Could you please stop shuffling the deck and play already?

32

3

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

Kevin Cruijssen

Posted 2019-03-11T10:19:29.470

Reputation: 67 575

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. – Olivier Grégoire – 2019-03-11T14:21:01.363

@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. – Kevin Cruijssen – 2019-03-11T14:27:23.683

My bad: I was sure the test case you mentioned was of size 8. But thanks for the extra test cases. – Olivier Grégoire – 2019-03-11T15:37:45.687

Is it ok to return false or other castable values instead of 0? – Alex – 2019-03-11T19:39:29.223

@Alex Yeah sure. I believe the Python answer already does so as well. – Kevin Cruijssen – 2019-03-12T07:38:32.800

1Question 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. – Steve – 2019-03-12T11:05:44.593

2@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. – Kevin Cruijssen – 2019-03-12T11:20:02.953

@IMSoP It was indeed supposed to be alternated, but I've edited it to interleaved now. – Kevin Cruijssen – 2019-03-13T11:00:02.397

There are two forms of riffle (faro) shuffle depending on whether the first card comes from the first half of the deck or from the second half. The first of these, which is the version described here is an out-shuffle. The second is an in-shuffle. It's fairly easy to see that an out-shuffle on n cards is equivalent to an in-shuffle on the n-2 inside cards, followed by replacing the first and last cards. – Alchymist – 2019-03-13T16:14:32.187

Answers

6

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

Jonathan Allan

Posted 2019-03-11T10:19:29.470

Reputation: 67 804

25

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)

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:

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

Arnauld

Posted 2019-03-11T10:19:29.470

Reputation: 111 334

12

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.

Erik the Outgolfer

Posted 2019-03-11T10:19:29.470

Reputation: 38 134

Save four bytes with f=lambda x:2!=x[1]and-~f(x[::2]+x[1::2]) – Jonathan Allan – 2019-03-11T14:48:39.147

@JonathanAllan Oh, of course! Well... != can be -. ;-) – Erik the Outgolfer – 2019-03-11T14:49:10.940

Ah, yeah caveat emptor :D (or just x[1]>2 I guess) – Jonathan Allan – 2019-03-11T14:50:07.713

5

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

Ven

Posted 2019-03-11T10:19:29.470

Reputation: 3 382

1Count the recursion depth for -3. – Erik the Outgolfer – 2019-03-11T13:26:20.633

@EriktheOutgolfer Much better, thanks! – Ven – 2019-03-11T13:29:41.987

1∧/2≤/⍵ -> ⍵≡⍳≢⍵ – ngn – 2019-03-12T15:58:28.197

@ngn didn't realize the array had no holes. Thanks! – Ven – 2019-03-12T16:59:41.960

5

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.

nwellnhof

Posted 2019-03-11T10:19:29.470

Reputation: 10 037

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.

Kirill L.

Posted 2019-03-11T10:19:29.470

Reputation: 6 693

Very nice! I was working on a similar approach but using a recursive function, which didn't work out as golfy. – user2390246 – 2019-03-12T11:55:30.290

4

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

Jo King

Posted 2019-03-11T10:19:29.470

Reputation: 38 234

3

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

Emigna

Posted 2019-03-11T10:19:29.470

Reputation: 50 798

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. :)

– Kevin Cruijssen – 2019-03-11T10:31:48.500

@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. – Emigna – 2019-03-11T11:18:35.803

3

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;}

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

Olivier Grégoire

Posted 2019-03-11T10:19:29.470

Reputation: 10 647

163 bytes by using two times x.clone() instead of A.copyOf(x,l). – Kevin Cruijssen – 2019-03-11T12:58:19.833

264 bytes – Arnauld – 2019-03-11T13:57:23.883

@Arnauld Thanks! I had a hard time figuring how to simplify that "length if odd else length - 1" – Olivier Grégoire – 2019-03-11T13:59:58.500

@Arnauld Oh! My new algorithm is actually the same as yours... And I spent half an hour figuring it out by myself... – Olivier Grégoire – 2019-03-11T14:04:39.020

More precisely, it's equivalent to an improvement over my original algorithm found by @nwellnhof. – Arnauld – 2019-03-11T14:10:25.067

I went away and thought about the algorithm over lunch and came up with the same idea... only to find you'd already discovered it. +1 – Neil – 2019-03-11T14:29:03.127

I've since found https://math.stackexchange.com/a/886088/30622 which explains why the value 2 is always in the largest cycle.

