Consistent cut

1. (aconsistent cutC)  (b happened before a) b  C If this is not true, then the cut is inconsistent Consistent cut A cut is a set of events. b g c a d P1 e m f P2 P3 k h i j Cut 1 Cut 2 (Not consistent) (Consistent)

2. Consistent snapshot The set of states immediately following a consistent cut forms a consistent snapshot of a distributed system. • A snapshot that is of practical interest is the most recent one. Let C1 and C2 be two consistent cuts and C1C2. Then C2 is more recent than C1. • Analyze why certain cuts in the one-dollar bank are inconsistent.

3. Consistent snapshot How to record a consistent snapshot? Note that 1. The recording must be non-invasive 2. Recording must be done on-the-fly. You cannot stop the system.

4. Works on a (1) strongly connected graph (2) each channel is FIFO. An initiator initiates the algorithm by sending out a marker ( ) Chandy-Lamport Algorithm

5. Initially every process is white. When a process receives a marker, it turns red if it has not already done so. Every action by a process, and every message sent by a process gets the color of that process. White and red processes

6. Step 1. In one atomic action, the initiator (a) Turns red (b) Records its own state (c) sends a marker along all outgoing channels Step 2. Every other process, upon receiving a marker for the first time (and before doing anything else) (a) Turns red (b) Records its own state (c) sends markers along all outgoing channels The algorithm terminates when (1) every process turns red, and (2) Every process has received a marker through each incoming channel. Two steps

7. Lemma 1. No red message is received in a white action. Why does it work?

8. Theorem. The global state recorded by Chandy-Lamport algorithm is equivalent to the ideal snapshot state SSS. Hint. A pair of actions (a, b) can be scheduled in any order, if there is no causal order between them, so (a; b) is equivalent to (b; a) Why does it work? All white All red SSS Easy conceptualization of the snapshot state

9. Why does it work? Let an observer observe the following actions: w[i] w[k] r[k] w[j] r[i] w[l] r[j] r[l] …  w[i] w[k] w[j] r[k] r[i] w[l] r[j] r[l] … [Lemma 1]  w[i] w[k] w[j] r[k] w[l] r[i] r[j] r[l] … [Lemma 1]  w[i] w[k] w[j] w[l] r[k] r[i] r[j] r[l] … [done!] Recorded state

10. Let us verify that Chandy-Lamport snapshot algorithm correctly counts the tokens circulating in the system Example 1. Count the tokens token C token no token A no token token no token B A no token token no token C B Are these consistent cuts? How to account for the channel states? Use sent and received variables for each process.

11. Another example of distributed snapshot: Communicating State Machines

12. Something unusual Let machine i start Chandy-lamport snapshot before it has sent M along ch1. Also, let machine j receive the marker after it sends out M’ along ch2. Observe that the snapshot state is down  up M’ Doesn’t this appear strange? This state was never reached during the computation!

13. Understanding snapshot

14. Understanding snapshot The observed state is a feasible state that is reachable from the initial configuration. It may not actually be visited during a specific execution. The final state of the original computation is always reachable from the observed state.

15. Discussions What good is a snapshot if that state has never been visited by the system? - It is relevant for the detection of stable predicates. - Useful for checkpointing.

16. Discussions What if the channels are not FIFO? Study how Lai-Yang algorithm works. It does not use any marker LY1. The initiator records its own state. When it needs to send a message m to another process, it sends a message (m, red). LY2. When a process receives a message (m, red), it records its state if it has not already done so, and then accepts the message m. Question 1. Why will it work? Question 1 Are there any limitations of this approach?

17. Questions Distributed snapshot = distributed read. Distributed reset = distributed write How difficult is distributed reset?