I like your version of all perms; it's even more elegant than
tap =: i.@! A. i.
Anyway, a quite elegant variant produces what I think you
require, here with a transpose to save lines:
|:(A.&i.~ -:&!)4
0 0 0 0 0 0 1 1 1 1 1 1
1 1 2 2 3 3 0 0 2 2 3 3
2 3 1 3 1 2 2 3 0 3 0 2
3 2 3 1 2 1 3 2 3 0 2 0
Any use?
Mike
On 15/06/2016 22:42, Louis de Forcrand wrote:
Hi,
How would you go about generating all unique permutations of a list,
while counting each perm and its reversal as one?
A perhaps clearer (but less efficient) way of doing this is generating
all permutations and then removing (reversed) duplicates:
allperms=: (A.&i.~ !) 4 NB. 4 is just an example
]those_I_want=: ([: {."2@~. (/:~@,: |.)"1) allperms
0 1 2 3
0 1 3 2
0 2 1 3
0 2 3 1
0 3 1 2
0 3 2 1
1 0 2 3
1 0 3 2
1 2 0 3
1 3 0 2
2 0 1 3
2 1 0 3
There is surely a better way to do this!
Thanks,
Louis
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