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Re: [cgal-discuss] degenerate beziers in an arrangement


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  • From: Fernando Cacciola <>
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  • Subject: Re: [cgal-discuss] degenerate beziers in an arrangement
  • Date: Wed, 3 Mar 2010 09:48:30 -0300
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Hi Ben,

>
> always cubics.  But...sometimes either C1==P1 or C2==P2.
>
I guess then that for your application the quadratic (P0,P1,P2) isn't
the same as the cubic (P0,P1,P2,P2)?
AFAICT they would both render the same (or maybe not? I never tried
such a degenerate cubic).

Anyway, in CGAL you just can't construct an N-degree bezier with only
M<N distinct control points. The degree is given by that count, so in
your case you're stuck with quadratics.

HTH

Fernando Cacciola
GeometryFactory



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