Trace a distributed snapshot across 3 nodes

Step through this trace the same way you did last chapter. N1 initiates now, which changes something structural: since N1 has no incoming marker to trigger its recording, BOTH of its incoming channels — from N0 and from N2 — start "recording" at the very first step, long before N0 or N2 have recorded anything of their own.

The rule is exactly the one from the previous chapter — reused here unchanged, via runSnapshot — applied to a new initiator and a new pair of in-flight messages.

• N1, the initiator this time — records its own state first, with both incoming channels open from the start.
• m1 — N2 → N0, $4, sent before the snapshot begins.
• m2 — N0 → N1, $6, sent after N1 has already started recording that very channel.

1. Two messages, two channels to watch

   Nothing about the algorithm changes with a second in-flight message — each channel is recorded independently. The only new work is tracking two channels' logs at once instead of one.

2. The initiator's early start matters

   Because N1 initiates, channel N0 → N1 opens for logging immediately — well before N0 has even recorded its own state. m2 only needs to arrive at N1 before N0's marker does on that same channel, and FIFO ordering guarantees that as long as N0 sent m2 before sending its marker.

Scrub through the full trace and watch all 6 channels at once. Two of them — N2 → N0 and N0 → N1 — pick up a logged message before they close; the other four close empty. Watch the running total climb from the recorded-states-only number up to the true total of 30 only once both messages are accounted for.

N0 = 10, N1 = 5, N2 = 15 (total 30). N2 → N0 sends $4 before the snapshot; N0 → N1 sends $6 while N1 is already recording that channel.

1. N1 initiates

   Records its own balance (5). Starts recording both incoming channels, N0→N1 and N2→N1, immediately — it has no triggering marker to exclude one of them.

2. N0 records on its first marker, after already sending m2

   N0 receives N1's marker and records its own balance. By this point it has already sent the $6 to N1, so N0's recorded balance already reflects that send: $10 - 6 = 4$.

3. N2 records on its first marker

   N2 receives N1's marker and records its own balance, which already reflects having sent the $4 to N0 earlier: $15 - 4 = 11$.

4. Both messages land in open channels

   • The $6 arrives at N1 before N0's marker closes channel N0→N1 — logged
   • The $4 arrives at N0 before N2's marker closes channel N2→N0 — logged
   • Every other channel closes with nothing in it
5. The totals check out

   Recorded states:

   • N0's $4$
   • N1's own $5$
   • N2's $11$

   Sum: $4 + 5 + 11 = 20$. Channel state:

   • N0→N1's $6$
   • N2→N0's $4$

   Sum: $6 + 4 = 10$. Together: $20 + 10 = 30$ — exactly the true starting total, split correctly between "already reflected in a node's balance" and "still crossing a channel."