Thanks!
It is not clear to me that the weeding step is straightforward to do
efficiently. Comparing using rev appears to me to be an operation on the order
of O(n^3).
I guess one way would be to include everything with all 1's in the first slot
(taking them out of consideration for future operations) and eliminate
everything with 1's in the last slot. Then go for the 2's and so on, perhaps
ending until nothing left. This looks like an O(n) operation, however I do not
know if I will be missing something.
Many thanks again and best wishes,
Ranjan
On Thu, 29 Mar 2018 23:26:12 -0700 Jeff Newmiller <jdnew...@dcn.davis.ca.us>
wrote:
> I don't know if this is more efficient than enumerating with distinct
> directions and weeding... it seems kind of heavyweight to me:
>
> #######
> library(gtools)
> directionless_circular_permutations <- function( n ) {
> v <- seq.int( n-1 )
> ix <- combinations( n-1, 2 )
> jx <- permutations( n-3, n-3 )
> x <- lapply( seq.int( nrow( ix ) )
> , function( i ) {
> cbind( ix[ i, 1 ]
> , permutations( n-3
> , n-3
> , v[ -ix[ i, ] ]
> )
> , ix[ i, 2 ]
> )
> }
> )
> cbind( do.call( rbind, x ), n )
> }
> #######
>
> For more speed, perhaps use Rcpp with [1]?
>
> [1] http://howardhinnant.github.io/combinations.html
>
> On Thu, 29 Mar 2018, Ranjan Maitra wrote:
>
> > Thanks!
> >
> > Yes, however, this seems a bit wasteful. Just wondering if there are
> > other, more efficient options possible.
> >
> > Best wishes,
> > Ranjan
> >
> >
> > On Thu, 29 Mar 2018 22:20:19 -0400 Boris Steipe <boris.ste...@utoronto.ca>
> > wrote:
> >
> >> If one is equal to the reverse of another, keep only one of the pair.
> >>
> >> B.
> >>
> >>
> >>
> >>> On Mar 29, 2018, at 9:48 PM, Ranjan Maitra <mai...@email.com> wrote:
> >>>
> >>> Dear friends,
> >>>
> >>> I would like to get all possible arrangements of n objects listed 1:n on
> >>> a circle.
> >>>
> >>> Now this is easy to do in R. Keep the last spot fixed at n and fill in
> >>> the rest using permuations(n-1, n-1) from the gtools package.
> >>>
> >>> However, what if clockwise or counterclockwise arrangements are the same?
> >>> I know that half of the above (n - 1)! arrangements are redundant.
> >>>
> >>> Is there an easy way to list these (n-1)!/2 arrangements?
> >>>
> >>> I thought of only listing the first half from a call to permuations(n -
> >>> 1, n - 1), but while this holds for n = 4, it does not for n = 5. So, I
> >>> am wondering if there is another function or tweak which would easily do
> >>> this.
> >>>
> >>> Many thanks in advance for any help. and best wishes,
> >>> Ranjan
> >>>
> >>> ______________________________________________
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> >>> and provide commented, minimal, self-contained, reproducible code.
> >>
> >> ______________________________________________
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> >>
> >
> >
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>
> ---------------------------------------------------------------------------
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