h3_bbox.c (82 lines of code) (raw):

/* * Copyright 2016-2021 Uber Technologies, Inc. * * Licensed under the Apache License, Version 2.0 (the "License"); * you may not use this file except in compliance with the License. * You may obtain a copy of the License at * * http://www.apache.org/licenses/LICENSE-2.0 * * Unless required by applicable law or agreed to in writing, software * distributed under the License is distributed on an "AS IS" BASIS, * WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied. * See the License for the specific language governing permissions and * limitations under the License. */ /** @file bbox.c * @brief Geographic bounding box functions */ #include "h3_bbox.h" #include <float.h> #include <math.h> #include <stdbool.h> #include "h3_constants.h" #include "h3_h3Index.h" #include "h3_latLng.h" /** * Whether the given bounding box crosses the antimeridian * @param bbox Bounding box to inspect * @return is transmeridian */ bool bboxIsTransmeridian(const BBox *bbox) { return bbox->east < bbox->west; } /** * Get the center of a bounding box * @param bbox Input bounding box * @param center Output center coordinate */ void bboxCenter(const BBox *bbox, LatLng *center) { center->lat = (bbox->north + bbox->south) / 2.0; // If the bbox crosses the antimeridian, shift east 360 degrees double east = bboxIsTransmeridian(bbox) ? bbox->east + M_2PI : bbox->east; center->lng = constrainLng((east + bbox->west) / 2.0); } /** * Whether the bounding box contains a given point * @param bbox Bounding box * @param point Point to test * @return Whether the point is contained */ bool bboxContains(const BBox *bbox, const LatLng *point) { return point->lat >= bbox->south && point->lat <= bbox->north && (bboxIsTransmeridian(bbox) ? // transmeridian case (point->lng >= bbox->west || point->lng <= bbox->east) : // standard case (point->lng >= bbox->west && point->lng <= bbox->east)); } /** * Whether two bounding boxes are strictly equal * @param b1 Bounding box 1 * @param b2 Bounding box 2 * @return Whether the boxes are equal */ bool bboxEquals(const BBox *b1, const BBox *b2) { return b1->north == b2->north && b1->south == b2->south && b1->east == b2->east && b1->west == b2->west; } /** * _hexRadiusKm returns the radius of a given hexagon in Km * * @param h3Index the index of the hexagon * @return the radius of the hexagon in Km */ double _hexRadiusKm(H3Index h3Index) { // There is probably a cheaper way to determine the radius of a // hexagon, but this way is conceptually simple LatLng h3Center; CellBoundary h3Boundary; H3_EXPORT(cellToLatLng)(h3Index, &h3Center); H3_EXPORT(cellToBoundary)(h3Index, &h3Boundary); return H3_EXPORT(greatCircleDistanceKm)(&h3Center, h3Boundary.verts); } /** * bboxHexEstimate returns an estimated number of hexagons that fit * within the cartesian-projected bounding box * * @param bbox the bounding box to estimate the hexagon fill level * @param res the resolution of the H3 hexagons to fill the bounding box * @param out the estimated number of hexagons to fill the bounding box * @return E_SUCCESS (0) on success, or another value otherwise. */ H3Error bboxHexEstimate(const BBox *bbox, int res, int64_t *out) { // Get the area of the pentagon as the maximally-distorted area possible H3Index pentagons[12] = {0}; H3Error pentagonsErr = H3_EXPORT(getPentagons)(res, pentagons); if (pentagonsErr) { return pentagonsErr; } double pentagonRadiusKm = _hexRadiusKm(pentagons[0]); // Area of a regular hexagon is 3/2*sqrt(3) * r * r // The pentagon has the most distortion (smallest edges) and shares its // edges with hexagons, so the most-distorted hexagons have this area, // shrunk by 20% off chance that the bounding box perfectly bounds a // pentagon. double pentagonAreaKm2 = 0.8 * (2.59807621135 * pentagonRadiusKm * pentagonRadiusKm); // Then get the area of the bounding box of the geoloop in question LatLng p1, p2; p1.lat = bbox->north; p1.lng = bbox->east; p2.lat = bbox->south; p2.lng = bbox->west; double d = H3_EXPORT(greatCircleDistanceKm)(&p1, &p2); double lngDiff = fabs(p1.lng - p2.lng); double latDiff = fabs(p1.lat - p2.lat); if (lngDiff == 0 || latDiff == 0) { return E_FAILED; } double length = fmax(lngDiff, latDiff); double width = fmin(lngDiff, latDiff); double ratio = length / width; // Derived constant based on: https://math.stackexchange.com/a/1921940 // Clamped to 3 as higher values tend to rapidly drag the estimate to zero. double a = d * d / fmin(3.0, ratio); // Divide the two to get an estimate of the number of hexagons needed double estimateDouble = ceil(a / pentagonAreaKm2); if (!isfinite(estimateDouble)) { return E_FAILED; } int64_t estimate = (int64_t)estimateDouble; if (estimate == 0) estimate = 1; *out = estimate; return E_SUCCESS; } /** * lineHexEstimate returns an estimated number of hexagons that trace * the cartesian-projected line * * @param origin the origin coordinates * @param destination the destination coordinates * @param res the resolution of the H3 hexagons to trace the line * @param out Out parameter for the estimated number of hexagons required to * trace the line * @return E_SUCCESS (0) on success or another value otherwise. */ H3Error lineHexEstimate(const LatLng *origin, const LatLng *destination, int res, int64_t *out) { // Get the area of the pentagon as the maximally-distorted area possible H3Index pentagons[12] = {0}; H3Error pentagonsErr = H3_EXPORT(getPentagons)(res, pentagons); if (pentagonsErr) { return pentagonsErr; } double pentagonRadiusKm = _hexRadiusKm(pentagons[0]); double dist = H3_EXPORT(greatCircleDistanceKm)(origin, destination); double distCeil = ceil(dist / (2 * pentagonRadiusKm)); if (!isfinite(distCeil)) { return E_FAILED; } int64_t estimate = (int64_t)distCeil; if (estimate == 0) estimate = 1; *out = estimate; return E_SUCCESS; }