The only difference between the correct and the incorrect result is a space before the second : . Should a space make a difference ever? In the correct result I can't understand what it is, what it does and why it does it.
/:~ :/:@": &.>a,b ------T-----┐ │01123│01123│ L-----+------ /:~:([:/:": )&.>a,b 0 1 Since I set out originally to remove "atop", how can I do it? /:~ :([: /: ":)&.>a,b ------T-----┐ │10123│32110│ L-----+------ Hope you can help. Thanks. Linda -----Original Message----- From: [email protected] [mailto:[email protected]] On Behalf Of Ian Clark Sent: Thursday, September 22, 2011 9:13 PM To: Programming forum Subject: Re: [Jprogramming] Permutations Off-the-cuff... a=: 10123 b=: 32110 c=: 32100 -:/ /:~ :/:@":&.> a,b 1 -:/ /:~ :/:@":&.> a,c 0 On Thu, Sep 22, 2011 at 10:41 PM, David Vaughan <[email protected]> wrote: > Hi, what is the best (fastest) way to check whether a number is a permutation of another number? > I've tried listing all permutations of one number and checking whether the other number is in the list of permutations. This seems pretty slow, so I was hoping for a faster method. > > Thanks. > ---------------------------------------------------------------------- > 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
