Re: raid6 and parity calculations

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>  -------Original Message-------
>  From: Neil Brown <neilb@xxxxxxx>
>  To: Michael Sallaway <michael@xxxxxxxxxxxx>
>  Cc: linux-raid@xxxxxxxxxxxxxxx
>  Subject: Re: raid6 and parity calculations
>  Sent: 15 Sep '10 10:26

>  It is best not to think to think to much about chunks.  Think about strips
>  (not stripes).
>  A strip is a set of blocks, one per device each at the same offset.
>  Think of page sizes blocks/strips.
>  Each strip has a P block and a Q block and a bunch of data blocks.  Which
>  is P and which is Q and which each data block is a function of the offset,
>  the layout and the chunk size.  Once you have used the chunksize to perform
>  that calculation, don't think about chunks any more - just blocks and strips.
>  

Aah, perfect -- that makes sense, thanks for that.

As a sort-of follow up question, would anyone know if the data size of a Q calculation affects the result at all? eg. if I do a 64kb Q calculation on 10 drives of data, would that be the same as doing 16x 4kb Q calculations on sequential blocks of the same data, then concatenating it together? (I can't remember what that operation property is called....?)

I've been reading the maths of RAID6 PDF (http://kernel.org/pub/linux/kernel/people/hpa/raid6.pdf), but I'm a bit too rusty to understand Galois fields, and if the data size matters. I presume the data ordering is also critical for a Q calculation, correct? (eg. drives have to be d0 -> d10 in order, not just random).

And, in contrast, for the P calculations, data size and input order makes no difference, correct? (since it's just a simple bitwise XOR of all the inputs).

>  
>  Ouch... good luck.

Thanks! I'm the only one to blame, though -- it happened in the month between "getting the new system set up" and "setting up backups for the new system". So it's the only copy of the data.... whoops. :-)

Thanks for the help/advice!

Cheers,
Michael
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