There was some discussion of this over at
http://code.google.com/p/sympy/issues/detail?id=2009.
Aaron Meurer
On Nov 24, 2010, at 9:04 PM, smichr wrote:
> sometimes solve returns None and sometimes it returns []. Is there any
> reason that a None would be preferred to []?
>
> --
> You received
sometimes solve returns None and sometimes it returns []. Is there any
reason that a None would be preferred to []?
--
You received this message because you are subscribed to the Google Groups
"sympy" group.
To post to this group, send email to sy...@googlegroups.com.
To unsubscribe from this gr
I agree. If anything, I think we may need to have separate classes to
represent a < b the inequality and a < b the boolean. One problem
with Le(3, 4) not returning True is that constructs like
if a < b:
do something
will break if a or b are sympified, because a < b returns Lt(a, b) in
that c
If I can contribute to the discussion, I support the simpler behavior
when Le(3,4) stays as a inequation until it has to be evaluated. Not
only the shortcut to True/False has now proven to bring in an error in
the solving routine, it is possible that in some situation one will
want to extract the l
Perhaps it has something to do with solve not taking into account that
Le() and friends evaluate themselves to a boolean in trivial cases:
In [1]: Le(3,4)
Out[1]: True
But the only way to really know for sure is to debug the code.
By the way, there's been some debate before on whether or not it
Now it is possible to calculate for example this:
In [451]: SolveFractionIeq((2*x-1)/(x-1), (2*x-1)/x, x)
Out[451]: [[1/2, 1], (-∞, 0]]
... which is almost correct if we omit the closedness of intervals
(somebody fixes?).
Note that the aforementioned UnboundLocalError in inequalities.py
prevents
I am working on a new solve() routine to fix the issue discussed in my
previous post.
However, I observed a bug (?) that sticks deep in the code. I am using
the "git clone https://matt...@github.com/mattpap/sympy-polys.git; cd
sympy-polys ; git checkout polys11" version.
In [388]: solve([Le(3,
I observe following behavior:
In [113]: solve([Le((-1 + x),(-2 + x+x**3)), Assume(x,Q.real)], x,
relational=False)
Out[113]: [[1, ∞)]
In [114]: solve([Le((-1 + x)/(-2 + x+x**3),1), Assume(x,Q.real)], x,
relational=False)
NotImplementedError (...)
But there is a workaround: one must put th
I observe following behavior:
In [113]: solve([Le((-1 + x),(-2 + x+x**3)), Assume(x,Q.real)], x,
relational=False)
Out[113]: [[1, ∞)]
In [114]: solve([Le((-1 + x)/(-2 + x+x**3),1), Assume(x,Q.real)], x,
relational=False)
NotImplementedError (...)
But there is a workaround: one must put th
