I think there are 8 solutions. Given one, minimal, solution, 0 1 3 2, how about this?
((,|."1)@:(|."0 1~i.@#))'0132' 0132 1320 3201 2013 2310 0231 1023 3102 Any use? Mike Sent from my iPad > On 25 May 2020, at 12:47, Brian Schott <schott.br...@gmail.com> wrote: > > I want all permutations of i. 4 for which 0 and 3 cannot be adjacent, nor > can 1 and 2. My idea is to create a as follows to describe my situation. > > ]a =. 0 1 1 0,:i. 4 > > 0 1 1 0 > > 0 1 2 3 > > > Next I created b and rotations on b to list the possibilities I can think > of. So I can think of 6 permutations. Are there more and is there a better > way to generate the real qualifying permutations? > ]b =. 0 1 3 2 ,:1 0 2 3 > 0 1 3 2 > 1 0 2 3 > 1|."1 b > 1 3 2 0 > 0 2 3 1 > _1|."1 b > 2 0 1 3 > 3 1 0 2 > -- > (B=) > ---------------------------------------------------------------------- > For information about J forums see http://www.jsoftware.com/forums.htm ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
