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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