Check a word given a generator matrix G












1












$begingroup$



Take the binary code C with generator matrix:
$$ left( begin{array}{ccccc|cccc}
1 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 0 \
0 & 1 & 0 & 0 & 0 & 0 & 1 & 1 & 1 \
0 & 0 & 1 & 0 & 0 & 1 & 0 & 1 & 1 \
0 & 0 & 0 & 1 & 0 & 1 & 1 & 0 & 1 \
0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 & 1 \
end{array} right) $$ And use it to correct the word that has at most one error: $011000011$.




Notice that this matrix is of the form $(I_5 | P)$, we obtain the parity matrix by transposition: $(P^T|I_{9-5})=(P^T|I_{4})$, this gives us the following parity matrix:
$$ left( begin{array}{ccccc|cccc}
1 & 0 & 1 & 1 & 1 & 1 & 0 & 0 & 0 \
1 & 1 & 0 & 1 & 1 & 0 & 1 & 0 & 0 \
1 & 1 & 1 & 0 & 1 & 0 & 0 & 1 & 0 \
0 & 1 & 1 & 1 & 1 & 0 & 0 & 0 & 1 \
end{array} right) $$
We obtain the four equations which each codeword must satisfy, where $m_i$ denotes the message bits and $r_i$ denotes the parity bit:
$$m_1 + m_3 +m_4+m_5 + r_1 = 0$$
$$m_1 + m_2 +m_4+m_5 + r_2 = 0$$
$$m_1 + m_2 +m_3+m_5 + r_3 = 0$$
$$m_2 + m_3 +m_4+m_5 + r_4 = 0$$
We will use these equations to check the word: $011000011= m_1 m_2 m_3 m_4 m_5 r_1 r_2 r_3 r_4$,
$$0 + 1 +0+0 + 0 not equiv 0$$
$$0 + 1 +0+0 + 0 not equiv 0$$
$$0 + 1 +1+0 + 1 not equiv 0$$
$$1 + 1 +0+0 + 1 not equiv 0$$



Here I ran into a problem because I do not gain any information from these equations as I cannot pinpoint the error. I've checked it several times but can't seem to find the error in my reasoning or how to proceed.






linear-algebra matrices coding-theory







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Just set $m_5$ to be $1$.
    $endgroup$
    – Berci
    Jan 10 at 14:07










  • $begingroup$
    Because it appears in every equation? so it must be the error?
    $endgroup$
    – Wesley Strik
    Jan 10 at 14:08






  • 2




    $begingroup$
    Yes, and because every equation is false originally.
    $endgroup$
    – Berci
    Jan 10 at 14:10
















1












$begingroup$



Take the binary code C with generator matrix:
$$ left( begin{array}{ccccc|cccc}
1 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 0 \
0 & 1 & 0 & 0 & 0 & 0 & 1 & 1 & 1 \
0 & 0 & 1 & 0 & 0 & 1 & 0 & 1 & 1 \
0 & 0 & 0 & 1 & 0 & 1 & 1 & 0 & 1 \
0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 & 1 \
end{array} right) $$ And use it to correct the word that has at most one error: $011000011$.




Notice that this matrix is of the form $(I_5 | P)$, we obtain the parity matrix by transposition: $(P^T|I_{9-5})=(P^T|I_{4})$, this gives us the following parity matrix:
$$ left( begin{array}{ccccc|cccc}
1 & 0 & 1 & 1 & 1 & 1 & 0 & 0 & 0 \
1 & 1 & 0 & 1 & 1 & 0 & 1 & 0 & 0 \
1 & 1 & 1 & 0 & 1 & 0 & 0 & 1 & 0 \
0 & 1 & 1 & 1 & 1 & 0 & 0 & 0 & 1 \
end{array} right) $$
We obtain the four equations which each codeword must satisfy, where $m_i$ denotes the message bits and $r_i$ denotes the parity bit:
$$m_1 + m_3 +m_4+m_5 + r_1 = 0$$
$$m_1 + m_2 +m_4+m_5 + r_2 = 0$$
$$m_1 + m_2 +m_3+m_5 + r_3 = 0$$
$$m_2 + m_3 +m_4+m_5 + r_4 = 0$$
We will use these equations to check the word: $011000011= m_1 m_2 m_3 m_4 m_5 r_1 r_2 r_3 r_4$,
$$0 + 1 +0+0 + 0 not equiv 0$$
$$0 + 1 +0+0 + 0 not equiv 0$$
$$0 + 1 +1+0 + 1 not equiv 0$$
$$1 + 1 +0+0 + 1 not equiv 0$$



