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- From: Fernando Cacciola <>
- To:
- 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
- [cgal-discuss] degenerate beziers in an arrangement, Ben Supnik, 03/02/2010
- Re: [cgal-discuss] degenerate beziers in an arrangement, Fernando Cacciola, 03/03/2010
- Re: [cgal-discuss] degenerate beziers in an arrangement, Ben Supnik, 03/03/2010
- Re: [cgal-discuss] degenerate beziers in an arrangement, Fernando Cacciola, 03/03/2010
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