How RAID-6 dual parity calculation works
igoro.comHere's someone doing the math to generate matrices for higher amounts of parity: http://www.spinics.net/lists/raid/msg44948.html
i would rather go with this : https://www.kernel.org/pub/linux/kernel/people/hpa/raid6.pdf for much better in depth coverage...
> This article is for computer engineers who would like to have a high-level understanding of how the dual parity calculation works, without diving into all of the mathematical details.
That is more in depth, but is a lot more math heavy, which the author specifically states they are trying to avoid.
I come from a different math background, and it always surprises me people don't explain erasure encodings in terms of polynomials.
We all know that a polynomial of degree n is basically defined by n points. So a line is defined by 2 points.
If I need to encode the 2 pieces of info [a,b] I can just generate a polynomial ax+b and sample it at 2 points to recover the info. If I sample it at 3 points I can lose any one point and still recover [a,b]. The concept generalizes to higher dimensions and discrete values.
afaik that's basically it.
From the article:
> This article is for computer engineers who would like to have a high-level understanding of how the dual parity calculation works, without diving into all of the mathematical details.
I believe that is the value proposition for reading: gaining an understanding of RAID-6 parity calculation while glossing over the math-y bits.
Why does it need separate parity formulas? It would be easier to put the same simple xor in both parity blocks.
Suppose you have four disks and lose two. Consider a set of blocks where you lost two data blocks and still have two parity blocks. The parity blocks would be identical and you would not be able to recover data for either of the two missing data blocks.