A game with coins












0












$begingroup$


You have 15 coins which are split to number of piles, in each step you must collect 1 coin from each pile and form a new pile and so on.



Prove that no matter how the arrangement was in the beginning you will end up in the stable case in which you have 5 piles of coins (1 with 1 coin, one with 2 coins .. till 1 with 5 coins).



It can be generalized to a number of coins N which equals N=m(m+1) where m is natural






elementary-number-theory







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$endgroup$












  • $begingroup$
    What have you tried so far?
    $endgroup$
    – Mindlack
    Jan 16 at 13:33










  • $begingroup$
    I only have time for a hint. A standard way to approach such kind of question is to construct certain invariant and keep track on them after each step. If the invariant is strictly monotone then you can claim that your operation stabilizes. For this question, the idea is to sort the piles in descending order and see how many coins are above the $5,4,3,2,1$-triangle. For instance, if the initial state is $3,3,3,2,2,2$, then $0+0+0+0+1+2=3$ coins are above the triangle. After the first step, it will give you $6,2,2,2,1,1,1$ and $1+0+0+0+0+1+1=3$ will be above the triangle.
    $endgroup$
    – Hw Chu
    Jan 16 at 14:59












  • $begingroup$
    Sadly the "above the triangle" invariant is only decreasing, but not strictly decreasing. So you need to get an auxiliary invariant to finish up, and that invariant is actually also something recording the coins above the triangle, but something more informative.
    $endgroup$
    – Hw Chu
    Jan 16 at 15:00












  • $begingroup$
    Like what? What more information?
    $endgroup$
    – Joshua Haim Mamou
    Jan 17 at 9:51
















0












$begingroup$


You have 15 coins which are split to number of piles, in each step you must collect 1 coin from each pile and form a new pile and so on.



Prove that no matter how the arrangement was in the beginning you will end up in the stable case in which you have 5 piles of coins (1 with 1 coin, one with 2 coins .. till 1 with 5 coins).



It can be generalized to a number of coins N which equals N=m(m+1) where m is natural






elementary-number-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    What have you tried so far?
    $endgroup$
    – Mindlack
    Jan 16 at 13:33










  • $begingroup$
    I only have time for a hint. A standard way to approach such kind of question is to construct certain invariant and keep track on them after each step. If the invariant is strictly monotone then you can claim that your operation stabilizes. For this question, the idea is to sort the piles in descending order and see how many coins are above the $5,4,3,2,1$-triangle. For instance, if the initial state is $3,3,3,2,2,2$, then $0+0+0+0+1+2=3$ coins are above the triangle. After the first step, it will give you $6,2,2,2,1,1,1$ and $1+0+0+0+0+1+1=3$ will be above the triangle.
    $endgroup$
    – Hw Chu
    Jan 16 at 14:59












  • $begingroup$
    Sadly the "above the triangle" invariant is only decreasing, but not strictly decreasing. So you need to get an auxiliary invariant to finish up, and that invariant is actually also something recording the coins above the triangle, but something more informative.
    $endgroup$
    – Hw Chu
    Jan 16 at 15:00












  • $begingroup$
    Like what? What more information?
    $endgroup$
    – Joshua Haim Mamou
    Jan 17 at 9:51














0












0








0


1



$begingroup$


You have 15 coins which are split to number of piles, in each step you must collect 1 coin from each pile and form a new pile and so on.



Prove that no matter how the arrangement was in the beginning you will end up in the stable case in which you have 5 piles of coins (1 with 1 coin, one with 2 coins .. till 1 with 5 coins).



It can be generalized to a number of coins N which equals N=m(m+1) where m is natural






elementary-number-theory







share|cite|improve this question











$endgroup$




You have 15 coins which are split to number of piles, in each step you must collect 1 coin from each pile and form a new pile and so on.



Prove that no matter how the arrangement was in the beginning you will end up in the stable case in which you have 5 piles of coins (1 with 1 coin, one with 2 coins .. till 1 with 5 coins).



