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 <[EMAIL PROTECTED]> 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 <[EMAIL PROTECTED]> 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 ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
