And for computing large Cantor sets really fast, you can use a recursive
method:
c0 =. 3 :', */&1 0 1^:y 1'
c1 =. 3 :', (*/~(3^y-2^l)&{.) , */~^:(l=.<.2^.y) 1 0 1'
(c0 -: c1) 18
1
10 (6!:2) 'c0 18'
1.22999
10 (6!:2) 'c1 18'
0.0728543
The J solution makes it easy to figure this out, because */ is
associative. With Mathematica, it's completely opaque.
Marshall
On Thu, Nov 30, 2017 at 10:20:43PM -0800, Roger Hui wrote:
> Cantor =: 3 : ', 1 0 1 */^:y ,1'
> SC =: 3 : '(3 3$4>i.5) ,./^:2@(*/)^:y ,.1'
>
> Recursive solutions using the Mathematica ReplaceAll (/.) idea are also
> possible, using indexing ({):
>
> Cantor1=: 3 : 'if. 0=y do. ,1 else. ,(Cantor1 y-1){0,:1 0 1 end.'
> SC1 =: 3 : 'if. 0=y do. ,.1 else. ,./^:2 (SC1 y-1){0,:3 3$4>i.5 end.'
>
> Checking that they give the same results:
>
> (Cantor -: Cantor1)"0 i.8
> 1 1 1 1 1 1 1 1
> (SC -: SC1)"0 i.8
> 1 1 1 1 1 1 1 1
>
> I claim the examples in my message unambiguously specify the extended H
> problem. More details (and solutions) can be found in
> http://code.jsoftware.com/wiki/Essays/Extended_H
>
>
>
>
> On Thu, Nov 30, 2017 at 9:33 PM, Dabrowski, Andrew John <
> [email protected]> wrote:
>
> > On 11/29/2017 11:40 PM, Roger Hui wrote:
> > > 2.5 Cantor Set
> > >
> > > Write a function to compute the Cantor set of order n, n>:0.
> > >
> > > Cantor 0
> > > 1
> > > Cantor 1
> > > 1 0 1
> > > Cantor 2
> > > 1 0 1 0 0 0 1 0 1
> > > Cantor 3
> > > 1 0 1 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 1 0 1
> > >
> >
> > In Mathematica:
> >
> > cantor[n_] := If[n == 0, {1},
> > cantor[n - 1] /. {0 -> Sequence[0, 0, 0], 1 -> Sequence[1, 0, 1]}]
> >
> > I doubt J could do substantially better, but I'll leave that to you
> > experts.
> >
> > > 2.6 Sierpinski Carpet
> > >
> > > Write a function to compute the Sierpinski Carpet of order n, n>:0.
> > >
> > > SC 0
> > > 1
> > > SC 1
> > > 1 1 1
> > > 1 0 1
> > > 1 1 1
> > > SC 2
> > > 1 1 1 1 1 1 1 1 1
> > > 1 0 1 1 0 1 1 0 1
> > > 1 1 1 1 1 1 1 1 1
> > > 1 1 1 0 0 0 1 1 1
> > > 1 0 1 0 0 0 1 0 1
> > > 1 1 1 0 0 0 1 1 1
> > > 1 1 1 1 1 1 1 1 1
> > > 1 0 1 1 0 1 1 0 1
> > > 1 1 1 1 1 1 1 1 1
> > I believe Mathematica has no built in tiling function, so I wrote one.
> >
> > tile[m_] := Join @@ ((Join @@@ #) & /@ (Transpose /@ m));
> >
> > hole = {{1, 1, 1}, {1, 0, 1}, {1, 1, 1}};
> > zeros = Table[0, {3}, {3}];
> >
> > sierpinski[n_] := If[n == 0, {{1}},
> > tile[sierpinski[n - 1] /. {1 -> hole, 0 -> zeros}]]
> >
> >
> > The tiling utilities in J are very nice.
> >
> >
> >
> > Could give a reference for the extend H algorithm? I get the idea, but
> > I'm a little unclear about the details.
> > ----------------------------------------------------------------------
> > For information about J forums see http://www.jsoftware.com/forums.htm
> ----------------------------------------------------------------------
> For information about J forums see http://www.jsoftware.com/forums.htm
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