///////////////////////////////////////////////////////////////////////////// // Name: polygon_test_point_inside.cpp ///////////////////////////////////////////////////////////////////////////// #include #include #include "PolyLine.h" using namespace std; /* this algo uses the the Jordan curve theorem to find if a point is inside or outside a polygon: * It run a semi-infinite line horizontally (increasing x, fixed y) * out from the test point, and count how many edges it crosses. * At each crossing, the ray switches between inside and outside. * If odd count, the test point is inside the polygon * This is called the Jordan curve theorem, or sometimes referred to as the "even-odd" test. */ /* 2 versions are given. * the second version is GPL (currently used) * the first version is for explanations and tests (used to test the second version) * both use the same algorithm. */ #if 0 /* This text and the algorithm come from http://www.ecse.rpi.edu/Homepages/wrf/Research/Short_Notes/pnpoly.html * * PNPOLY - Point Inclusion in Polygon Test * W. Randolph Franklin (WRF) * * Table of Contents * * 1. The C Code <#The C Code> * 2. The Method <#The Method> * 3. Originality <#Originality> * 4. The Inequality Tests are Tricky <#The Inequality Tests are Tricky> * 5. C Semantics <#C Semantics> * 6. Point on a (Boundary) Edge <#Point on an Edge> * 7. Multiple Components and Holes <#Listing the Vertices> * 8. Testing Which One of Many Polygons Contains the Point <#Testing a * Point Against Many Polygons> * 9. Explanation of /"for (i = 0, j = nvert-1; i < nvert; j = i++)"/ * <#Explanation> * 10. Fortran Code for the Point in Polygon Test <#Fortran Code for the * Point in Polygon Test> * 11. Converting the Code to All Integers <#Converting the Code to All * Integers> * 12. License to Use <#License to Use> * * The C Code * * Here is the code, for reference. Excluding lines with only braces, there * are only /7 lines/ of code. * * int pnpoly(int nvert, float *vertx, float *verty, float ref_pointX, float ref_pointY) * { * int i, j, c = 0; * for (i = 0, j = nvert-1; i < nvert; j = i++) { * if ( ((verty[i]>ref_pointY) != (verty[j]>ref_pointY)) && * (ref_pointX < (vertx[j]-vertx[i]) * (ref_pointY-verty[i]) / (verty[j]-verty[i]) + vertx[i]) ) * c = !c; * } * return c; * } * * Argument Meaning * nvert Number of vertices in the polygon. Whether to repeat the first * vertex at the end is discussed below. * vertx, verty Arrays containing the x- and y-coordinates of the * polygon's vertices. * ref_pointX, ref_pointY X- and y-coordinate of the test point. * * * The Method * * I run a semi-infinite ray horizontally (increasing x, fixed y) out from * the test point, and count how many edges it crosses. At each crossing, * the ray switches between inside and outside. This is called the /Jordan * curve theorem/. * * The case of the ray going thru a vertex is handled correctly via a * careful selection of inequalities. Don't mess with this code unless * you're familiar with the idea of /Simulation of Simplicity/. This * pretends to shift the ray infinitesimally to one side so that it either * clearly intersects, or clearly doesn't touch. Since this is merely a * conceptual, infinitesimal, shift, it never creates an intersection that * didn't exist before, and never destroys an intersection that clearly * existed before. * * The ray is tested against each edge thus: * * 1. Is the point in the half-plane below the extended edge? and * 2. Is the point's X coordinate within the edge's X-range? * * Handling endpoints here is tricky. * * * Originality * * I make no claim to having invented the idea. However in 1970, I did * produce the Fortran code given below on my own, and include it in a * package of cartographic SW publicly-distributed by David Douglas, Dept * of Geography, Simon Fraser U and U of Ottawa. * * Earlier implementations of point-in-polygon testing presumably exist, * tho the code might never have been released. Pointers to prior art, * especially publicly available code, are welcome. One early publication, * which doesn't handle the point on an edge, and has a typo, is this: * * M Shimrat, "Algorithm 112, Position of Point Relative to Polygon", * /Comm. ACM/ 5(8), Aug 1962, p 434. * * A well-written recent summary is this: * * E Haines, /Point in Polygon Strategies/, * http://www.acm.org/pubs/tog/editors/erich/ptinpoly/, 1994. * * * The Inequality Tests are Tricky * * If translating the program to another language, be sure to get the * inequalities in the conditional correct. They were carefully chosen to * make the program work correctly when the point is vertically below a vertex. * * Several people have thought that my program was wrong, when really * /they/ had gotten the inequalities wrong. * * * C Semantics * * My code uses the fact that, in the C language, when executing the code |a&&b|, if |a| is false, then |b| must not be evaluated. If your * compiler doesn't do this, then it's not implementing C, and you will get * a divide-by-zero, i.a., when the test point is vertically in line with a * vertical edge. When translating this code to another language with * different semantics, then you must implement this test explicitly. * * * Point on a (Boundary) Edge * * PNPOLY partitions the plane into points inside the polygon and points * outside the polygon. Points that are on the boundary are classified as * either inside or