I think the CRT is related ( https://en.wikipedia.org/wiki/Chinese_remainder_theorem ) in the sense that it is about modulo arithmetic and congruences, but I am not sure it has much to do with permutations fo the moduli. It states that given some pairwise coprime x1 x2 x3 ... xk, then if N = */ x1 x2 x3 ... xk and we choose any a1 a2 a3 ... ak then there must be some X such that X = (x1|a1), X = (x2|a2) ,....
e.g. k = 3 5 7 (all mutually coprime) k=: 3 5 7 a=: 9 2 11 NB. arbitrary a's N=: */k KA=: k | a I. KA -:"1 1 |: k |"0 _[ i. 1000 NB. look for X up to 1000 102 207 312 417 522 627 732 837 942 can confirm that k | 102 is the same as k | a Doesnt work if the k's are not coprime (e.g. k =: 3 6 11) On Wednesday, January 29, 2020, 12:10:28 PM GMT+9, 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 ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
