#14498: trees and binary trees
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Reporter: elixyre | Owner: sage-combinat
Type: enhancement | Status: needs_review
Priority: major | Milestone: sage-5.11
Component: combinatorics | Resolution:
Keywords: trees, binary trees, latex | Work issues:
Report Upstream: N/A | Reviewers:
Authors: Jean-Baptiste Priez | Merged in:
Dependencies: #8703 | Stopgaps:
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Comment (by darij):
Hi Jean-Baptiste, thanks for the changes (but aren't the "An other" and
"explorer" typos still there?).
In the q-hook-length formula, are you sure you want to work in the
symbolic ring? That is, you really want not to do any cancellations? For
example, your doctested example
{{{
sage: BinaryTree([[],[]]).q_hook_length_formula()
(((q + 2)*q + 2)*q + 1)*q/((q + 1)*q + 1)
}}}
simplifies to {{{q^2 + q}}}. I have not done any speed tests, but I would
be surprised if the fraction field of a univariate polynomial ring wasn't
faster than the symbolic ring (also I don't trust the symbolic ring, but
this might be unfounded). Alternatively, you can make a recursive
computation with q-binomial coefficients that never leaves the polynomial
ring; but not sure if this is very efficient (q-binomial coefficient might
be dense).
I think the formula in the docstring of q_hook_length_formula() doesn't
match the algorithm. In the formula, there is a positive power of q in the
denominator (which should give a Laurent polynomial, not a polynomial),
while in the algorithm there is a negative power of q in the denominator
(which gives a polynomial divisible by some power of q in the end). I'm
not sure why the power of q is there at all...
I hate to say I am still confused by the input syntax of trees:
{{{
sage: b = BinaryTree([[None,[]],[[[],[]],[]]]).canonical_labelling()
sage: parent(b)
Labelled binary trees
sage: b
3[1[., 2[., .]], 7[5[4[., .], 6[., .]], 8[., .]]]
sage: OrderedTree(b)
[[[], [[], []]], [[[[], []], [[], []]], [[], []]]]
sage: b.node_number()
8
sage: OrderedTree(b).node_number()
17
}}}
Apparently the coercion(!) from b to an ordered tree replaces every leaf
by a node with two children because for some reason leaves of binary trees
are stored as nodes-with-two-children internally. This is all defensible
from some viewpoint, but I want it doced. Once somebody clears this up I
would volunteer to review #14498, but so far I don't want to spread my own
confusion...
--
Ticket URL: <http://trac.sagemath.org/sage_trac/ticket/14498#comment:12>
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