Swapping registers in an old calculator












5














I came up with this problem inspired by the limitations of an old non-scientific calculator I owned years ago (the two registers were the display, and an internal memory for an additional number).



We have a primitive calculator with only two registers $R_1$ and $R_2$, and the following four operations:




  • $R_1+R_2to R_2$ (add the content of register $R_1$ to register $R_2$.)


  • $-R_1+R_2to R_2$ (subtract the content of register $R_1$ from register $R_2$.)


  • $R_1+R_2to R_1$ (add the content of register $R_2$ to register $R_1$.)


  • $R_1-R_2to R_1$ (subtract the content of register $R_2$ from register $R_1$.)



For instance, if $R_1=x$ (register $R_1$ contains number $x$) and
$R_2=y$ ($R_2$ contains $y$), after applying the operation $R_1+R_2to
R_2$ we end up with $R_1=x$ and $R_2=x+y$.



Assume that initially we have $R_1=x$ and $R_2=y$, where $x$ and $y$ are arbitrary numbers. For each of the following tasks describe a sequence of operations that would allow us to perform it, or prove that the task cannot be performed (task 1 is very easy, the real challenge is about task 2):




  • Task 1. Swap the contents of registers $R_1$ and $R_2$ changing the sign of $y$ in the process, so we would end up with $R_1=-y$, $R_2=x$.


  • Task 2. Swap the contents of registers $R_1$ and $R_2$, so that we would end up with $R_1=y$, $R_2=x$.







reachability







share|improve this question



























    5














    I came up with this problem inspired by the limitations of an old non-scientific calculator I owned years ago (the two registers were the display, and an internal memory for an additional number).



    We have a primitive calculator with only two registers $R_1$ and $R_2$, and the following four operations:




    • $R_1+R_2to R_2$ (add the content of register $R_1$ to register $R_2$.)


    • $-R_1+R_2to R_2$ (subtract the content of register $R_1$ from register $R_2$.)


    • $R_1+R_2to R_1$ (add the content of register $R_2$ to register $R_1$.)


    • $R_1-R_2to R_1$ (subtract the content of register $R_2$ from register $R_1$.)



    For instance, if $R_1=x$ (register $R_1$ contains number $x$) and
    $R_2=y$ ($R_2$ contains $y$), after applying the operation $R_1+R_2to
    R_2$     we end up with $R_1=x$ and $R_2=x+y$.



    Assume that initially we have $R_1=x$ and $R_2=y$, where $x$ and $y$ are arbitrary numbers. For each of the following tasks describe a sequence of operations that would allow us to perform it, or prove that the task cannot be performed (task 1 is very easy, the real challenge is about task 2):




    • Task 1. Swap the contents of registers $R_1$ and $R_2$ changing the sign of $y$ in the process, so we would end up with $R_1=-y$, $R_2=x$.


    • Task 2. Swap the contents of registers $R_1$ and $R_2$, so that we would end up with $R_1=y$, $R_2=x$.







    reachability







    share|improve this question

























      5












      5








      5







      I came up with this problem inspired by the limitations of an old non-scientific calculator I owned years ago (the two registers were the display, and an internal memory for an additional number).



      We have a primitive calculator with only two registers $R_1$ and $R_2$, and the following four operations:




      • $R_1+R_2to R_2$ (add the content of register $R_1$ to register $R_2$.)


      • $-R_1+R_2to R_2$ (subtract the content of register $R_1$ from register $R_2$.)


      • $R_1+R_2to R_1$ (add the content of register $R_2$ to register $R_1$.)


      • $R_1-R_2to R_1$ (subtract the content of register $R_2$ from register $R_1$.)



      For instance, if $R_1=x$ (register $R_1$ contains number $x$) and
      $R_2=y$ ($R_2$ contains $y$), after applying the operation $R_1+R_2to
      R_2$       we end up with $R_1=x$ and $R_2=x+y$.



      Assume that initially we have $R_1=x$ and $R_2=y$, where $x$ and $y$ are arbitrary numbers. For each of the following tasks describe a sequence of operations that would allow us to perform it, or prove that the task cannot be performed (task 1 is very easy, the real challenge is about task 2):




      • Task 1. Swap the contents of registers $R_1$ and $R_2$ changing the sign of $y$ in the process, so we would end up with $R_1=-y$, $R_2=x$.


      • Task 2. Swap the contents of registers $R_1$ and $R_2$, so that we would end up with $R_1=y$, $R_2=x$.







      reachability







      share|improve this question













      I came up with this problem inspired by the limitations of an old non-scientific calculator I owned years ago (the two registers were the display, and an internal memory for an additional number).



      We have a primitive calculator with only two registers $R_1$ and $R_2$, and the following four operations:




      • $R_1+R_2to R_2$ (add the content of register $R_1$ to register $R_2$.)


