5 ms·
As a post script, I should add that I had a thought that the RS(1,3) was less redundant than copies=4 as I think that RS(1,3) requires more than 1 of its four s
by compressedgas 9mo ago
As a post script, I should add that I had a thought that the RS(1,3) was less redundant than copies=4 as I think that RS(1,3) requires more than 1 of its four symbols to be present.
Based on https://en.wikipedia.org/wiki/Reed%E2%80%93Solomon_error_correction https://en.wikipedia.org/wiki/Reed%E2%80%93Solomon_error_cor... which gives (n=1 - k=3)/2 is 1 as the answer to the question of the number of missing symbols that RS(1,3) can recover which means that RS(1,3) is not a worse way of storing four copies but a worse way of storing 2 copies. It takes the space of four copies to store what only has the redundancy of two copies.