kicad/polygon/polygon_test_point_inside.cpp

/*
 * This program source code file is part of KiCad, a free EDA CAD application.
 *
 * Copyright (C) 2007-2014 Jean-Pierre Charras, jp.charras at wanadoo.fr
 * Copyright (C) 2007-2014 KiCad Developers, see CHANGELOG.TXT for contributors.
 *
 * This program is free software; you can redistribute it and/or
 * modify it under the terms of the GNU General Public License
 * as published by the Free Software Foundation; either version 2
 * of the License, or (at your option) any later version.
 *
 * This program is distributed in the hope that it will be useful,
 * but WITHOUT ANY WARRANTY; without even the implied warranty of
 * MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
 * GNU General Public License for more details.
 *
 * You should have received a copy of the GNU General Public License
 * along with this program; if not, you may find one here:
 * http://www.gnu.org/licenses/old-licenses/gpl-2.0.html
 * or you may search the http://www.gnu.org website for the version 2 license,
 * or you may write to the Free Software Foundation, Inc.,
 * 51 Franklin Street, Fifth Floor, Boston, MA  02110-1301, USA
 */

/**
 * @file polygon_test_point_inside.cpp
 */

#include <cmath>
#include <vector>

#include "polygon_test_point_inside.h"

/* 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.
 * Take care to starting and ending points of segments outlines, when the horizontal line crosses a segment outline
 * exactly on an ending point:
 * Because the starting point of a segment is also the ending point of the previous, only one must be used.
 * And we do no use twice the same segment, so we do NOT use both starting and ending points of these 2 segments.
 * So we must use only one ending point of each segment when calculating intersections
 * but it cannot be always the starting or the ending point. This depends on relative position of 2 consectutive segments
 * Here, the ending point above the Y reference position is used
 *   and the ending point below or equal the Y reference position is NOT used
 * Obviously, others cases are irrelevant because there is not intersection.
 */

#define OUTSIDE false
#define INSIDE true


/* Function TestPointInsidePolygon (overlaid)
 * same as previous, but use wxPoint and aCount corners
 */
bool TestPointInsidePolygon( const wxPoint *aPolysList, int aCount, const wxPoint &aRefPoint )
{
    // count intersection points to right of (refx,refy). If odd number, point (refx,refy) is inside polyline
    int ics, ice;
    int count = 0;
    // find all intersection points of line with polyline sides
    for( ics = 0, ice = aCount-1; ics < aCount; ice = ics++ )
    {
        int seg_startX = aPolysList[ics].x;
        int seg_startY = aPolysList[ics].y;
        int seg_endX   = aPolysList[ice].x;
        int seg_endY   = aPolysList[ice].y;

        /* Trivial cases: skip if ref above or below the segment to test */
        if( ( seg_startY > aRefPoint.y ) && (seg_endY > aRefPoint.y ) )
            continue;

        // segment below ref point, or one of its ends has the same Y pos as the ref point: skip
        // So we eliminate one end point of 2 consecutive segments.
        // Note: also we skip horizontal segments if ref point is on this horizontal line
        // So reference points on horizontal segments outlines always are seen as outside the polygon
        if( ( seg_startY <= aRefPoint.y ) && (seg_endY <= aRefPoint.y ) )
            continue;

        /* refy is between seg_startY and seg_endY.
         * note: here: horizontal segments (seg_startY == seg_endY) are skipped,
         * either by the first test or by the second test
         * see if an horizontal semi infinite line from refx is intersecting the segment
         */

        // calculate the x position of the intersection of this segment and the semi infinite line
        // this is more easier if we move the X,Y axis origin to the segment start point:
        seg_endX -= seg_startX;
        seg_endY -= seg_startY;
        double newrefx = (double) (aRefPoint.x - seg_startX);
        double newrefy = (double) (aRefPoint.y - seg_startY);

        // Now calculate the x intersection coordinate of the line from (0,0) to (seg_endX,seg_endY)
        // with the horizontal line at the new refy position
        // the line slope  = seg_endY/seg_endX;
        // and the x pos relative to the new origin is intersec_x = refy/slope
        // Note: because horizontal segments are skipped, 1/slope exists (seg_endY never == O)
        double intersec_x = (newrefy * seg_endX) / seg_endY;
        if( newrefx < intersec_x )  // Intersection found with the semi-infinite line from refx to infinite
            count++;
    }

    return count & 1 ? INSIDE : OUTSIDE;
}