Reindexing two lists / flattening indices












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If I have two lists/indiced that run from -N to N, call them A and B. How can I create a third index, C such that C goes from 0 to $(2N+1)^2-1$?










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    $begingroup$


    If I have two lists/indiced that run from -N to N, call them A and B. How can I create a third index, C such that C goes from 0 to $(2N+1)^2-1$?










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      1





      $begingroup$


      If I have two lists/indiced that run from -N to N, call them A and B. How can I create a third index, C such that C goes from 0 to $(2N+1)^2-1$?










      share|cite|improve this question









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      If I have two lists/indiced that run from -N to N, call them A and B. How can I create a third index, C such that C goes from 0 to $(2N+1)^2-1$?







      combinatorics






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      asked Jan 17 at 20:15









      yankeefan11yankeefan11

      1,0041920




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          $begingroup$

          Call $i = -N, -N + 1, cdots, N$ the index of $A$, and $j = -N, -N + 1, cdots, N$ the index of $B$. The index



          $$ bbox[5px,border:2px solid blue]
          {
          k = (i + N)(2N + 1) + (j + N)
          }
          $$



          goes from $k = 0$ for $(i, j) = (-N, -N)$ to $k = 2N(2N + 1) + 2N = (2N + 1)^2 - 1$ for $(i, j) = (N, N)$






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            1 Answer
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            active

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            oldest

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            active

            oldest

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            2












            $begingroup$

            Call $i = -N, -N + 1, cdots, N$ the index of $A$, and $j = -N, -N + 1, cdots, N$ the index of $B$. The index



            $$ bbox[5px,border:2px solid blue]
            {
            k = (i + N)(2N + 1) + (j + N)
            }
            $$



            goes from $k = 0$ for $(i, j) = (-N, -N)$ to $k = 2N(2N + 1) + 2N = (2N + 1)^2 - 1$ for $(i, j) = (N, N)$






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              Call $i = -N, -N + 1, cdots, N$ the index of $A$, and $j = -N, -N + 1, cdots, N$ the index of $B$. The index



              $$ bbox[5px,border:2px solid blue]
              {
              k = (i + N)(2N + 1) + (j + N)
              }
              $$



              goes from $k = 0$ for $(i, j) = (-N, -N)$ to $k = 2N(2N + 1) + 2N = (2N + 1)^2 - 1$ for $(i, j) = (N, N)$






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                Call $i = -N, -N + 1, cdots, N$ the index of $A$, and $j = -N, -N + 1, cdots, N$ the index of $B$. The index



                $$ bbox[5px,border:2px solid blue]
                {
                k = (i + N)(2N + 1) + (j + N)
                }
                $$



                goes from $k = 0$ for $(i, j) = (-N, -N)$ to $k = 2N(2N + 1) + 2N = (2N + 1)^2 - 1$ for $(i, j) = (N, N)$






                share|cite|improve this answer









                $endgroup$



                Call $i = -N, -N + 1, cdots, N$ the index of $A$, and $j = -N, -N + 1, cdots, N$ the index of $B$. The index



                $$ bbox[5px,border:2px solid blue]
                {
                k = (i + N)(2N + 1) + (j + N)
                }
                $$



                goes from $k = 0$ for $(i, j) = (-N, -N)$ to $k = 2N(2N + 1) + 2N = (2N + 1)^2 - 1$ for $(i, j) = (N, N)$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 17 at 20:34









                caveraccaverac

                14.8k31130




                14.8k31130






























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