#10776: poset top() function breaks when top element has boolean value False
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Reporter: niles | Owner: sage-combinat
Type: defect | Status: new
Priority: major | Milestone:
Component: combinatorics | Keywords: beginner, poset, top
Author: | Upstream: N/A
Reviewer: | Merged:
Work_issues: |
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Posets determine whether there is a top element by converting the element
returned by `._hasse_diagram.top()` to a boolean. Instead, the function
should check whether the returned value is `None`.
Here is the current definition of `.top()`:
{{{
def top(self):
"""
Returns the top element of the poset, if it exists.
EXAMPLES::
sage: P = Poset({0:[3],1:[3],2:[3],3:[4,5],4:[],5:[]})
sage: P.top() is None
True
sage: Q = Poset({0:[1],1:[]})
sage: Q.top()
1
"""
hasse_top = self._hasse_diagram.top()
if hasse_top:
return self._vertex_to_element(hasse_top)
else:
return None
}}}
And here is an example showing how it fails:
{{{
sage: p = Poset([[0],[]]); p
Finite poset containing 1 elements
sage: p.list()
[0]
sage: p.top()
sage: p.top() is None
True
sage: p._hasse_diagram.top()
0
sage: p._hasse_diagram.top() is None
False
}}}
--
Ticket URL: <http://trac.sagemath.org/sage_trac/ticket/10776>
Sage <http://www.sagemath.org>
Sage: Creating a Viable Open Source Alternative to Magma, Maple, Mathematica,
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