Could you please stop shuffling the deck and play already?

Question

Challenge:

Input: A list of distinct positive integers within the range \$[1, \text{list-size}]\$.

Output: An integer: the amount of times the list is riffle-shuffled. For a list, this means the list is split in two halves, and these halves are interleaved (i.e. riffle-shuffling the list [1,2,3,4,5,6,7,8,9,10] once would result in [1,6,2,7,3,8,4,9,5,10], so for this challenge the input [1,6,2,7,3,8,4,9,5,10] would result in 1).

Challenge rules:

• You can assume the list will only contain positive integers in the range \$[1, \text{list-size}]\$ (or \$[0, \text{list-size}-1]\$ if you choose to have 0-indexed input-lists).
• You can assume all input-lists will either be a valid riffle-shuffled list, or a sorted list which isn't shuffled (in which case the output is 0).
• You can assume the input-list will contain at least three values.

Step-by-step example:

Input: [1,3,5,7,9,2,4,6,8]

Unshuffling it once becomes: [1,5,9,4,8,3,7,2,6], because every even 0-indexed item comes first [1, ,5, ,9, ,4, ,8], and then all odd 0-indexed items after that [ ,3, ,7, ,2, ,6, ].
The list isn't ordered yet, so we continue:

Unshuffling the list again becomes: [1,9,8,7,6,5,4,3,2]
Again becomes: [1,8,6,4,2,9,7,5,3]
Then: [1,6,2,7,3,8,4,9,5]
And finally: [1,2,3,4,5,6,7,8,9], which is an ordered list, so we're done unshuffling.

We unshuffled the original [1,3,5,7,9,2,4,6,8] five times to get to [1,2,3,4,5,6,7,8,9], so the output is 5 in this case.

General rules:

• This is code-golf, so shortest answer in bytes wins.
  Don't let code-golf languages discourage you from posting answers with non-codegolfing languages. Try to come up with an as short as possible answer for 'any' programming language.
• Standard rules apply for your answer with default I/O rules, so you are allowed to use STDIN/STDOUT, functions/method with the proper parameters and return-type, full programs. Your call.
• Default Loopholes are forbidden.
• If possible, please add a link with a test for your code (i.e. TIO).
• Also, adding an explanation for your answer is highly recommended.

Test cases:

Input                                                   Output

[1,2,3]                                                 0
[1,2,3,4,5]                                             0
[1,3,2]                                                 1
[1,6,2,7,3,8,4,9,5,10]                                  1
[1,3,5,7,2,4,6]                                         2
[1,8,6,4,2,9,7,5,3,10]                                  2
[1,9,8,7,6,5,4,3,2,10]                                  3
[1,5,9,4,8,3,7,2,6,10]                                  4
[1,3,5,7,9,2,4,6,8]                                     5
[1,6,11,5,10,4,9,3,8,2,7]                               6
[1,10,19,9,18,8,17,7,16,6,15,5,14,4,13,3,12,2,11,20]    10
[1,3,5,7,9,11,13,15,17,19,2,4,6,8,10,12,14,16,18,20]    17
[1,141,32,172,63,203,94,234,125,16,156,47,187,78,218,109,249,140,31,171,62,202,93,233,124,15,155,46,186,77,217,108,248,139,30,170,61,201,92,232,123,14,154,45,185,76,216,107,247,138,29,169,60,200,91,231,122,13,153,44,184,75,215,106,246,137,28,168,59,199,90,230,121,12,152,43,183,74,214,105,245,136,27,167,58,198,89,229,120,11,151,42,182,73,213,104,244,135,26,166,57,197,88,228,119,10,150,41,181,72,212,103,243,134,25,165,56,196,87,227,118,9,149,40,180,71,211,102,242,133,24,164,55,195,86,226,117,8,148,39,179,70,210,101,241,132,23,163,54,194,85,225,116,7,147,38,178,69,209,100,240,131,22,162,53,193,84,224,115,6,146,37,177,68,208,99,239,130,21,161,52,192,83,223,114,5,145,36,176,67,207,98,238,129,20,160,51,191,82,222,113,4,144,35,175,66,206,97,237,128,19,159,50,190,81,221,112,3,143,34,174,65,205,96,236,127,18,158,49,189,80,220,111,2,142,33,173,64,204,95,235,126,17,157,48,188,79,219,110,250]
                                                        45

