Opened 9 years ago

Closed 9 years ago

# Two algorithms for k-charge do not give same answer

Reported by: Owned by: aschilling major sage-6.1 combinatorics tableaux, charge sage-combinat, zabrocki Anne Schilling Mike Zabrocki N/A public/combinat/k-charge-15444 8df647454982f5799a4267712551a78989f08992

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.

### comment:1 follow-up:  2 Changed 9 years ago by zabrocki

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
```

### comment:2 in reply to:  1 Changed 9 years ago by aschilling

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 9 years ago by aschilling

Authors: → Anne Schilling → public/combinat/k-charge-15444 → 8df647454982f5799a4267712551a78989f08992 new → needs_review

New commits:

 ​8df6474 Fixed bug in k-charge implementation for J-algorithm

### comment:4 follow-up:  5 Changed 9 years ago by aschilling

Description: modified (diff)

### comment:5 in reply to:  4 Changed 9 years ago by aschilling

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 9 years ago by zabrocki

Status: needs_review → 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 9 years ago by aschilling

Thank you for the swift review! Anne

### comment:8 Changed 9 years ago by jdemeyer

Milestone: sage-5.13 → sage-6.0

### comment:9 Changed 9 years ago by aschilling

Reviewers: → Mike Zabrocki

### comment:10 Changed 9 years ago by vbraun_spam

Milestone: sage-6.0 → sage-6.1

### comment:11 Changed 9 years ago by vbraun

Resolution: → fixed positive_review → closed
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