lower code version, of Raul's.
I may not have understood problem, but I get 2492 solutions (oops... now I
understand that each digit needs to be used just once.
I =: ]F.: NB. insert using fold for shape flexibility
#(>:i.9) ([ (] #~ 0 = |) (i.10) +("1 0)(,@:) 10 * ])I 9 8 7 6 5 4 3 2
# (>:i.9) ([ (] #~ 0 = |) (i.10) +("1 0)(,@:) 10 * ])I 9 8 7 6 5 4 3 2
2492
5 {. (>:i.9) ([ (] #~ 0 = |) (i.10) +("1 0)(,@:) 10 * ])I 9 8 7 6 5 4 3 2
102000564 102006162 102006801 102006882 102054402
with spec, but 0 allowed to be included in distinct digits
(#~ 9 = 10&#.inv(#@~."1@:)) (>:i.9) ([ (] #~ 0 = |) (i.10) +("1 0)(,@:) 10 *
])I 9 8 7 6 5 4 3 2
381654720 381654729 783204165 801654723
only 1 above doesn't include 0 (the 1 solution sought)
timespacex '(#~ 9 = 10&#.inv(#@~."1@:)) (>:i.9) ([ (] #~ 0 = |) (i.10) +("1
0)(,@:) 10 * ])I 9 8 7 6 5 4 3 2'
0.000899 603168
On Sunday, September 10, 2023 at 06:09:30 p.m. EDT, 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
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