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andras_gerlits | 1 year ago
https://medium.com/p/5e397cb12e63#7df1
Considering that this is maybe the 12th time I'm linking the explanation, I now think you're not looking for a discussion, you're here for a fight. Please let me know if you find any problems with the article in its reasoning. It was vetted by a lot of people before, so I really need to notify them too if you do.
sausagefeet|1 year ago
Redundancy doesn't fix any problems, it just makes them less likely to occur. Again this not only does nothing to address the CAP theorem but is done for a wide array of problems already, there is nothing new here.
andras_gerlits|1 year ago
https://medium.com/p/5e397cb12e63#373c
There are three separate mitigation-solutions that go into how total-order strong consistency can keep marching on even if a specific child-group is intermittently isolated.
The first one is redundancy by deterministic replication, so there will always be many replicas which aren't just shards, but full copies of the _consistency ledger_. It's not a database, not a cache, just the thing that establishes the consistency between nodes. These instances all "race each other" to provide the first value of the outputs to other nodes.
The second one is the latency-mitigation we talked about earlier, I don't think we need to waste more breath on that.
The third one is that since the consistency-mechanism requires an explicit association-instruction to interleave the children's versions into its parent's (so that these versions can be exposed to nodes observing it from afar). If the child goes AWOL, it won't be able to associate its versions to its parent, so it won't keep up everyone else either. In this case the total-order won't be affected by the child-group's local-order, which is still allowed to make progress, as long as it's not trying to update any record that is distant to it.