
13 Feb
2008
13 Feb
'08
9:27 a.m.
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.