<Dd> Q = ⨁ i g i D i = g 0 D 0 ⊕ g 1 D 1 ⊕ g 2 D 2 ⊕...⊕ g n − 1 D n − 1 (\ displaystyle \ mathbf (Q) = \ bigoplus _ (i) (g ^ (i) D_ (i)) = g ^ (0) \ mathbf (D) _ (0) \; \ oplus \; g ^ (1) \ mathbf (D) _ (1) \; \ oplus \; g ^ (2) \ mathbf (D) _ (2) \; \ oplus \;...\; \ oplus \; g ^ (n - 1) \ mathbf (D) _ (n - 1)) </Dd> <P> For a computer scientist, a good way to think about this is that ⊕ (\ displaystyle \ oplus) is a bitwise XOR operator and g i (\ displaystyle g ^ (i)) is the action of a linear feedback shift register on a chunk of data . Thus, in the formula above, the calculation of P is just the XOR of each stripe . This is because addition in any characteristic two finite field reduces to the XOR operation . The computation of Q is the XOR of a shifted version of each stripe . </P> <P> Mathematically, the generator is an element of the field such that g i (\ displaystyle g ^ (i)) is different for each nonnegative i (\ displaystyle i) satisfying i <m − 1 (\ displaystyle i <m - 1). </P> <P> If one data drive is lost, the data can be recomputed from P just like with RAID 5 . If two data drives are lost or a data drive and the drive containing P are lost, the data can be recovered from P and Q or from just Q, respectively, using a more complex process . The details can be computed using field theory; suppose that D i (\ displaystyle D_ (i)) and D j (\ displaystyle D_ (j)) are the lost values with i ≠ j (\ displaystyle i \ neq j), then, using the other values of D (\ displaystyle D), we find constants A (\ displaystyle A) and B (\ displaystyle B): </P>

Name the raid level that provides the best redundancy