[I thought I had sent this message after Mike's first, but apparently forgot to "Send"]
Hauke, I could not figure out your approach. Mike, I think your solution is correct. I tested it with the following approach. First I went to NuVoc and looked up A. to find the following line of code. n=: !N=: 4 NB. number of distinct permutations on N=4 points I then executed the following line which gave me the 26 permutations of both i.4 and of 0 1 1 0. ((i.n) A. i. N);(i.n) A. 0 1 1 0 From there in my editor I manually deleted the offending cases, leaving only the 8 below. Which I think correspond to your 8. ┌───────┬───────┐ │0 1 3 2│0 1 0 1│ │0 2 3 1│0 1 0 1│ │1 0 2 3│1 0 1 0│ │1 3 2 0│1 0 1 0│ │2 0 1 3│1 0 1 0│ │2 3 1 0│1 0 1 0│ │3 1 0 2│0 1 0 1│ │3 2 0 1│0 1 0 1│ └───────┴───────┘ On Mon, May 25, 2020 at 8:15 AM 'Mike Day' via Programming < programm...@jsoftware.com> wrote: > I think there are 8 solutions. > Given one, minimal, solution, 0 1 3 2, how about this? > > NN((,|."1)@:(|."0 1~i.@#))'0132' > 0132 > 1320 > 3201 > 2013 > 2310 > 0231 > 1023 > 3102 > > Any use? > > Mike > > Sent from my iPad > > > ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
