#14641: Does the "promotion" method for tableaux really compute Schuetzenberger
promotion?
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Reporter: darij |
Owner: tbd
Type: defect |
Status: new
Priority: major |
Milestone: sage-5.10
Component: PLEASE CHANGE |
Resolution:
Keywords: tableaux, partitions, combinat, jeu de taquin, promotion |
Work issues:
Report Upstream: N/A |
Reviewers:
Authors: |
Merged in:
Dependencies: |
Stopgaps:
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Comment (by jessicapalencia):
Hi Darij,
So I think the 'n' in this promotion operator specifies that the highest
entry value we are allowing in the tableau is 'n+1'. Now for standard
tableaux, we want 'n' to be the number of boxes minus 1. But when you're
working with semistandard tableaux, you really do need to specify this,
since it may be that no 'n+1's are present in the tableaux, in which case
promotion just increments all the entries.
It would be good to change the code so that if no 'n' is specified, it
takes n = (# boxes - 1) by default.
I agree that the code is confusing in the case of rectangular tableaux,
but it seems to be working correctly in the rectangular examples I've
tried. It's likely just a computational shortcut.
Also, I think the code isn't designed to make sense with 'n' less than the
largest entry minus 1. So it gives nonsensical answers. Perhaps the code
should check that 'n' is at least as big as the largest entry minus 1.
Jessica
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Ticket URL: <http://trac.sagemath.org/sage_trac/ticket/14641#comment:2>
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