1. Hwajung Lee

2. -- How many messages are in transit on the internet? --What is the global state of a distributed system of N processes? How do we compute these?

3. Let a $1 coin circulate in a network of a million banks. How can someone count the total $ in circulation? If not counted “properly,” then one may think the total $ in circulation to be one million.

4. Major uses in - deadlock detection - termination detection - rollback recovery - global predicate computation

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

6.  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.

7. 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.

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

9. 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.

10. 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.

11. Lemma 1. No red message is received in a white action.

12. 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) SSS Easy conceptualization of the snapshot state All whiteAll red

13. Let an observer see 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

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

16. 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  upM’ Doesn’t this appear strange? This state was never reached during the computation!

18. 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.

19. 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.

20. 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?