Could you please stop shuffling the deck and play already?Unflatten an ArrayLisp Extraction MissionLeaving the NestCalculate the sum of ILDPost-determined Array SortingFind the “Recursive Size” of a ListCould you make me a hexagon please?Visit and exit an arrayCodegolf Rainbow : Sorting Colors with ReflectionLongest Repeating Subsequence of a Single Digit

How is the Swiss post e-voting system supposed to work, and how was it wrong?

What is this large pipe coming out of my roof?

Sword in the Stone story where the sword was held in place by electromagnets

It's a yearly task, alright

Instead of Universal Basic Income, why not Universal Basic NEEDS?

Is it normal that my co-workers at a fitness company criticize my food choices?

What is the greatest age difference between a married couple in Tanach?

Why does Deadpool say "You're welcome, Canada," after shooting Ryan Reynolds in the end credits?

How do I hide Chekhov's Gun?

Why are the outputs of printf and std::cout different

Converting Functions to Arrow functions

Be in awe of my brilliance!

Will a pinhole camera work with instant film?

Informing my boss about remarks from a nasty colleague

Co-worker team leader wants to inject his friend's awful software into our development. What should I say to our common boss?

Do I need life insurance if I can cover my own funeral costs?

Why doesn't using two cd commands in bash script execute the second command?

Bash: What does "masking return values" mean?

Why did it take so long to abandon sail after steamships were demonstrated?

An Accountant Seeks the Help of a Mathematician

Is it possible to upcast ritual spells?

Is it possible that AIC = BIC?

Define, (actually define) the "stability" and "energy" of a compound

Make a transparent 448*448 image



Could you please stop shuffling the deck and play already?


Unflatten an ArrayLisp Extraction MissionLeaving the NestCalculate the sum of ILDPost-determined Array SortingFind the “Recursive Size” of a ListCould you make me a hexagon please?Visit and exit an arrayCodegolf Rainbow : Sorting Colors with ReflectionLongest Repeating Subsequence of a Single Digit













28












$begingroup$


Challenge:



Input: A list of distinct positive integers within the range $[1, textlist-size]$.



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 $[1, textlist-size]$ (or $[0, textlist-size-1]$ 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









share|improve this question











$endgroup$











  • $begingroup$
    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.
    $endgroup$
    – Olivier Grégoire
    Mar 11 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.
    $endgroup$
    – Kevin Cruijssen
    Mar 11 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.
    $endgroup$
    – Olivier Grégoire
    Mar 11 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
    2 days ago






  • 2




    $begingroup$
    @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.
    $endgroup$
    – Kevin Cruijssen
    2 days ago















28












$begingroup$


Challenge:



Input: A list of distinct positive integers within the range $[1, textlist-size]$.



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 $[1, textlist-size]$ (or $[0, textlist-size-1]$ 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









share|improve this question











$endgroup$











  • $begingroup$
    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.
    $endgroup$
    – Olivier Grégoire
    Mar 11 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.
    $endgroup$
    – Kevin Cruijssen
    Mar 11 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.
    $endgroup$
    – Olivier Grégoire
    Mar 11 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
    2 days ago






  • 2




    $begingroup$
    @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.
    $endgroup$
    – Kevin Cruijssen
    2 days ago













28












28








28


2



$begingroup$


Challenge:



Input: A list of distinct positive integers within the range $[1, textlist-size]$.



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 $[1, textlist-size]$ (or $[0, textlist-size-1]$ 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









share|improve this question











$endgroup$




Challenge:



Input: A list of distinct positive integers within the range $[1, textlist-size]$.



