#18987: Parallel computation for TilingSolver.number_of_solutions
-------------------------+-------------------------------------------------
Reporter: | Owner:
slabbe | Status: needs_review
Type: | Milestone: sage-6.9
enhancement | Resolution:
Priority: major | Merged in:
Component: | Reviewers: Vincent Delecroix
combinatorics | Work issues:
Keywords: | Commit:
Authors: | 0c752d4038a7419285a1c1fa9b0c21842b593a2e
Sébastien Labbé | Stopgaps:
Report Upstream: N/A |
Branch: |
public/18987 |
Dependencies: |
-------------------------+-------------------------------------------------
Comment (by slabbe):
> Ok, now I understand why I needed it that way.
Wait. I really need the quotient itself finally. I was just lucky that the
transformation keeping some the pentamino invariant was in the group.
{{{
sage: from sage.combinat.tiling import ncube_isometry_group
sage: from sage.games.quantumino import pentaminos
sage: L = ncube_isometry_group(3)
sage: f = lambda p : [m for m in L[1:] if (m*p).canonical() ==
p.canonical()]
sage: [(i, f(p)) for i,p in enumerate(pentaminos) if f(p)]
[(6, [
[ 0 0 -1]
[ 0 -1 0]
[-1 0 0]
]),
(7, [
[ 0 0 1]
[ 0 -1 0]
[ 1 0 0]
]),
(12, [
[-1 0 0]
[ 0 0 -1]
[ 0 -1 0]
]),
(13, [
[ 0 0 -1]
[ 0 -1 0]
[-1 0 0]
]),
(16, [
[ 0 0 -1]
[ 0 -1 0]
[-1 0 0]
])]
}}}
Above, I get a problem with pentamino number 7 because it is invariant
under a transformation that is not in the subgroup isomorphic to the
quotient. So I really need to consider the quotient with all of the
elements in each coset. Chosing a representative won't work even if it is
well chosen.
Give me more time. I'll update my branch.
--
Ticket URL: <http://trac.sagemath.org/ticket/18987#comment:33>
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