3 ms·
Post author here. Hmm. The invariant we're trying to maintain is that any write in 'good' should have its transactional "siblings" in either 'good' or 'pending'
by pbailis 13y ago
Post author here. Hmm. The invariant we're trying to maintain is that any write in 'good' should have its transactional "siblings" in either 'good' or 'pending' on their respective servers. So if we are trying to write x=1 and y=1, then, if x=1 is in the x server's 'good', then y=1 should be in the y server's 'good' or 'pending'. But in the example under "not okay", y is not present in either.
- dscrd 13y agoOk, then I just did not understand that part yet.