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Cycles on the torus


Decompose a permutation into cyclesMoore IterationGoogle Code Jam - New Lottery GameCo-primality and the number piRotate every row and column in a matrixNumber of cycles of a permutationFire propagation simulator2D Array Middle PointPrint a Quinella TablePartitioning the grid into triangles













6












$begingroup$


Challenge



This challenge will have you write a program that takes in two integers n and m and outputs the number non-intersecting loops on the n by m torus made by only taking steps up and to the right. You can think of torus as the grid with wraparound both at the top and the bottom.



This is code-golf so fewest bytes wins.



Example



For example, if the input is n=m=5, one valid walk is



(0,0) -> (0,1) -> (0,2) -> (1,2) -> (2,2) -> (2,3) -> (2,4) -> 
(2,0) -> (3,0) -> (4,0) -> (4,1) -> (4,2) -> (4,3) ->
(0,3) -> (1,3) -> (1,4) ->
(1,0) -> (1,1) -> (2,1) -> (3,1) -> (3,2) -> (3,3) -> (3,4) -> (4,4) -> (0,4) -> (0,0)


as shown in the graphic.



A loop on the torus.



Some example input/outputs



f(1,1) = 2 (up or right)
f(1,2) = 2 (up or right-right)
f(2,2) = 4 (up-up, up-right-up-right, right-right, right-up-right-up)
f(2,3) = 7
f(3,3) = 22
f(2,4) = 13
f(3,4) = 66
f(4,4) = 258









share|improve this question









$endgroup$

















    6












    $begingroup$


    Challenge



    This challenge will have you write a program that takes in two integers n and m and outputs the number non-intersecting loops on the n by m torus made by only taking steps up and to the right. You can think of torus as the grid with wraparound both at the top and the bottom.



    This is code-golf so fewest bytes wins.



    Example



    For example, if the input is n=m=5, one valid walk is



    (0,0) -> (0,1) -> (0,2) -> (1,2) -> (2,2) -> (2,3) -> (2,4) -> 
    (2,0) -> (3,0) -> (4,0) -> (4,1) -> (4,2) -> (4,3) ->
    (0,3) -> (1,3) -> (1,4) ->
    (1,0) -> (1,1) -> (2,1) -> (3,1) -> (3,2) -> (3,3) -> (3,4) -> (4,4) -> (0,4) -> (0,0)


    as shown in the graphic.



    A loop on the torus.



    Some example input/outputs



    f(1,1) = 2 (up or right)
    f(1,2) = 2 (up or right-right)
    f(2,2) = 4 (up-up, up-right-up-right, right-right, right-up-right-up)
    f(2,3) = 7
    f(3,3) = 22
    f(2,4) = 13
    f(3,4) = 66
    f(4,4) = 258









    share|improve this question









    $endgroup$















      6












      6








      6





      $begingroup$


      Challenge



      This challenge will have you write a program that takes in two integers n and m and outputs the number non-intersecting loops on the n by m torus made by only taking steps up and to the right. You can think of torus as the grid with wraparound both at the top and the bottom.



      This is code-golf so fewest bytes wins.



      Example



      For example, if the input is n=m=5, one valid walk is



      (0,0) -> (0,1) -> (0,2) -> (1,2) -> (2,2) -> (2,3) -> (2,4) -> 
      (2,0) -> (3,0) -> (4,0) -> (4,1) -> (4,2) -> (4,3) ->
      (0,3) -> (1,3) -> (1,4) ->
      (1,0) -> (1,1) -> (2,1) -> (3,1) -> (3,2) -> (3,3) -> (3,4) -> (4,4) -> (0,4) -> (0,0)


      as shown in the graphic.



      A loop on the torus.



      Some example input/outputs



      f(1,1) = 2 (up or right)
      f(1,2) = 2 (up or right-right)
      f(2,2) = 4 (up-up, up-right-up-right, right-right, right-up-right-up)
      f(2,3) = 7
      f(3,3) = 22
      f(2,4) = 13
      f(3,4) = 66
      f(4,4) = 258









      share|improve this question









      $endgroup$




      Challenge



      This challenge will have you write a program that takes in two integers n and m and outputs the number non-intersecting loops on the n by m torus made by only taking steps up and to the right. You can think of torus as the grid with wraparound both at the top and the bottom.



