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by pbailis
4770 days ago
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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. |
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