It can be generalized to a number of coins N which equals N=m(m+1) where m is natural






elementary-number-theory




elementary-number-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 16 at 14:04







Joshua Haim Mamou

















asked Jan 16 at 13:25









Joshua Haim MamouJoshua Haim Mamou

245




245












  • $begingroup$
    What have you tried so far?
    $endgroup$
    – Mindlack
    Jan 16 at 13:33










  • $begingroup$
    I only have time for a hint. A standard way to approach such kind of question is to construct certain invariant and keep track on them after each step. If the invariant is strictly monotone then you can claim that your operation stabilizes. For this question, the idea is to sort the piles in descending order and see how many coins are above the $5,4,3,2,1$-triangle. For instance, if the initial state is $3,3,3,2,2,2$, then $0+0+0+0+1+2=3$ coins are above the triangle. After the first step, it will give you $6,2,2,2,1,1,1$ and $1+0+0+0+0+1+1=3$ will be above the triangle.
    $endgroup$
    – Hw Chu
    Jan 16 at 14:59












  • $begingroup$
    Sadly the "above the triangle" invariant is only decreasing, but not strictly decreasing. So you need to get an auxiliary invariant to finish up, and that invariant is actually also something recording the coins above the triangle, but something more informative.
    $endgroup$
    – Hw Chu
    Jan 16 at 15:00












  • $begingroup$
    Like what? What more information?
    $endgroup$
    – Joshua Haim Mamou
    Jan 17 at 9:51


















  • $begingroup$
    What have you tried so far?
    $endgroup$
    – Mindlack
    Jan 16 at 13:33










  • $begingroup$
    I only have time for a hint. A standard way to approach such kind of question is to construct certain invariant and keep track on them after each step. If the invariant is strictly monotone then you can claim that your operation stabilizes. For this question, the idea is to sort the piles in descending order and see how many coins are above the $5,4,3,2,1$-triangle. For instance, if the initial state is $3,3,3,2,2,2$, then $0+0+0+0+1+2=3$ coins are above the triangle. After the first step, it will give you $6,2,2,2,1,1,1$ and $1+0+0+0+0+1+1=3$ will be above the triangle.
    $endgroup$
    – Hw Chu
    Jan 16 at 14:59












  • $begingroup$
    Sadly the "above the triangle" invariant is only decreasing, but not strictly decreasing. So you need to get an auxiliary invariant to finish up, and that invariant is actually also something recording the coins above the triangle, but something more informative.
    $endgroup$
    – Hw Chu
    Jan 16 at 15:00












  • $begingroup$
    Like what? What more information?
    $endgroup$
    – Joshua Haim Mamou
    Jan 17 at 9:51
















$begingroup$
What have you tried so far?
$endgroup$
– Mindlack
Jan 16 at 13:33




$begingroup$
What have you tried so far?
$endgroup$
– Mindlack
Jan 16 at 13:33












$begingroup$
I only have time for a hint. A standard way to approach such kind of question is to construct certain invariant and keep track on them after each step. If the invariant is strictly monotone then you can claim that your operation stabilizes. For this question, the idea is to sort the piles in descending order and see how many coins are above the $5,4,3,2,1$-triangle. For instance, if the initial state is $3,3,3,2,2,2$, then $0+0+0+0+1+2=3$ coins are above the triangle. After the first step, it will give you $6,2,2,2,1,1,1$ and $1+0+0+0+0+1+1=3$ will be above the triangle.
$endgroup$
– Hw Chu
Jan 16 at 14:59






$begingroup$
I only have time for a hint. A standard way to approach such kind of question is to construct certain invariant and keep track on them after each step. If the invariant is strictly monotone then you can claim that your operation stabilizes. For this question, the idea is to sort the piles in descending order and see how many coins are above the $5,4,3,2,1$-triangle. For instance, if the initial state is $3,3,3,2,2,2$, then $0+0+0+0+1+2=3$ coins are above the triangle. After the first step, it will give you $6,2,2,2,1,1,1$ and $1+0+0+0+0+1+1=3$ will be above the triangle.
$endgroup$
– Hw Chu
Jan 16 at 14:59














$begingroup$
Sadly the "above the triangle" invariant is only decreasing, but not strictly decreasing. So you need to get an auxiliary invariant to finish up, and that invariant is actually also something recording the coins above the triangle, but something more informative.
$endgroup$
– Hw Chu
Jan 16 at 15:00






$begingroup$
Sadly the "above the triangle" invariant is only decreasing, but not strictly decreasing. So you need to get an auxiliary invariant to finish up, and that invariant is actually also something recording the coins above the triangle, but something more informative.
$endgroup$
– Hw Chu
Jan 16 at 15:00














$begingroup$
Like what? What more information?
$endgroup$
– Joshua Haim Mamou
Jan 17 at 9:51




$begingroup$
Like what? What more information?
$endgroup$
– Joshua Haim Mamou
Jan 17 at 9:51










2 Answers
2






active

oldest

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0












$begingroup$

generale case :



First you can show that because the number of pile is constant, exactly one pile should have one coin at each step.



