Here’s a generator matrix,

A parity check matrix for this, is one where (hence the product of with a codeword is also 0, though an error word generates a non-zero output).

Two candidates for this present themselves (are there others? I can’t see them):

Taking , there are distinct messages (codewords), where , but he generator can create different outputs. So we have non-zero code words and possible outputs, the number of detectable errors is:

The most likely error is the one with the lowest Hamming weight:

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H’ works as a parity check matrix, but I think H” does not: H” times (1,1,1,1) yields (0,0), but (1,1,1,1) is not generated by G.

OK, you are the professional, but I don’t follow your argument.

The maths is modulo 2, so surely each row of is zero?

True, but that’s only half the battle. Don’t you want H” x non-zero (meaning at least one component is nonzero, not that all components are nonzero) for x a non-word? In other words, shouldn’t a non-word generate at least one nonzero bit when checked?

Yes – you have a point! I’ll have to edit that/update it.

As a lark, I wrote a program to compute all minimum-dimension parity check matrices for a given generator matrix. Charitably assuming that I understood what I’m doing, there are two for your generator. H’ is one. The other has rows [1,1,1,1] and [0,0,1,0]. (Note that permuting the rows of a parity check matrix gives another parity check matrix, and adding an additional row x to a parity check matrix is fine as long as x^TG = 0.)

Cheers! I bought a book – was just about to blog about it.

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