... and here’s an explicit function to generate the minimal solution, and
indeed the
minimal solutions for each starting value:
rp =: 3 : 0
NB. generate minimal restricted perm where adjacent pairs' sums .ne. 3
cur =. p =. 4 | y NB. suitable seed
l =. cur -.~ i. 4
while. #l do.
p =. p, cur =. cur {.@(] #~ 3 ~: +) l
l =. l -. cur
end.
p
)
rp 0
0 1 3 2
rp"0 i.4 NB. the lower perms for each starting value
0 1 3 2
1 0 2 3
2 0 1 3
3 1 0 2
Just append the reverse of every row for all solutions,
$(,|."1)rp"0 i.4
8 4
Cheers,
Mike
Sent from my iPad
> On 25 May 2020, at 13:15, Mike Day <mike_liz....@tiscali.co.uk> wrote:
>
> 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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