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Re: Polygon cutter

  • To: mathgroup at smc.vnet.net
  • Subject: [mg87982] Re: Polygon cutter
  • From: carlos at Colorado.EDU
  • Date: Mon, 21 Apr 2008 06:40:28 -0400 (EDT)
  • References: <200804200349.XAA11201@smc.vnet.net> <fuhf9s$ib1$1@smc.vnet.net>

On Apr 21, 1:21 am, "W_Craig Carter" <ccar... at mit.edu> wrote:
> Hello Carlos,
> I believe the interpretation of good style is a bit like ethics--it's
> easy to agree in extreme cases, but the in-between cases are
> subjective.
>
> I think this depends on whether you intend to modify the code in the
> future. I believe V1 is easier to read, and if you think that someone
> else may wish to modify it, then I recommend sticking with that.
> However,"Which" will probably be more efficient, and rank the order of
> tests with decreasing frequency.
>
> V2 looks cool--some people like that.  There is probably  a way to
> improve the coolness even more, and remove the need to return the
> impossible "Null"s.  Creating V2 probably hones your skills and is
> more interesting to do, which is similarly important; but probably has
> a direct impact on a broader interpretation of efficiency.
>
> On Sat, Apr 19, 2008 at 11:49 PM,  <car... at colorado.edu> wrote:
> > A programming style question.  The code fragments below come
> >  Version 1. Straight translation from Fortran:
>
> >  p={Null};
> >  If [tb>0&&tb==tt,  p={P1,P2,P4,P3}];
>
> :
> :
>
>
>
> >  poly=Graphics[Polygon[p]];
>
> >  Version 2. Table driven variant of above:
>
> >  p={{{},{Pc1,Pc2,Pc3},{Pc1,P4,P3},{Pc1,Pc2,P3,P4}},
> >    {{Null},{},{Null},{Null}},
> >    {{Null},{Pc3,Pc2,P1,P2},{P1,P2,P4,P3},{Pc2,P1,P2,P4,P3}},
> >    {{Null},{Pc3,P1,P2},{Null},{P1,P2,P4,P3}}} [[tb+2,tt+2]];
> >  poly=Graphics[Polygon[p]];
>
> >  Question: which version is preferable in Mathematica, or is
> >  there a better one? Both run roughly at the same speed
> >  (about 25 microsec under 5.2 on an Intel MacBook Pro).
> >  Since this  loop is traversed once for each polygon,
> >  efficiency is important.
>
> --
> W. Craig Carter

Your suggestion of looking at the Which construct was an interesting
one -
didn't know it existed. So I tried it as

  p={Null};
  Which [tb>0&&tb==tt,  p={P1,P2,P4,P3}, tb==1&&tt==2,
p={Pc2,P1,P2,P4,P3},
             tb==-1&&tt==-1,p={},            tb==-1&&tt==0=
,
p={Pc1,Pc2,Pc3},
	     tb==-1&&tt==1, p={Pc1,P4,P3},   tb==-1&&tt==2,
p={Pc1,Pc2,P3,P4},
	     tb==0&&tt==0,  p={},            tb==1&&tt==0,
p={Pc3,Pc2,P1,P2},
	     tb==2&&tt==0,  p={Pc3,P1,P2}];
   poly=Graphics[Polygon[p]];

The two most common cases are the first two.  I expected this to be
quicker than V1
but it wasnt.  On a MacBook Pro under 5.2:

            V1          25 microsec/polygon
            V2          24
            Which     30

Similar rankings for 4.1 and 4.2. Havent tested it on 6.0.



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