I only looked at this very quickly:
I get
>>> from sympy import *
>>> pg00 = Symbol('pg00')
>>> pg01 = Symbol('pg01')
>>> eq1 = sympify(pg01**2/pg00**2)
>>> eq2 = sympify(1.0 - 2*pg01/pg00)
>>> eq3 = sympify(pg01*(1.0 - pg01/pg00))
>>> eq = [eq1,eq2,eq3]
>>> print eq
[pg01**2/pg00**2, 1.0 - 2*pg01/pg00, pg01*(1.0 - pg01/pg00)]
>>> (red,rep) = cse(eq)
>>> red
[(x0, -pg01/pg00)]
>>> rep
[pg01**2/pg00**2, 1.0 - 2*pg01/pg00, pg01*(x0 + 1.0)]
after changing ndiv to be
def ndiv(a, b):
"""if b divides a in an extractive way (like 1/4 divides 1/2
but not vice versa, and 2/5 does not divide 1/3) then return
the integer number of times it divides, else return 0.
"""
if a not in (S.Infinity, S.NegativeInfinity) and \
(not b.q % a.q or not a.q % b.q):
return int(a/b)
return 0
But it could be that the better fix is to not have infinity pulled out
-- by breakup?
Hopefully that can get someone started.
/c
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