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 $[1, textlist-size]$ (or $[0, textlist-size-1]$ 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






code-golf number array-manipulation integer






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited yesterday







Kevin Cruijssen

















asked Mar 11 at 10:19









Kevin CruijssenKevin Cruijssen

40.5k565210




40.5k565210











  • $begingroup$
    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.
    $endgroup$
    – Olivier Grégoire
    Mar 11 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.
    $endgroup$
    – Kevin Cruijssen
    Mar 11 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.
    $endgroup$
    – Olivier Grégoire
    Mar 11 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
    2 days ago






  • 2




    $begingroup$
    @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.
    $endgroup$
    – Kevin Cruijssen
    2 days ago
















  • $begingroup$
    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.
    $endgroup$
    – Olivier Grégoire
    Mar 11 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.
    $endgroup$
    – Kevin Cruijssen
    Mar 11 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.
    $endgroup$
    – Olivier Grégoire
    Mar 11 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
    2 days ago






  • 2




    $begingroup$
    @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.
    $endgroup$
    – Kevin Cruijssen
    2 days ago















$begingroup$
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.
$endgroup$
– Olivier Grégoire
Mar 11 at 14:21




$begingroup$
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.
$endgroup$
– Olivier Grégoire
Mar 11 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.
$endgroup$
– Kevin Cruijssen
Mar 11 at 14:27





$begingroup$
@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.
$endgroup$
– Kevin Cruijssen
Mar 11 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.
$endgroup$
– Olivier Grégoire
Mar 11 at 15:37




$begingroup$
My bad: I was sure the test case you mentioned was of size 8. But thanks for the extra test cases.
$endgroup$
– Olivier Grégoire
Mar 11 at 15:37




1




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
2 days ago




$begingroup$
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
2 days ago




2




2




$begingroup$
@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.
$endgroup$
– Kevin Cruijssen
2 days ago




$begingroup$
@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.
$endgroup$
– Kevin Cruijssen
2 days ago










21 Answers
21






active

oldest

votes


















24












$begingroup$

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:



$$x_n=begincases
2^nbmod L&textif Ltext is odd\
2^nbmod (L-1)&textif Ltext is even\
endcases$$



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:



$$c_2equivleft(frack+12right)^npmod k$$



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



$$k=begincases
L&textif Ltext is odd\
L-1&textif Ltext is even\
endcases$$






share|improve this answer











$endgroup$




















    10












    $begingroup$


    Python 2, 39 bytes





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


    Try it online!



    -4 thanks to Jonathan Allan.






    share|improve this answer











    $endgroup$












    • $begingroup$
      Save four bytes with f=lambda x:2!=x[1]and-~f(x[::2]+x[1::2])
      $endgroup$
      – Jonathan Allan
      Mar 11 at 14:48










    • $begingroup$
      @JonathanAllan Oh, of course! Well... != can be -. ;-)
      $endgroup$
      – Erik the Outgolfer
      Mar 11 at 14:49











    • $begingroup$
      Ah, yeah caveat emptor :D (or just x[1]>2 I guess)
      $endgroup$
      – Jonathan Allan
      Mar 11 at 14:50



















    5












    $begingroup$


    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





    share|improve this answer











    $endgroup$




















      5












      $begingroup$


      Perl 6, 34 32 bytes



      -2 bytes thanks to Jo King





      (.[(2 X**^$_)X%$_-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.






      share|improve this answer











      $endgroup$




















        4












        $begingroup$


        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





        share|improve this answer











        $endgroup$




















          4












          $begingroup$


          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.






          share|improve this answer











          $endgroup$












          • $begingroup$
            Very nice! I was working on a similar approach but using a recursive function, which didn't work out as golfy.
            $endgroup$
            – user2390246
            2 days ago


















          4












          $begingroup$


          APL (Dyalog Unicode), 35 26 23 22 bytesSBCS





          ⍳⍴⍵]


          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:



          ⍳⍴⍵]
          ⍳⍴⍵] ⍝ 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.



