Brian Schott wrote:
> John,
>
> I don't understand how you have identified hh as the
> inverse of h. I see that it does accomplish the desired or
> simpler result of using { rather that }, but in general how
> can an inverse like hh be identified from its dual (is dual
> the right terminology)? That is, I thought I could use h
> with amend, but if I can always find hh from h and then use
> {, that could be superior. Also, could you demonstrate that
> h and hh are really inverses. My understanding says that an
> inverse reverses the effect of its dual, but I don't see hh
> reversing h. Are you saying that { and } are inverse
> (functors -- I don't even know what a functor is, but this
> sounds like it might be right) and if { and } are
> inverses, then that is a revelation to me.
>
Brian:
Here are a couple of ways of finding the inverse of a permutation.
This is the inverse in a group-theoretic sense, where the group
multiplication is (non-commutative) composition of permutations. If
the permutations are expressed in direct form, composition is { .
h=:1 2 0 3
hh=:2 0 1 3
h { hh
0 1 2 3
hh { h
0 1 2 3
The easiest "mathematical" way of thinking about inverses of
permutations is the following:
h=:1 2 0 3
In 2-row notation this is
0 1 2 3
1 2 0 3
You get the inverse by flipping the rows,
1 2 0 3
0 1 2 3
sorting the columns,
0 1 2 3
2 0 1 3
and taking the last row
2 0 1 3
This corresponds to the J verb
pinv=:3 : '{: /:~&.|: |.(i.#y),:y'
pinv h
2 0 1 3
However, you can do this in one:
/: h
2 0 1 3
Alternatively, you can find the inverse by reversing each cycle when
the permutation is expressed in disjoint cycle form.
Convert h to cycle form,
+-----+-+
|2 0 1|3|
+-----+-+
reverse each cycle,
+-----+-+
|1 0 2|3|
+-----+-+
convert back to direct form
2 0 1 3
This corresponds to
pinv3=:|.&.>&.C.
pinv3 h
2 0 1 3
Obviously this method is most useful when you start and finish with
the permutation in cycle form, and you find the inverse with |.&.> .
Finally, { and } and (kind of) inverses with respect to permutations,
but this is really the result of the inverse permutation being given
by /: . This is what I was getting at:
f=:3 : 'y { i.#y' NB. identity
f h
1 2 0 3
g=:3 : '(i.#y) y } y' NB. inverse: only shape of RHA to } matters
g h
2 0 1 3
Best wishes,
John
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