      • $-R_1+R_2to R_2$ (subtract the content of register $R_1$ from register $R_2$.)


      • $R_1+R_2to R_1$ (add the content of register $R_2$ to register $R_1$.)


      • $R_1-R_2to R_1$ (subtract the content of register $R_2$ from register $R_1$.)



      For instance, if $R_1=x$ (register $R_1$ contains number $x$) and
      $R_2=y$ ($R_2$ contains $y$), after applying the operation $R_1+R_2to
      R_2$       we end up with $R_1=x$ and $R_2=x+y$.



      Assume that initially we have $R_1=x$ and $R_2=y$, where $x$ and $y$ are arbitrary numbers. For each of the following tasks describe a sequence of operations that would allow us to perform it, or prove that the task cannot be performed (task 1 is very easy, the real challenge is about task 2):




      • Task 1. Swap the contents of registers $R_1$ and $R_2$ changing the sign of $y$ in the process, so we would end up with $R_1=-y$, $R_2=x$.


      • Task 2. Swap the contents of registers $R_1$ and $R_2$, so that we would end up with $R_1=y$, $R_2=x$.







      reachability




      reachability






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked 4 hours ago









      mlerma54

      1434




      1434






















          2 Answers
          2






          active

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          2














          Task 1




          1. $R_1 - R_2to R_1$ (now $R_1 = x-y$, $R_2 = y$)

          2. $R_1 + R_2to R_2$ (now $R_1 = x-y$, $R_2 = x$)

          3. $R_1 - R_2to R_1$ (now $R_1 = -y$, $R_2 = x$)







          share|improve this answer



















          • 2




            Task 2 the second operation isn't in the list of allowed operations
            – Dr Xorile
            4 hours ago






          • 2




            @DrXorile Oops, so it seems...
            – jafe
            4 hours ago



















          1














          Task 1 was answered. Task 2:




          Not possible. We always have $begin{bmatrix}R_1 \ R_2end{bmatrix} = Mbegin{bmatrix}x \ yend{bmatrix}$ for some matrix $M$ that depends only on the sequence of operations. Initially $M = begin{bmatrix}1 & 0 \ 0 & 1end{bmatrix}$, and the given operations cause $M$ to be left-multiplied by $begin{bmatrix}1 & 0 \ 1 & 1end{bmatrix}$, $begin{bmatrix}1 & 0 \ -1 & 1end{bmatrix}$, $begin{bmatrix}1 & 1 \ 0 & 1end{bmatrix}$, and $begin{bmatrix}1 & -1 \ 0 & 1end{bmatrix}$ respectively. All of these matrices have determinant $1$, so we will always have $det M = 1$, which makes it impossible to reach the desired $begin{bmatrix}0 & 1 \ 1 & 0end{bmatrix}$ with determinant $-1$.






          share





















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            2 Answers
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            2 Answers
            2






            active

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            active

            oldest

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            active

            oldest

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            2














            Task 1




            1. $R_1 - R_2to R_1$ (now $R_1 = x-y$, $R_2 = y$)

            2. $R_1 + R_2to R_2$ (now $R_1 = x-y$, $R_2 = x$)

            3. $R_1 - R_2to R_1$ (now $R_1 = -y$, $R_2 = x$)







            share|improve this answer



















            • 2




              Task 2 the second operation isn't in the list of allowed operations
              – Dr Xorile
              4 hours ago






            • 2




              @DrXorile Oops, so it seems...
              – jafe
              4 hours ago
















            2














            Task 1




            1. $R_1 - R_2to R_1$ (now $R_1 = x-y$, $R_2 = y$)

            2. $R_1 + R_2to R_2$ (now $R_1 = x-y$, $R_2 = x$)

            3. $R_1 - R_2to R_1$ (now $R_1 = -y$, $R_2 = x$)







            share|improve this answer



















            • 2




              Task 2 the second operation isn't in the list of allowed operations
              – Dr Xorile
              4 hours ago






            • 2




              @DrXorile Oops, so it seems...
              – jafe
              4 hours ago














            2












            2








            2






            Task 1




            1. $R_1 - R_2to R_1$ (now $R_1 = x-y$, $R_2 = y$)

            2. $R_1 + R_2to R_2$ (now $R_1 = x-y$, $R_2 = x$)

            3. $R_1 - R_2to R_1$ (now $R_1 = -y$, $R_2 = x$)







            share|improve this answer














            Task 1




            1. $R_1 - R_2to R_1$ (now $R_1 = x-y$, $R_2 = y$)

            2. $R_1 + R_2to R_2$ (now $R_1 = x-y$, $R_2 = x$)

            3. $R_1 - R_2to R_1$ (now $R_1 = -y$, $R_2 = x$)








            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited 4 hours ago

























            answered 4 hours ago









            jafe

            16.8k243168




            16.8k243168








            • 2




              Task 2 the second operation isn't in the list of allowed operations
              – Dr Xorile
              4 hours ago






