Oleg - when I saw the length of "2 burg 10", I thought maybe I had
mis-remembered the length of my long-ago solution.
However upon inspection, I see that your solution has unnecessary entries:
#2 burg 10
110
2 burg 10
0 0 0 1 0 2 0 3 0 4 0 5 0 6 0 7 0 8 0 9 1 1 1 2 1 3 1 4 1 5 1 6 1 7 1 8 1 9
2 2 2 3 2 4 2 5 2 6 2 7 2 8 2 9 3 3 3 4 3 5 3 6 3 7 3 8 3 9 4 4 4 5 4 6 4 7
4 8 4 9 5 5 5 6 5 7 5 8 5 9 6 6 6 7 6 8 6 9 7 7 7 8 7 9 8 8 8 9 9 0
NB. Removing dupes by hand:
#0 0 1 0 2 0 3 0 4 0 5 0 6 0 7 0 8 0 9 1 1 2 1 3 1 4 1 5 1 6 1 7 1 8 1
9 2 2 3 2 4 2 5 2 6 2 7 2 8 2 9 3 3 4 3 5 3 6 3 7 3 8 3 9 4 4 5 4
6 4 7 4 8 4 9 5 5 6 5 7 5 8 5 9 6 6 7 6 8 6 9 7 7 8 7 9 8 8 9 9 0
101
Regards,
Devon
On 6/24/08, Oleg Kobchenko <
olegykj@...> wrote:
>
> One simple way, not optimal, but not terribly bad, I think,
> for burglars anyway ...
>
> NB. 4 burg 5
> burg=: 4 : 0
> B=. x#y
> L=. ''
> for_i. i.y^x do.
> p=. B#:i
> if. -.p +./@E. L do. L=. L,p end.
> end.
> L
> )
>
> ...
> --- On Tue, 6/24/08, Devon McCormick <
devonmcc@...> wrote:
> > Dan - I solved a similar problem a number of years ago ...
>
> > that I came up with a 101-digit "universal" answering machine
>
>
> #2 burg 10
> 110
> htm
>
...
--
Devon McCormick, CFA
^me^ at acm.
org is my
preferred e-mail
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