NetNews Usenet Archive 1992 #19/NN_1992_19.iso/spool/comp/programm/2527

Text File | 1992-08-29 | 1.8 KB | 37 lines

Newsgroups: comp.programming
Path: sparky!uunet!zaphod.mps.ohio-state.edu!darwin.sura.net!cs.ucf.edu!schnitzi
From: [e-mail] (Mark Schnitzius)
Subject: Re: How to decide if a point is inside a polygon?
Message-ID: <[e-mail]>
Sender: [e-mail] (News system)
Organization: University of Central Florida
References: <[e-mail]> <[e-mail]> <[e-mail]> <[e-mail]> <[e-mail]>
Date: Fri, 28 Aug 1992 20:02:22 GMT
Lines: 25

[e-mail] (Alexander Berlin) writes:
>| I write:
>|> Well, one solution to this problem is this: consider each vertice of the
>|> polygon as belonging to only _one_ of the line segments it is connected to.
>|> So in the above example,  only consider point C to be on the line segment
>|> BC and not on AC.  Then if the line you're drawing goes through point C,
>|> it is only considered to have gone through line segment BC and not AC.
>|> This simple fix should clear up any problems.

>No, it doesn't. I didn't say that the line drawn from point X has one or two
>crossings. My point is that X-line and Y-line have the same number of crossings
>(be it 1 or 2) while X is outside and Y is inside. The fix you suggest doesn't
>distinguish X- and Y- cases in my example.
>The rule that would work is to check if AC and CB segments are on the same
>side of the line (2 or 0 crossings, case X) or on the different sides
>(1 crossing, case Y). But I think it is much better to just avoid these
>confusing cases.

You are right; sorry, I misunderstod your earlier point.  So how can we
get this to work for all cases?  Just find a direction to extend the
infinite line that doesn't intersect any vertices?

Mark Schnitzius
[e-mail]
University of Central Florida