ASKSAGE: Sage Q&A Forum - Latest question feedhttps://ask.sagemath.org/questions/Q&A Forum for SageenCopyright Sage, 2010. Some rights reserved under creative commons license.Sat, 31 Mar 2018 14:49:26 -0500Combinatorial Species of Phylogenetic Treeshttps://ask.sagemath.org/question/41852/combinatorial-species-of-phylogenetic-trees/I would like to study the combinatorial class ${\cal H}$ of labelled phylogenetic trees, defined as rooted trees, whose internal nodes are unlabelled
and are constrained to have outdegree $\geq 2$, while their leaves have labels attached to them.
According to Flajolet, Sedgewick:
*Analytic Combinatorics*, p. 128, this class satisfies the recursive specification
$${\cal H}={\cal Z}+\mbox{Set}_{\geq 2}({\cal H})$$
I defined their [species](https://en.wikipedia.org/wiki/Combinatorial_species) $H$ by the code
Z=species.SingletonSpecies()
E2=species.SetSpecies(min=2)
G=CombinatorialSpecies()
H=CombinatorialSpecies()
G.define(E2(Z+G))
H=Z+G
The number of labelled structures is given by the [integer sequence](http://oeis.org/A000311).
In particular I tried to calculate (for a small cardinal number)
1. a list of such structures.
2. a generating series of this structures.
3. the number of such structures.
4. isomorphism-types of such structures.
5. a generating series of the isomorphism-types of such structures.
6. a random such structure.
7. automorphism-groups of of such structures.
using the [following](http://doc.sagemath.org/html/en/reference/combinat/sage/combinat/tutorial.html#section-generic-species) code:
1. H.structures([1,2,3]).list()
2. g = H.generating_series()
3. g.counts(3)
4. H.isotypes([1,2,3])
5. g=H.isotype_generating_series()
6. r=H.structures([1,2,3]).random_element()
7. r.automorphism_group()
Only 2. seems to work.
If my code ist correct, I wonder if some of these (i.e. composition species)
are not implemented and when (whether) they will be.Frank ZenterSat, 31 Mar 2018 14:49:26 -0500https://ask.sagemath.org/question/41852/