In the process of RS code, if our code for n, the number of information element for k, then the number of redundant code element is n - k, you can correct (n - k)/2, but in the current distributed storage system to six data block codes into three redundant pieces, for any three of the nine pieces are missing then recovered by calculation from the remaining 6 pieces, I don't see how this is done, the RS coding in n - k redundant code can correct (n - k)/2, the three redundancy should be only restore l=1.5 blocks to ah, why can resume 3 piece?