Answer 1

JavaScript (ES6), 44 bytes

Shorter version suggested by @nwellnhof

Expects a deck with 1-indexed cards as input.

f=(a,x=1)=>a[x]-2&&1+f(a,x*2%(a.length-1|1))

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Given a deck \$[c_0,\ldots,c_{L-1}]\$ of length \$L\$, we define:

$$x_n=\begin{cases} 2^n\bmod L&\text{if }L\text{ is odd}\\ 2^n\bmod (L-1)&\text{if }L\text{ is even}\\ \end{cases}$$

And we look for \$n\$ such that \$c_{x_n}=2\$.

---

JavaScript (ES6),  57 52  50 bytes

Expects a deck with 0-indexed cards as input.

f=(a,x=1,k=a.length-1|1)=>a[1]-x%k&&1+f(a,x*-~k/2)

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How?

Since JS is lacking native support for extracting array slices with a custom stepping, simulating the entire riffle-shuffle would probably be rather costly (but to be honest, I didn't even try). However, the solution can also be found by just looking at the 2nd card and the total number of cards in the deck.

Given a deck of length \$L\$, this code looks for \$n\$ such that:

$$c_2\equiv\left(\frac{k+1}{2}\right)^n\pmod k$$

where \$c_2\$ is the second card and \$k\$ is defined as:

$$k=\begin{cases} L&\text{if }L\text{ is odd}\\ L-1&\text{if }L\text{ is even}\\ \end{cases}$$

Answer 2

Python 2, 39 bytes

f=lambda x:x[1]-2and-~f(x[::2]+x[1::2])

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-4 thanks to Jonathan Allan.

Answer 3

Jelly, 8 bytes

ŒœẎ$ƬiṢ’

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How?

ŒœẎ$ƬiṢ’ - Link: list of integers A
    Ƭ    - collect up until results are no longer unique...
   $     -   last two links as a monad:
Œœ       -     odds & evens i.e. [a,b,c,d,...] -> [[a,c,...],[b,d,...]]
  Ẏ      -     tighten                         -> [a,c,...,b,d,...]
     Ṣ   - sort A
    i    - first (1-indexed) index of sorted A in collected shuffles
      ’  - decrement

Answer 4

APL (Dyalog Unicode), 35 26 23 22 bytes^SBCS

{⍵≡⍳≢⍵:0⋄1+∇⍵[⍒2|⍳⍴⍵]}

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Thanks to Adám for the help, Erik the Outgolfer for -3 and ngn for -1.

The TIO link contains two test cases.

Explanation:

{⍵≡⍳≢⍵:0⋄1+∇⍵[⍒2|⍳⍴⍵]}
{⍵≡⍳≢⍵:0⋄1+∇⍵[⍒2|⍳⍴⍵]} ⍝ function takes one argument: ⍵, the array
 ⍵≡⍳≢⍵                 ⍝ if the array is sorted:
 ⍵≡⍳≢⍵                 ⍝ array = 1..length(array)
      :0               ⍝ then return 0
        ⋄              ⍝ otherwise
         1+            ⍝ increment
           ∇           ⍝ the value of the recursive call with this argument:
            ⍵[      ]  ⍝ index into the argument with these indexes:
                 ⍳⍴⍵   ⍝ - generate a range from 1 up to the size of ⍵
               2|      ⍝ - %2: generate a binary mask like [1 0 1 0 1 0]
              ⍒        ⍝ - grade (sorts but returns indexes instead of values), so we have the indexes of all the 1s first, then the 0s.

¹

Answer 5

R, 58 55 45 bytes

a=scan();while(a[2]>2)a=matrix(a,,2,F<-F+1);F

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Simulates the sorting process. Input is 1-indexed, returns FALSE for 0.

