[R] getting all circular arrangements without accounting for order

[Ranjan Maitra]

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 <jdnewmil at 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.steipe at 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 <maitra at 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.
> >>
> >> ______________________________________________
> >> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
> >> https://stat.ethz.ch/mailman/listinfo/r-help
> >> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> >> and provide commented, minimal, self-contained, reproducible code.
> >>
> >
> >
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> > ______________________________________________
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> >
> 
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