A nice explanation of the algorithm; thanks.
However, this part isn't quite right.

`> # @p is the position of each element in @o,
> # that is, @o = (0..$#e)[@p]
> my @p= (0..$#e);
> # Note that it is also true that @p = (0..$#e)[@o].
`

In the language of group theory,

`(0..$#e)` is the
identity, while

`@o, @p` are inverses of each other. Thus

`@p[@o]=@o[@p]=(0..$#e)`.
For example,

` # an arbitrary permutation
@o = @permutation = (1,2,0);
# position of 0 is in $o[2], so $p[0]=2.
# This leads to
@p = @positions = (2,0,1);
# And then
print " o[p] = (@o[@p]) = p[o] = (@p[@o]) \n";
`

which outputs

` o[p] = (0 1 2) = p[o] = (0 1 2)
`

Regards,

barrachois

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