Perfect riffle shuffles












7












$begingroup$


Take a deck of cards (with indexed position from 1 (top) to 52 (bottom)) and perform a perfect riffle shuffle, such that the top card (1) is still on top and the bottom (52) is still on the bottom.



Amazingly, if you perform 8 such riffle shuffles you will return to where you started.



Obviously, cards 1 and 52 do not change position. Most of the cards will go through some cycle and land back where they started only after 8 riffles.



But two cards will simply swap position back and forth each shuffle.




What are they?




Bonus question:




If you throw in the two jokers, you will have 54 cards. How many riffle shuffles will it take to get this deck back to the starting positions?











share|improve this question









$endgroup$

















    7












    $begingroup$


    Take a deck of cards (with indexed position from 1 (top) to 52 (bottom)) and perform a perfect riffle shuffle, such that the top card (1) is still on top and the bottom (52) is still on the bottom.



    Amazingly, if you perform 8 such riffle shuffles you will return to where you started.



    Obviously, cards 1 and 52 do not change position. Most of the cards will go through some cycle and land back where they started only after 8 riffles.



    But two cards will simply swap position back and forth each shuffle.




    What are they?




    Bonus question:




    If you throw in the two jokers, you will have 54 cards. How many riffle shuffles will it take to get this deck back to the starting positions?











    share|improve this question









    $endgroup$















      7












      7








      7





      $begingroup$


      Take a deck of cards (with indexed position from 1 (top) to 52 (bottom)) and perform a perfect riffle shuffle, such that the top card (1) is still on top and the bottom (52) is still on the bottom.



      Amazingly, if you perform 8 such riffle shuffles you will return to where you started.



      Obviously, cards 1 and 52 do not change position. Most of the cards will go through some cycle and land back where they started only after 8 riffles.



      But two cards will simply swap position back and forth each shuffle.




      What are they?




      Bonus question:




      If you throw in the two jokers, you will have 54 cards. How many riffle shuffles will it take to get this deck back to the starting positions?











      share|improve this question









      $endgroup$




      Take a deck of cards (with indexed position from 1 (top) to 52 (bottom)) and perform a perfect riffle shuffle, such that the top card (1) is still on top and the bottom (52) is still on the bottom.



      Amazingly, if you perform 8 such riffle shuffles you will return to where you started.



      Obviously, cards 1 and 52 do not change position. Most of the cards will go through some cycle and land back where they started only after 8 riffles.



      But two cards will simply swap position back and forth each shuffle.




      What are they?




      Bonus question:




      If you throw in the two jokers, you will have 54 cards. How many riffle shuffles will it take to get this deck back to the starting positions?








      cards






      share|improve this question













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      share|improve this question










      asked 10 hours ago









      Dr XorileDr Xorile

      13.8k32975




      13.8k32975






















          1 Answer
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          10












          $begingroup$


          Cards in the top half besides 1 always increase from $k$ to $2k-1$ (because it will be the first card of the $k$th pair). Cards in the bottom half besides 52 always decrease from $m$ to $2(m-26)$ (because it will be the second card of the $(m-26)$th pair). Then to return to the original position after two shuffles, the cards must swap between the halves.


          If $k$ is the position of the card in the top half, then after one shuffle it will move to position $m=2k-1$. If it is now in the bottom half ($m>26$), it will move to position $2(m-26)=2(2k-1-26)=4k-54$ after the second shuffle. To have returned to its initial position, we must have $4k-54=kiff3k=54iff k=18$, so that $m=2k-1=35>26$ holds. The two cards are at positions $boxed{18text{ and }35}$.




          Bonus:




          It can be shown using group theory that $k$ shuffles will restore a deck of size $n$ if $n-1$ divides $2^k-1$. The sequence of least such $k$ for every $n$ is in the OEIS, which gives the answer for 54 cards as $boxed{52text{ shuffles}}$.







          share|improve this answer









          $endgroup$









          • 1




            $begingroup$
            These answers are the reason i feel guilty for not going back to continues learning.
            $endgroup$
            – Alex
            10 hours ago











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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          10












          $begingroup$


          Cards in the top half besides 1 always increase from $k$ to $2k-1$ (because it will be the first card of the $k$th pair). Cards in the bottom half besides 52 always decrease from $m$ to $2(m-26)$ (because it will be the second card of the $(m-26)$th pair). Then to return to the original position after two shuffles, the cards must swap between the halves.


          If $k$ is the position of the card in the top half, then after one shuffle it will move to position $m=2k-1$. If it is now in the bottom half ($m>26$), it will move to position $2(m-26)=2(2k-1-26)=4k-54$ after the second shuffle. To have returned to its initial position, we must have $4k-54=kiff3k=54iff k=18$, so that $m=2k-1=35>26$ holds. The two cards are at positions $boxed{18text{ and }35}$.