Answer 6

Perl 6, 34 32 bytes

-2 bytes thanks to Jo King

{(.[(2 X**^$_)X%$_-1+|1]...2)-1}

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Similar to Arnauld's approach. The index of the second card after n shuffles is 2**n % k with k defined as in Arnauld's answer.

Answer 7

Java (JDK), 59 bytes

a->{int c=0;for(;a[(1<<c)%(a.length-1|1)]>2;)c++;return c;}

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Works reliably only for arrays with a size less than 31 or solutions with less than 31 iterations. For a more general solution, see the following solution with 63 bytes:

a->{int i=1,c=0;for(;a[i]>2;c++)i=i*2%(a.length-1|1);return c;}

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Explanation

In a riffle, the next position is the previous one times two modulo either length if it's odd or length - 1 if it's even.

So I'm iterating over all indices using this formula until I find the value 2 in the array.

Credits

• -8 bytes thanks to Kevin Cruijssen. (Previous algorithm, using array)
• -5 bytes thanks to Arnauld.

Answer 8

Perl 6, 36 34 32 bytes

-2 bytes thanks to nwellnhof

$!={.[1]-2&&$!(.sort:{$++%2})+1}

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Reverse riffle shuffles by sorting by the index modulo 2 until the list is sorted, then returns the length of the sequence.

It's funny, I don't usually try the recursive approach for Perl 6, but this time it ended up shorter than the original.

Explanation:

$!={.[1]-2&&$!(.sort:{$++%2})+1}
$!={                           }   # Assign the anonymous code block to $!
    .[1]-2&&                       # While the list is not sorted
            $!(             )      # Recursively call the function on
               .sort:{$++%2}       # It sorted by the parity of each index
                             +1    # And return the number of shuffles

Answer 9

05AB1E (legacy), 9 bytes

[DāQ#ι˜]N

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Explanation

[   #  ]     # loop until
  ā          # the 1-indexed enumeration of the current list
 D Q         # equals a copy of the current list
     ι˜      # while false, uninterleave the current list and flatten
        N    # push the iteration index N as output

Answer 10

J, 28 26 bytes

-2 bytes thanks to Jonah!

 1#@}.(\:2|#\)^:(2<1{])^:a:

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Inspired be Ven's APL solution.

Explanation:

               ^:       ^:a:   while 
                 (2<1{])       the 1-st (zero-indexed) element is greater than 2   
     (        )                do the following and keep the intermediate results
          i.@#                 make a list form 0 to len-1
        2|                     find modulo 2 of each element
      /:                       sort the argument according the list of 0's and 1's
1  }.                          drop the first row of the result
 #@                            and take the length (how many rows -> steps)     

K (ngn/k), 25 bytes

Thanks to ngn for the advice and for his K interpreter!

{#1_{~2=x@1}{x@<2!!#x}\x}

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Answer 11

Husk, 11 bytes

L↑S≠ŀ¡ȯΣTC2

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Answer 12

APL(NARS), chars 49, bytes 98

{0{∧/¯1↓⍵≤1⌽⍵:⍺⋄(⍺+1)∇⍵[d],⍵[i∼d←↑¨i⊂⍨2∣i←⍳≢⍵]}⍵}

why use in the deepest loop, one algo that should be nlog(n), when we can use one linear n? just for few bytes more? [⍵≡⍵[⍋⍵] O(nlog n) and the confront each element for see are in order using ∧/¯1↓⍵≤1⌽⍵ O(n)]test:

  f←{0{∧/¯1↓⍵≤1⌽⍵:⍺⋄(⍺+1)∇⍵[d],⍵[i∼d←↑¨i⊂⍨2∣i←⍳≢⍵]}⍵}
  f ,1
0
  f 1 2 3
0
  f 1,9,8,7,6,5,4,3,2,10
3
  f 1,3,5,7,9,11,13,15,17,19,2,4,6,8,10,12,14,16,18,20
17

Answer 13

Ruby, 42 bytes

f=->d,r=1{d[r]<3?0:1+f[d,r*2%(1|~-d.max)]}

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How:

Search for number 2 inside the array: if it's in second position, the deck hasn't been shuffled, otherwise check the positions where successive shuffles would put it.

