> -----Original Message----- > From: linux-raid-owner@vger.kernel.org > [mailto:linux-raid-owner@vger.kernel.org]On Behalf Of Stephenson, Dale > Sent: Tuesday, March 26, 2002 12:03 PM > To: 'Ross Vandegrift'; Marcel > Cc: linux-raid@vger.kernel.org > Subject: RE: SW RAID6 ? > > You can get parity sets of the appropriate size (and necessary > non-multiple > intersection) by calculating parity vertically as well as horizontally. > Evenodd utilizes this with fixed parity drives, IIRC. The illustration of > RAID6 at http://www.acnc.com/04_01_06.html implies this, though the > illustration they show will not work (losing the first and last > drive would > lose data block A0 as well as parity A and parity 0. Rule of thumb -- a > chunk that is a member of a two parity sets cannot have both parity blocks > on the same drive). However, there ARE schemes that will work with simple > encoding. The original RAID 6 implementation did not use simple encoding, > IIRC it used Huffman codes. I've been wracking my brain, but am completely unable to come up with how you would avoid this. Any ideas? - To unsubscribe from this list: send the line "unsubscribe linux-raid" in the body of a message to majordomo@vger.kernel.org More majordomo info at http://vger.kernel.org/majordomo-info.html
