Talk:Inversion (discrete mathematics)

Inversion vectors

@Watchduck:When calculating inversion vectors, different methods are used to obtain the two sets of the inversion pairs (i,j):

$\operatorname {inv} (\pi )\ =\ \{\ (1,6),\ (2,3),\ (2,5),\ (2,6),\ (3,5),\ (3,6),\ (4,5),\ (4,6),\ (5,6)\}$

$\operatorname {inv} (\pi ^{-1})\ =\ \{\ (1,2),\ (1,3),\ (1,4),\ (1,5),\ (1,6),\ (3,4),\ (3,5),\ (3,6),\ (4,5)\}$

which can be represented by a triangle with the 15 possible pairs as can be seen from the two figures File:Inversion example; invset p-b 1.svg and File:Inversion example; invset p-b 2.svg. The vectors obtained in the right calculation are denoted:

$\operatorname {r} _{(!)}\ =\ 132210_{(!)}\quad \operatorname {v} _{(!)}\ =\ 503100_{(!)}$

while those for the left calculation, of which the bibliography reports the first, are:

$\operatorname {l} _{(!)}\ =\ 001035_{(!)}\quad \operatorname {v'} _{(!)}\ =\ 011232_{(!)}$

In the first image we represent the vectors $l,\ r$ obtained in the calculation performed on $\pi =254631$ but those of the second image come from different pairs related to the inverse permutation $\pi ^{-1}=615324$ and it makes no sense to call them the same as the previous ones.

The software Mathematica cites one of the two as (see the image) $\operatorname {v} _{(!)}\ =\ 503100_{(!)}$ the other to distinguish it could be denoted, as already said $\operatorname {v'} _{(!)}\ =\ 011232_{(!)}$ that's why the two symbols give rise to the thought that the same vectors are always obtained $l,\ r$, and should be changed to $v',\ v$. Grasso Luigi (discuss • contribs) 20:41, 12 October 2024 (UTC)

@Grasso Luigi: You are aware, that each of the two permutations has its own set of ten files, right? This section shows them both together.

And in each set, the permutation is just called $\pi$, and the three vectors are just called $l$, $r$ and $v$.

I have shortened the heading of this discussion. Yours included: $\pi \ =\ 254631\quad \pi ^{-1}\ =\ 615324$

But that is misleading, when you talk about files from both sets.

I hope indices will make it clear:

$\pi _{1}=\pi _{2}^{-1}={\begin{pmatrix}1&2&3&4&5&6\\2&5&4&6&3&1\end{pmatrix}}$            $\pi _{2}=\pi _{1}^{-1}={\begin{pmatrix}1&2&3&4&5&6\\6&1&5&3&2&4\end{pmatrix}}$

$r_{1}=v_{2}=(1,3,2,2,1,0)$            $r_{2}=v_{1}=(5,0,3,1,0,0)$

I do not care about the fourth vector, because no-one else does. (Anyway, I have shown it as u in Inversions in sequences.) --Watchduck (quack) 21:38, 12 October 2024 (UTC)