      This is code-golf so fewest bytes wins.



      Example



      For example, if the input is n=m=5, one valid walk is



      (0,0) -> (0,1) -> (0,2) -> (1,2) -> (2,2) -> (2,3) -> (2,4) -> 
      (2,0) -> (3,0) -> (4,0) -> (4,1) -> (4,2) -> (4,3) ->
      (0,3) -> (1,3) -> (1,4) ->
      (1,0) -> (1,1) -> (2,1) -> (3,1) -> (3,2) -> (3,3) -> (3,4) -> (4,4) -> (0,4) -> (0,0)


      as shown in the graphic.



      A loop on the torus.



      Some example input/outputs



      f(1,1) = 2 (up or right)
      f(1,2) = 2 (up or right-right)
      f(2,2) = 4 (up-up, up-right-up-right, right-right, right-up-right-up)
      f(2,3) = 7
      f(3,3) = 22
      f(2,4) = 13
      f(3,4) = 66
      f(4,4) = 258






      code-golf combinatorics grid






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked 4 hours ago









      Peter KageyPeter Kagey

      878518




      878518






















          1 Answer
          1






          active

          oldest

          votes


















          3












          $begingroup$


          Python 2, 87 bytes





          f=lambda m,n,z=0,l=[]:z==0if z in l else sum(f(m,n,(z+d)%m%(n*1j),l+[z])for d in(1,1j))


          Try it online!



          The interesting thing here is using a complex number z to store the coordinate of the current position. We can move up by adding 1 and move right by adding 1j. To my surprise, modulo works on complex numbers in a way that lets us handle the wrapping for each dimension separately: doing %m acts on the real part, and %(n*1j) acts on the imaginary part.






          share|improve this answer









          $endgroup$













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            1 Answer
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            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

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            3












            $begingroup$


            Python 2, 87 bytes





            f=lambda m,n,z=0,l=[]:z==0if z in l else sum(f(m,n,(z+d)%m%(n*1j),l+[z])for d in(1,1j))


            Try it online!



            The interesting thing here is using a complex number z to store the coordinate of the current position. We can move up by adding 1 and move right by adding 1j. To my surprise, modulo works on complex numbers in a way that lets us handle the wrapping for each dimension separately: doing %m acts on the real part, and %(n*1j) acts on the imaginary part.






            share|improve this answer









            $endgroup$


















              3












              $begingroup$


              Python 2, 87 bytes





              f=lambda m,n,z=0,l=[]:z==0if z in l else sum(f(m,n,(z+d)%m%(n*1j),l+[z])for d in(1,1j))


              Try it online!



              The interesting thing here is using a complex number z to store the coordinate of the current position. We can move up by adding 1 and move right by adding 1j. To my surprise, modulo works on complex numbers in a way that lets us handle the wrapping for each dimension separately: doing %m acts on the real part, and %(n*1j) acts on the imaginary part.






              share|improve this answer









              $endgroup$
















                3












                3








                3





                $begingroup$


                Python 2, 87 bytes





                f=lambda m,n,z=0,l=[]:z==0if z in l else sum(f(m,n,(z+d)%m%(n*1j),l+[z])for d in(1,1j))


                Try it online!



                The interesting thing here is using a complex number z to store the coordinate of the current position. We can move up by adding 1 and move right by adding 1j. To my surprise, modulo works on complex numbers in a way that lets us handle the wrapping for each dimension separately: doing %m acts on the real part, and %(n*1j) acts on the imaginary part.






                share|improve this answer









                $endgroup$




                Python 2, 87 bytes





                f=lambda m,n,z=0,l=[]:z==0if z in l else sum(f(m,n,(z+d)%m%(n*1j),l+[z])for d in(1,1j))


                Try it online!



                The interesting thing here is using a complex number z to store the coordinate of the current position. We can move up by adding 1 and move right by adding 1j. To my surprise, modulo works on complex numbers in a way that lets us handle the wrapping for each dimension separately: doing %m acts on the real part, and %(n*1j) acts on the imaginary part.







                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered 2 hours ago









                xnorxnor

                91.8k18187444




                91.8k18187444






























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                    If this is an answer to a challenge…




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                    • …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).





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