[cconroy]

this proof is sound (for this simple structure) but does it imply you can never have all three (CAP) in some scheme where you can draw from many nodes and edges (not infinite) to handle partition failure while always being available and consistent? Perhaps a proof that you need to draw from infinite nodes and edges to achieve CAP would be interesting.

disclaimer: i am not well educated on the literature in Distributed systems :(

[legacynl]

I was thinking the same thing. The ability to handle partition failure comes from having more nodes. This proves this is not possible when haviing just two nodes. Kind of makes sense as it stops being a distributed system if you disconnect all nodes.

[y4mi]

The CAP Theorem thinks about guarantees, not probabilities.

There are various ways to reduce the impact and probability of partitions, but increasing the quantity of nodes does not make partitions impossible -- in fact, you now have to worry about n possible partitions (n being the number of nodes).

the only way to guarantee partition tolerance is by majority votum. If you do that, you can no longer guarantee availability. For example, a cluster of 11 nodes might be partitioned thrice. (2 + 4 + 5) None of the nodes are allowed to answer, breaking the availability guarantee.

[gweinberg]

If for any two nodes there is some path by which the nodes can communicate, the network is not partitioned. That's what partitioning means.

[zzzcpan]

In asynchronous networks you never have that. You never know whether you can communicate with the other nodes, until you try and get the response back. You can only reason whether you were able to communicate with other nodes about communications that already took place.