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On Mon, Aug 20, 2012 at 2:02 AM, Mike S. <maikxlx@gmail.com> wrote:For example, starting with ((t1 t2)(t3 t4)):
/\
/ \
/ \
/\ /\
t1 t2 t3 t4
t3
/ \
t1 t4
\
t2
o Voilà, we're done.If I start with (((t1 t2) t3) t4) instead:/\
/ \
/\ \/ \ \/\ \ \t1 t2 t3 t4I seem to end up with the same raised tree you ended up with:I first raise t1:/\
/ \
/\ \t1 \ \\ \ \t2 t3 t4then t3 (since t1 already has a right branch):/\
t3 \
/ \t1 \\ \t2 t4And t3 again:t3
/ \
t1 \\ \
t2 \\t4
If two different trees end up having the same raised tree, then encoding the raised tree won't be enough to encode any tree.mu'o mi'e xorxes
--
Até mais!
Leonardo
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