Then by induction you can show that for each $i$ less or equal to the number of pile. There is exactly one pile with $i$ coins.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    How is the number of piles constant ? If the initial state is $3,3,3,3,3$ then the next state is $2,2,2,2,2,5$ and the number of piles has increased from $5$ to $6$ surely ?
    $endgroup$
    – gandalf61
    Jan 16 at 15:15










  • $begingroup$
    @levgeni That works if you can show that game always eventually reaches a single stable state. But how do you know it does not repeat a cycle of two (or more) states ? For example with $4$ coins $(1,3)$, $(2,2)$ and $(1,1,2)$ form a repeating cycle with three states.
    $endgroup$
    – gandalf61
    Jan 16 at 15:34










  • $begingroup$
    You have right, it misses something in my reasoning...
    $endgroup$
    – Ievgeni
    Jan 16 at 15:45



















0












$begingroup$

As Levgeni shows above, a single fixed state (as opposed to a cycle which loops through multiple states) must have the form $(1,2,3,dots,n)$.



I can also show that any state that differs from a $(1,2,3,dots,n)$ state by $1$ in one or more places will be part of cycle of length $n+1$. So for $16$ coins, for example, there is a cycle



$(1,2,3,4,6) rightarrow (1,2,3,5,5) rightarrow (1,2,4,4,5) rightarrow (1,3,3,4,5) rightarrow (2,2,3,4,5) rightarrow (1,1,2,3,4,5) rightarrow (1,2,3,4,6)$



and for $17$ coins there are two distinct $6$-cycles:



$(1,2,3,5,6) rightarrow (1,2,4,5,5) rightarrow (1,3,4,4,5) rightarrow (2,3,3,4,5) rightarrow (1,2,2,3,4,5) rightarrow (1,1,2,3,4,6) rightarrow (1,2,3,5,6)$



and



$(1,2,4,4,6) rightarrow (1,3,3,5,5) rightarrow (2,2,4,4,5) rightarrow (1,1,3,3,4,5) rightarrow (2,2,3,4,6) rightarrow (1,1,2,3,5,5) rightarrow (1,2,4,4,6)$



I cannot yet prove that these are the only possible cycles.






share|cite|improve this answer









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    2 Answers
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    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

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    active

    oldest

    votes









    0












    $begingroup$

    generale case :



    First you can show that because the number of pile is constant, exactly one pile should have one coin at each step.



    Then by induction you can show that for each $i$ less or equal to the number of pile. There is exactly one pile with $i$ coins.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      How is the number of piles constant ? If the initial state is $3,3,3,3,3$ then the next state is $2,2,2,2,2,5$ and the number of piles has increased from $5$ to $6$ surely ?
      $endgroup$
      – gandalf61
      Jan 16 at 15:15










    • $begingroup$
      @levgeni That works if you can show that game always eventually reaches a single stable state. But how do you know it does not repeat a cycle of two (or more) states ? For example with $4$ coins $(1,3)$, $(2,2)$ and $(1,1,2)$ form a repeating cycle with three states.
      $endgroup$
      – gandalf61
      Jan 16 at 15:34










    • $begingroup$
      You have right, it misses something in my reasoning...
      $endgroup$
      – Ievgeni
      Jan 16 at 15:45
















    0












    $begingroup$

    generale case :



    First you can show that because the number of pile is constant, exactly one pile should have one coin at each step.



    Then by induction you can show that for each $i$ less or equal to the number of pile. There is exactly one pile with $i$ coins.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      How is the number of piles constant ? If the initial state is $3,3,3,3,3$ then the next state is $2,2,2,2,2,5$ and the number of piles has increased from $5$ to $6$ surely ?
      $endgroup$
      – gandalf61
      Jan 16 at 15:15










    • $begingroup$
      @levgeni That works if you can show that game always eventually reaches a single stable state. But how do you know it does not repeat a cycle of two (or more) states ? For example with $4$ coins $(1,3)$, $(2,2)$ and $(1,1,2)$ form a repeating cycle with three states.
      $endgroup$
      – gandalf61
      Jan 16 at 15:34










    • $begingroup$
      You have right, it misses something in my reasoning...
      $endgroup$
      – Ievgeni
      Jan 16 at 15:45














    0












    0








    0





    $begingroup$

    generale case :



    First you can show that because the number of pile is constant, exactly one pile should have one coin at each step.



