Jeff,
I wanted to let you know that your function is faster than generating the
directional circular permutations and weeding.
Here is the time for n = 10. I compared with just doing the permutations, there
is no point in proceeding further with the weeding since it is slower at the
start itself.
system.time(directionless_circular_permutations(10))
user system elapsed
1.576 0.000 1.579
system.time(permutations(9,9))
user system elapsed
3.030 0.000 3.037
Repeats keep the values similar. All computations on a 10-core processor
@3.1GHz with 20 threads (using only one thread) and running the Fedora 27
with 4.15.9 kernel.
I have to say that I was surprised to see the difference at n = 10 itself.
Thanks again!
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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