Re: Issue 978 in sympy: extract (-1) from all expressions canonically

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Comment #2 on issue 978 by smi...@gmail.com: extract (-1) from all  
expressions canonically
http://code.google.com/p/sympy/issues/detail?id=978

This issue should really be fixed. Without it, zeros can hide in  
expressions and lead to undue simplification difficulties. Consider:

m = Matrix([[ 0,    -1,     0,     0,     0,  0,    -1,  0,     0],
... [-1, x,    -1,     0,     0,  0,     0, -1,     0],
... [ 0,    -1, x,     0,     0,  0,     0,  0,    -1],
... [ 0,     0,     0, x,    -1,  0,    -1,  0,     0],
... [ 0,     0,     0,    -1, x, -1,     0, -1,     0],
... [ 0,     0,     0,     0,    -1,  0,     0,  0,    -1],
... [-1,     0,     0,    -1,     0,  0, x, -1,     0],
... [ 0,    -1,     0,     0,    -1,  0,    -1,  0,    -1],
... [ 0,     0,    -1,     0,     0, -1,     0, -1, x]])
r=m.row_join(eye(9)).rref()

When you simplify (with cse) some of the terms in the "inverted" matrix  
(which is singular) you will find that they are NaN, but if you factor such  
expressions you come up with some interesting results:

for j in range(9):
...  for k in range(9,18):
...   rjk=r[0][j,k]
...   rep,e=cse(rjk)
...   if e[0] is S.NaN:
...     print j,k,factor(rjk)
...
0 11 -1/(2*x)
0 13 nan
0 14 oo*(x - 1)**2*(x + 1)**2
0 15 -1/2
0 17 0
1 9 -1/2
etc...


Let's look at the first one:

z=r[0][0,11]

In its current form it is not NaN:

z.count_ops()
611

I've picked one of the arguments of z that is infinity (another one is a  
symbolic zero and that, I suppose, is why cse is able to determine that z  
is NaN).

z1=z.args[1].args[2]
1/(-(-x + 1/x)/(x - 1/x)**2 - (-(-x + 1/x)/(x*(x - 1/x)**2) - 1/(x*(x -  
1/x)))**
2/(2*x - 2/x - (-x + 1/x)/(x**2*(x - 1/x)**2) - 1/(x**2*(x - 1/x))) - 1/(x  
- 1/x
))

If this argument is sympified the infinity shows up:

S(str(_))
oo

How is a symbolic infinity getting into an expression? I ask somewhat  
rhetorically. If the expression had been sign simplified (made canonical  
wrt sign) then perhaps this situation wouldn't have arisen.

zz=z.args[1].args[2]
zz.subs([(a,Mul(-1,a,evaluate=False) if a.could_extract_minus_sign()  
else a)
 for a in zz.atoms(Add)])
oo

But an interesting thing is that factor gets into trouble (giving -1/(2*x))  
as the factorization of this expression that cse can determine quickly is  
nan...and simplify takes about two minutes to do nothing to the expression:

t=time();s=simplify(z);print time()-t
124.861000061
s==z
True

Here's a little simpler expression that doesn't respond well to simple  
attempts at simplification:

n
-1 + 1/x
z=n/x/(-n)**2-1/n/x
z.expand()
-1/(x*(1 - 2/x + x**(-2))) - 1/(x*(-1 + 1/x)) + 1/(x**2*(1 - 2/x + x**(-2)))
z.diff(x).expand()
1/(x**2*(1 - 2/x + x**(-2))) + 1/(x**2*(-1 + 1/x)) + 2/(x**3*(1 - 3/x +  
3/x**2 -
 1/x**3)) - 3/(x**3*(1 - 2/x + x**(-2))) - 2/(x**4*(1 - 3/x + 3/x**2 -  
1/x**3))
factor_terms(z)
((-x + 1)**2 - (x - 1)**2)/((-x + 1)*(x - 1)**2)
cancel(z)
0

Again, sign simplification quickly resolves this expression:

z.subs([(a,Mul(-1,a,evaluate=False) if a.could_extract_minus_sign()  
else a) for a in z.atoms(Add)])
0



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