Allan Sandfeld wrote:
>
> I just looked at it. It is possible allright and the diagram looks ok.
>
> If you have 3 disks A,B and C the parity is calculated by dividing the diskw
> into typical lines, in this example I use 3 like they use on the diagram. We
> then have a parity per line and one per disk. You can only regenerate one
> block per parity, but since you have two full independ parities you can
> replace any two.
>
> A1 B1 C1 P1 (P1 = A1^B1^C1)
> A2 B2 C2 P2
> A3 B3 C3 P3
> PA PB PC
> (PA=A1^A2^A3)
>
> As you can see if you wish to chech the parity for one read line(eg.A1-C1),
> you can check directly against the horizontal parity P1. But if you wish to
> check the horizontal parity you need to read the entire diskarray!
I don't think I got that one. How many disks would you use, and
how would you distribute the above fields across the disks? If
you put each column on a disk you can handle any two lost
blocks, but not two lost disks. Or would you use a total of 15
disks? Or do you have some way of placing them on 5 disks with
3 blocks on each disk?
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