Time, clocks & ordering

Step through nine events across three processes. Every process keeps a counter of its own — bumped by one for every event — except a receive, which first jumps up to catch the sender's counter before continuing. Watch each process's three-number vector grow as messages cross between them.

A Lamport clock gives every event a single number: bump your own counter, and on a receive, jump to $\max(\text{own}, \text{sender's})$ first.

A vector clock keeps one counter per process instead of one shared number: bump only your own component, but on receive take the componentwise max with the sender's stamped vector first.

• $L_p$ — process $p$'s own Lamport counter, just before this event.
• $L_{msg}$ — the Lamport timestamp attached to the message being received.
• $a \le b$ — every component of vector $a$ is $\le$ the matching component of $b$.

1. Comparing two vector clocks

   $a$ happened-before $b$ exactly when $a \le b$ componentwise and $a \neq b$. If neither vector dominates the other, the events are concurrent — no message chain links them in either direction.

2. Why a vector, not just a number

   A single Lamport number can always be compared to any other — but that comparison doesn't mean one event caused the other. A vector clock's componentwise structure is what actually distinguishes "definitely before" from "no relation at all."

Pick any two of the nine events. compareVectorClocks reads their vectors directly and reports before, after, concurrent, or equal — the same rule, applied to whichever pair you choose.

Trace the first four events by hand: e1 (P0, internal), e2 (P0, sends m1), e3 (P0, internal), e4 (P1, receives m1).

1. e1 — P0's first event

   P0 bumps its own counter: Lamport = 1, vector = [1, 0, 0].

2. e2 — P0 sends m1

   Another internal bump on P0: Lamport = 2, vector = [2, 0, 0]. This vector rides along inside m1.

3. e3 — P0 does more local work

   Bump again: Lamport = 3, vector = [3, 0, 0].

4. e4 — P1 receives m1

   P1 first takes the componentwise max of its own [0,0,0] and m1's stamped [2,0,0] → [2,0,0], then bumps its own component: vector = [2, 1, 0]. Lamport jumps to $\max(0, 2) + 1 = 3$ — coincidentally equal to e3's Lamport number, even though e3 and e4 are on different processes entirely.