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Equivalence of codes defines a symmetric relation


Binary codes behaving differently from other codes?Knuth equivalence of unshufflesCycles “converging” to an infinite cycle?Permutation that gives a sequence of non-negative partial sumsBlocks in permutation group theory (D&F)Factorizations in the symmetric groupHow do I form a group of permutations and determine the multiplication table for the group?Generator polynomial of the sum of cyclic codesNumber of deranged indicesAn algorithm on $S_n$. What is the final permutation?













2












$begingroup$


Two codes $C_1, C_2 subseteq A^n$ are called equivalent (notation: $sim$) if there are permutations $pi in Sym(A)$ and $sigma_1, dots, sigma_n in S_n$ such that



$$C_2 = (sigma_1(a_pi(1)), dots, sigma_n(a_pi(n))) mid (a_1, dots, a_n ) in C_1$$



Intuitively, this means that we can permute the order of the symbols in our code words, and subsequently permute the alphabet $A$ on every single position on everyy word.



The text I'm reading mentions without proof that $sim$ defines an equivalence relation, but I do not succeed in proving this.



In particular, I'm stuck at showing that



$$C_1 sim C_2 implies C_2 sim C_1$$



I tried with a little example:



if we have the code $C_1 := 00000,01101,10110,11011 subseteq 0,1^5$ and we perform $pi = (12)(345)$ and $sigma_3 = sigma_4 = 1$, $sigma_1 = sigma_2 = sigma_5 neq 1$, then we get the code $C_2 = 11001,01111,10010,00100$



To get back to $C_1$, we have to apply $pi' = (21543)$ and $sigma_2' = sigma_3' = 1$ and $sigma_1' = sigma_4' = sigma_5' neq 1$. I don't see a relation between the permutations $pi, sigma_i$ and $pi', sigma_j$ that can help me.



How can I show that this relation is symmetric?










share|cite|improve this question











$endgroup$
















    2












    $begingroup$


    Two codes $C_1, C_2 subseteq A^n$ are called equivalent (notation: $sim$) if there are permutations $pi in Sym(A)$ and $sigma_1, dots, sigma_n in S_n$ such that



    $$C_2 = (sigma_1(a_pi(1)), dots, sigma_n(a_pi(n))) mid (a_1, dots, a_n ) in C_1$$



    Intuitively, this means that we can permute the order of the symbols in our code words, and subsequently permute the alphabet $A$ on every single position on everyy word.



    The text I'm reading mentions without proof that $sim$ defines an equivalence relation, but I do not succeed in proving this.



    In particular, I'm stuck at showing that



    $$C_1 sim C_2 implies C_2 sim C_1$$



    I tried with a little example:



    if we have the code $C_1 := 00000,01101,10110,11011 subseteq 0,1^5$ and we perform $pi = (12)(345)$ and $sigma_3 = sigma_4 = 1$, $sigma_1 = sigma_2 = sigma_5 neq 1$, then we get the code $C_2 = 11001,01111,10010,00100$



    To get back to $C_1$, we have to apply $pi' = (21543)$ and $sigma_2' = sigma_3' = 1$ and $sigma_1' = sigma_4' = sigma_5' neq 1$. I don't see a relation between the permutations $pi, sigma_i$ and $pi', sigma_j$ that can help me.



    How can I show that this relation is symmetric?










    share|cite|improve this question











    $endgroup$














      2












      2








      2





      $begingroup$


      Two codes $C_1, C_2 subseteq A^n$ are called equivalent (notation: $sim$) if there are permutations $pi in Sym(A)$ and $sigma_1, dots, sigma_n in S_n$ such that



      $$C_2 = (sigma_1(a_pi(1)), dots, sigma_n(a_pi(n))) mid (a_1, dots, a_n ) in C_1$$



      Intuitively, this means that we can permute the order of the symbols in our code words, and subsequently permute the alphabet $A$ on every single position on everyy word.



      The text I'm reading mentions without proof that $sim$ defines an equivalence relation, but I do not succeed in proving this.



