I sometimes don’t get the order of thoght and speech right.
Sorry for shouting 16 prior to thinking well about it.
For each first number, the third is fixed, so there are only
two possible ways (permute the other 2). In total, that’s 8.
Am 25.05.20 um 14:15 schrieb 'Mike Day' via Programming:
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=)
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