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# min. distance of product code and incomplete product code

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#### zzungboy

##### Newbie level 3
There are two linear block code C1 and C2.
C1 is (n1, k1, d1) where d1 is min. distance. C2 is (n2, k2, d2).

The min. distance for product code C1C2 is d1d2.
And, the min. distance for incomplete product code C1C2 is d1+d2-1.

Can you prove these?
If you know any materials or book about proof of these, please let me know.

Consider a row of an information block only has one nonzero bit and all other positions in the block are zero, then for product code, the row weight after row-wise encoding for this row is d1. In addition, since there are d1 nonzero positions in a row, each nonzero position will produce d2 column weight after column-wise encoding. The min. distance is then d1*d2.

For imcomplete product code, we also consider the case that only one nonzero position is in the information block, then the row weight will be d1 for this row after row-wise encoding. Since this in an incomplete product code, the check on check part is ignored. So, after column encoding, there is only d1 column weight produced by the nonzero information bit. The min. weight is d1+d2-1.

### zzungboy

Points: 2
Re: min. distance of product code and incomplete product cod

Thank you for the response.

Regarding min. distance of imcomplete product code,
first, we think the row-wise. the min. distance of information part is 1 (since it is all k-tuple digits) and the min distance of parity check is d1-1.
similarly, we also think column-wise. the min. distance of information part is also 1. and the min distance of parity check is d2-1.

Therefore, if we ignore checks on checks for incomplete product code, the min. distance of checks on checks is reduced.
So, d1d2-(d1-1)(d2-1)=d1+d2-1.

It's a my thinking. is it also correct?

Re: min. distance of product code and incomplete product cod

There are many ways to prove the min. distance.
I think this is also right...

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