36
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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 , 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
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9
  • \$\begingroup\$ One or two test cases with an odd length and an output greater than 0 would be nice. It's easy to mess the riffle in such cases if you have to write the riffle code by yourself instead of relying on builtins. \$\endgroup\$ Mar 11, 2019 at 14:21
  • \$\begingroup\$ @OlivierGrégoire The [1,3,5,7,9,2,4,6,8] is of length 9, but I will add a few more for lengths 7 and 11 perhaps. EDIT: Added the test cases [1,3,5,7,2,4,6] = 2 (length 7) and [1,6,11,5,10,4,9,3,8,2,7] = 6 (length 11). Hope that helps. \$\endgroup\$ Mar 11, 2019 at 14:27
  • \$\begingroup\$ My bad: I was sure the test case you mentioned was of size 8. But thanks for the extra test cases. \$\endgroup\$ Mar 11, 2019 at 15:37
  • 1
    \$\begingroup\$ Question as currently formulated seems "wrong"... a single riffle shuffle should result in the first and last cards changing, unless you're pulling some kind of con trick! i.e. [6,1,7,2,8,3,9,4,10,5] after a single shuffle of 10 cards. \$\endgroup\$
    – Steve
    Mar 12, 2019 at 11:05
  • 3
    \$\begingroup\$ @Steve I guess you're kinda right. Riffle-shuffling in general simply interleaves two halves, so both [1,6,2,7,3,8,4,9,5,10] or [6,1,7,2,8,3,9,4,10,5] are possible. In my challenge it does mean that the top card will always remain the top card, so it's indeed a bit of a con-trick.. I've never seen someone irl use only riffle-shuffles to shuffle a deck of cards however. Usually they also use other type of shuffles in between. Anyway, it's too late to change the challenge now, so for the sake of this challenge the top card will always remain the top card after a riffle-shuffle. \$\endgroup\$ Mar 12, 2019 at 11:20

27 Answers 27

25
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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))

Try it online!

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)

Try it online!

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:

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

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0
12
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Python 2, 39 bytes

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

Try it online!

-4 thanks to Jonathan Allan.

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3
  • \$\begingroup\$ Save four bytes with f=lambda x:2!=x[1]and-~f(x[::2]+x[1::2]) \$\endgroup\$ Mar 11, 2019 at 14:48
  • \$\begingroup\$ @JonathanAllan Oh, of course! Well... != can be -. ;-) \$\endgroup\$ Mar 11, 2019 at 14:49
  • \$\begingroup\$ Ah, yeah caveat emptor :D (or just x[1]>2 I guess) \$\endgroup\$ Mar 11, 2019 at 14:50
7
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Jelly, 8 bytes

ŒœẎ$ƬiṢ’

Try it online!

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
\$\endgroup\$
0
6
+100
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APL (Dyalog Unicode), 35 26 23 22 bytesSBCS

{⍵≡⍳≢⍵:0⋄1+∇⍵[⍒2|⍳⍴⍵]}

Try it 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

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4
  • 1
    \$\begingroup\$ Count the recursion depth for -3. \$\endgroup\$ Mar 11, 2019 at 13:26
  • \$\begingroup\$ @EriktheOutgolfer Much better, thanks! \$\endgroup\$
    – Ven
    Mar 11, 2019 at 13:29
  • 1
    \$\begingroup\$ ∧/2≤/⍵ -> ⍵≡⍳≢⍵ \$\endgroup\$
    – ngn
    Mar 12, 2019 at 15:58
  • \$\begingroup\$ @ngn didn't realize the array had no holes. Thanks! \$\endgroup\$
    – Ven
    Mar 12, 2019 at 16:59
5
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R, 58 55 45 bytes

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

Try it online!

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

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1
  • \$\begingroup\$ Very nice! I was working on a similar approach but using a recursive function, which didn't work out as golfy. \$\endgroup\$ Mar 12, 2019 at 11:55
5
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Perl 6, 34 32 bytes

-2 bytes thanks to Jo King

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

Try it online!

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.

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0
4
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Java (JDK), 59 bytes

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

Try it online!

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;}

Try it online!

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

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10
  • \$\begingroup\$ 163 bytes by using two times x.clone() instead of A.copyOf(x,l). \$\endgroup\$ Mar 11, 2019 at 12:58
  • 2
    \$\begingroup\$ 64 bytes \$\endgroup\$
    – Arnauld
    Mar 11, 2019 at 13:57
  • \$\begingroup\$ @Arnauld Thanks! I had a hard time figuring how to simplify that "length if odd else length - 1" \$\endgroup\$ Mar 11, 2019 at 13:59
  • \$\begingroup\$ @Arnauld Oh! My new algorithm is actually the same as yours... And I spent half an hour figuring it out by myself... \$\endgroup\$ Mar 11, 2019 at 14:04
  • \$\begingroup\$ More precisely, it's equivalent to an improvement over my original algorithm found by @nwellnhof. \$\endgroup\$
    – Arnauld
    Mar 11, 2019 at 14:10
4
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Perl 6, 36 34 32 bytes

-2 bytes thanks to nwellnhof

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

Try it online!

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
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0
3
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05AB1E (legacy), 9 bytes

[DāQ#ι˜]N

Try it online!

