lineVector.z * t2 + z0; + for (final Membership bound : bounds) { + if (!bound.isWithin(point2X, point2Y, point2Z)) { + return false; + } + } + for (final Membership bound : moreBounds) { + if (!bound.isWithin(point2X, point2Y, point2Z)) { + return false; + } + } + return true; + } else { + //System.err.println(" no solutions - no intersection"); + return false; + } } /**