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

[David Brown <david@westcontrol.com>]

On 18/04/12 20:22, Piergiorgio Sartor wrote:
> Hi David,
>
> On Tue, Apr 17, 2012 at 10:18:55PM +0200, David Brown wrote:
> [...]
>
>> For quad parity, we can try g3 = 8 as the obvious next choice in the
>> pattern.  Unfortunately, we start hitting conflicts.  To recover
>
> you should not use 8, because this is not a generator
> of GF(256) with polynomial 285, the standard for the
> RAID-5/6 setup.
>
> This means than 8^k does not cover the complete field
> for k in [0 254], thus having cycles and, consequently,
> creating conflicts.
>
> Some generators could be:
>
> 2, 4, 6, 9 13, 14, 16...
>
> but not 32 nor 64.
>
> I know that powers of two are nice, but if you want to
> have generic RAID, you must use other values.
>

I know that 8 is not a generator, and therefore you cannot expect to get 
a full set of (256 - noOfParities) disks.  But picking another generator 
(such as 16) is not enough to guarantee you the full range - it is a 
requirement, but not sufficient.  The generators need to be independent 
of each other, in the sense that all the simultaneous equations for all 
the combinations of failed disks need to be soluble.

It turns out that if you pick 16 as the forth parity generator here (1, 
2, 4, 16), then you can only have 5 data disks.  In fact, there are no 
other values for g3 that give significantly more than 21 data disks in 
combination with (1, 2, 4), whether or not they happen to be a generator 
for all of GF(2⁸).

> The log/exp tables, are, of course, always valid.
>
> BTW, the GF(256) with polynomial 285 has exactly 128
> generators, so it would be possible to have up to 129
> parity disk (1 is not a generator), for, I guess, a
> max of 256 disks (or maybe 255?).
>
> Hope this helps,
>
> bye,
>

When I started out with this, I thought it was as simple as you are 
suggesting.  But it is not - picking a set of generators for GF(2⁸) is 
not enough.  You have to check that all the solution matrices are 
invertible for all combinations of failed disks.

In fact, it is a little surprising that (1, 2, 4) works so well for 
triple parity.  I don't know whether it is just luck, or genius in Peter 
Anvin's choice of the multiply operation.

mvh.,

David
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