            • 2




              @DrXorile Oops, so it seems...
              – jafe
              4 hours ago














            • 2




              Task 2 the second operation isn't in the list of allowed operations
              – Dr Xorile
              4 hours ago






            • 2




              @DrXorile Oops, so it seems...
              – jafe
              4 hours ago








            2




            2




            Task 2 the second operation isn't in the list of allowed operations
            – Dr Xorile
            4 hours ago




            Task 2 the second operation isn't in the list of allowed operations
            – Dr Xorile
            4 hours ago




            2




            2




            @DrXorile Oops, so it seems...
            – jafe
            4 hours ago




            @DrXorile Oops, so it seems...
            – jafe
            4 hours ago











            1














            Task 1 was answered. Task 2:




            Not possible. We always have $begin{bmatrix}R_1 \ R_2end{bmatrix} = Mbegin{bmatrix}x \ yend{bmatrix}$ for some matrix $M$ that depends only on the sequence of operations. Initially $M = begin{bmatrix}1 & 0 \ 0 & 1end{bmatrix}$, and the given operations cause $M$ to be left-multiplied by $begin{bmatrix}1 & 0 \ 1 & 1end{bmatrix}$, $begin{bmatrix}1 & 0 \ -1 & 1end{bmatrix}$, $begin{bmatrix}1 & 1 \ 0 & 1end{bmatrix}$, and $begin{bmatrix}1 & -1 \ 0 & 1end{bmatrix}$ respectively. All of these matrices have determinant $1$, so we will always have $det M = 1$, which makes it impossible to reach the desired $begin{bmatrix}0 & 1 \ 1 & 0end{bmatrix}$ with determinant $-1$.






            share


























              1














              Task 1 was answered. Task 2:




              Not possible. We always have $begin{bmatrix}R_1 \ R_2end{bmatrix} = Mbegin{bmatrix}x \ yend{bmatrix}$ for some matrix $M$ that depends only on the sequence of operations. Initially $M = begin{bmatrix}1 & 0 \ 0 & 1end{bmatrix}$, and the given operations cause $M$ to be left-multiplied by $begin{bmatrix}1 & 0 \ 1 & 1end{bmatrix}$, $begin{bmatrix}1 & 0 \ -1 & 1end{bmatrix}$, $begin{bmatrix}1 & 1 \ 0 & 1end{bmatrix}$, and $begin{bmatrix}1 & -1 \ 0 & 1end{bmatrix}$ respectively. All of these matrices have determinant $1$, so we will always have $det M = 1$, which makes it impossible to reach the desired $begin{bmatrix}0 & 1 \ 1 & 0end{bmatrix}$ with determinant $-1$.






              share
























                1












                1








                1






                Task 1 was answered. Task 2:




                Not possible. We always have $begin{bmatrix}R_1 \ R_2end{bmatrix} = Mbegin{bmatrix}x \ yend{bmatrix}$ for some matrix $M$ that depends only on the sequence of operations. Initially $M = begin{bmatrix}1 & 0 \ 0 & 1end{bmatrix}$, and the given operations cause $M$ to be left-multiplied by $begin{bmatrix}1 & 0 \ 1 & 1end{bmatrix}$, $begin{bmatrix}1 & 0 \ -1 & 1end{bmatrix}$, $begin{bmatrix}1 & 1 \ 0 & 1end{bmatrix}$, and $begin{bmatrix}1 & -1 \ 0 & 1end{bmatrix}$ respectively. All of these matrices have determinant $1$, so we will always have $det M = 1$, which makes it impossible to reach the desired $begin{bmatrix}0 & 1 \ 1 & 0end{bmatrix}$ with determinant $-1$.






                share












                Task 1 was answered. Task 2:




                Not possible. We always have $begin{bmatrix}R_1 \ R_2end{bmatrix} = Mbegin{bmatrix}x \ yend{bmatrix}$ for some matrix $M$ that depends only on the sequence of operations. Initially $M = begin{bmatrix}1 & 0 \ 0 & 1end{bmatrix}$, and the given operations cause $M$ to be left-multiplied by $begin{bmatrix}1 & 0 \ 1 & 1end{bmatrix}$, $begin{bmatrix}1 & 0 \ -1 & 1end{bmatrix}$, $begin{bmatrix}1 & 1 \ 0 & 1end{bmatrix}$, and $begin{bmatrix}1 & -1 \ 0 & 1end{bmatrix}$ respectively. All of these matrices have determinant $1$, so we will always have $det M = 1$, which makes it impossible to reach the desired $begin{bmatrix}0 & 1 \ 1 & 0end{bmatrix}$ with determinant $-1$.







                share











                share


                share










                answered 7 mins ago









                Anders Kaseorg

                40625




                40625






























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