          Bonus:




          It can be shown using group theory that $k$ shuffles will restore a deck of size $n$ if $n-1$ divides $2^k-1$. The sequence of least such $k$ for every $n$ is in the OEIS, which gives the answer for 54 cards as $boxed{52text{ shuffles}}$.







          share|improve this answer









          $endgroup$









          • 1




            $begingroup$
            These answers are the reason i feel guilty for not going back to continues learning.
            $endgroup$
            – Alex
            10 hours ago
















          10












          $begingroup$


          Cards in the top half besides 1 always increase from $k$ to $2k-1$ (because it will be the first card of the $k$th pair). Cards in the bottom half besides 52 always decrease from $m$ to $2(m-26)$ (because it will be the second card of the $(m-26)$th pair). Then to return to the original position after two shuffles, the cards must swap between the halves.


          If $k$ is the position of the card in the top half, then after one shuffle it will move to position $m=2k-1$. If it is now in the bottom half ($m>26$), it will move to position $2(m-26)=2(2k-1-26)=4k-54$ after the second shuffle. To have returned to its initial position, we must have $4k-54=kiff3k=54iff k=18$, so that $m=2k-1=35>26$ holds. The two cards are at positions $boxed{18text{ and }35}$.




          Bonus:




          It can be shown using group theory that $k$ shuffles will restore a deck of size $n$ if $n-1$ divides $2^k-1$. The sequence of least such $k$ for every $n$ is in the OEIS, which gives the answer for 54 cards as $boxed{52text{ shuffles}}$.







          share|improve this answer









          $endgroup$









          • 1




            $begingroup$
            These answers are the reason i feel guilty for not going back to continues learning.
            $endgroup$
            – Alex
            10 hours ago














          10












          10








          10





          $begingroup$


          Cards in the top half besides 1 always increase from $k$ to $2k-1$ (because it will be the first card of the $k$th pair). Cards in the bottom half besides 52 always decrease from $m$ to $2(m-26)$ (because it will be the second card of the $(m-26)$th pair). Then to return to the original position after two shuffles, the cards must swap between the halves.


          If $k$ is the position of the card in the top half, then after one shuffle it will move to position $m=2k-1$. If it is now in the bottom half ($m>26$), it will move to position $2(m-26)=2(2k-1-26)=4k-54$ after the second shuffle. To have returned to its initial position, we must have $4k-54=kiff3k=54iff k=18$, so that $m=2k-1=35>26$ holds. The two cards are at positions $boxed{18text{ and }35}$.




          Bonus:




          It can be shown using group theory that $k$ shuffles will restore a deck of size $n$ if $n-1$ divides $2^k-1$. The sequence of least such $k$ for every $n$ is in the OEIS, which gives the answer for 54 cards as $boxed{52text{ shuffles}}$.







          share|improve this answer









          $endgroup$




          Cards in the top half besides 1 always increase from $k$ to $2k-1$ (because it will be the first card of the $k$th pair). Cards in the bottom half besides 52 always decrease from $m$ to $2(m-26)$ (because it will be the second card of the $(m-26)$th pair). Then to return to the original position after two shuffles, the cards must swap between the halves.


          If $k$ is the position of the card in the top half, then after one shuffle it will move to position $m=2k-1$. If it is now in the bottom half ($m>26$), it will move to position $2(m-26)=2(2k-1-26)=4k-54$ after the second shuffle. To have returned to its initial position, we must have $4k-54=kiff3k=54iff k=18$, so that $m=2k-1=35>26$ holds. The two cards are at positions $boxed{18text{ and }35}$.




          Bonus:




          It can be shown using group theory that $k$ shuffles will restore a deck of size $n$ if $n-1$ divides $2^k-1$. The sequence of least such $k$ for every $n$ is in the OEIS, which gives the answer for 54 cards as $boxed{52text{ shuffles}}$.








          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered 10 hours ago









          noednenoedne

          7,60212160




          7,60212160








          • 1




            $begingroup$
            These answers are the reason i feel guilty for not going back to continues learning.
            $endgroup$
            – Alex
            10 hours ago














          • 1




            $begingroup$
            These answers are the reason i feel guilty for not going back to continues learning.
            $endgroup$
            – Alex
            10 hours ago








          1




          1




          $begingroup$
          These answers are the reason i feel guilty for not going back to continues learning.
          $endgroup$
          – Alex
          10 hours ago




          $begingroup$
          These answers are the reason i feel guilty for not going back to continues learning.
          $endgroup$
          – Alex
          10 hours ago


















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