Tantangan:

Input: Daftar bilangan bulat positif yang berbeda dalam kisaran $[1, \text{list-size}]$ .

Keluaran: Bilangan bulat: berapa kali daftar diacak-acak . Untuk daftar, ini berarti daftar ini dibagi menjadi dua bagian, dan bagian ini saling terkait (yaitu mengacak-acak daftar yang [1,2,3,4,5,6,7,8,9,10]akan dihasilkan sekali [1,6,2,7,3,8,4,9,5,10], jadi untuk tantangan ini input [1,6,2,7,3,8,4,9,5,10]akan menghasilkan 1).

Aturan tantangan:

• Anda dapat mengasumsikan bahwa daftar hanya akan berisi bilangan bulat positif dalam kisaran $[1, \text{list-size}]$ (atau $[0, \text{list-size}-1]$ jika Anda memilih untuk membuat daftar input yang diindeks 0).
• Anda dapat mengasumsikan semua daftar input akan menjadi daftar acak-acak, atau daftar diurutkan yang tidak dikocok (dalam hal ini hasilnya adalah 0 ).
• Anda dapat mengasumsikan bahwa daftar input akan mengandung setidaknya tiga nilai.

Contoh langkah demi langkah:

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

Unshuffling sekali menjadi:, [1,5,9,4,8,3,7,2,6]karena setiap item bahkan diindeks 0 datang terlebih dahulu [1, ,5, ,9, ,4, ,8], dan kemudian semua item 0-diindeks aneh setelah itu [ ,3, ,7, ,2, ,6, ].
Daftar ini belum dipesan, jadi kami melanjutkan:

Membatalkan daftar lagi menjadi: [1,9,8,7,6,5,4,3,2]
Sekali lagi menjadi: [1,8,6,4,2,9,7,5,3]
Lalu: [1,6,2,7,3,8,4,9,5]
Dan akhirnya:[1,2,3,4,5,6,7,8,9] :, yang merupakan daftar yang diurutkan, jadi kita selesai mengacak.

Kami melepas paket aslinya [1,3,5,7,9,2,4,6,8]lima kali [1,2,3,4,5,6,7,8,9], jadi hasilnya adalah5 dalam kasus ini.

Aturan umum:

• Ini adalah kode-golf , jadi jawaban tersingkat dalam byte menang.
  Jangan biarkan bahasa kode-golf mencegah Anda memposting jawaban dengan bahasa non-codegolf. Cobalah untuk memberikan jawaban sesingkat mungkin untuk bahasa pemrograman 'apa saja'.
• Aturan standar berlaku untuk jawaban Anda dengan aturan I / O default , sehingga Anda diizinkan untuk menggunakan STDIN / STDOUT, fungsi / metode dengan parameter yang tepat dan tipe pengembalian, program penuh. Panggilanmu.
• Celah Default tidak diperbolehkan.
• Jika memungkinkan, silakan tambahkan tautan dengan tes untuk kode Anda (yaitu TIO ).
• Juga, menambahkan penjelasan untuk jawaban Anda sangat dianjurkan.

Kasus uji:

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

Jelly , 8 byte

ŒœẎ$ƬiṢ’

Cobalah online!

Bagaimana?

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

JavaScript (ES6), 44 byte

Versi lebih pendek disarankan oleh @nwellnhof

Mengharapkan dek dengan kartu 1-diindeks sebagai input.

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

Cobalah online!

Diberi setumpuk $[c_0,\ldots,c_{L-1}]$ dengan panjang $L$ , kami mendefinisikan:

Dan kami mencari $n$ sedemikian rupa sehingga $c_{x_n}=2$ .

JavaScript (ES6),  57 52  50 byte

Mengharapkan dek dengan kartu 0-diindeks sebagai input.

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

Cobalah online!

Bagaimana?

Karena JS kurang memiliki dukungan asli untuk mengekstraksi irisan array dengan loncatan kustom, mensimulasikan seluruh riffle-shuffle mungkin akan agak mahal (tapi jujur, saya bahkan tidak mencoba). Namun, solusinya juga dapat ditemukan dengan hanya melihat kartu ke-2 dan jumlah total kartu di dek.

Diberi setumpuk panjang $L$ , kode ini mencari $n$ sedemikian rupa sehingga:

di mana $c_2$ adalah kartu kedua dan $k$ didefinisikan sebagai:

Python 2 , 39 byte

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

Cobalah online!

-4 Terima kasih kepada Jonathan Allan .

R , 58 55 45 byte

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

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Mensimulasikan proses penyortiran. Input diindeks 1, pengembalian FALSEuntuk 0.

Perl 6 , 34 32 byte

-2 byte terima kasih kepada Jo King

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

Cobalah online!

Mirip dengan pendekatan Arnauld . Indeks kartu kedua setelah n mengocok adalah 2**n % kdengan k didefinisikan seperti dalam jawaban Arnauld.

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

{⍵≡⍳≢⍵:0⋄1+∇⍵[⍒2|⍳⍴⍵]}

Cobalah online!

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

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

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

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.

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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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

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.

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.

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
    }
}

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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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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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.

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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Perl 5 -pa, 77 bytes

map{push@{$_%2},$_}0..$#F;$_=0;++$_,@s=sort{$a-$b}@F=@F[@0,@1]while"@F"ne"@s"

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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

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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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

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 =)