Hi Oleg,
it's not immediately clear (to me at least) how efficient your method
will be in "practice". Any method based on common sub-expression
elimination surely must inspect every node in the flattened graph. In
the worst case, an acyclic graph containing n nodes could have 2^n
nodes when flattened to a tree:
tricky 0 = constant 0
tricky d = add g g
where
g = tricky (d-1)
Of course, in "practice" the worst case may not occur. Also, my
mental model is big circuits, which may be different to yours and
Tom's.
(Sorry if I'm just pointing out the obvious.)
Matt.
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