Answer 14

R, 70 72 bytes

x=scan();i=0;while(any(x>sort(x))){x=c(x[y<-seq(x)%%2>0],x[!y]);i=i+1};i

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Now handles the zero shuffle case.

Answer 15

C (GCC) 64 63 bytes

-1 byte from nwellnhof

i,r;f(c,v)int*v;{for(i=r=1;v[i]>2;++r)i=i*2%(c-1|1);return~-r;}

This is a drastically shorter answer based on Arnauld's and Olivier Grégoire's answers. I'll leave my old solution below since it solves the slightly more general problem of decks with cards that are not contiguous.

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---

C (GCC) 162 bytes

a[999],b[999],i,r,o;f(c,v)int*v;{for(r=0;o=1;++r){for(i=c;i--;(i&1?b:a)[i/2]=v[i])o=(v[i]>v[i-1]|!i)&o;if(o)return r;for(i+=o=c+1;i--;)v[i]=i<o/2?a[i]:b[i-o/2];}}

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a[999],b[999],i,r,o; //pre-declare variables
f(c,v)int*v;{ //argument list
    for(r=0;o=1;++r){ //major loop, reset o (ordered) to true at beginning, increment number of shuffles at end
        for(i=c;i--;(i&1?b:a)[i/2]=v[i]) //loop through v, split into halves a/b as we go
            o=(v[i]>v[i-1]|!i)&o; //if out of order set o (ordered) to false
        if(o) //if ordered
            return r; //return number of shuffles
        //note that i==-1 at this point
        for(i+=o=c+1;i--;)//set i=c and o=c+1, loop through v
            v[i]=i<o/2?a[i]:b[i-o/2];//set first half of v to a, second half to b
    }
}

Answer 16

R, 85 bytes

s=scan();u=sort(s);k=0;while(any(u[seq(s)]!=s)){k=k+1;u=as.vector(t(matrix(u,,2)))};k

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Explanation

Stupid (brute force) method, much less elegant than following the card #2.

Instead of unshuffling the input s we start with a sorted vector u that we progressively shuffle until it is identical with s. This gives warnings (but shuffle counts are still correct) for odd lengths of input due to folding an odd-length vector into a 2-column matrix; in that case, in R, missing data point is filled by recycling of the first element of input.

The loop will never terminate if we provide a vector that cannot be unshuffled.

Addendum: you save one byte if unshuffling instead. Unlike the answer above, there is no need to transpose with t(), however, ordering is byrow=TRUE which is why T appears in matrix().

R, 84 bytes

s=scan();u=sort(s);k=0;while(any(s[seq(u)]!=u)){k=k+1;s=as.vector(matrix(s,,2,T))};k

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Answer 17

PowerShell, 116 114 108 84 78 bytes

-24 bytes thanks to Erik the Outgolfer's solution.

-6 bytes thanks to mazzy.

param($a)for(;$a[1]-2){$n++;$t=@{};$a|%{$t[$j++%2]+=,$_};$a=$t.0+$t.1;$j=0}+$n

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Answer 18

Lua (LuaJIT), 119 96 79 bytes

n=t;while 2~=n[2]do n={}for a=1,#t*2,2 do x=a-#t n[#n+1]=t[a]or t[x+x%2]end;t=n

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Answer 19

MATLAB, 70 bytes

function s(l),x=max(l);find(find(l==2)==1+mod(2.^(1:x),x-1))*(l(2)~=2)

explanation:

every nth shuffle, 2 will be pushed n^2 indices down from its previous position, wrapping around when it reaches the last position. That means that the function for index(n) is

1+mod(2^n,list-size-1)

for a list-size of 10, then, the indices are:

• index(1) = 3 -> [1 6 2 7 3 8 4 9 5 10]
• index(2) = 5 -> [1 8 6 4 2 9 7 5 3 10]
• index(3) = 9 -> [1 9 8 7 6 5 4 3 2 10]
• index(4) = 17 => 8 -> [1 5 9 4 8 3 7 2 6 10]

etc.

