Opened 6 years ago
Closed 6 years ago
#15444 closed defect (fixed)
Two algorithms for k-charge do not give same answer
Reported by: | aschilling | Owned by: | |
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Priority: | major | Milestone: | sage-6.1 |
Component: | combinatorics | Keywords: | tableaux, charge |
Cc: | sage-combinat, zabrocki | Merged in: | |
Authors: | Anne Schilling | Reviewers: | Mike Zabrocki |
Report Upstream: | N/A | Work issues: | |
Branch: | public/combinat/k-charge-15444 (Commits) | Commit: | 8df647454982f5799a4267712551a78989f08992 |
Dependencies: | Stopgaps: |
Description (last modified by )
Currently, the two implementations of k-charge do not give the same answer:
sage: T = WeakTableaux(4,[4,3,2,1],[2,2,2,2,1,1],representation='bounded') sage: for t in T: print t.k_charge(), t.k_charge(algorithm='J') ....: 9 10 10 10 8 8 9 9 10 10 8 9 11 11
Comparing against the expansion of Hall-Littlewood symmetric functions in terms of k-Schur functions, it seems that the I-implementation is correct
sage: Sym = SymmetricFunctions(QQ['t']) sage: Qp = Sym.hall_littlewood().Qp() sage: ks = Sym.kschur(4) sage: ks(Qp[2,2,2,2,1,1])[Partition([4,3,2,1])] t^11 + 2*t^10 + 2*t^9 + 2*t^8
Compared to the book http://arxiv.org/abs/1301.3569 pg. 84 the bug seems to be in the method _height_of_restricted_subword in k_tableau.py.
Change History (11)
comment:1 follow-up: ↓ 2 Changed 6 years ago by
comment:2 in reply to: ↑ 1 Changed 6 years ago by
Yes, I think the implementation that Avi and Nate did is not quite correct. If you look at the second standard subword of your example above, then the program computes the height of the restricted subword incorrectly for the letter r=3.
comment:3 Changed 6 years ago by
- Branch set to public/combinat/k-charge-15444
- Commit set to 8df647454982f5799a4267712551a78989f08992
- Status changed from new to needs_review
New commits:
8df6474 | Fixed bug in k-charge implementation for J-algorithm |
comment:4 follow-up: ↓ 5 Changed 6 years ago by
- Description modified (diff)
comment:5 in reply to: ↑ 4 Changed 6 years ago by
I ran the following code for k=3 and 4 and now the two implementations seem to agree:
sage: for n in range(10): for la in Partitions(n,max_part=k): for mu in Partitions(n,max_part=k): T = WeakTableaux(k,la,mu,representation='bounded') if not all(t.k_charge() == t.k_charge(algorithm="J") for t in T): print la,mu ....:
comment:6 follow-up: ↓ 7 Changed 6 years ago by
- Status changed from needs_review to positive_review
Looks good to me. I tested it on much larger examples and everything seems to be correct now.
Thanks for fixing it.
comment:7 in reply to: ↑ 6 Changed 6 years ago by
Thank you for the swift review! Anne
comment:8 Changed 6 years ago by
- Milestone changed from sage-5.13 to sage-6.0
comment:9 Changed 6 years ago by
- Reviewers set to Mike Zabrocki
comment:10 Changed 6 years ago by
- Milestone changed from sage-6.0 to sage-6.1
comment:11 Changed 6 years ago by
- Resolution set to fixed
- Status changed from positive_review to closed
Proposition 3.15 on p. 84 of our book states that the k-charge with the two algorithms are equal. Am I correct to assume that since you opened a ticket that the error is in the algorithm and not in that proposition?
To clarify