From: Joe Lewis Wilkins <PepeToo@...>

Date: Thu, 08 Jun 2000 08:49:37 -0700

Date: Thu, 08 Jun 2000 08:49:37 -0700

Chris, I started to suggest the same thing, but thought in my ignorance that I was probably overlooking something. One of the things that has been bothering me, about all of this C to FB stuff that has been flittering about on the list, is that there is almost nothing that needs to be done that cannot be done with a ToolBox call if you know which ones to use. Apple has done a very thorough job, and most of it is faster, now that we have access to PPC, than any C code anyone can come up with, so why not spend all of this energy exploring the enormous potential of the Toolbox, and leave the "convoluted" C thinking where it belongs - on PCs. IMHO, Joe Wilkins Chris Stasny wrote: > Being math ignorant, I would have approached this problem from the > toolbox. Conversationally... > > Open region > Draw poly > Close region > Use PTINRGN > Dispose region > > 5 lines of code to do the same thing. :) > > >Hi All, > > > >Given a point polygon p and a point (x,y) NOT ON p, we are deciding > >whether (x,y) lies inside of p -- by suitably counting edge > >intersections with the ray going right (east) from (x,y). In fact, > >points (x,y) too near to p may give a wrong answer or even cause a > >zero-divide error. > > > >I happen to have looked at this problem recently for loops that are > >piecewise b'ezier (or worse) rather than piecewise linear. So I cannot > >resist commenting on the C-code. (I'll explain the b'ezize curve case > >privately if someone is really interested.) > > > >Randolph Franklin original C code looks clean and solid to me. > >He does miss one optimization that Paul D. Bourke > >uses. Namely bypass the guts of the for loop in case the max of xp[i] > >and xp[j] is less than x. That is because the edge p[j] to p[i] > >could then not possibly hit the ray. > > > >Franklin's C code revised is: > > > > int pnpoly(int npol, float *xp, float *yp, float x, float y) > > {int i, j, c = 0; > > for (i = 0, j = npol-1; i < npol; j = i++) > > {if (xp[i]<=x && xp[j]<=x) \* the added condition is "if..."; LS *\ > > {if ((((yp[i] <= y) && (y < yp[j])) || > > ((yp[j] <= y) && (y < yp[i]))) && > > (x < (xp[j] - xp[i]) * (y - yp[i]) / (yp[j] - yp[i]) + xp[i])) > > c = !c; > > } > > } > > return c; > > } > > > > Paul C. Bourke's code (below) on the other hand seems to contain > >a a redundancy and a glitch. > > > >The redundancy is Bourke's condition: > > > > if (p1.y != p2.y) > > > >At the point in question (p1.y != p2.y) is necessarily true. > > > >The possible glitch is Bourke's > > > > if (p1.x == p2.x || p.x <= xinters) > > > >I think that line should become: > > > > if (p.x <= xinters) > > > >My reason: Bourke seems to be counting all *vertical* edges* that hit > >the horizontal line through (x,y) and protrude below it. That seems > >wrong. > > > >I may be confused myself, but I believe Bourke's code would report > >that (x,y)=(.5,.5) is not inside the unit square! > > > >Sorry to address the meaning rather than the FB translation. > >But I believe translation in general must pass via meaning. > > > > Cheers > > > > Larry S > > > >PS. Here is Bourke's code: > > > > > Content: futurebasic_18774.ezm > > > Date: Tue, 6 Jun 2000 22:55:14 -0400 > > > From: Robert Covington <t88@...> > > >... > > > From Paul D. Bourke: > > > > > > The following C function returns INSIDE or OUTSIDE indicating > >the status of > > > a point P with respect to a polygon with N points. > > > > > > #define MIN(x,y) (x < y ? x : y) > > > #define MAX(x,y) (x > y ? x : y) > > > #define INSIDE 0 > > > #define OUTSIDE 1 > > > > > > typedef struct { > > > double x,y; > > > } Point; > > > > > > int InsidePolygon(Point *polygon,int N,Point p) > > > { > > > DIM counter as int > > > counter =0 > > > int i as int > > > DIM xinters as double > > > DIM p1 as point ,p2 as point > > > > > > p1 = polygon[0]; > > > for (i=1;i<=N;i++) { > > > p2 = polygon[i % N]; > > > if (p.y > MIN(p1.y,p2.y)) { > > > if (p.y <= MAX(p1.y,p2.y)) { > > > if (p.x <= MAX(p1.x,p2.x)) { > > > if (p1.y != p2.y) { > > > xinters = (p.y-p1.y)*(p2.x-p1.x)/(p2.y-p1.y)+p1.x; > > > if (p1.x == p2.x || p.x <= xinters) > > > counter++; > > > } > > > } > > > } > > > } > > > p1 = p2 > > > } > > > > > > if (counter % 2 == 0) > > > return(OUTSIDE); > > > else > > > return(INSIDE); > > > > > >-- > >To unsubscribe, send ANY message to <futurebasic-unsubscribe@...> > > Best, > > -STAZ ~)~ > > 800.348.2623 Orders http://www.stazsoftware.com > 228.255.7086 FAX mailto:staz@... > > -- > To unsubscribe, send ANY message to <futurebasic-unsubscribe@...>