How RAID-6 dual parity calculation works

How RAID-6 dual parity calculation works(igoro.com)

55 points by c0ff 11 years ago | 7 comments

sp332 11 years ago |

Here's someone doing the math to generate matrices for higher amounts of parity: http://www.spinics.net/lists/raid/msg44948.html

signa11 11 years ago |

i would rather go with this : https://www.kernel.org/pub/linux/kernel/people/hpa/raid6.pdf for much better in depth coverage...

level 11 years ago | |

> 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.

shanemhansen 11 years ago |

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.

dmiladinov 11 years ago |

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.

oofabz 11 years ago |

Why does it need separate parity formulas? It would be easier to put the same simple xor in both parity blocks.

aYiZtr 11 years ago | |

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.