          Old solution:
          i⊣i+←1⋄⍵[⍒2⍣∧/2≤/⍵⍵⊣i←¯1



          ¹






          share|improve this answer











          $endgroup$








          • 1




            $begingroup$
            Count the recursion depth for -3.
            $endgroup$
            – Erik the Outgolfer
            Mar 11 at 13:26










          • $begingroup$
            @EriktheOutgolfer Much better, thanks!
            $endgroup$
            – Ven
            Mar 11 at 13:29










          • $begingroup$
            ∧/2≤/⍵ -> ⍵≡⍳≢⍵
            $endgroup$
            – ngn
            2 days ago










          • $begingroup$
            @ngn didn't realize the array had no holes. Thanks!
            $endgroup$
            – Ven
            2 days ago


















          3












          $begingroup$


          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





          share|improve this answer









          $endgroup$












          • $begingroup$
            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. :)
            $endgroup$
            – Kevin Cruijssen
            Mar 11 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 at 11:18


















          3












          $begingroup$


          Java (JDK), 59 bytes





          a->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->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



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

          • -5 bytes thanks to Arnauld.





          share|improve this answer











          $endgroup$












          • $begingroup$
            163 bytes by using two times x.clone() instead of A.copyOf(x,l).
            $endgroup$
            – Kevin Cruijssen
            Mar 11 at 12:58







          • 2




            $begingroup$
            64 bytes
            $endgroup$
            – Arnauld
            Mar 11 at 13:57










          • $begingroup$
            @Arnauld Thanks! I had a hard time figuring how to simplify that "length if odd else length - 1"
            $endgroup$
            – Olivier Grégoire
            Mar 11 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...
            $endgroup$
            – Olivier Grégoire
            Mar 11 at 14:04










          • $begingroup$
            More precisely, it's equivalent to an improvement over my original algorithm found by @nwellnhof.
            $endgroup$
            – Arnauld
            Mar 11 at 14:10


















          2












          $begingroup$

          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





          share|improve this answer











          $endgroup$












          • $begingroup$
            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!
            $endgroup$
            – Kerndog73
            Mar 11 at 21:31










          • $begingroup$
            @Kerndog73 All is number, but in APL think characters are not numbers... (they seems element in AV array)
            $endgroup$
            – RosLuP
            Mar 11 at 23:41


















          2












          $begingroup$


          Ruby, 42 bytes





          f=->d,r=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.






          share|improve this answer











          $endgroup$




















            2












            $begingroup$


            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.






            share|improve this answer











            $endgroup$








            • 1




              $begingroup$
              @user2390246 fair point. Adjusted accordingly
              $endgroup$
              – Nick Kennedy
              2 days ago


















            2












            $begingroup$

            C (GCC) 64 63 bytes



            -1 byte from nwellnhof



            i,r;f(c,v)int*v;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)!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)!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







            share|improve this answer











            $endgroup$




















              1












              $begingroup$


              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.






              share|improve this answer









              $endgroup$




















                1












                $begingroup$


                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!






                share|improve this answer











                $endgroup$




















                  1












                  $begingroup$


                  Perl 5 -pa, 77 bytes





                  mappush@$_%2,$_0..$#F;$_=0;++$_,@s=sort$a-$b@F=@F[@0,@1]while"@F"ne"@s"


                  Try it online!






                  share|improve this answer









                  $endgroup$




















                    1












                    $begingroup$


                    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 == sum(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






                    share|improve this answer











                    $endgroup$




















                      1












                      $begingroup$


                      Japt, 13 11 bytes



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



                      g1 Í©ÒßUñÏu


                      Try it or run all test cases



                      g1 Í©ÒßUñÏu :Implicit input of integer array U
                      g1 :Get the element at 0-based index 1
                      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





                      share|improve this answer











                      $endgroup$








                      • 1




                        $begingroup$
                        This interpreter looks super cool.
                        $endgroup$
                        – recursive
                        Mar 11 at 23:51


















                      1












                      $begingroup$


                      J, 28 bytes



                      1#@}.(/:2|i.@#)^:(2<1 












                      • $begingroup$
                        Very nice! I was working on a similar approach but using a recursive function, which didn't work out as golfy.
                        $endgroup$
                        – user2390246
                        2 days ago















