Algebra

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sara15

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Feb 4, 2011
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consider the binary encoding function that sends (a1,a2,a3) into (a1,a2,a3,a1+a2, a2+a3, a1+a3,a1+a2+a3). First , how can I give a generator matrix of this code
. then how can I provide the parity check matrix , giving n,k and the minimum distance of the code. Finall, how to decode m1=1100010 and m2= 0111010
please help me
 
Just write it down...

First column: 1 0 0
Second Column: 0 1 0
Third Column: 0 0 1
Fourth Column: 1 1 0
You're not going to make me do ALL the work, are you?
 
I want a simple explanation, because did not study the principles of algebra
 
Well, it can't be a whole lot simpler than giving you half the answer.

If you REALLY don't know enough algebra to put together row (a1 a2 a3) and column (1 1 1) to come up with a1+a2+a3, then I have to say you are in the wrong class. You don't have the prerequisites to survive this class.

Feel free to prove me wrong.
 
Hi
I respect your altitude but sorry you are wrong , because if I asked this question this doesn not meant that I do not know any thing in Algebra. since I have taken many course in Algebra and number theory but my courses were theoritical . and long time I did not review these basic things.I just need a remainder.
Thanks
 
Lovely argument. Like I said, feel free to prove it.

I still gave you half the answer. Tell me what the next column is.

First column: 1 0 0
Second Column: 0 1 0
Third Column: 0 0 1
Fourth Column: 1 1 0
Fifth Column: ? ? ?
 
I like your spunk. Willing to fight for it.

No. You need 2 and 3

Fifth Column: 0 1 1
Sixth Column: ? ? ? -- Hint: You need 1 and 3

Can you multiply these matrices?

\(\displaystyle \begin{pmatrix} 1& 2 \\ 3& 4 \end{pmatrix} \cdot \begin{pmatrix} a& b \\ c& d \end{pmatrix}\)
 
The sixth is 101
sure I can multiply them
ad get first column a+2c 3a+4c
second column b+2d 3b+4d
 
Excellent. That's all we're doing, here. Your task, in the original problem, is to multiply two matrices.

1) 1x3 (a1 a2 a3)
2) 3x7 (most of which we have defined already.)

Way to hang in there. I stand corrected. Good work!
 
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