Has no one mentioned "A." yet? On Tue, Jan 28, 2020 at 10:10 PM Henry Rich <henryhr...@gmail.com> wrote:
> This is the Chinese Remainder Theorem, no? > > Henry Rich > > On 1/28/2020 9:49 PM, 'Jon Hough' via Programming wrote: > > See Modulo Multiplication Group: > > https://en.wikipedia.org/wiki/Multiplicative_group_of_integers_modulo_n > > http://mathworld.wolfram.com/ModuloMultiplicationGroup.html > > > > > > Since 11 is prime, its modulo multiplication group has 10 elements (all > the numbers from 1 to 10). i.e. multiplying the elements by any > > number in 1 to 10 will permute the elements. > > > > If you used 8, say, instead. 8 is only coprime to 1,3,5,7. So its Modulo > multiplication group has only 4 elements. > > If you do > > 8 | 2 * i.8 > > you do not get a permutation of the numbers of i.8, because 2 and 8 are > not coprime. > > > > A long time ago I wrote a script to calculate the modulo multiplication > group for arbitrary integers. > > https://github.com/jonghough/PermuJ/blob/master/modulomultiplication.ijs > > > > On Wednesday, January 29, 2020, 10:20:12 AM GMT+9, 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 > > ---------------------------------------------------------------------- > > For information about J forums see http://www.jsoftware.com/forums.htm > > > > ---------------------------------------------------------------------- > > For information about J forums see http://www.jsoftware.com/forums.htm > > ---------------------------------------------------------------------- > For information about J forums see http://www.jsoftware.com/forums.htm > -- Devon McCormick, CFA Quantitative Consultant ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