                      $begingroup$
                      Very nice! I was working on a similar approach but using a recursive function, which didn't work out as golfy.
                      $endgroup$
                      – user2390246
                      2 days ago




                      $begingroup$
                      Very nice! I was working on a similar approach but using a recursive function, which didn't work out as golfy.
                      $endgroup$
                      – user2390246
                      2 days ago












                      1












                      $begingroup$


                      J, 28 bytes



                      1#@.(/:2|i.@#)^:(2<1improve this answer











                      $endgroup$












                      • $begingroup$
                        converge-iterate, then drop one, and count - this leads to shorter code
                        $endgroup$
                        – ngn
                        2 days ago










                      • $begingroup$
                        @ngn. So, similar to my J solution - I'll try it later, thanks!
                        $endgroup$
                        – Galen Ivanov
                        2 days ago









                      share.(/:2|i.@#)^:(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. 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@1x@<2!!#xx


                      Try it online!







                      share|improve this answer














                      share|improve this answer



                      share|improve this answer








                      edited yesterday

























                      answered Mar 11 at 20:24









                      Galen IvanovGalen Ivanov

                      7,12211034




                      7,12211034











                      • $begingroup$
                        converge-iterate, then drop one, and count - this leads to shorter code
                        $endgroup$
                        – ngn
                        2 days ago










                      • $begingroup$
                        @ngn. So, similar to my J solution - I'll try it later, thanks!
                        $endgroup$
                        – Galen Ivanov
                        2 days ago
















                      • $begingroup$
                        converge-iterate, then drop one, and count - this leads to shorter code
                        $endgroup$
                        – ngn
                        2 days ago










                      • $begingroup$
                        @ngn. So, similar to my J solution - I'll try it later, thanks!
                        $endgroup$
                        – Galen Ivanov
                        2 days ago















                      $begingroup$
                      converge-iterate, then drop one, and count - this leads to shorter code
                      $endgroup$
                      – ngn
                      2 days ago




                      $begingroup$
                      converge-iterate, then drop one, and count - this leads to shorter code
                      $endgroup$
                      – ngn
                      2 days ago












                      $begingroup$
                      @ngn. So, similar to my J solution - I'll try it later, thanks!
                      $endgroup$
                      – Galen Ivanov
                      2 days ago




                      $begingroup$
                      @ngn. So, similar to my J solution - I'll try it later, thanks!
                      $endgroup$
                      – Galen Ivanov
                      2 days ago











                      1












                      $begingroup$

                      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!






                      share|improve this answer










                      New contributor




                      Volare is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                      Check out our Code of Conduct.






                      $endgroup$












                      • $begingroup$
                        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! :)
                        $endgroup$
                        – Kevin Cruijssen
                        yesterday
















                      1












                      $begingroup$

                      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!






                      share|improve this answer










                      New contributor




                      Volare is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                      Check out our Code of Conduct.






                      $endgroup$












                      • $begingroup$
                        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! :)
                        $endgroup$
                        – Kevin Cruijssen
                        yesterday














                      1












                      1








                      1





                      $begingroup$

                      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!






                      share|improve this answer










                      New contributor




                      Volare is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                      Check out our Code of Conduct.






                      $endgroup$



                      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!







                      share|improve this answer










                      New contributor




                      Volare is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                      Check out our Code of Conduct.









                      share|improve this answer



                      share|improve this answer








                      edited yesterday









                      Stephen

                      7,49323397




                      7,49323397






                      New contributor




                      Volare is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                      Check out our Code of Conduct.









                      answered yesterday









                      VolareVolare

                      112




                      112




                      New contributor




                      Volare is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                      Check out our Code of Conduct.





                      New contributor





                      Volare is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                      Check out our Code of Conduct.






                      Volare is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                      Check out our Code of Conduct.











