Not so elegant, then!
Would have been nice....
M
On 16/06/2016 00:33, Roger Hui wrote:
Your version is incorrect?
(g allperms 4) -: (A.&i.~ -:&!)4
0
(g allperms 4) ; (A.&i.~ -:&!)4
┌───────┬───────┐
│0 1 2 3│0 1 2 3│
│0 1 3 2│0 1 3 2│
│0 2 1 3│0 2 1 3│
│0 2 3 1│0 2 3 1│
│0 3 1 2│0 3 1 2│
│0 3 2 1│0 3 2 1│
│1 0 2 3│1 0 2 3│
│1 0 3 2│1 0 3 2│
│1 2 0 3│1 2 0 3│
│1 3 0 2│1 2 3 0│
│2 0 1 3│1 3 0 2│
│2 1 0 3│1 3 2 0│
└───────┴───────┘
- Both 0 2 3 1 and its reversal occurs in your result
- Neither 2 0 1 3 nor its reversal occurs in your result
On Wed, Jun 15, 2016 at 4:16 PM, Mike Day <[email protected]>
wrote:
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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