Oops, sorry, my bad, just realized that the function assumes the tree is
unrooted, so this is not a bug!
Leonid
Leonid Chindelevitch,
Assistant Professor,
School of Computing Science,
Simon Fraser University.
Office: TASC 1 9425,
8888 University Avenue,
Burnaby, BC V5A 1S6.
Phone: 778-782-4973.
2016-11-12 15:29 GMT-08:00 Leonid Chindelevitch <leonard...@gmail.com>:
> It looks like sometimes, the splits produced are not all distinct, namely:
>
> > Tr=rtree(10,rooted=TRUE)
> > Br=bitsplits(Tr)[[1]]
> > Br
> [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
> [1,] f3 fc f0 f0 c0 e0 f0 f0
> [2,] c0 c0 c0 40 00 00 00 00
>
> As you can see, the last two columns are identical. I unfortunately did
> not save the tree which gave me this output, but I believe it should be
> possible to test all the options that have those splits to find the
> offender.
>
> Thanks,
>
> Leonid Chindelevitch
>
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