Here I ran into a problem because I do not gain any information from these equations as I cannot pinpoint the error. I've checked it several times but can't seem to find the error in my reasoning or how to proceed.






linear-algebra matrices coding-theory







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Just set $m_5$ to be $1$.
    $endgroup$
    – Berci
    Jan 10 at 14:07










  • $begingroup$
    Because it appears in every equation? so it must be the error?
    $endgroup$
    – Wesley Strik
    Jan 10 at 14:08






  • 2




    $begingroup$
    Yes, and because every equation is false originally.
    $endgroup$
    – Berci
    Jan 10 at 14:10














1












1








1





$begingroup$



Take the binary code C with generator matrix:
$$ left( begin{array}{ccccc|cccc}
1 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 0 \
0 & 1 & 0 & 0 & 0 & 0 & 1 & 1 & 1 \
0 & 0 & 1 & 0 & 0 & 1 & 0 & 1 & 1 \
0 & 0 & 0 & 1 & 0 & 1 & 1 & 0 & 1 \
0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 & 1 \
end{array} right) $$ And use it to correct the word that has at most one error: $011000011$.




Notice that this matrix is of the form $(I_5 | P)$, we obtain the parity matrix by transposition: $(P^T|I_{9-5})=(P^T|I_{4})$, this gives us the following parity matrix:
$$ left( begin{array}{ccccc|cccc}
1 & 0 & 1 & 1 & 1 & 1 & 0 & 0 & 0 \
1 & 1 & 0 & 1 & 1 & 0 & 1 & 0 & 0 \
1 & 1 & 1 & 0 & 1 & 0 & 0 & 1 & 0 \
0 & 1 & 1 & 1 & 1 & 0 & 0 & 0 & 1 \
end{array} right) $$
We obtain the four equations which each codeword must satisfy, where $m_i$ denotes the message bits and $r_i$ denotes the parity bit:
$$m_1 + m_3 +m_4+m_5 + r_1 = 0$$
$$m_1 + m_2 +m_4+m_5 + r_2 = 0$$
$$m_1 + m_2 +m_3+m_5 + r_3 = 0$$
$$m_2 + m_3 +m_4+m_5 + r_4 = 0$$
We will use these equations to check the word: $011000011= m_1 m_2 m_3 m_4 m_5 r_1 r_2 r_3 r_4$,
$$0 + 1 +0+0 + 0 not equiv 0$$
$$0 + 1 +0+0 + 0 not equiv 0$$
$$0 + 1 +1+0 + 1 not equiv 0$$
$$1 + 1 +0+0 + 1 not equiv 0$$



Here I ran into a problem because I do not gain any information from these equations as I cannot pinpoint the error. I've checked it several times but can't seem to find the error in my reasoning or how to proceed.






linear-algebra matrices coding-theory







share|cite|improve this question











$endgroup$





Take the binary code C with generator matrix:
$$ left( begin{array}{ccccc|cccc}
1 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 0 \
0 & 1 & 0 & 0 & 0 & 0 & 1 & 1 & 1 \
0 & 0 & 1 & 0 & 0 & 1 & 0 & 1 & 1 \
0 & 0 & 0 & 1 & 0 & 1 & 1 & 0 & 1 \
0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 & 1 \
end{array} right) $$ And use it to correct the word that has at most one error: $011000011$.




Notice that this matrix is of the form $(I_5 | P)$, we obtain the parity matrix by transposition: $(P^T|I_{9-5})=(P^T|I_{4})$, this gives us the following parity matrix:
$$ left( begin{array}{ccccc|cccc}
1 & 0 & 1 & 1 & 1 & 1 & 0 & 0 & 0 \
1 & 1 & 0 & 1 & 1 & 0 & 1 & 0 & 0 \
1 & 1 & 1 & 0 & 1 & 0 & 0 & 1 & 0 \
0 & 1 & 1 & 1 & 1 & 0 & 0 & 0 & 1 \
end{array} right) $$
We obtain the four equations which each codeword must satisfy, where $m_i$ denotes the message bits and $r_i$ denotes the parity bit:
$$m_1 + m_3 +m_4+m_5 + r_1 = 0$$
$$m_1 + m_2 +m_4+m_5 + r_2 = 0$$
$$m_1 + m_2 +m_3+m_5 + r_3 = 0$$
$$m_2 + m_3 +m_4+m_5 + r_4 = 0$$
We will use these equations to check the word: $011000011= m_1 m_2 m_3 m_4 m_5 r_1 r_2 r_3 r_4$,
$$0 + 1 +0+0 + 0 not equiv 0$$
$$0 + 1 +0+0 + 0 not equiv 0$$
$$0 + 1 +1+0 + 1 not equiv 0$$
$$1 + 1 +0+0 + 1 not equiv 0$$