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
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3
  • \$\begingroup\$ I didn't even knew it was possible to output the index outside the loop in the legacy. I thought it would be 0 again at that point, just like in the new 05AB1E version. Nice answer! Shorter than my 10-byter using the unshuffle-builtin Å≠ that inspired this challenge. :) \$\endgroup\$ Mar 11, 2019 at 10:31
  • \$\begingroup\$ @KevinCruijssen: Interesting. I didn't know there was an unshuffle. In this instance it's the same as my version, but unshuffle maintains dimensions on 2D arrays. \$\endgroup\$
    – Emigna
    Mar 11, 2019 at 11:18
  • \$\begingroup\$ Alternative 9-byter that works in both versions: [DāQD–#ι˜ \$\endgroup\$
    – ovs
    Jan 17, 2021 at 14:13
3
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J, 28 26 bytes

-2 bytes thanks to Jonah!

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

Try it online!

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}

Try it online!

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4
  • \$\begingroup\$ converge-iterate, then drop one, and count - this leads to shorter code \$\endgroup\$
    – ngn
    Mar 12, 2019 at 16:28
  • \$\begingroup\$ @ngn. So, similar to my J solution - I'll try it later, thanks! \$\endgroup\$ Mar 12, 2019 at 18:16
  • 1
    \$\begingroup\$ 1#@}.(\:2|#\)^:(2<1{])^:a: for 26 bytes \$\endgroup\$
    – Jonah
    Apr 22, 2019 at 3:29
  • \$\begingroup\$ @Jonah Thank you! \$\endgroup\$ Apr 22, 2019 at 3:54
3
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Husk, 11 bytes

L↑S≠ŀ¡ȯΣTC2

Try it online!

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2
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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
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2
  • \$\begingroup\$ That’s the first time I’ve seen someone differentiate between characters and bytes 👍. It always bugs me when I see Unicode characters and they claim that it’s one byte per character. This 😠 is not one byte! \$\endgroup\$ Mar 11, 2019 at 21:31
  • \$\begingroup\$ @Kerndog73 All is number, but in APL think characters are not numbers... (they seems element in AV array) \$\endgroup\$
    – user58988
    Mar 11, 2019 at 23:41
2
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Ruby, 42 bytes

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

Try it online!

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.

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2
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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

Try it online!

Now handles the zero shuffle case.

\$\endgroup\$
1
  • 1
    \$\begingroup\$ @user2390246 fair point. Adjusted accordingly \$\endgroup\$ Mar 12, 2019 at 12:03
2
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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.

Try it online


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];}}

Try it online

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
    }
}
\$\endgroup\$
0
2
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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

Try it online.

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

Try it online!

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1
  • \$\begingroup\$ I took the liberty of fixing your title and adding a TIO-link for the test cases (based on the other R answer), and also verified your answer works as intended, so +1 from me and welcome to PPCG! :) \$\endgroup\$ Mar 13, 2019 at 10:52
2
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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

Try it online!

\$\endgroup\$
2
  • \$\begingroup\$ you can save a bit more: Try it online! \$\endgroup\$
    – mazzy
    Mar 18, 2019 at 4:49
  • \$\begingroup\$ @muzzy, you're right again :) thanks \$\endgroup\$ Mar 18, 2019 at 7:41
2
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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

Try it online!

\$\endgroup\$
2
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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.

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1
  • \$\begingroup\$ Welcome to CGCC! Pretty smart approach, so +1 from me. :) MATLAB doesn't have an online compiler, right? Since I don't see it in the tio.run list. Could you in that case perhaps add a screenshot as verification for the test cases? :) \$\endgroup\$ Mar 4, 2021 at 21:49
2
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Perl 5 -pa, 77 68 bytes

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

Try it online!

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1
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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

Try it online!

Explanation

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

\$\endgroup\$
1
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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]

Try it online!

\$\endgroup\$
1
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Pyth, 18 bytes

L?SIb0hys%L2>Bb1
y

Try it online!

-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

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1
  • \$\begingroup\$ 14 bytes \$\endgroup\$
    – hakr14
    Jan 17, 2021 at 20:25
1
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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

Try it online!

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1
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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
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1
  • 1
    \$\begingroup\$ This interpreter looks super cool. \$\endgroup\$
    – recursive
    Mar 11, 2019 at 23:51
1
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Stax, 14 bytes

ü≈:☻‼Xí┌ùß♦▲▬á

Run and debug it

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0
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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

Try it online!

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

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4
  • 1
    \$\begingroup\$ -2 by removing the f=. No need to assign the lambda to a variable, it is allowed without it; and it's still possible to call it. The "invisible" variable _ is set to the last evaluated value, so lambda x:x[1]-2and 1+f(x[::2]+x[1::2]) would set _ to that exact lambda. \$\endgroup\$
    – Makonede
    Jan 18, 2021 at 17:17
  • \$\begingroup\$ Oh, this is brilliant! I will certainly have fun (and debugging profit) playing with it \$\endgroup\$
    – Alex
    Jan 18, 2021 at 21:01
  • \$\begingroup\$ No way to demonstrate it though: it only works in REPL mode \$\endgroup\$
    – Alex
    Jan 18, 2021 at 21:17
  • \$\begingroup\$ Never mind, I only just realized that you use recursion and need it assigned to f. However, for answers that don't use recursion you still don't need it. \$\endgroup\$
    – Makonede
    Feb 26, 2021 at 21:55

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