Using this, I find the index where 2 is, and find which power of 2 that is along the array. That power corresponds the the number of shuffles. To account for 0 shuffles, the whole thing is multiplied by the boolean value of l(2)~=2 to make sure that it returns 0 when 2 is in the right place, which only happens for an unshuffled array.

Answer 20

Perl 5 -pa, 77 68 bytes

map{push@{$_%2},$_}0..$#F;++$\,@F=@F[@0,@1]while"@F"ne"@{[1..@F]}"}{

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Answer 21

Wolfram Language (Mathematica), 62 bytes

c=0;While[Sort[a]!=a,a=a[[1;;-1;;2]]~Join~a[[2;;-1;;2]];c++];c

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Explanation

The input list is a . It is unriffled and compared with the sorted list until they match.

Answer 22

Red, 87 79 78 bytes

func[b][c: 0 while[b/2 > 2][c: c + 1 b: append extract b 2 extract next b 2]c]

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Answer 23

Pyth, 18 bytes

L?SIb0hys%L2>Bb1
y

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-2 thanks to @Erik the Outgolfer.

The script has two line: the first one defines a function y, the second line calls y with the implicit Q (evaluated stdin) argument.

L?SIb0hys%L2>Bb1
L                function y(b)
 ?               if...
  SIb            the Invariant b == sort(b) holds
     0           return 0
      h          otherwise increment...
       y         ...the return of a recursive call with:
             B   the current argument "bifurcated", an array of:
              b   - the original argument
            >  1  - same with the head popped off
          L      map...
         % 2     ...take only every 2nd value in each array
        s         and concat them back together

¹

Answer 24

PowerShell, 62 71 70 66 bytes

+9 bytes when Test cases with an even number of elements added.

-1 byte with splatting.

-4 bytes: wrap the expression with $i,$j to a new scope.

for($a=$args;$a[1]-2;$a=&{($a|?{++$j%2})+($a|?{$i++%2})}){$n++}+$n

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Answer 25

Japt, 13 11 10 bytes

Taking my shiny, new, very-work-in-progress interpreter for a test drive.

ÅÎÍ©ÒßUñÏu

Try it or run all test cases

ÅÎÍ©ÒßUñÏu     :Implicit input of integer array U
Å              :Slice the first element off U
 Î             :Get the first element
  Í            :Subtract from 2
   ©           :Logical AND with
    Ò          :  Negation of bitwise NOT of
     ß         :  A recursive call to the programme with input
      Uñ       :    U sorted
        Ï      :    By 0-based indices
         u     :    Modulo 2

Answer 26

Stax, 14 bytes

ü≈:☻‼Xí┌ùß♦▲▬á

Run and debug it

Answer 27

Vyxal 3, 7 bytes

ϩUJiᶤᵞS

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ϩUJiᶤᵞS­⁡​‎‎⁡⁠⁡‏⁠⁠⁠‎⁡⁠⁤‏‏​⁡⁠⁡‌⁢​‎‎⁡⁠⁢‏⁠‎⁡⁠⁣‏‏​⁡⁠⁡‌⁣​‎‎⁡⁠⁢⁡‏⁠‎⁡⁠⁢⁢‏⁠‎⁡⁠⁢⁣‏‏​⁡⁠⁡‌­
ϩ  i     # ‎⁡Apply and collect until value is no longer unique, including initial value
 UJ      # ‎⁢Uninterleave and append
    ᶤᵞS  # ‎⁣first index where the list is sorted (invariant after sort)
💎

Created with the help of Luminespire.

Answer 28

Python 3, 40 bytes

f=lambda x:x[1]-2and 1+f(x[::2]+x[1::2])  # 1-based
f=lambda x:x[1]-1and 1+f(x[::2]+x[1::2])  # 0-based

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I need to refresh the page more frequently: missed Erik the Outgolfer's edit doing a similar trick =)