Is there a function for mapping 1 to 1, 2 to 2…n to n, 1,2 to n+1, 1,3 to n+2…1,n-1 to 2n-1 for n...












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So I'm programming something and I need to create this mapping. This is my rough idea of how it will work.



I have a list of variables, $x_1$ to $x_n$.



$x_1$ maps to $1$, $x_1$ maps to $2$, so on until $x_n$ which maps to $n$. This has $n$ terms.



For 2 variables together, I need to map $x_1x_2$ to $n+1$, all the way to $x_1x_n$ which will map to $2n-1$. This has $n-1$ terms. Note that I'm not considering $x_1x_1$.



As I go along to $x_2x_3$ to $x_2x_n$, I will be mapping it to the additional $n-2$ terms. For 2 variables, I should have a total of $frac{(n)(n+1)}{2}-n$ terms mapped to.



For 3 variables and beyond, I know the number of terms that it needs to get mapped to. It is $n$ choose 3, and so on for 4 variables. But is there a function which perhaps takes in the indicies of $x$ and spits out where it should be along this long chain? I can't seem to find it.



Thanks.






combinatorics generating-functions







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    1












    $begingroup$


    So I'm programming something and I need to create this mapping. This is my rough idea of how it will work.



    I have a list of variables, $x_1$ to $x_n$.



    $x_1$ maps to $1$, $x_1$ maps to $2$, so on until $x_n$ which maps to $n$. This has $n$ terms.



    For 2 variables together, I need to map $x_1x_2$ to $n+1$, all the way to $x_1x_n$ which will map to $2n-1$. This has $n-1$ terms. Note that I'm not considering $x_1x_1$.



    As I go along to $x_2x_3$ to $x_2x_n$, I will be mapping it to the additional $n-2$ terms. For 2 variables, I should have a total of $frac{(n)(n+1)}{2}-n$ terms mapped to.



    For 3 variables and beyond, I know the number of terms that it needs to get mapped to. It is $n$ choose 3, and so on for 4 variables. But is there a function which perhaps takes in the indicies of $x$ and spits out where it should be along this long chain? I can't seem to find it.



    Thanks.






    combinatorics generating-functions







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      1












      1








      1





      $begingroup$


      So I'm programming something and I need to create this mapping. This is my rough idea of how it will work.



      I have a list of variables, $x_1$ to $x_n$.



      $x_1$ maps to $1$, $x_1$ maps to $2$, so on until $x_n$ which maps to $n$. This has $n$ terms.



      For 2 variables together, I need to map $x_1x_2$ to $n+1$, all the way to $x_1x_n$ which will map to $2n-1$. This has $n-1$ terms. Note that I'm not considering $x_1x_1$.



      As I go along to $x_2x_3$ to $x_2x_n$, I will be mapping it to the additional $n-2$ terms. For 2 variables, I should have a total of $frac{(n)(n+1)}{2}-n$ terms mapped to.



      For 3 variables and beyond, I know the number of terms that it needs to get mapped to. It is $n$ choose 3, and so on for 4 variables. But is there a function which perhaps takes in the indicies of $x$ and spits out where it should be along this long chain? I can't seem to find it.



      Thanks.






      combinatorics generating-functions







      share|cite|improve this question









      $endgroup$




      So I'm programming something and I need to create this mapping. This is my rough idea of how it will work.



      I have a list of variables, $x_1$ to $x_n$.



      $x_1$ maps to $1$, $x_1$ maps to $2$, so on until $x_n$ which maps to $n$. This has $n$ terms.



      For 2 variables together, I need to map $x_1x_2$ to $n+1$, all the way to $x_1x_n$ which will map to $2n-1$. This has $n-1$ terms. Note that I'm not considering $x_1x_1$.



      As I go along to $x_2x_3$ to $x_2x_n$, I will be mapping it to the additional $n-2$ terms. For 2 variables, I should have a total of $frac{(n)(n+1)}{2}-n$ terms mapped to.



      For 3 variables and beyond, I know the number of terms that it needs to get mapped to. It is $n$ choose 3, and so on for 4 variables. But is there a function which perhaps takes in the indicies of $x$ and spits out where it should be along this long chain? I can't seem to find it.



      Thanks.






      combinatorics generating-functions




      combinatorics generating-functions






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      asked Jan 7 at 8:31









      Yip Jung HonYip Jung Hon

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          $begingroup$

          The following bijection takes a set of $k$ distinct indices $a_1<a_2<dots<a_k$ and returns a number between $0$ and $binom{n}k-1$, so that every set gets a different number.



          $$
          S={a_1<a_2<dots<a_k}implies f(S)=binom{n-a_1}k+binom{n-a_2}{k-1}+dots+binom{n-a_k}1
          $$
          However, this is exactly the opposite of the order you want, so you should instead use $binom{n}k-f(S)$.



          Therefore, given an arbitrary set $S$ of $k$ indices, your desired indexing operation is
          $$
          g(S) = binom{n}1+dots+binom{n}{k-1}+binom{n}k-f(S)
          $$
          For example, when $n=10$ and your set of indices is $x_2x_5x_7$, the place in the list is
          $$
          g({2,5,7})=binom{10}1+binom{10}2+binom{10}3-binom{10-2}3-binom{10-5}2-binom{10-7}1=
          $$
          As a sanity check, the place of $x_1x_n$ in the list is $binom{n}1+binom{n}2-binom{n-1}2-binom{n-n}1=n+(n-1)+0=2n-1$, which agrees with your example. The position of $x_1x_2$ is $binom{n}1+binom{n}2-binom{n-1}2-binom{n-2}1=n+1$.






