#14019: equality is broken for Posets
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Reporter: ncohen | Owner: sage-combinat
Type: defect | Status: needs_work
Priority: major | Milestone: sage-6.4
Component: combinatorics | Resolution:
Keywords: posets | Merged in:
Authors: Travis Scrimshaw, | Reviewers: Travis Scrimshaw,
Anne Schilling | Anne Schilling
Report Upstream: N/A | Work issues:
Branch: | Commit:
public/combinat/poset/fix_equality-14019|
8dddb16ee9bce94fab03989228cf83dd567f8ac0
Dependencies: #17059 | Stopgaps: #14185
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Comment (by ncohen):
> That condition is fine, but the labeling of the "canonical" labeling is
still arbitrary and not unique (as you can see, it has changed).
No, it is unique. And it is always integers from 0 to n-1.
Try it in Sage: take a graph `D1` labelled on whatever you want, then
compute its canonical representative `R`. Then relabel the vertices of
`D1` into a graph `D2` in any way you like: the representatie of `D2` will
be equal to `R`.
> Why? Your isomorphism property is still satisfied:
Two pairs "Poset, linear_extension" are "isomorphic" if there is an
isomorphism of the two posets that sends the first linear extension on the
second linear extension. But we never check this second part (and
similarly for the canonical representatives).
For instance, if `P` is a poset on three points 0,1,2 where the cover
relations are only (1<2), then the pair "Poset, linear extension" equal to
`P,[0,1,2]` is not isomorphic to `P,[1,0,2]`.
> No, it checks that the posets are isomorphic (which has nothing to do
with the linear extension chosen).
Indeed, and to me that is incorrect. It is quite simple: you cannot say
that two "posets with linear extension" `P1,P2` are isomorphic unless
there is a relabel function `f` such that `P1.relabel(f)==P2`.
As it is implemented right now this does not hold.
Nathann
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Ticket URL: <http://trac.sagemath.org/ticket/14019#comment:66>
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