# Maths - linear Hamming codes

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Thread starter 7 years ago
#1
I have attached both the questions and the 'answers'.
I don't see how finding a check matrix shows 3a) and I have no clue about 3b onwards.

In Q4 I thought you did n=(2^r)-1 and r=3 therefore n=7 but in the answers they have n as 14. I am v confused.
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7 years ago
#2
(Original post by CammieInfinity)
I have attached both the questions and the 'answers'.
I don't see how finding a check matrix shows 3a) and I have no clue about 3b onwards.

In Q4 I thought you did n=(2^r)-1 and r=3 therefore n=7 but in the answers they have n as 14. I am v confused.
I'm just learning this myself so I could be wrong..

Since there is a weight 7 codeword, the existence of weight 1 and 2 codewords implies the existence of weight 6 and 5 codewords respectively.

A weight 1 or 2 codeword is not possible since minimum weight for Hamming code is 3.

As for question 4, this one isn't binary. You need

Edit:...and shouldn't it be 13?
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7 years ago
#3
Your parity check matrix for Ham(3,3) can have as its columns all non-zero vectors in V(3,3) which have 1 as their first non-zero entry.

001, 010, 011, 012, 100, etc..
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Thread starter 7 years ago
#4
Yes you are right I meant 13. Apologies. I see it now. I'll just ask my tutor when I get back about the other bit because I do not see the relevance of the check matrix.

I am also struggling with 1c) (new attachment)

I don't know how to find the magnitude of C? the base included 3 words if that helps. and for S(c,1) why is 6 choose 1 multiplied by 1? As in why 1? What does the 1 represent? if it was S(c,2) would it be S(c,1) + 6 choose 2 *1?
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7 years ago
#5
(Original post by CammieInfinity)
I'll just ask my tutor when I get back about the other bit because I do not see the relevance of the check matrix.
Which other bit do you mean?

Wasn't the check matrix required?
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Thread starter 7 years ago
#6
As in you said use check matrix to show you can't have a weight of 1
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7 years ago
#7
(Original post by CammieInfinity)
As in you said use check matrix to show you can't have a weight of 1
The check matrix was used to show that 1111111 is a codeword.

You can use the check matrix to show that no codeword has weight one but when I mentioned that, I neglected to use the fact that the minimum weight of non-zero codewords of a Hamming code is 3.
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