top | item 29354802 (no title) cdtwigg | 4 years ago In this formulation I believe the last bit would also be repeated 3 times, and you’d have to read those three bits to detect corruption. discuss order hn newest dataflow|4 years ago Yes but how would you know that without reading all 6 bits first? netizen-936824|4 years ago You wouldn't, as long as the bits are received sequentially. I think the idea is that, depending on where the corruption is, you only need to read up to two bits past the corrupted one
dataflow|4 years ago Yes but how would you know that without reading all 6 bits first? netizen-936824|4 years ago You wouldn't, as long as the bits are received sequentially. I think the idea is that, depending on where the corruption is, you only need to read up to two bits past the corrupted one
netizen-936824|4 years ago You wouldn't, as long as the bits are received sequentially. I think the idea is that, depending on where the corruption is, you only need to read up to two bits past the corrupted one
dataflow|4 years ago
netizen-936824|4 years ago