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Coding Theory Homework Question

 
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Jan22-06, 06:29 PM   #1
 

Coding Theory Homework Question


Construct a [4, 7^2, 3] code. I know it exists because 7 is prime, so there are 6 MOLS. However, I am not quite sure how to go about constructing this code.
 
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Jan22-06, 06:36 PM   #2
 
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You sure you have that right? You can't encode 49 symbols in a codeword that's only 4 symbols long.

I don't know the acronym `MOL' either.


This sounds like a homework question -- you can usually metareason these out: it would probably be done using a code (or a technique) that you learned recently.
 
Jan22-06, 07:25 PM   #3
 
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MOLS=mutually orthogonal latin squares

I'm not positive what the (4,7^2,3) notation refers to, I'm guessing a 3-error correcting code with 7^2 codewords of length 4 (7 symbols each position?).

The construction I have in mind would make codewords of length 8 though. This is a pretty standard construction, and assuming it's what you're trying to do: are you having problems producing the MOLS or coming up with the code given a set of 6 MOLS? (or both?)
 
Jan26-06, 09:47 AM   #4
 

Coding Theory Homework Question


Quote by shmoe
MOLS=mutually orthogonal latin squares

I'm not positive what the (4,7^2,3) notation refers to, I'm guessing a 3-error correcting code with 7^2 codewords of length 4 (7 symbols each position?).

The construction I have in mind would make codewords of length 8 though. This is a pretty standard construction, and assuming it's what you're trying to do: are you having problems producing the MOLS or coming up with the code given a set of 6 MOLS? (or both?)
Yes, MOLS is mutually orthogonal latin squares. the notation (4, 7^2, 3) referes to an [n, M, d] code where n is the length of each vector in the code, M is the number of vectors, and d is the minimum distance between them.

Anyway, I figured it our, so it's all good. Thanks for the responses.
 
Jan26-06, 10:57 AM   #5
 
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Quote by SomeRandomGuy
Yes, MOLS is mutually orthogonal latin squares. the notation (4, 7^2, 3) referes to an [n, M, d] code where n is the length of each vector in the code, M is the number of vectors, and d is the minimum distance between them.

Anyway, I figured it our, so it's all good. Thanks for the responses.
ahh, 3 being minimum distance and not the level of error correction would explain how length 4 code words will be sufficient. Much simpler to construct the code when you only need 2 MOLS and not 6!
 
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