#15480: Words.__eq__ returns wrong answers
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Reporter: ncohen | Owner:
Type: defect | Status: needs_review
Priority: major | Milestone: sage-5.13
Component: combinatorics | Resolution:
Keywords: | Merged in:
Authors: Nathann Cohen | Reviewers:
Report Upstream: N/A | Work issues:
Branch: u/ncohen/15480 | Commit:
Dependencies: | Stopgaps:
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Changes (by ncohen):
* status: new => needs_review
* branch: => u/ncohen/15480
Old description:
> Right now equality between sets of words only compare the alphabets
>
> {{{
> sage: Words(3, 10) == Words(3,900)
> True
> sage: Words(2, finite=False) == Words(2)
> True
> sage: Words(2) == Words(2,30)
> True
> }}}
New description:
Right now equality between sets of words only compare the alphabets
{{{
sage: Words(3, 10) == Words(3,900)
True
sage: Words(2, finite=False) == Words(2)
True
sage: Words(2) == Words(2,30)
True
sage: Words(10,0) == Words(20,0)
False
}}}
I am not proud of this patch's code, but I see NO other way to all these
wrongs answers without rewriting the class hierarchy. The fact that a
class of finite words is an instance (i.e. it inherits) from a class of
infinite words makes it really hard to implement `O_o`
{{{
sage: isinstance(Words(4,30), type(Words(3)))
True
}}}
You cannot be sure, when implementing the `__eq__` method of `Words_all`,
that self really represents an infinite class of words.
Besides, the old code read :
{{{
if isinstance(other, Words_all):
return self.alphabet() == other.alphabet()
else:
return NotImplemented
}}}
I haven't been able to guess when `not (type(self) is type(other))` does
not mean that the two classes should not be reported as equal.
Though the old code seems to imply that this can happen.
If this can happen we need to add a doctest somewhere.
Nathann
--
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Ticket URL: <http://trac.sagemath.org/ticket/15480#comment:1>
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