分布式系统和算法：Distributed Snapshot

1 .Distributed Snapshot 

2 . Think about these -- 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 . One-dollar bank 2 (2,0) (1,2) 1 0 (0,1) 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 .Importance of snapshots Major uses in - deadlock detection - termination detection - rollback recovery - global predicate computation 

5 . Consistent cut A cut is a set of events. If a cut C is consistent then (a ∈C)∧(bp a) ⇒ b∈C If this is not true, then the cut C is inconsistent time 

6 . Consistent snapshot The set of states immediately following the events (actions) in 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 . 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. 

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

9 . White and red processes 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. So, white action = action by a white process red action = action by a red process white message = message sent by a white process red message = message sent by a red process 

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

12 . Why does it work? All white All red SSS Easy conceptualization of the snapshot state 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) 

13 . 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 

14 . Example 1: Count the tokens Let us verify that Chandy-Lamport snapshot algorithm correctly counts the tokens circulating in the system 2 token no token token C A no token token no token B A no token token no token C 1 3 B Are these consistent cuts? How to account for the channel states? Use sent and received variables for each process. 

15 . Example 2: Communicating State Machines up up ch1 send receive send receive M i j M' M' M ch2 down down state machine state machine i j global state i ch1 j ch2 S0 down φ down φ S1 up M down φ S2 up M up M' S3 down M up φ 

16 . 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! 

17 .Understanding snapshot S0 j sends M' i sends M recorded state SSS S1 S1' j receives M j sends M' i receives M' S2' i sends M S2 j receives M i receives M' S3' S3 i receives M' j receives M S0 

18 . 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. 

19 . 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. 

20 . 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? 

21 . Food for thought Distributed snapshot = distributed read. Distributed reset = distributed write How difficult is distributed reset? 

22 .Distributed debugging (Marzullo and Neiger, 1991) e, VC(e) observer Distributed system 

23 . Distributed debugging Sij is a global state after the ith action by process 0 and the jth action by process 1 

24 . Distributed debugging Possibly ϕ: At least one consistent global state S is reachable from the initial global state, such that φ(S) = true. Definitely ϕ: All computations pass through some consistent global state S such that φ(S) = true. Never ϕ: No computation passes through some consistent global state S such that φ(S) = true. Definitely ϕ ⇒Possibly ϕ Possibly ϕ 

25 . Examples ϕ = x+y =12 (true at S21) Possibly ϕ ϕ = x+y > 15 (true at S31) Definitely ϕ ϕ = x=y=5 (true at S40 and S22) Never ϕ *Neither S40 nor S22 are consistent states*