I'm studying Distributed Systems and synchronization and I didn't catch this solution of totally ordered multicast with Lamport timestamps. I read that it doesn't need ack to deliver a message to the application, but
"It is sufficient to multicast any other type of message, as long as that message has a timestamp larger than the received message. The condition for delivering a message m to the application, is that another message has been received from each other process with a large timestamp. This guarantees that there are no more messages underway with a lower timestamp."
This is a definition from a book. I tried to apply this definition to an example but I guess that something is wrong.
There are 4 processes and they multicast the following messages (the second number in parentheses is timestamp) :
P1 multi-casts (m11, 5); (m12, 12); (m13, 14);
P2 multi-casts (m21, 6); (m22, 14);
P3 multi-casts (m31, 5); (m32, 7); (m33, 11);
P4 multi-casts (m41, 8); (m42, 15); (m43, 19).
Supposing that there are no acknoledgments, can I guess which messages can be delivered and which not? Based on definition, my guess is that only m11 and m31 can be delivered to the application, because all the other messages received will have a timestamp greater, but this seems very strange, and I think I didn't understand the delivery condition very well. I have an exam on Tuesday about this, but anyway I would understand this mechanism.