#14126: Count Number of Linear Extensions of a Poset
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Reporter: csar | Owner: sage-combinat
Type: enhancement | Status: needs_work
Priority: major | Milestone: sage-6.4
Component: combinatorics | Resolution:
Keywords: days45 | Merged in:
Authors: Jori Mäntysalo | Reviewers:
Report Upstream: N/A | Work issues:
Branch: u/jmantysalo | Commit:
/count-lin-ext | f908696aa7c63cefbc382af326cfe085e0dfa522
Dependencies: | Stopgaps:
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Comment (by jmantysalo):
I don't quite understand. AFAIK it is proved that counting linear
extensions is in `#P`, and according to
http://link.springer.com/chapter/10.1007%2F11537311_39 there is an
algorithm that lists them in linear time.
We can trivially reduce time by doing series-parallel decomposition first.
It won't help on average.
So do you have some good algorithm in mind? Being in `#P` does not mean
that `k` could not be improved in `O(k^n)`.
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Ticket URL: <https://trac.sagemath.org/ticket/14126#comment:30>
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