This is a beautiful list! Thank you for sharing.
Regarding your 3) and 6) versions with modulo arithmetic, happily, this is a
standard result from number theory. You should be able to find the relevant
background and theorem itself near the beginning of any introductory Number
Theory book. IIRC, my class back in the day used Andrews[0] which is a Dover
publication, so extremely affordable.
For a taste, in standard parlance we call the set {0..n-1} a "(least) residue
system" and the subset of coprimes, "reduced residue system". A residue system
endowed with addition modulo n we then call the "additive group (of integers)
modulo n". With that terminology out of the way, the phenomenon you noticed is
often stated as:
Each m in a reduced residue system modulo n generates the attitive group of
integers modulo n.
Anyway, have fun exploring some number theory!
[0]:https://store.doverpublications.com/0486682528.html
Jimmy Gauvin <jimmy.gau...@gmail.com> wrote:
> Hi,
>
> I am looking for some reference texts on permutations and modular
> arithmetic.
>
> I recently stumbled on some interesting properties of card shuffles.
> For example, using a deck of 11 cards labeled 0 through 10 and shuffling
> them to obtain this layout :
>
> 0 6 1 7 2 8 3 9 4 10
>
> There are several ways to find out which position each card occupies .
> 1) index of
> 0 6 1 7 2 8 3 9 4 10 i. i.11
> 0 2 4 6 8 10 1 3 5 7 9
> 2) grading
> /: 0 6 1 7 2 8 3 9 4 10
> 0 2 4 6 8 10 1 3 5 7 9
> 3) and computing the positions with modulo
> 11 | 2*i.11
> 0 2 4 6 8 10 1 3 5 7 9
>
> Going from the positions to the card layout can also be done several ways :
> 4) assignment
> (i.11) ( 0 2 4 6 8 10 1 3 5 7 9 ) } 11$0
> 0 6 1 7 2 8 3 9 4 10
> 5) grading
> /: 0 2 4 6 8 10 1 3 5 7 9
> 0 6 1 7 2 8 3 9 4 10
> 6) and, this is the kicker for me, modulo with the right multiplier
> 11 | 6*i.11
> 0 6 1 7 2 8 3 9 4 10
>
> While 3) is obvious, I find 6) disconcerting. And it seems to work for all
> cases where the number of cards and the interval between cards are coprime.
>
> I know this must be explained somewhere but I can't find the relevant
> material.
>
> Thanks for your assistance,
>
> Jimmy
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