Re: layout of far blocks in raid10

[Neil Brown <neilb@suse.de>]

On Tue, 11 May 2010 13:13:06 -0400
Aryeh Gregor <Simetrical+list@gmail.com> wrote:

> On Tue, May 11, 2010 at 11:12 AM, Keld Simonsen <keld@keldix.com> 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. 

I'm not quite sure how to respond to this...  As a mathematician I would
expect you to understand the important of precision in choosing words, yet
you use the word "know" for something that is exactly wrong.  Either you mean
"guess" or you have been seriously misinformed.  If it is the latter, then
please let me know where this misinformation came from so I can see about
getting it corrected.

md/raid10 uses a simple cyclic layout in all cases.  It does so because this
layout is completely general and works for all numbers of devices and copies.

So you can only survive multiple device failures where are most N-1 are
adjacent where N is the number of copies, and the first and last devices are
treated as adjacent.

NeilBrown

>                            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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