Re: layout of far blocks in raid10

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On Tue, May 11, 2010 at 11:12 AM, Keld Simonsen <keld@xxxxxxxxxx> wrote:
> There is a quesition on block layout in the raid10 far layout,
> that I would like to know more about.
> For 4 drives, and with 2 copies (-n 4 -p n2)  I see several
> possible layouts, 3 of them are, showing the beginning of each raid0 section:

There are only two layouts possible here: cyclic, and
double-transposition.  The first can be summarized in cycle notation
<http://en.wikipedia.org/wiki/Cycle_notation> as (abcd), where two
letters are adjacent if the extra copy of the first letter is on the
same disk as the second letter, and it's assumed the letters wrap
around in the parentheses (so the extra copy of d is on the same disk
as a).  The second is (ab)(cd).  So for instance, your example 1 is
(1432), example 2 is (13)(24), and example 3 is (1234).  For larger
numbers you have more possibilities, like (abc)(def) or (abcd)(ef) for
six drives.  The exact number of possibilities is the number of
partitions of the number of drives
<http://en.wikipedia.org/wiki/Partition_(number_theory)> that don't
include 1.

As far as I know (hopefully someone will correct me if I'm wrong),
RAID10 in mdadm stores data like (ab)(cd)(ef)..., at least if you have
an even number of drives.  Thus one disk out of every pair can fail
and you'll still have your data, where the pairs are determined by the
order you specify on the command line.  I don't know if this behavior
is guaranteed, but you can verify it by leaving some devices missing
-- trying to create a RAID10 with "/dev/sda1 /dev/sdb1 missing
missing" will fail, but "/dev/sda1 missing /dev/sdb1 missing" will
succeed, at least in my limited experience.

I don't know what mdadm does if there are an odd number of drives --
perhaps something like (ab)(cd)(efg), perhaps something more
complicated.  I know more about mathematics than about mdadm.  :)
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