# What happens when two different nodes have the same Lamport clock ID

With Lamport clocks, each node keeps its own counter. Before sending a message, a node increments its counter by one: LC(A)=LC(A)+1, and sends {1,msg} to B. Upon receiving a message, B updates its counter using the following formula: LC(B)=max(LC(B),LC(msg))+1=max(0,1)+1=2.

This works as expected when A and B take turns: A sends a message to B and B reacts by sending a message to A (or another node entirely).

However, this is not always the case. For example, in a typical TCP connection:

• A sends a message to B ("SYN")

• B replies to A ("SYN/ACK")

• A replies to B ("ACK") and then sends an actual request, such as an HTTP GET.

What happens (assuming both A and B start with their logical clocks set to zero):

• A increments its counter: LC(A)=1

• A sends {1,SYN} to B

• B updates its counter: LC(B)=max(0,1)+1=2

• B increments its counter: LC(B)=2+1=3

• B sends {3,SYN/ACK} to A

• A updates its counter: LC(A)=max(1,3)+1=4

• A increments its counter: LC(A)=4+1=5

• A sends {5,ACK} to B

• B updates its counter: LC(B)=max(3,5)+1=6

• At this point, A has to send another message (e.g. "GET /"). It increments its counter: LC(A)=5+1=6 and sends {6, GET /} to B.

As a result, both nodes have the same logical clock ID: LC(A)=LC(B)=6.

Is this a valid result? What happens next?

Though there can be a case where both nodes have the same clock id for the key and if so we will have to rely on some external mechanism to resolve the conflict.

Source

But I'm not clear on the details. What kind of external mechanism could be adopted, if it is even needed?

To break ties, we use any arbitrary total ordering $$\prec$$ of the processes.