nunosempere.github.io/maths-prog/Outshuffles.md

8 perfect outshuffles leave a deck of 52 cards unchanged

Here is some code in R which demonstrates this:

c(1:52) -> z

outshuffle <- function(z){  
     NULL -> l  
     for(i in c(1:26)){  
         c(l,z[i],z[26+i]) -> l  
         }  
     l->> z  
}  

for(i in c(1:8)){  
    outshuffle(z)  
}  

And the results are:

[1] "Step 1"
 [1]  1 27  2 28  3 29  4 30  5 31  6 32  7 33  8 34  9 35 10 36 11 37 12 38 13
[26] 39 14 40 15 41 16 42 17 43 18 44 19 45 20 46 21 47 22 48 23 49 24 50 25 51
[51] 26 52
[1] "Step 2"
 [1]  1 14 27 40  2 15 28 41  3 16 29 42  4 17 30 43  5 18 31 44  6 19 32 45  7
[26] 20 33 46  8 21 34 47  9 22 35 48 10 23 36 49 11 24 37 50 12 25 38 51 13 26
[51] 39 52
[1] "Step 3"
 [1]  1 33 14 46 27  8 40 21  2 34 15 47 28  9 41 22  3 35 16 48 29 10 42 23  4
[26] 36 17 49 30 11 43 24  5 37 18 50 31 12 44 25  6 38 19 51 32 13 45 26  7 39
[51] 20 52
[1] "Step 4"
 [1]  1 17 33 49 14 30 46 11 27 43  8 24 40  5 21 37  2 18 34 50 15 31 47 12 28
[26] 44  9 25 41  6 22 38  3 19 35 51 16 32 48 13 29 45 10 26 42  7 23 39  4 20
[51] 36 52
[1] "Step 5"
 [1]  1  9 17 25 33 41 49  6 14 22 30 38 46  3 11 19 27 35 43 51  8 16 24 32 40
[26] 48  5 13 21 29 37 45  2 10 18 26 34 42 50  7 15 23 31 39 47  4 12 20 28 36
[51] 44 52
[1] "Step 6"
 [1]  1  5  9 13 17 21 25 29 33 37 41 45 49  2  6 10 14 18 22 26 30 34 38 42 46
[26] 50  3  7 11 15 19 23 27 31 35 39 43 47 51  4  8 12 16 20 24 28 32 36 40 44
[51] 48 52
[1] "Step 7"
 [1]  1  3  5  7  9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49
[26] 51  2  4  6  8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44 46 48
[51] 50 52
[1] "Step 8"
 [1]  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
[26] 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
[51] 51 52

q.e.d.