Hamming(7,4) problem : Check my answer is correct or incorrect.












-1












$begingroup$


Find a data to be transmitted with Hamming(7,4) if given data bits : 1010.



$a_1=a_3oplus a_5oplus a_7=1oplus0oplus0=1$



$a_2=a_3oplus a_6oplus a_7=1oplus1oplus0=0$



$a_3=a_5oplus a_6oplus a_7=0oplus1oplus0=1$



begin{eqnarray}
a_1&a_2&a_3&a_4&a_5&a_6&a_7\
1&0&1&1&0&1&0
end{eqnarray}



Is it correct answer?










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$endgroup$








  • 2




    $begingroup$
    Depends on the concrete generator matrix.
    $endgroup$
    – Wuestenfux
    Dec 16 '18 at 12:30
















-1












$begingroup$


Find a data to be transmitted with Hamming(7,4) if given data bits : 1010.



$a_1=a_3oplus a_5oplus a_7=1oplus0oplus0=1$



$a_2=a_3oplus a_6oplus a_7=1oplus1oplus0=0$



$a_3=a_5oplus a_6oplus a_7=0oplus1oplus0=1$



begin{eqnarray}
a_1&a_2&a_3&a_4&a_5&a_6&a_7\
1&0&1&1&0&1&0
end{eqnarray}



Is it correct answer?










share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    Depends on the concrete generator matrix.
    $endgroup$
    – Wuestenfux
    Dec 16 '18 at 12:30














-1












-1








-1





$begingroup$


Find a data to be transmitted with Hamming(7,4) if given data bits : 1010.



$a_1=a_3oplus a_5oplus a_7=1oplus0oplus0=1$



$a_2=a_3oplus a_6oplus a_7=1oplus1oplus0=0$



$a_3=a_5oplus a_6oplus a_7=0oplus1oplus0=1$



begin{eqnarray}
a_1&a_2&a_3&a_4&a_5&a_6&a_7\
1&0&1&1&0&1&0
end{eqnarray}



Is it correct answer?










share|cite|improve this question









$endgroup$




Find a data to be transmitted with Hamming(7,4) if given data bits : 1010.



$a_1=a_3oplus a_5oplus a_7=1oplus0oplus0=1$



$a_2=a_3oplus a_6oplus a_7=1oplus1oplus0=0$



$a_3=a_5oplus a_6oplus a_7=0oplus1oplus0=1$



begin{eqnarray}
a_1&a_2&a_3&a_4&a_5&a_6&a_7\
1&0&1&1&0&1&0
end{eqnarray}



Is it correct answer?







coding-theory






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asked Dec 16 '18 at 11:39









Ongky Denny WijayaOngky Denny Wijaya

3798




3798








  • 2




    $begingroup$
    Depends on the concrete generator matrix.
    $endgroup$
    – Wuestenfux
    Dec 16 '18 at 12:30














  • 2




    $begingroup$
    Depends on the concrete generator matrix.
    $endgroup$
    – Wuestenfux
    Dec 16 '18 at 12:30








2




2




$begingroup$
Depends on the concrete generator matrix.
$endgroup$
– Wuestenfux
Dec 16 '18 at 12:30




$begingroup$
Depends on the concrete generator matrix.
$endgroup$
– Wuestenfux
Dec 16 '18 at 12:30










1 Answer
1






active

oldest

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1












$begingroup$

Well, it all depends on the generator matrix, say
$$G =left(begin{array}{ccccccc}
1 & 0 & 0 & 0 & 1 & 1 & 1 \
0 & 1 & 0 & 0 & 0 & 1 & 1 \
0 & 0 & 1 & 0 & 1 & 0 & 1 \
0 & 0 & 0 & 1 & 1 & 1 & 0 \
end{array}right).$$

Then $(a,b,c) G = (a,b,c,d,a+c+d,a+b+c,a+b+c)$, in particular $(1,0,1,0)G =(1,0,1,0,0,1,0)$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you so much, @Wuestenfux
    $endgroup$
    – Ongky Denny Wijaya
    Dec 17 '18 at 3:04











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1 Answer
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oldest

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1 Answer
1






active

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active

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active

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1












$begingroup$

Well, it all depends on the generator matrix, say
$$G =left(begin{array}{ccccccc}
1 & 0 & 0 & 0 & 1 & 1 & 1 \
0 & 1 & 0 & 0 & 0 & 1 & 1 \
0 & 0 & 1 & 0 & 1 & 0 & 1 \
0 & 0 & 0 & 1 & 1 & 1 & 0 \
end{array}right).$$

Then $(a,b,c) G = (a,b,c,d,a+c+d,a+b+c,a+b+c)$, in particular $(1,0,1,0)G =(1,0,1,0,0,1,0)$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you so much, @Wuestenfux
    $endgroup$
    – Ongky Denny Wijaya
    Dec 17 '18 at 3:04
















1












$begingroup$

Well, it all depends on the generator matrix, say
$$G =left(begin{array}{ccccccc}
1 & 0 & 0 & 0 & 1 & 1 & 1 \
0 & 1 & 0 & 0 & 0 & 1 & 1 \
0 & 0 & 1 & 0 & 1 & 0 & 1 \
0 & 0 & 0 & 1 & 1 & 1 & 0 \
end{array}right).$$

Then $(a,b,c) G = (a,b,c,d,a+c+d,a+b+c,a+b+c)$, in particular $(1,0,1,0)G =(1,0,1,0,0,1,0)$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you so much, @Wuestenfux
    $endgroup$
    – Ongky Denny Wijaya
    Dec 17 '18 at 3:04














1












1








1





$begingroup$

Well, it all depends on the generator matrix, say
$$G =left(begin{array}{ccccccc}
1 & 0 & 0 & 0 & 1 & 1 & 1 \
0 & 1 & 0 & 0 & 0 & 1 & 1 \
0 & 0 & 1 & 0 & 1 & 0 & 1 \
0 & 0 & 0 & 1 & 1 & 1 & 0 \
end{array}right).$$

Then $(a,b,c) G = (a,b,c,d,a+c+d,a+b+c,a+b+c)$, in particular $(1,0,1,0)G =(1,0,1,0,0,1,0)$.






share|cite|improve this answer









$endgroup$



Well, it all depends on the generator matrix, say
$$G =left(begin{array}{ccccccc}
1 & 0 & 0 & 0 & 1 & 1 & 1 \
0 & 1 & 0 & 0 & 0 & 1 & 1 \
0 & 0 & 1 & 0 & 1 & 0 & 1 \
0 & 0 & 0 & 1 & 1 & 1 & 0 \
end{array}right).$$

Then $(a,b,c) G = (a,b,c,d,a+c+d,a+b+c,a+b+c)$, in particular $(1,0,1,0)G =(1,0,1,0,0,1,0)$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 16 '18 at 15:36









WuestenfuxWuestenfux

4,9391513




4,9391513












  • $begingroup$
    Thank you so much, @Wuestenfux
    $endgroup$
    – Ongky Denny Wijaya
    Dec 17 '18 at 3:04


















  • $begingroup$
    Thank you so much, @Wuestenfux
    $endgroup$
    – Ongky Denny Wijaya
    Dec 17 '18 at 3:04
















$begingroup$
Thank you so much, @Wuestenfux
$endgroup$
– Ongky Denny Wijaya
Dec 17 '18 at 3:04




$begingroup$
Thank you so much, @Wuestenfux
$endgroup$
– Ongky Denny Wijaya
Dec 17 '18 at 3:04


















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