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On Mon, 30 Aug 1999 07:52:45 -0500, Mark Gordon wrote:
>Nieminen Juha wrote:
>>
>> Mark Gordon <mtg### [at] mailbag com> wrote:
>> : If the polygons are not convex, recursively lop off pointy bits as
>> : triangles until they are.
>>
>> If you take any three points of the polygon, how can you be sure that
>> the triangle defined by them is completely inside the polygon?
>
>I made implicit reference to a "pointy bit test" ... yeah, that's right.
>;-)
>
>Assume the polygon is defined as a sequence of points on its periphery.
>A triplet of points on the edge of the polygon forms a "pointy bit" (my
>jargon) if the interior angle formed at the second point (as measured in
>the plane defined by the three points) is less than 180 degrees. If the
>interior angle formed at the second point is greater than 180 degrees
>it's an, um, "anti-pointy bit". :-) If it's exactly 180, you may as
>well remove the middle point as redundant.
Unfortunately, it's not quite that easy. You can't lop of just any
pointy bit. Take a simple isosceles triangle. Add a new vertex at
the center of the base. Move that vertex upward, so what you get looks
like a rather angular version of the Federation emblem from Star Trek.
Note that the angle surrounding the topmost vertex qualifies as a
"pointy bit" yet the resulting triangle is neither inside nor outside
the polygon.
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