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MathGroup Archive 2007

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Re: [Mathematica 6]3 lines of code which crashes Mathematica every time

  • To: mathgroup at smc.vnet.net
  • Subject: [mg76538] Re: [mg76449] [Mathematica 6]3 lines of code which crashes Mathematica every time
  • From: John Fultz <jfultz at wolfram.com>
  • Date: Wed, 23 May 2007 05:41:18 -0400 (EDT)
  • Reply-to: jfultz at wolfram.com

On Tue, 22 May 2007 02:57:02 -0400 (EDT), Nasser Abbasi wrote:
>
>
> (* WARNING, SAVE ALL YOUR WORK BEFORE RUNNING THIS *)
>
> m = 100; n = 70;
> A = Table[Table[RandomReal[{0, 100.}], {n}], {m}];
> ListPlot3D[A, Mesh -> None, InterpolationOrder -> 0, ColorFunction ->
> "SouthwestColors", Filling -> Axis]
>
>
> (*Now Mathematica will spin for sometime then disappear.*)
>
> Nasser

The killer here is Filling->Axis.  Using Filling introduces more polygons,=
 which 
doesn't help, but the big problem is that the default FillingStyle=
 introduces 
transparency into the graphic.  Removing the Filling, or keeping the Filling=
 but 
not using a transparent FillingStyle, e.g.,

m = 70; n = 100;
A = Table[Table[RandomReal[{0, 100.}], {n}], {m}];
ListPlot3D[A, Mesh -> None, InterpolationOrder -> 0, 
 ColorFunction -> "SouthwestColors", Filling -> Axis, 
 FillingStyle -> Red ] 

will make this work much better.

The transparency issue is a general one in 3D graphics.  It arises because=
 it's 
algorithmically much more difficult to sort polygons (i.e. to know which 
polygons are in front of which other polygons) when some of those polygons=
 are 
transparent.  If you're at all familiar with polygon sorting algorithms,=
 here's 
a brief description of what we're doing.

We use Z-buffering to sort polygons when no transparency exists.  Not only=
 is it 
simple and fast, modern video cards have hardware that speeds it up.  If 
transparent polygons are present, Z-buffering will produce bad results, so=
 we 
sort the polygons using a BSP tree, which is very efficient once created,=
 but 
the initial tree creation has a high cost.  Also, the resulting BSP tree has=
 
many more polygons because of the necessity of splitting existing polygons.

In your example, there are so many polygons (particularly the ones that are=
 
created by splitting existing polygons) that simply creating the BSP tree 
exhausts the available memory.  With smaller values of m and n, the BSP tree=
 
could be created, but the rendering will still be much slower.

Sincerely,
 
John Fultz
jfultz at wolfram.com
User Interface Group
Wolfram Research, Inc.




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