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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