                      • $begingroup$
                        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! :)
                        $endgroup$
                        – Kevin Cruijssen
                        yesterday

















                      • $begingroup$
                        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! :)
                        $endgroup$
                        – Kevin Cruijssen
                        yesterday
















                      $begingroup$
                      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! :)
                      $endgroup$
                      – Kevin Cruijssen
                      yesterday





                      $begingroup$
                      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! :)
                      $endgroup$
                      – Kevin Cruijssen
                      yesterday












                      0












                      $begingroup$

                      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 =)






                      share|improve this answer











                      $endgroup$

















                        0












                        $begingroup$

                        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 =)






                        share|improve this answer











                        $endgroup$















                          0












                          0








                          0





                          $begingroup$

                          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 =)






                          share|improve this answer











                          $endgroup$



                          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 =)







                          share|improve this answer














                          share|improve this answer



                          share|improve this answer








                          edited Mar 11 at 19:32

























                          answered Mar 11 at 19:22









                          AlexAlex

                          3263




                          3263



























                              draft saved

                              draft discarded
















































                              If this is an answer to a challenge…



                              • …Be sure to follow the challenge specification. However, please refrain from exploiting obvious loopholes. Answers abusing any of the standard loopholes are considered invalid. If you think a specification is unclear or underspecified, comment on the question instead.


                              • …Try to optimize your score. For instance, answers to code-golf challenges should attempt to be as short as possible. You can always include a readable version of the code in addition to the competitive one.
                                Explanations of your answer make it more interesting to read and are very much encouraged.


                              • …Include a short header which indicates the language(s) of your code and its score, as defined by the challenge.


                              More generally…



                              • …Please make sure to answer the question and provide sufficient detail.


                              • …Avoid asking for help, clarification or responding to other answers (use comments instead).




                              draft saved


                              draft discarded














                              StackExchange.ready(
                              function ()
                              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcodegolf.stackexchange.com%2fquestions%2f181299%2fcould-you-please-stop-shuffling-the-deck-and-play-already%23new-answer', 'question_page');

                              );

                              Post as a guest















                              Required, but never shown





















































                              Required, but never shown














                              Required, but never shown












                              Required, but never shown







                              Required, but never shown

































                              Required, but never shown














                              Required, but never shown












                              Required, but never shown







                              Required, but never shown







                              Popular posts from this blog

                              How should I support this large drywall patch? Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?How do I cover large gaps in drywall?How do I keep drywall around a patch from crumbling?Can I glue a second layer of drywall?How to patch long strip on drywall?Large drywall patch: how to avoid bulging seams?Drywall Mesh Patch vs. Bulge? To remove or not to remove?How to fix this drywall job?Prep drywall before backsplashWhat's the best way to fix this horrible drywall patch job?Drywall patching using 3M Patch Plus Primer

                              random experiment with two different functions on unit interval Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Random variable and probability space notionsRandom Walk with EdgesFinding functions where the increase over a random interval is Poisson distributedNumber of days until dayCan an observed event in fact be of zero probability?Unit random processmodels of coins and uniform distributionHow to get the number of successes given $n$ trials , probability $P$ and a random variable $X$Absorbing Markov chain in a computer. Is “almost every” turned into always convergence in computer executions?Stopped random walk is not uniformly integrable

                              Lowndes Grove History Architecture References Navigation menu32°48′6″N 79°57′58″W / 32.80167°N 79.96611°W / 32.80167; -79.9661132°48′6″N 79°57′58″W / 32.80167°N 79.96611°W / 32.80167; -79.9661178002500"National Register Information System"Historic houses of South Carolina"Lowndes Grove""+32° 48' 6.00", −79° 57' 58.00""Lowndes Grove, Charleston County (260 St. Margaret St., Charleston)""Lowndes Grove"The Charleston ExpositionIt Happened in South Carolina"Lowndes Grove (House), Saint Margaret Street & Sixth Avenue, Charleston, Charleston County, SC(Photographs)"Plantations of the Carolina Low Countrye