How does paxos algorithm handle partial failures of accept messages? I've been reading the basic paxos algorithm from the "Paxos Made Simple" paper. I can't understand how paxos algorithm maintains the safety properties under partial failures of the accept messages, meaning cases where the proposer receives prepare responses from a majority, but the subsequent accept message does not reach all the acceptors. It seems to me this could lead in a violation of the safety properties, where there will be no majority that has agreed on the same value.

I created a contrived example of the use-case I am thinking about. In every round, the proposer sends a prepare message to a majority of nodes, gets a response from all of them, but then the accept message arrives only to one of them (the other ones get lost). More specifically, as shown in the image below:


in round 1, the proposer sends a propose (1) message. The acceptors haven't received any prepare message yet, so they all reply with a promise. The proposer gets a majority in phase 1, so sends an accept(1, A) message. This message reaches the first node that accepts it.
in round 2, the proposer sends a propose (2) message to a majority, which does not contain the node that has accepted a value. As a result, the proposes gets a majority and sends an accept(2, B) message, which reaches only the second node.
the round 3 is similar, where none of the selected acceptors during the first phase have accepted any value yet. As a result, the proposer picks a value and sends an accept(3, C) message, which reaches only the third node.
in round 4, the majority has to contain at least one of the acceptors that has accepted a value. Let's say this is the first node that has accepted the value A, which returns the number of the proposal (1) as a reply to the prepare(4) message the proposer sent. Since there's only one accepted value, the proposer uses this one and sends an accept(4, A) message, which reaches the fourth node.
in round 5, the proposer sends a propose(5) message to a majority that contains the first 2 nodes, which have accepted the values A and B. They both return the associated proposal numbers (1 and 2). The proposer picks the value corresponding to the highest one, B. The proposer then sends an accept(5, B) message, which only reaches the fifth node.




At this point, there is no way to satisfy the safety property, since we would need a majority (6/2 + 1 = 4) of acceptors to have accepted the same value.

Where is the flaw in my thinking ?
The paper's specification for the phase 2(b) of the algorithm, which is how an acceptor responds to an accept message is the following:


  If an acceptor receives an accept request for a proposal numbered n,
  it accepts the proposal unless it has already responded to a prepare
  request having a number greater than n.


This means that acceptors are allowed to accept multiple proposals (even with different values) in a single instance of Paxos, as long as they have not responded to a prepare request with a number greater than the number of the accept message. As a result, even after the last row of the example, a proposer could bring the system to an agreement, by making some of the nodes accept a different value.

For example, the proposer could send a subsequent prepare(6) message to nodes 1,2,5,6, followed by an accept(6, B) that reaches everyone, thus reaching a majority and concluding the consensus. The important thing to note is that after this point, it's impossible to form a majority with a different value, because any potential majority quorum would contain at least one node returning (6, B) during the prepare phase.