Re: Is this enough for us to have triple-parity RAID?

["H. Peter Anvin" <hpa@zytor.com>]

On 04/20/2012 11:58 AM, David Brown wrote:
> Hi,
> 
> Yes, being a generator for GF(2^8) is a requirement for a parity
> generator (sorry for the confusing terminology here - if anyone has a
> better suggestion, please say) to be part of a 255 data disk system.
> However, being a GF generator is necessary but not sufficient - using
> parity generators (1, 2, 4, 16) will /not/ give quad parity for 255 data
> disks, even though individually each of 1, 2, 4 and 16 are generators
> for GF.
> 
> 255 data disks is the theoretical limit for GF(2⁸).  But it is a
> theoretical limit of the algorithms - I don't know whether Linux md raid
> actually supports that many disks.  I certainly doubt if it is useful.
> 
> It might well be that a 21 data disk limit quad parity is useful - or at
> least, as useful as quad parity ever would be.  It would fit well within
> a typical large chassis with 24 disk slots.  And then it doesn't matter
> that 8 is not a generator for GF(2⁸) - it becomes the best choice
> because of the easiest implementation.
> 

It is also worth noting that there is nothing magical about GF(2^8).  It
is just a reasonable tradeoff when tables are needed.

There are hardware tricks one can play to do efficient operation of
wider fields, too.

But It sounds like {04} or {8e} are particular interesting generators of
the existing GF(2^8) field for an efficient second field, giving
triple-parity RAID again at a reasonable cost.

	-hpa

-- 
H. Peter Anvin, Intel Open Source Technology Center
I work for Intel.  I don't speak on their behalf.

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