5 ms·
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.
by cdtwigg 5y 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.
- dataflow 5y agoYes but how would you know that without reading all 6 bits first?
- netizen-936824 5y agoYou 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