Two algorithms for k-charge do not give same answer

[anneschilling]

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.

CC: @sagetrac-sage-combinat @zabrocki

Component: combinatorics

Keywords: tableaux, charge

Author: Anne Schilling

Branch/Commit: public/combinat/k-charge-15444 @ 8df6474

Reviewer: Mike Zabrocki

Issue created by migration from https://trac.sagemath.org/ticket/15444

[zabrocki]

comment:1

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

sage: T = WeakTableau([[1, 1, 2, 3], [2, 4, 4], [3, 6], [5]],4,representation='bounded')
sage: T.k_charge()
9
sage: T.k_charge(algorithm='J')
10

[anneschilling]

comment:2

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.

[anneschilling]

Commit: 8df6474

[anneschilling]

Branch: public/combinat/k-charge-15444

[anneschilling]

New commits:

[8df6474](https://github.com/sagemath/sagetrac-mirror/commit/8df6474)

Fixed bug in k-charge implementation for J-algorithm

[anneschilling]

Author: Anne Schilling

[anneschilling]

comment:5

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
....:         

[zabrocki]

comment:6

Looks good to me. I tested it on much larger examples and everything seems to be correct now.

Thanks for fixing it.

[anneschilling]

comment:7

Thank you for the swift review! Anne

[anneschilling]

Reviewer: Mike Zabrocki