Here I ran into a problem because I do not gain any information from these equations as I cannot pinpoint the error. I've checked it several times but can't seem to find the error in my reasoning or how to proceed.






linear-algebra matrices coding-theory




linear-algebra matrices coding-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 10 at 13:52







Wesley Strik

















asked Jan 10 at 13:45









Wesley StrikWesley Strik

1,665423




1,665423








  • 2




    $begingroup$
    Just set $m_5$ to be $1$.
    $endgroup$
    – Berci
    Jan 10 at 14:07










  • $begingroup$
    Because it appears in every equation? so it must be the error?
    $endgroup$
    – Wesley Strik
    Jan 10 at 14:08






  • 2




    $begingroup$
    Yes, and because every equation is false originally.
    $endgroup$
    – Berci
    Jan 10 at 14:10














  • 2




    $begingroup$
    Just set $m_5$ to be $1$.
    $endgroup$
    – Berci
    Jan 10 at 14:07










  • $begingroup$
    Because it appears in every equation? so it must be the error?
    $endgroup$
    – Wesley Strik
    Jan 10 at 14:08






  • 2




    $begingroup$
    Yes, and because every equation is false originally.
    $endgroup$
    – Berci
    Jan 10 at 14:10








2




2




$begingroup$
Just set $m_5$ to be $1$.
$endgroup$
– Berci
Jan 10 at 14:07




$begingroup$
Just set $m_5$ to be $1$.
$endgroup$
– Berci
Jan 10 at 14:07












$begingroup$
Because it appears in every equation? so it must be the error?
$endgroup$
– Wesley Strik
Jan 10 at 14:08




$begingroup$
Because it appears in every equation? so it must be the error?
$endgroup$
– Wesley Strik
Jan 10 at 14:08




2




2




$begingroup$
Yes, and because every equation is false originally.
$endgroup$
– Berci
Jan 10 at 14:10




$begingroup$
Yes, and because every equation is false originally.
$endgroup$
– Berci
Jan 10 at 14:10










1 Answer
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$begingroup$

For such a simple code, the simple way to proceed is the following:



You have 32 code words. You can calculate them and check the weight of each word: if each weight is at least three, the code being linear, this implies that each pair of two different code words has a Hamming distance of at least three.



Then, for correcting the word $y = 011000011$: simply calculate its Hamming distance from each code word. You should find one at distance zero or one.



To correct from the syndrome ($H,y$): by considering all the error positions, you can establish a correspondence (a table) between error positions and the syndrome.






share|cite|improve this answer











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    $begingroup$

    For such a simple code, the simple way to proceed is the following:



    You have 32 code words. You can calculate them and check the weight of each word: if each weight is at least three, the code being linear, this implies that each pair of two different code words has a Hamming distance of at least three.



    Then, for correcting the word $y = 011000011$: simply calculate its Hamming distance from each code word. You should find one at distance zero or one.



    To correct from the syndrome ($H,y$): by considering all the error positions, you can establish a correspondence (a table) between error positions and the syndrome.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      For such a simple code, the simple way to proceed is the following:



      You have 32 code words. You can calculate them and check the weight of each word: if each weight is at least three, the code being linear, this implies that each pair of two different code words has a Hamming distance of at least three.



      Then, for correcting the word $y = 011000011$: simply calculate its Hamming distance from each code word. You should find one at distance zero or one.



      To correct from the syndrome ($H,y$): by considering all the error positions, you can establish a correspondence (a table) between error positions and the syndrome.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        For such a simple code, the simple way to proceed is the following:



        You have 32 code words. You can calculate them and check the weight of each word: if each weight is at least three, the code being linear, this implies that each pair of two different code words has a Hamming distance of at least three.



        Then, for correcting the word $y = 011000011$: simply calculate its Hamming distance from each code word. You should find one at distance zero or one.



        To correct from the syndrome ($H,y$): by considering all the error positions, you can establish a correspondence (a table) between error positions and the syndrome.






        share|cite|improve this answer











        $endgroup$



        For such a simple code, the simple way to proceed is the following:



        You have 32 code words. You can calculate them and check the weight of each word: if each weight is at least three, the code being linear, this implies that each pair of two different code words has a Hamming distance of at least three.



        Then, for correcting the word $y = 011000011$: simply calculate its Hamming distance from each code word. You should find one at distance zero or one.



        To correct from the syndrome ($H,y$): by considering all the error positions, you can establish a correspondence (a table) between error positions and the syndrome.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 10 at 14:07

























        answered Jan 10 at 13:59









        DamienDamien

        60714




        60714






























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