Hi!
http://trac.sagemath.org/sage_trac/ticket/12536
is now ready for review. Is there any volunteer?
On 2/19/12 3:39 PM, Nicolas M. Thiery wrote:
> On Sat, Feb 18, 2012 at 01:39:31PM -0800, Anne Schilling wrote:
>> Having this as an input in Poset was my very first suggestion.
>
> Was it? I had understood you wanted to change the linear extension of
> a poset, and I had (mis?)interpreted that relabeling a poset would be
> a building block of the algorithmic, which would create a bunch of
> copies of the same poset with a different linear extension. This did
> not feel reasonable. On the other hand, specifying the default linear
> extension for a Poset is certainly a natural feature!
Relabeling a poset using a linear extension does not necessarily give the same
poset:
sage: P = Poset(([1,2,3,4], [[1,2],[1,4],[2,3]]), cover_relations = True)
sage: P.linear_extensions()
[[1, 2, 3, 4], [1, 2, 4, 3], [1, 4, 2, 3]]
sage: p = LinearExtensionOfPoset([1,2,3,4],P)
sage: p.to_poset() == P
True
sage: p = LinearExtensionOfPoset([1,4,2,3],P)
sage: p.to_poset() == P
False
But they are isomorphic, perhaps this is what you mean?
In fact, for what I am doing, I only need the algorithmic on finite posets with
labels 1,2,...,n and then there is a natural order 1<2<...<n which is now used
in LinearExtensionOfPoset. So the order problem has now disappeared.
So P.linear_extensions() as before gives a list of permutations of its vertices.
LinearExtensionOfPoset is only implemented for posets with vertices 1,2,...,n
and the elements are permutations of these numbers.
> Just to make sure: here you mean permutation as in PermutationGroup,
> right?
Now they are just lists of integer from 1,2,...,n. But mathematically they
can be interpreted as permutations in one-line notation.
> This is an interesting point, though fairly independent on the
> previous one, since we can now have permutations of any set (thanks
> Mike!). Does the multiplicative structure of permutations play a role?
\pi^{-1} gives the labels of the new poset. Other than that not as far as I
know.
Well, the tau_i acts as simple transpositions in certain cases and as the
identity
in others. But since they do not satisfy the braid relations, it is not really
the
symmetric group.
Cheers,
Anne
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