It depends if you mean "can multiple values be fixed" in a given round of Paxos or if you mean "can any given node accept more than one value".
As per the accepted answer by @Rakis you cannot get more than one value that is fixed into a location. Yet due to message loss, you can have a node accept more than one value into a location before the location is fixed.
To understand how this can happen we can image arbitrary messages getting lost arbitrary nodes crashing. For example nodes {A, B, C, D, E}
with A
the leader. It issues a prepare, gets a majority response, issues an accept(Va)
then crashes and never recovers. The accept message is only seen by node B
due to lost messages. Node E
times out and issues a prepare. Due to message loss, this is only seen by nodes C
, D
and E
. Node E
has a majority response but does not know about the value Va
sent by node A
and accepted by node B
. It is free to choose its own value and issues accept(Ve)
. This time all messages get through. Node B
accepts Ve
along with nodes C
, D
and E
.
In that scenario, the only value fixed by the algorithm was Ve
. Yet node B
accepted two different values Va
and Vb
.
To answer your question is this multipaxos the answer is no. See this answer for the explanation about mutltipaxos is. Note it isn't helpful to call things Paxos
and Multi-Paxos
. In his 2019 lectures, the inventor talks about the Single Synod Algorithm
for the minimal algorithm to fix a single value. Then the actual algorithm that is practically useful is the extension to choose many values from a stable leader. See this post about that. If you are sitting exams of the topic you may need to stick with the old confusing terminology.