#12993: Bug in computing the rank function a poset
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Reporter: saliola | Owner:
sage-combinat
Type: defect | Status:
needs_review
Priority: major | Milestone: sage-5.1
Component: combinatorics | Resolution:
Keywords: poset, combinat | Work issues:
Report Upstream: N/A | Reviewers:
Authors: Darij Grinberg, Anne Schilling | Merged in:
Dependencies: | Stopgaps:
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Description changed by saliola:
Old description:
> This should return True:
> {{{
> sage: P = Poset(([1,2,3,4],[[1,4],[2,3],[3,4]]), facade = True)
> sage: P.is_graded()
> False
> }}}
>
> Anne's suggestion from this [https://groups.google.com/d/topic/sage-
> combinat-devel/XsMrxKdvl1A/discussion thread] on sage-combinat-devel:
>
> For a finite poset perhaps the easiest would be to
>
> - start with a random element in the poset, assign rank 0
> - look at all covers and cocovers and assign the rank according to
> the recurrence rank(x) = rank(y) + 1 if x covers y.
> - repeat with the new elements
> - if at any point an element is reached again and is assigned a
> different value from before, the poset is not graded; otherwise continue
> with new elements
New description:
This should return True:
{{{
sage: P = Poset(([1,2,3,4],[[1,4],[2,3],[3,4]]), facade = True)
sage: P.is_ranked()
False
}}}
Anne's suggestion from this [https://groups.google.com/d/topic/sage-
combinat-devel/XsMrxKdvl1A/discussion thread] on sage-combinat-devel:
For a finite poset perhaps the easiest would be to
- start with a random element in the poset, assign rank 0
- look at all covers and cocovers and assign the rank according to the
recurrence rank(x) = rank(y) + 1 if x covers y.
- repeat with the new elements
- if at any point an element is reached again and is assigned a
different value from before, the poset is not graded; otherwise continue
with new elements
--
--
Ticket URL: <http://trac.sagemath.org/sage_trac/ticket/12993#comment:4>
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