# Point in a poly

point winding number lastpt int orig thispt wind quad

Robert's suggestion is a good one. The sum of the angles about the test point is known as the winding number. For non intersecting polygons if the winding number is non-zero then the test point is inside the polygon. It works just fine for convex and concave polygon's. Intersecting polygon's give reasonable results too. A figure 8 will give a negitive winding number for a test point in one of the loops and a positive winding number for the other loop, with all points outside having a winding number of 0. Other advantages of the winding number calculation are that it does not suffer from the confusion of the infinite ray algorithm when the ray crosses a vertex or is colinear with an edge.

Here is my version of a point in poly routine using a quadrant granularity winding number. The basic idea is to total the angle changes for a wiper arm with its origin at the point and whos end follows the polygon points. If the angle change is 0 then you are outside, otherwise you are in some sense inside. It is not necessary to compute exact angles, resolution to a quadrant is sufficient, greatly simplifying the calculations.

I pulled this out of some other code and hopefully didn't break it in doing so. There is some ambiguity in this version as to whether a point lying on the polygon is inside or out. This can be fairly easily detected though, so you can do what you want in that case.

decwrl!sci!kenm

Here is my version of a point in poly routine using a quadrant granularity winding number. The basic idea is to total the angle changes for a wiper arm with its origin at the point and whos end follows the polygon points. If the angle change is 0 then you are outside, otherwise you are in some sense inside. It is not necessary to compute exact angles, resolution to a quadrant is sufficient, greatly simplifying the calculations.

I pulled this out of some other code and hopefully didn't break it in doing so. There is some ambiguity in this version as to whether a point lying on the polygon is inside or out. This can be fairly easily detected though, so you can do what you want in that case.

----------------------------------------------------------------- /* * Quadrants: * 1 | 0 * ----- * 2 | 3 */ typedef struct { int x,y; } point; pointinpoly(pt,cnt,polypts) point pt; /* point to check */ int cnt; /* number of points in poly */ point *polypts; /* array of points, */ /* last edge from polypts[cnt-1] to polypts[0] */ { int oldquad,newquad; point thispt,lastpt; int a,b; int wind; /* current winding number */ wind = 0; lastpt = polypts[cnt-1]; oldquad = whichquad(lastpt,pt); /* get starting angle */ for(i=0;i thispt = polypts[i]; newquad = whichquad(thispt,pt); if(oldquad!=newquad) { /* adjust wind */ /* * use mod 4 comparsions to see if we have * advanced or backed up one quadrant */ if(((oldquad+1)&3)==newquad) wind++; else if((((newquad+1)&3)==oldquad) wind--; else { /* * upper left to lower right, or * upper right to lower left. Determine * direction of winding by intersection * with x==0. */ a = lastpt.y - thispt.y; a *= (pt.x - lastpt.x); b = lastpt.x - thispt.x; a += lastpt.y * b; b *= pt.y; if(a > b) wind += 2; else wind -= 2; } } lastpt = thispt; } return(wind); /* non zero means point in poly */ } /* * Figure out which quadrent pt is in with respect to orig */ int whichquad(pt,orig) point pt; point orig; { if(pt.x < orig.x) { if(pt.y < orig.y) quad = 2; else quad = 1; } else { if(pt.y < orig.y) quad = 3; else quad = 0; } }Ken McElvain

decwrl!sci!kenm

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