Path: utzoo!utgpu!news-server.csri.toronto.edu!rpi!crdgw1!uunet!mcsun!hp4nl!alchemy!ptavoly
From: ptavoly@cs.ruu.nl (Peter Tavoly)
Newsgroups: comp.sys.amiga.programmer
Subject: Re: Clicking on irregular shapes?
Message-ID: <1991Apr09.164000.1998@cs.ruu.nl>
Date: 9 Apr 91 16:40:00 GMT
Organization: Utrecht University, Dept. of Computer Science
Lines: 96


This post contains reactions to three other posts in the same thread.

1.

umueller@iiic.ethz.ch (Urban Dominik Mueller) writes:
>>        /\
>>       /  \    /\
>>      / O  \  /  \                                                   X
>>     /      \/ ___\
>>    /_______  /
>>            \/
>>where you draw the horizontal line from the 'O' to the 'X', now, it intersects
>>an odd number of boundaries, yet it is OUTside the shape. Or am I mistaking
>>something here?
>
>Am *I* mistaking *you*? The way I see it, the 'O' is clearly inside, as the
>algorithm states. Or are you talking about the 'X'? 'X' must be a point that's
>sure to be outside, but 'O' is the point in question. The polygon can be as
>complex as you want.

You wrote, that you explained it the other way around; I guess that's why I
did it that way too :)

>>       /\
>>      /  \    /
>>     / O  \  /                                                       X
>>    /      \/
>>   /_______/
>>Here, an even number of boundaries is crossed. (The second intersection is a
>>single line)
>
>The polygon must be closed for this algorithm to work. To make it closed,
>you have to draw a second line over that single line already drawn, there-
>fore you get a double interection at that line, so the total number of inter-
>sections is odd again -> inside.

Yes, that is the problem, what if it is not closed? You are bound to come
across one of those at some point. 

BTW, I was thinking of a bitmap indeed, not a polygon, or maybe an imaginary
polygon only, which is not visible but is the internal representation of
the boundaries and then a bitmap inside for the graphical stuff.

>
>BTW I just discovered that I explained the problem with the imaginary line
>going through a vertex exactly the wrong way round. Count the intersection
>only if it's *no* local maximum or minimun, otherwise don't (or do it twice).
>My apologies if that caused problems to anyone.

No problem.

> |          Urban Mueller         |      / / |    Urban Mueller    |
> | INTERNET:umueller@iiic.ethz.ch | __  / /  |    Schulhausstr. 83 |
> | FIDONET: 2:302/906 (AUGL)      | \ \/ /   | CH-6312 Steinhausen |
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2.

nj@magnolia.Berkeley.EDU (Narciso Jaramillo) writes:
>In article <1991Apr05.104141.2118@cs.ruu.nl> ptavoly@cs.ruu.nl (Peter Tavoly) writes:
[shapes and stuff deleted]
>
>I just realized that you may be assuming that this test works with raw
>bitmaps.  In general, it only completely works if you have an internal
>representation of your polygons.  If you're only working with a bitmap
>image, this won't work so well in degenerate cases.  It's probably
>better to go ahead and represent your polygons internally--a simple
>array of line endpoints won't take up much room.

Exactly that's what I was thinking. Thank you for the suggestions.

>nj

3.

dillon@overload.Berkeley.CA.US (Matthew Dillon) writes:
>    I think the best solution is to simply have a grainy bitmap of the
>    object and just compare the coordinate against the grainy bitmap (e.g.
>    a 16x16 bitmap = 32 bytes)
>
>    (for this case)

Yes, I think I might be lazy and do this :^)

>				-Matt

 -Thomas T.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
  "Help, I've been attacked by a Prime terminal at 1200 baud!!!"  -Thomas T.
  
  Did you know that IBM have a secret factory with an infinite number of
  monkeys? That's how they invented MS-DOS...      -ptavoly@praxis.cs.ruu.nl
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