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


Leonid Chindelevitch

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