Jeff,
Thanks! This one is much faster, no doubt!
system.time(directionless_circular_permutations2(12))
user system elapsed
5.331 0.745 6.089
system.time(directionless_circular_permutations(12))
user system elapsed
173.659 0.890 174.946
However, from n = 13, things start bec
New function below is a bit faster due to more efficent memory handling.
for-loop FTW!
directionless_circular_permutations2 <- function( n ) {
n1 <- n - 1L
v <- seq.int( n1 )
ix <- combinations( n1, 2L )
jx <- permutations( n-3L, n-3L )
jxrows <- nrow( jx )
jxoffsets <- seq.int( jxro
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 its
> On Mar 29, 2018, at 6:48 PM, Ranjan Maitra 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 pa
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 ope
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 )
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
wrote:
> If one is equal to the reverse of another, keep only one of the pair.
>
> B.
>
>
>
> > On Mar
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 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
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
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