Polygons & Nodal Points...

schoon at amgt.com schoon at amgt.com
Mon Jul 16 12:55:55 CDT 2001

```~sdpm~
Hey, thanks! Yes, I did think of this approach - simply check to see if
my node falls on a line... This will work if all lines are orthoganal,
but I can't be guaranteed this will be the case. The approach I'm
thinking of is like this:

All polygons start at the origin.

Start at origin X=0, Y=0;
If X == vertice, check for max/min Y and increment Y from min to max.
Increment X check max/min Y, increment Y from min to max....

My thoughts are this would handle angled line segments. What I'm doing
is creating a mesh within the polygon.... This is discussed somewhat in
the Mastering Algorithms book, but solving for the case of a node on a
vertice was out of the books scope.

.mark

>----------
>From: 	cabney[SMTP:cabney at cyberpass.net]
>Sent: 	Sunday, July 15, 2001 11:42 PM
>To: 	Mark Schoonover
>Cc: 	san-diego-pm-list at happyfunball.pm.org
>Subject: 	RE: Polygons & Nodal Points...
>
>~sdpm~
>On Sun, 15 Jul 2001 schoon at amgt.com wrote:
>
>> Yes, this is true, using a ray... But it fails when a point is right on
>> a boundary, and the polygon is odd shaped which would make an even
>> amount of crossings... Still a valid point, but fails the test...
>
>K, i have to admit... you need to handle when you are leaving the
>polyon and your y-value is on the edge:
>
>sub in_poly
>{
>	# @isect are x,y quads from the polygon that cross at x
>	my (\$x, \$y, @isect) = @_;
>
>	return "hit!" if is_node(\$x,\$y, at isect);
>	my \$c = 0;
>
>	# slice gives you y at x for each x,y quad
>	my @ladder = slice  (\$x, @isect);
>
>	{
>		my \$rung = shift @ladder;
>		if ( \$y == \$rung ) {
>			unless (\$c % 2) { \$c-- }
>			last;
>		}
>		last if ( \$y <  \$rung );
>		\$c++;
>	}
>	return (\$c % 2) ? "hit!" : "miss!";
>}
>
>CA
>--
>There was a time
>A wind that blew so young
>For this could be the biggest sky
>And I could have the faintest idea
>
>~sdpm~
>
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~sdpm~

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```