Hamming(7,4) problem : Check my answer is correct or incorrect.
$begingroup$
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?
coding-theory
asked Dec 16 '18 at 11:39
Ongky Denny WijayaOngky Denny Wijaya
3798
$endgroup$
add a comment |
$begingroup$
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?
coding-theory
asked Dec 16 '18 at 11:39
Ongky Denny WijayaOngky Denny Wijaya
3798
$endgroup$
2
$begingroup$
Depends on the concrete generator matrix.
$endgroup$
– Wuestenfux
Dec 16 '18 at 12:30
add a comment |
$begingroup$
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?
coding-theory
asked Dec 16 '18 at 11:39
Ongky Denny WijayaOngky Denny Wijaya
3798
$endgroup$
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
coding-theory
asked Dec 16 '18 at 11:39
Ongky Denny WijayaOngky Denny Wijaya
3798
asked Dec 16 '18 at 11:39
Ongky Denny WijayaOngky Denny Wijaya
3798
asked Dec 16 '18 at 11:39
Ongky Denny WijayaOngky Denny Wijaya
3798
asked Dec 16 '18 at 11:39
Ongky Denny WijayaOngky Denny Wijaya
3798
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
add a comment |
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
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
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 $(1,0,1,0)G =(1,0,1,0,0,1,0)$.
answered Dec 16 '18 at 15:36
WuestenfuxWuestenfux
4,9391513
$endgroup$
$begingroup$
Thank you so much, @Wuestenfux
$endgroup$
– Ongky Denny Wijaya
Dec 17 '18 at 3:04
add a comment |
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1 Answer
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active
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1 Answer
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active
oldest
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$begingroup$
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 $(1,0,1,0)G =(1,0,1,0,0,1,0)$.
answered Dec 16 '18 at 15:36
WuestenfuxWuestenfux
4,9391513
$endgroup$
$begingroup$
Thank you so much, @Wuestenfux
$endgroup$
– Ongky Denny Wijaya
Dec 17 '18 at 3:04
add a comment |
$begingroup$
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 $(1,0,1,0)G =(1,0,1,0,0,1,0)$.
answered Dec 16 '18 at 15:36
WuestenfuxWuestenfux
4,9391513
$endgroup$
$begingroup$
Thank you so much, @Wuestenfux
$endgroup$
– Ongky Denny Wijaya
Dec 17 '18 at 3:04
add a comment |
$begingroup$
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 $(1,0,1,0)G =(1,0,1,0,0,1,0)$.
answered Dec 16 '18 at 15:36
WuestenfuxWuestenfux
4,9391513
$endgroup$
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 $(1,0,1,0)G =(1,0,1,0,0,1,0)$.
answered Dec 16 '18 at 15:36
WuestenfuxWuestenfux
4,9391513
answered Dec 16 '18 at 15:36
WuestenfuxWuestenfux
4,9391513
answered Dec 16 '18 at 15:36
WuestenfuxWuestenfux
4,9391513
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
Depends on the concrete generator matrix.
$endgroup$
– Wuestenfux
Dec 16 '18 at 12:30