On 12/01/2017 01:20 AM, Roger Hui wrote:
> Cantor =: 3 : ', 1 0 1 */^:y ,1'
> SC =: 3 : '(3 3$4>i.5) ,./^:2@(*/)^:y ,.1'
Nice! The first algorithm can easily be implemented in Mathematica with
cantor2[n_] := Nest[Flatten@Outer[Times, #, {1, 0, 1}] &, {1}, n]
Mathematica also allows many solutions to a problem.
I don't want to sound like a shill for Mathematica, but it's a great
package with a great ecosystem; it's just obscenely expensive. J might
be a nice free alternative.
SC confuses me. I would have thought that
(3 3$4>i.5) (*/)^:y ,.1
would produce a noun of rank 2y+2, and that each application of ,./
would reduce the rank by 1, so we would have to apply it 2y times; yet
you only apply it twice.
>
> 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
>
So the result should be a matrix in which each entry is either a number,
a "-" or "|", or ""?
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