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 t4
I seem to end up with the same raised tree you ended up with:
I first raise t1:
/\
/ \
/\ \
t1 \ \
\ \ \
t2 t3 t4
then t3 (since t1 already has a right branch):
/\
t3 \
/ \
t1 \
\ \
t2 t4
And 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