Hi,
On 27 November 2011 05:59, Aaron Meurer <[email protected]> wrote:
> This is an interesting question. Definitely is_zero should work.
>
> Actually, it's not so bad. It's just not implemented enough to know
> that zero + zero == zero:
>
> In [4]: a.is_zero
> Out[4]: True
>
> In [3]: var('a b',positive=False,negative=False,real=True, zero=False)
> AssertionError: ('inconsistency between facts', {'real': True,
> 'prime': False, 'comparable': False, 'commutative': True, 'composite':
> False, 'positive': False, 'negative': False, 'zero': False, 'complex':
> True, 'imaginary': False, 'nonzero': True}, 'real', False)
>
> I think this should be considered in the context of the new
> assumptions system, because that is what we eventually use. There,
> the refine() function simplifies expressions based on assumptions.
> Right now it doesn't work for this:
>
> In [7]: refine(a, ~Q.positive(a)&~Q.negative(a)&Q.real(a))
> Out[7]: a
>
> But I think what we should look at is making refine smart enough to do
> this, rather than trying to fix the old system to be smart about these
> things, because, again, the old system is going away. Not only that,
> but there's a reason that we're going with a model where you call
> refine rather than things trying to simplify automatically.
>
With new assumptions adding such feature is very simple, e.g.:
~/repo/git/sympy$ git diff
diff --git a/sympy/assumptions/refine.py b/sympy/assumptions/refine.py
index ab7c34e..b7a0c9b 100644
--- a/sympy/assumptions/refine.py
+++ b/sympy/assumptions/refine.py
@@ -153,8 +153,13 @@ def refine_exp(expr, assumptions):
elif ask(Q.odd(coeff + S.Half), assumptions):
return S.ImaginaryUnit
+def refine_Symbol(expr, assumptions):
+ if ask(~Q.nonzero(expr), assumptions):
+ return S.Zero
+
handlers_dict = {
'Abs' : refine_abs,
'Pow' : refine_Pow,
'exp' : refine_exp,
+ 'Symbol' : refine_Symbol,
}
~/repo/git/sympy$ bin/isympy -q
IPython console for SymPy 0.7.1-git (Python 2.7.2-64-bit) (ground types:
gmpy)
In [1]: refine(x, ~Q.nonzero(x))
Out[1]: 0
In [2]: refine(x, ~Q.negative(x) & ~Q.positive(x) & Q.real(x))
Out[2]: 0
I'm just curious why there is Q.nonzero and not Q.zero. Also, I'm not sure
about the following:
In [3]: refine(x, ~Q.negative(x) & ~Q.positive(x))
Out[3]: 0
In [4]: refine(x, ~Q.negative(x))
Out[4]: x
Shouldn't assumption in [3] be understood as zero and all complex numbers
with non-zero imaginary part? In this case the result should be x.
>
> Aaron Meurer
>
> On Thu, Nov 24, 2011 at 9:16 PM, smichr <[email protected]> wrote:
> >>>> var('a b',positive=False,negative=False,real=True)
> > (a, b)
> >>>> a+b
> > a + b
> >>>> _.is_zero
> >
> > Should such symbols be allowed or should 0 have been returned instead?
> >
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>
Mateusz
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