outside. * * 1. * * Any particular point is always classified consistently the same * way. In the following figure, consider what PNPOLY would say when * the red point, /P/, is tested against the two triangles, /T_L / * and /T_R /. Depending on internal roundoff errors, PNPOLY may say * that /P/ is in /T_L / or in /T_R /. However it will always give * the same answer when /P/ is tested against those triangles. That * is, if PNPOLY finds that /P/ is in /T_L /, then it will find that * /P/ is not /T_R /. If PNPOLY finds that /P/ is not in /T_L /, then * it will find that /P/ is in /T_R /. * * 2. If you want to know when a point is exactly on the boundary, you * need another program. This is only one of many functions that * PNPOLY lacks; it also doesn't predict tomorrow's weather. You are * free to extend PNPOLY's source code. * * 3. The first reason for this is the numerical analysis position that * you should not be testing exact equality unless your input is * exact. Even then, computational roundoff error would often make * the result wrong. * * 4. The second reason is that, if you partition a region of the plane * into polygons, i.e., form a planar graph, then PNPOLY will locate * each point into exactly one polygon. In other words, PNPOLY * considers each polygon to be topologically a semi-open set. This * makes things simpler, i.e., causes fewer special cases, if you use * PNPOLY as part of a larger system. Examples of this include * locating a point in a planar graph, and intersecting two planar * graphs. * * * Explanation of /"for (i = 0, j = nvert-1; i < nvert; j = i++)"/ * * The intention is to execute the loop for each i from 0 to nvert-1. For * each iteration, j is i-1. However that wraps, so if i=0 then j=nvert-1. * Therefore the current edge runs between verts j and i, and the loop is * done once per edge. In detail: * * 1. Start by setting i and j: * i = 0 * j = nvert-1 * 2. If i You may use my material for non-profit research * and education, provided that you credit me, and link back to my home page. * http://www.ecse.rpi.edu/Homepages/wrf/Research/Short_Notes/pnpoly.html, * 05/20/2008 20:36:42 */ bool TestPointInsidePolygon( std::vector aPolysList, int istart, int iend, int refx, int refy ) /** Function TestPointInsidePolygon * test if a point is inside or outside a polygon. * @param aPolysList: the list of polygons * @param istart: the starting point of a given polygon in m_FilledPolysList. * @param iend: the ending point of the polygon in m_FilledPolysList. * @param refx, refy: the point coordinate to test * @return true if the point is inside, false for outside */ { double ref_pointX = refx; double ref_pointY = refy; bool inside = false; for( int ii = istart, jj = iend; ii <= iend; jj = ii++ ) { double seg_startX, seg_startY; // starting point for the segment to test seg_startX = aPolysList[ii].x; seg_startY = aPolysList[ii].y; double seg_endX, seg_endY; // ending point for the segment to test seg_endX = aPolysList[jj].x; seg_endY = aPolysList[jj].y; if( ( ( seg_startY > ref_pointY ) != (seg_endY > ref_pointY ) ) && (ref_pointX < (seg_endX - seg_startX) * (ref_pointY - seg_startY) / (seg_endY - seg_startY) + seg_startX) ) inside = not inside; } return inside; } #else bool TestPointInsidePolygon( std::vector aPolysList, int istart, int iend, int refx, int refy ) /** Function TestPointInsidePolygon * test if a point is inside or outside a polygon. * if a point is on a outline segment, it is considered outside the polygon * the polygon must have only lines (not arcs) for outlines. * Use TestPointInside or TestPointInsideContour for more complex polygons * @param aPolysList: the list of polygons * @param istart: the starting point of a given polygon in m_FilledPolysList. * @param iend: the ending point of the polygon in m_FilledPolysList. * @param refx,refy: the point coordinate to test * @return true if the point is inside, false for outside */ { #define OUTSIDE_IF_ON_SIDE 0 // = 1 if we consider point on a side outside the polygon // define line passing through (x,y), with slope = 0 (horizontal line) // get intersection points // count intersection points to right of (x,y), if odd (x,y) is inside polyline int xx, yy; double slope = 0; // Using an horizontal line. double a = refy - slope * refx; int ics, ice; bool inside = false; // find all intersection points of line with polyline sides for( ics = istart, ice = iend; ics <= iend; ice = ics++ ) { double intersectx1, intersecty1, intersectx2, intersecty2; int ok; ok = FindLineSegmentIntersection( a, slope, aPolysList[ics].x, aPolysList[ics].y, aPolysList[ice].x, aPolysList[ice].y, CPolyLine::STRAIGHT, &intersectx1, &intersecty1, &intersectx2, &intersecty2 ); /* FindLineSegmentIntersection() returns 0, 1 or 2 coordinates (ok = 0, 1, 2) * for straight line segments, only 0 or 1 are possible * (2 intersections points are possible only with arcs */ if( ok ) // Intersection found { xx = (int) intersectx1; yy = (int) intersecty1; /* if the intersection point is on the start point of the current segment, * do not count it, * because it was already counted, as ending point of the previous segment */ if( xx == aPolysList[ics].x && yy == aPolysList[ics].y ) continue; #if OUTSIDE_IF_ON_SIDE if( xx == refx && yy == refy ) return false; // (x,y) is on a side, call it outside else #endif if( xx > refx ) inside = not inside; } } return inside; } #endif