Since 7 is a prime number, this is an example of a finite field. One simple approach that works is to pick a prime number M, and change the meaning of +, вЂ“, and Г— so that values 'wrap around' at M. Thankfully, this is a well-studied problem in algebra. WeвЂ™d like to redefine the arithmetic operators +, вЂ“, Г—, / such that our equations still work, but the values of P and Q stay within the ranges of the input values. For example, if you are dealing with 1-byte blocks, it would be ideal if P and Q were 1 byte as well. But in that case, the Q value can reach 3825! Ideally, youвЂ™d want P and Q to have the same range as the data values themselves. For example, if the data block is 1 byte, its value ranges from 0 to 255. But, integers are inconvenientвЂ¦Д¬alculating P and Q from the formulas above will allow you to recover up to two lost data blocks.Д«ut, there is an annoying technical problem: the calculated values of P and Q can be significantly larger than the original data values. In fact, if you look at the original matrix, any five rows would have been linearly independent, so we could solve the equations regardless of which two devices were missing. This system of equations is solvable because the matrix is invertible: the five rows are all linearly independent. Effectively, we lost two rows in the matrix and the two matching rows in the result vector: In our example, we lose blocks c and d and end up with 5 equations of 5 variables. We have 7 linear equations of 5 variables. We can express the relationship between a, b, c, d, e, P, Q as the following matrix-vector multiplication: To explain why we can always solve the equations to recover two missing blocks, we need to take a short detour into matrix algebra. How come you can always solve the equations? If you lost c and d, you can recover them by solving the original equations. But, what if you want protection against the loss off two devices? Can we extend the parity idea to protect the data even in that scenario? Single parity protects data against the loss of a single storage device.

Parity blocks gets more overwrites than data blocks, so it makes sense to distribute them among the devices. RAID-5 uses the exclusive-or parity approach, except that the placement of parity is rotated among the storage devices. Exclusive-or parity is commonly used in storage systems as RAID-5 configuration: Given a, b, c, e and P, it is easy to reconstruct the missing block d: If the first parity bit is 0, you know that the number of 1 bits among the first bits of a, b, c, d, e is even.If any one device fails, you can still recompute the block from the failed device, and so you havenвЂ™t lost any data.ДЄ simple way to calculate the parity P is just to combine the values a, b, c, d, e with the exclusive-or (XOR) operator:Д®ffectively, every parity bit protects the corresponding bits of a, b, c, d, e. You can put the 5 data blocks on 5 different storage devices, and then store a parity block on a 6th device. As an example, say that you need to reliably store 5 data blocks of a fixed size (e.g., 4kB). LetвЂ™s start with the simple case вЂ“ single parity.

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. Parity protection is a common technique for reliable data storage on mediums that may fail (HDDs, SSDs, storage servers, etc.) Calculation of parity uses tools from algebra in very interesting ways, in particular in the dual parity case.

What is data parity