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maths-prog/Outshuffles.md
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maths-prog/Outshuffles.md
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# 8 perfect outshuffles leave a deck of 52 cards unchanged
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Here is some code in R which demonstrates this:
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~~~~
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c(1:52) -> z
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outshuffle <- function(z){
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NULL -> l
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for(i in c(1:26)){
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c(l,z[i],z[26+i]) -> l
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}
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l->> z
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}
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for(i in c(1:8)){
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outshuffle(z)
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}
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~~~~
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And the results are:
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~~~~
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[1] "Step 1"
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[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
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[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
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[51] 26 52
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[1] "Step 2"
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[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
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[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
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[51] 39 52
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[1] "Step 3"
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[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
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[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
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[51] 20 52
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[1] "Step 4"
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[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
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[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
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[51] 36 52
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[1] "Step 5"
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[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
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[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
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[51] 44 52
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[1] "Step 6"
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[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
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[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
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[51] 48 52
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[1] "Step 7"
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[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
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[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
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[51] 50 52
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[1] "Step 8"
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[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
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[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
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[51] 51 52
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~~~~
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q.e.d.
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