If we do a bit of work by ourselves? Consider that the 5 has to be in fifth
position (starting from 1), and that there must be even digits (four of
them) on the even positions. The other four odd digits go to the remaining
odd positions. Then there are only 576 possibilities left.
odds =: (i.24) A. '1379' NB. there's a better trick, is there?
evens =: (i.24) A. '2468'
all =: ,/ odds {{ }. x 1 3 7 9 } y 2 4 6 8 } '0oeoe5eoeo' }}"1/ evens
all #~ (1+i.9) {{*./ 0 = x | ". x {."0 1 y }}"1 all
381654729
Direct Definition rules!
Greetings,
Ben
On Mon, 11 Sept 2023 at 08:09, Raul Miller <rauldmil...@gmail.com> wrote:
> I guess to simplify it, I would filter the possibilities at each step.
>
> first digit divisible by 1:
> N1=: 1+i.9
>
> Second digit divisible by 2:
> N2=: ;(10*N1)+each (<2*1+i.4)-.each N
>
> Third digit divisible by 3:
> digits=: 10&#.inv
> N3=: (#~ 0=3|]);(10*N2)+each (<1+i.9) (-. digits) each N2
>
> And, so on... and since we're basically doing the same thing at each
> step, we could encapsulate and parameterize this step as a function.
>
> F=: {{(#~ 0=x|]);(10*y)+each (<N1) (-. digits) each y}}
> 9 F 8 F 7 F 6 F 5 F 4 F 3 F 2 F N1
> 381654729
>
> timespacex '9 F 8 F 7 F 6 F 5 F 4 F 3 F 2 F N1'
> 0.0006679 38400
>
> So that took about a millisecond of machine time on this little laptop.
>
> Further simplifications are possible, but since this is a one use
> calculation this is probably good enough.
>
> --
> Raul
>
> On Sun, Sep 10, 2023 at 2:02 PM 'Skip Cave' via Programming
> <programm...@jsoftware.com> wrote:
> >
> > Quora Question: Can you arrange the digits 1-9 to make a nine-digit
> number
> > such that the first digit is divisible by one, the first two digits are
> > divisible by two, the first three divisible by three and so on?
> >
> > My solution:
> > ea=.&.>
> >
> > at=.>10#.ea(a=.362880 9$1 to 9){.ea{n=.>:perm 9
> >
> > {:"1 at#~*./"1[0=a|"1 at
> >
> > 381654729
> >
> >
> > The answer is 381654729.
> >
> >
> > Check:
> >
> > ]m=.>10#.ea(;/1 to 9){.ea{sep 381654729
> >
> > 3 38 381 3816 38165 381654 3816547 38165472 381654729
> >
> > (1 to 9)| m
> >
> > 0 0 0 0 0 0 0 0 0
> >
> >
> > There is only one answer. Is there a way to simplify or minimize the big
> > repetitive array 'a' to make the two J code lines more memory efficient &
> > succinct?
> >
> >
> > Skip Cave
> > Cave Consulting LLC
> > ----------------------------------------------------------------------
> > For information about J forums see http://www.jsoftware.com/forums.htm
> ----------------------------------------------------------------------
> For information about J forums see http://www.jsoftware.com/forums.htm
>
----------------------------------------------------------------------
For information about J forums see http://www.jsoftware.com/forums.htm