    Then by induction you can show that for each $i$ less or equal to the number of pile. There is exactly one pile with $i$ coins.






    share|cite|improve this answer









    $endgroup$



    generale case :



    First you can show that because the number of pile is constant, exactly one pile should have one coin at each step.



    Then by induction you can show that for each $i$ less or equal to the number of pile. There is exactly one pile with $i$ coins.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 16 at 14:23









    IevgeniIevgeni

    1




    1












    • $begingroup$
      How is the number of piles constant ? If the initial state is $3,3,3,3,3$ then the next state is $2,2,2,2,2,5$ and the number of piles has increased from $5$ to $6$ surely ?
      $endgroup$
      – gandalf61
      Jan 16 at 15:15










    • $begingroup$
      @levgeni That works if you can show that game always eventually reaches a single stable state. But how do you know it does not repeat a cycle of two (or more) states ? For example with $4$ coins $(1,3)$, $(2,2)$ and $(1,1,2)$ form a repeating cycle with three states.
      $endgroup$
      – gandalf61
      Jan 16 at 15:34










    • $begingroup$
      You have right, it misses something in my reasoning...
      $endgroup$
      – Ievgeni
      Jan 16 at 15:45


















    • $begingroup$
      How is the number of piles constant ? If the initial state is $3,3,3,3,3$ then the next state is $2,2,2,2,2,5$ and the number of piles has increased from $5$ to $6$ surely ?
      $endgroup$
      – gandalf61
      Jan 16 at 15:15










    • $begingroup$
      @levgeni That works if you can show that game always eventually reaches a single stable state. But how do you know it does not repeat a cycle of two (or more) states ? For example with $4$ coins $(1,3)$, $(2,2)$ and $(1,1,2)$ form a repeating cycle with three states.
      $endgroup$
      – gandalf61
      Jan 16 at 15:34










    • $begingroup$
      You have right, it misses something in my reasoning...
      $endgroup$
      – Ievgeni
      Jan 16 at 15:45
















    $begingroup$
    How is the number of piles constant ? If the initial state is $3,3,3,3,3$ then the next state is $2,2,2,2,2,5$ and the number of piles has increased from $5$ to $6$ surely ?
    $endgroup$
    – gandalf61
    Jan 16 at 15:15




    $begingroup$
    How is the number of piles constant ? If the initial state is $3,3,3,3,3$ then the next state is $2,2,2,2,2,5$ and the number of piles has increased from $5$ to $6$ surely ?
    $endgroup$
    – gandalf61
    Jan 16 at 15:15












    $begingroup$
    @levgeni That works if you can show that game always eventually reaches a single stable state. But how do you know it does not repeat a cycle of two (or more) states ? For example with $4$ coins $(1,3)$, $(2,2)$ and $(1,1,2)$ form a repeating cycle with three states.
    $endgroup$
    – gandalf61
    Jan 16 at 15:34




    $begingroup$
    @levgeni That works if you can show that game always eventually reaches a single stable state. But how do you know it does not repeat a cycle of two (or more) states ? For example with $4$ coins $(1,3)$, $(2,2)$ and $(1,1,2)$ form a repeating cycle with three states.
    $endgroup$
    – gandalf61
    Jan 16 at 15:34












    $begingroup$
    You have right, it misses something in my reasoning...
    $endgroup$
    – Ievgeni
    Jan 16 at 15:45




    $begingroup$
    You have right, it misses something in my reasoning...
    $endgroup$
    – Ievgeni
    Jan 16 at 15:45











    0












    $begingroup$

    As Levgeni shows above, a single fixed state (as opposed to a cycle which loops through multiple states) must have the form $(1,2,3,dots,n)$.



    I can also show that any state that differs from a $(1,2,3,dots,n)$ state by $1$ in one or more places will be part of cycle of length $n+1$. So for $16$ coins, for example, there is a cycle



    $(1,2,3,4,6) rightarrow (1,2,3,5,5) rightarrow (1,2,4,4,5) rightarrow (1,3,3,4,5) rightarrow (2,2,3,4,5) rightarrow (1,1,2,3,4,5) rightarrow (1,2,3,4,6)$



    and for $17$ coins there are two distinct $6$-cycles:



    $(1,2,3,5,6) rightarrow (1,2,4,5,5) rightarrow (1,3,4,4,5) rightarrow (2,3,3,4,5) rightarrow (1,2,2,3,4,5) rightarrow (1,1,2,3,4,6) rightarrow (1,2,3,5,6)$



    and



    $(1,2,4,4,6) rightarrow (1,3,3,5,5) rightarrow (2,2,4,4,5) rightarrow (1,1,3,3,4,5) rightarrow (2,2,3,4,6) rightarrow (1,1,2,3,5,5) rightarrow (1,2,4,4,6)$



    I cannot yet prove that these are the only possible cycles.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      As Levgeni shows above, a single fixed state (as opposed to a cycle which loops through multiple states) must have the form $(1,2,3,dots,n)$.