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            1 Answer
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            1 Answer
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            active

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            1












            $begingroup$

            The following bijection takes a set of $k$ distinct indices $a_1<a_2<dots<a_k$ and returns a number between $0$ and $binom{n}k-1$, so that every set gets a different number.



            $$
            S={a_1<a_2<dots<a_k}implies f(S)=binom{n-a_1}k+binom{n-a_2}{k-1}+dots+binom{n-a_k}1
            $$
            However, this is exactly the opposite of the order you want, so you should instead use $binom{n}k-f(S)$.



            Therefore, given an arbitrary set $S$ of $k$ indices, your desired indexing operation is
            $$
            g(S) = binom{n}1+dots+binom{n}{k-1}+binom{n}k-f(S)
            $$
            For example, when $n=10$ and your set of indices is $x_2x_5x_7$, the place in the list is
            $$
            g({2,5,7})=binom{10}1+binom{10}2+binom{10}3-binom{10-2}3-binom{10-5}2-binom{10-7}1=
            $$
            As a sanity check, the place of $x_1x_n$ in the list is $binom{n}1+binom{n}2-binom{n-1}2-binom{n-n}1=n+(n-1)+0=2n-1$, which agrees with your example. The position of $x_1x_2$ is $binom{n}1+binom{n}2-binom{n-1}2-binom{n-2}1=n+1$.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              The following bijection takes a set of $k$ distinct indices $a_1<a_2<dots<a_k$ and returns a number between $0$ and $binom{n}k-1$, so that every set gets a different number.



              $$
              S={a_1<a_2<dots<a_k}implies f(S)=binom{n-a_1}k+binom{n-a_2}{k-1}+dots+binom{n-a_k}1
              $$
              However, this is exactly the opposite of the order you want, so you should instead use $binom{n}k-f(S)$.



              Therefore, given an arbitrary set $S$ of $k$ indices, your desired indexing operation is
              $$
              g(S) = binom{n}1+dots+binom{n}{k-1}+binom{n}k-f(S)
              $$
              For example, when $n=10$ and your set of indices is $x_2x_5x_7$, the place in the list is
              $$
              g({2,5,7})=binom{10}1+binom{10}2+binom{10}3-binom{10-2}3-binom{10-5}2-binom{10-7}1=
              $$
              As a sanity check, the place of $x_1x_n$ in the list is $binom{n}1+binom{n}2-binom{n-1}2-binom{n-n}1=n+(n-1)+0=2n-1$, which agrees with your example. The position of $x_1x_2$ is $binom{n}1+binom{n}2-binom{n-1}2-binom{n-2}1=n+1$.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                The following bijection takes a set of $k$ distinct indices $a_1<a_2<dots<a_k$ and returns a number between $0$ and $binom{n}k-1$, so that every set gets a different number.



                $$
                S={a_1<a_2<dots<a_k}implies f(S)=binom{n-a_1}k+binom{n-a_2}{k-1}+dots+binom{n-a_k}1
                $$
                However, this is exactly the opposite of the order you want, so you should instead use $binom{n}k-f(S)$.



                Therefore, given an arbitrary set $S$ of $k$ indices, your desired indexing operation is
                $$
                g(S) = binom{n}1+dots+binom{n}{k-1}+binom{n}k-f(S)
                $$
                For example, when $n=10$ and your set of indices is $x_2x_5x_7$, the place in the list is
                $$
                g({2,5,7})=binom{10}1+binom{10}2+binom{10}3-binom{10-2}3-binom{10-5}2-binom{10-7}1=
                $$
                As a sanity check, the place of $x_1x_n$ in the list is $binom{n}1+binom{n}2-binom{n-1}2-binom{n-n}1=n+(n-1)+0=2n-1$, which agrees with your example. The position of $x_1x_2$ is $binom{n}1+binom{n}2-binom{n-1}2-binom{n-2}1=n+1$.






                share|cite|improve this answer









                $endgroup$



                The following bijection takes a set of $k$ distinct indices $a_1<a_2<dots<a_k$ and returns a number between $0$ and $binom{n}k-1$, so that every set gets a different number.



                $$
                S={a_1<a_2<dots<a_k}implies f(S)=binom{n-a_1}k+binom{n-a_2}{k-1}+dots+binom{n-a_k}1
                $$
                However, this is exactly the opposite of the order you want, so you should instead use $binom{n}k-f(S)$.



                Therefore, given an arbitrary set $S$ of $k$ indices, your desired indexing operation is
                $$
                g(S) = binom{n}1+dots+binom{n}{k-1}+binom{n}k-f(S)
                $$
                For example, when $n=10$ and your set of indices is $x_2x_5x_7$, the place in the list is
                $$
                g({2,5,7})=binom{10}1+binom{10}2+binom{10}3-binom{10-2}3-binom{10-5}2-binom{10-7}1=
                $$
                As a sanity check, the place of $x_1x_n$ in the list is $binom{n}1+binom{n}2-binom{n-1}2-binom{n-n}1=n+(n-1)+0=2n-1$, which agrees with your example. The position of $x_1x_2$ is $binom{n}1+binom{n}2-binom{n-1}2-binom{n-2}1=n+1$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 7 at 19:41









                Mike EarnestMike Earnest

                21.8k12051




                21.8k12051






























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