      In particular, I'm stuck at showing that



      $$C_1 sim C_2 implies C_2 sim C_1$$



      I tried with a little example:



      if we have the code $C_1 := 00000,01101,10110,11011 subseteq 0,1^5$ and we perform $pi = (12)(345)$ and $sigma_3 = sigma_4 = 1$, $sigma_1 = sigma_2 = sigma_5 neq 1$, then we get the code $C_2 = 11001,01111,10010,00100$



      To get back to $C_1$, we have to apply $pi' = (21543)$ and $sigma_2' = sigma_3' = 1$ and $sigma_1' = sigma_4' = sigma_5' neq 1$. I don't see a relation between the permutations $pi, sigma_i$ and $pi', sigma_j$ that can help me.



      How can I show that this relation is symmetric?










      share|cite|improve this question











      $endgroup$




      Two codes $C_1, C_2 subseteq A^n$ are called equivalent (notation: $sim$) if there are permutations $pi in Sym(A)$ and $sigma_1, dots, sigma_n in S_n$ such that



      $$C_2 = (sigma_1(a_pi(1)), dots, sigma_n(a_pi(n))) mid (a_1, dots, a_n ) in C_1$$



      Intuitively, this means that we can permute the order of the symbols in our code words, and subsequently permute the alphabet $A$ on every single position on everyy word.



      The text I'm reading mentions without proof that $sim$ defines an equivalence relation, but I do not succeed in proving this.



      In particular, I'm stuck at showing that



      $$C_1 sim C_2 implies C_2 sim C_1$$



      I tried with a little example:



      if we have the code $C_1 := 00000,01101,10110,11011 subseteq 0,1^5$ and we perform $pi = (12)(345)$ and $sigma_3 = sigma_4 = 1$, $sigma_1 = sigma_2 = sigma_5 neq 1$, then we get the code $C_2 = 11001,01111,10010,00100$



      To get back to $C_1$, we have to apply $pi' = (21543)$ and $sigma_2' = sigma_3' = 1$ and $sigma_1' = sigma_4' = sigma_5' neq 1$. I don't see a relation between the permutations $pi, sigma_i$ and $pi', sigma_j$ that can help me.



      How can I show that this relation is symmetric?







      permutations equivalence-relations coding-theory






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Mar 16 at 20:07







      Math_QED

















      asked Mar 16 at 19:58









      Math_QEDMath_QED

      7,71131454




      7,71131454




















          1 Answer
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          0












          $begingroup$

          Answering my own question:



          It is easy to check that (show that the right side is contained in the left side and then use that both sides have equal finite order)



          $$C_1 = (sigma^-1_pi^-1(1)(a_pi^-1(1))), dots, (sigma^-1_pi^-1(a_pi^-1(n)))mid (a_1, dots, a_n) in C_2$$



          and the symmetry follows.






          share|cite|improve this answer









          $endgroup$












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            1 Answer
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            active

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            0












            $begingroup$

            Answering my own question:



            It is easy to check that (show that the right side is contained in the left side and then use that both sides have equal finite order)



            $$C_1 = (sigma^-1_pi^-1(1)(a_pi^-1(1))), dots, (sigma^-1_pi^-1(a_pi^-1(n)))mid (a_1, dots, a_n) in C_2$$



            and the symmetry follows.






            share|cite|improve this answer









            $endgroup$

















              0












              $begingroup$

              Answering my own question:



              It is easy to check that (show that the right side is contained in the left side and then use that both sides have equal finite order)



              $$C_1 = (sigma^-1_pi^-1(1)(a_pi^-1(1))), dots, (sigma^-1_pi^-1(a_pi^-1(n)))mid (a_1, dots, a_n) in C_2$$



              and the symmetry follows.






              share|cite|improve this answer









              $endgroup$















                0












                0








                0





                $begingroup$

                Answering my own question:



                It is easy to check that (show that the right side is contained in the left side and then use that both sides have equal finite order)



                $$C_1 = (sigma^-1_pi^-1(1)(a_pi^-1(1))), dots, (sigma^-1_pi^-1(a_pi^-1(n)))mid (a_1, dots, a_n) in C_2$$



                and the symmetry follows.






                share|cite|improve this answer









                $endgroup$



                Answering my own question:



                It is easy to check that (show that the right side is contained in the left side and then use that both sides have equal finite order)



                $$C_1 = (sigma^-1_pi^-1(1)(a_pi^-1(1))), dots, (sigma^-1_pi^-1(a_pi^-1(n)))mid (a_1, dots, a_n) in C_2$$



                and the symmetry follows.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 9 hours ago









                Math_QEDMath_QED

                7,71131454




                7,71131454



























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