      I can also show that any state that differs from a $(1,2,3,dots,n)$ state by $1$ in one or more places will be part of cycle of length $n+1$. So for $16$ coins, for example, there is a cycle



      $(1,2,3,4,6) rightarrow (1,2,3,5,5) rightarrow (1,2,4,4,5) rightarrow (1,3,3,4,5) rightarrow (2,2,3,4,5) rightarrow (1,1,2,3,4,5) rightarrow (1,2,3,4,6)$



      and for $17$ coins there are two distinct $6$-cycles:



      $(1,2,3,5,6) rightarrow (1,2,4,5,5) rightarrow (1,3,4,4,5) rightarrow (2,3,3,4,5) rightarrow (1,2,2,3,4,5) rightarrow (1,1,2,3,4,6) rightarrow (1,2,3,5,6)$



      and



      $(1,2,4,4,6) rightarrow (1,3,3,5,5) rightarrow (2,2,4,4,5) rightarrow (1,1,3,3,4,5) rightarrow (2,2,3,4,6) rightarrow (1,1,2,3,5,5) rightarrow (1,2,4,4,6)$



      I cannot yet prove that these are the only possible cycles.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        As Levgeni shows above, a single fixed state (as opposed to a cycle which loops through multiple states) must have the form $(1,2,3,dots,n)$.



        I can also show that any state that differs from a $(1,2,3,dots,n)$ state by $1$ in one or more places will be part of cycle of length $n+1$. So for $16$ coins, for example, there is a cycle



        $(1,2,3,4,6) rightarrow (1,2,3,5,5) rightarrow (1,2,4,4,5) rightarrow (1,3,3,4,5) rightarrow (2,2,3,4,5) rightarrow (1,1,2,3,4,5) rightarrow (1,2,3,4,6)$



        and for $17$ coins there are two distinct $6$-cycles:



        $(1,2,3,5,6) rightarrow (1,2,4,5,5) rightarrow (1,3,4,4,5) rightarrow (2,3,3,4,5) rightarrow (1,2,2,3,4,5) rightarrow (1,1,2,3,4,6) rightarrow (1,2,3,5,6)$



        and



        $(1,2,4,4,6) rightarrow (1,3,3,5,5) rightarrow (2,2,4,4,5) rightarrow (1,1,3,3,4,5) rightarrow (2,2,3,4,6) rightarrow (1,1,2,3,5,5) rightarrow (1,2,4,4,6)$



        I cannot yet prove that these are the only possible cycles.






        share|cite|improve this answer









        $endgroup$



        As Levgeni shows above, a single fixed state (as opposed to a cycle which loops through multiple states) must have the form $(1,2,3,dots,n)$.



        I can also show that any state that differs from a $(1,2,3,dots,n)$ state by $1$ in one or more places will be part of cycle of length $n+1$. So for $16$ coins, for example, there is a cycle



        $(1,2,3,4,6) rightarrow (1,2,3,5,5) rightarrow (1,2,4,4,5) rightarrow (1,3,3,4,5) rightarrow (2,2,3,4,5) rightarrow (1,1,2,3,4,5) rightarrow (1,2,3,4,6)$



        and for $17$ coins there are two distinct $6$-cycles:



        $(1,2,3,5,6) rightarrow (1,2,4,5,5) rightarrow (1,3,4,4,5) rightarrow (2,3,3,4,5) rightarrow (1,2,2,3,4,5) rightarrow (1,1,2,3,4,6) rightarrow (1,2,3,5,6)$



        and



        $(1,2,4,4,6) rightarrow (1,3,3,5,5) rightarrow (2,2,4,4,5) rightarrow (1,1,3,3,4,5) rightarrow (2,2,3,4,6) rightarrow (1,1,2,3,5,5) rightarrow (1,2,4,4,6)$



        I cannot yet prove that these are the only possible cycles.







        share|cite|improve this answer












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        answered Jan 21 at 8:56









        gandalf61gandalf61

        9,209825




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