– Neil – 2019-03-12T00:51:15.303

@Neil My original idea was to "follow" a value. For the sake of simplicity I chose the value next to 1, which is 2 in the sorted array. That's how I noticed the pattern and made the formula. – Olivier Grégoire – 2019-03-12T08:45:26.420

I was just making sure that the formula works for all inputs. Obviously 2 is the simplest value to check, but it is also never one of the cases where the algorithm fails. (A example failure case would be checking for 4 in a shuffle of size 10; since 4 and 7 just swap places, it would always return 0 or 1.) – Neil – 2019-03-12T09:29:58.200

@Neil Nice to know: I hadn't noticed the swap. Thanks :) – Olivier Grégoire – 2019-03-12T09:31:44.207

3

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!

Galen Ivanov

Posted 2019-03-11T10:19:29.470

Reputation: 13 815

converge-iterate, then drop one, and count - this leads to shorter code – ngn – 2019-03-12T16:28:28.593

@ngn. So, similar to my J solution - I'll try it later, thanks! – Galen Ivanov – 2019-03-12T18:16:26.413

11#@}.(\:2|#\)^:(2<1{])^:a: for 26 bytes – Jonah – 2019-04-22T03:29:49.683

@Jonah Thank you! – Galen Ivanov – 2019-04-22T03:54:57.650

2

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

RosLuP

Posted 2019-03-11T10:19:29.470

Reputation: 3 036

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! – Indiana Kernick – 2019-03-11T21:31:45.813

@Kerndog73 All is number, but in APL think characters are not numbers... (they seems element in AV array) – RosLuP – 2019-03-11T23:41:26.123

2

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
    }
}

rtpax

Posted 2019-03-11T10:19:29.470

Reputation: 411

2

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.

Nick Kennedy

Posted 2019-03-11T10:19:29.470

Reputation: 11 829

1@user2390246 fair point. Adjusted accordingly – Nick Kennedy – 2019-03-12T12:03:55.750

2

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.

G B

Posted 2019-03-11T10:19:29.470

Reputation: 11 099

2

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!

Volare

Posted 2019-03-11T10:19:29.470

Reputation: 21

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! :)

– Kevin Cruijssen – 2019-03-13T10:52:55.763

2

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!

Andrei Odegov

Posted 2019-03-11T10:19:29.470

Reputation: 939

you can save a bit more: Try it online!

– mazzy – 2019-03-18T04:49:41.267

@muzzy, you're right again :) thanks – Andrei Odegov – 2019-03-18T07:41:17.473

1

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

Shaggy

Posted 2019-03-11T10:19:29.470

Reputation: 24 623

1This interpreter looks super cool. – recursive – 2019-03-11T23:51:55.863

1

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!

Galen Ivanov

Posted 2019-03-11T10:19:29.470

Reputation: 13 815

1

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

Ven

Posted 2019-03-11T10:19:29.470

Reputation: 3 382

1

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.

Rainer Glüge

Posted 2019-03-11T10:19:29.470

Reputation: 131

1

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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Xcali

Posted 2019-03-11T10:19:29.470

Reputation: 7 671

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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mazzy

Posted 2019-03-11T10:19:29.470

Reputation: 4 832

0

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 =)

Alex

Posted 2019-03-11T10:19:29.470

Reputation: 326

Asked: 2019-03-11T10:19:29.470

Viewed: 7 779 times

Active: 2019-08-26T15:35:27.283