Re: DelaunayTriangulation[] output
- To: mathgroup at smc.vnet.net
- Subject: [mg83218] Re: DelaunayTriangulation[] output
- From: Frank Iannarilli <frankeye at cox.net>
- Date: Thu, 15 Nov 2007 05:39:04 -0500 (EST)
- References: <fhegn1$m6p$1@smc.vnet.net>
On Nov 14, 4:55 am, Dominick <pd20012... at yahoo.com> wrote:
Hi Dominick,
I used the following functions and command sequence (they worked in
Mathematica 5.2, I presume they'll still work in 6). In below,
"facet"=triangle.
deltri=DelaunayTriangulation[.....];
The following function works (for all but the 1st vertex, which yields
bogus facet {1, deltri[[1,2,-1]], 2})
makeFacets[record_List]:=Module[{return={},base=record[[1]],list=record[[2]]},
For[i=1,i<=Length[list],i++,If[(list[[1]]> base &&list[[2]]> base),
return=Append[return,{base,list[[1]],list[[2]]}]];list=RotateLeft[list,
1]
];
return]
A quick test of makeFacets[]:
In: makeFacets[{20,{43,19,1,21,44}}]
Out: {{20,21,44},{20,44,43}}
facets=Partition[Flatten[Map[(makeFacets[#])&,deltri]],3];
facets=DeleteCases[facets,{1,deltri[[1,2,-1]],2}];
"facets" will be a list of triangles (each a list of 3 vertex indices)
Good luck