#18987: Parallel computation for TilingSolver.number_of_solutions
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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: |
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Comment (by slabbe):
> > So the quotient is just the signed permutation matrices with either
`0` or `1` coefficient `-1` (all others being `1`). Isn't it?
>
> Maybe. When the dimension is odd, the quotient is the signed permutation
matrices where the signs is either all positive or all negative with
determinant 1. That is the formula that I was using.
Ok, now I understand why I needed it that way. I need well-chosen
representatives for the quotient. By this, I mean that the chosen
representatives of the cosets form a group itself.
For example, these representatives do not form a group:
{{{
sage: n = 3
sage: c = identity_matrix(n)
sage: c[0,0] = -1
sage: L = [w.matrix() for w in WeylGroup(['A', n-1])]
sage: L = [(w if w.det() == 1 else c*w) for w in L]
[
[1 0 0] [ 0 0 -1] [0 0 1] [ 0 -1 0] [0 1 0] [-1 0 0]
[0 1 0] [ 0 1 0] [1 0 0] [ 1 0 0] [0 0 1] [ 0 0 1]
[0 0 1], [ 1 0 0], [0 1 0], [ 0 0 1], [1 0 0], [ 0 1 0]
]
sage: MatrixGroup(L).cardinality()
24
}}}
But these representatives forms a group:
{{{
sage: L = [w.matrix() for w in WeylGroup(['A', n-1])]
sage: L = [m.det() * m for m in L]
sage: L
[
[1 0 0] [ 0 0 -1] [0 0 1] [ 0 -1 0] [0 1 0] [-1 0 0]
[0 1 0] [ 0 -1 0] [1 0 0] [-1 0 0] [0 0 1] [ 0 0 -1]
[0 0 1], [-1 0 0], [0 1 0], [ 0 0 -1], [1 0 0], [ 0 -1 0]
]
sage: MatrixGroup(L).cardinality()
6
}}}
And I still don't know how to construct the quotient when n is even.
--
Ticket URL: <http://trac.sagemath.org/ticket/18987#comment:31>
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