1. CPSC 668Set 11: Asynchronous Consensus1 CPSC 668 Distributed Algorithms and Systems Fall 2009 Prof. Jennifer Welch

2. CPSC 668Set 11: Asynchronous Consensus2 Impossibility of Asynchronous Consensus 1.Show impossible in read/write shared memory with n processors and n - 1 faults prove directly: not hard since so many faults implies there is no 2-proc algorithm for 1 fault 2.Show impossible in r/w shared memory with n processors and 1 fault. Two approaches: Reduction: use a hypothetical n-proc algorithm for 1 fault as a subroutine to design a 2-proc algorithm for 1 fault Direct proof: Use similar ideas to n-1 failures case

3. CPSC 668Set 11: Asynchronous Consensus3 Impossibility of Asynchronous Consensus 3Show impossible in message passing with n processors and 1 fault. Two approaches: Reduction: Use a hypothetical message passing algorithm for n procs and 1 fault as a subroutine to design a shared memory algorithm for n procs and 1 fault. This would contradict previous result. Direct approach: Use similar ideas to shared memory case, augmented to handle messages. (Historically, this was the first version that was proven.)

4. CPSC 668Set 11: Asynchronous Consensus4 Modeling Asynchronous Systems with Crash Failures Let f be the maximum number of faulty processors. For both SM and MP: All but f of the processors must take an infinite number of steps in an admissible execution. For MP: Also require that all messages sent must eventually be delivered, except for those sent by a faulty processor in its last step, which might or might not be delivered.

5. CPSC 668Set 11: Asynchronous Consensus5 Wait-Free Algorithms An algorithm for n processors is wait-free if it can tolerate n - 1 failures. Intuition is that a nonfaulty processor does not wait for other processors to do something: it cannot, because it might be the only processor left alive. First result is to show that there is no wait- free consensus algorithm in the asynchronous r/w shared memory model.

6. CPSC 668Set 11: Asynchronous Consensus6 Impossibility of Wait-Free Consensus Suppose in contradiction there is an n- processor algorithm for n - 1 faults in the asynchronous read/write shared memory model. Proof is similar to that showing f + 1 rounds are necessary in the synchronous message passing model. bivalent initial config bivalent config bivalent config bivalent config bivalent config …

7. CPSC 668Set 11: Asynchronous Consensus7 Modified Notion of Bivalence In the synchronous round lower bound proof, valency referred to which decisions are reachable in failure- sparse admissible executions. For this proof, we are concerned with which decisions are reachable in any execution, as long as it is admissible (for the asynchronous shared memory model with up to n - 1 failures).

8. CPSC 668Set 11: Asynchronous Consensus8 Univalent Similarity Lemma (5.15): If C 1 and C 2 are both univalent and they are similar w.r.t. p i, then they have the same valency. Proof: C 1 v-valent p i -only  p i decides v C 2 w-valent  p i decides v

9. CPSC 668Set 11: Asynchronous Consensus9 Bivalent Initial Configuration Lemma (5.16): There exists a bivalent initial configuration Proof is a simpler version of what we did for the synchronous f + 1 round lower bound proof.

10. CPSC 668Set 11: Asynchronous Consensus10 Critical Processors Def: If C is bivalent and i(C) (result of p i taking one step) is univalent, then p i is critical in C. Lemma (5.17): If C is bivalent, then at least one processor is not critical in C, i.e., there is a bivalent extension. Proof: Suppose in contradiction all processors are critical. C bival. j(C) 1-val. i(C) 0-val. pipi pjpj Rest of proof is case analysis of what p i and p j do in their two steps

11. CPSC 668Set 11: Asynchronous Consensus11 Critical Processors Case 1: p i and p j access different registers. j(C) 1-val. pjpj C bival. i(C) 0-val. pipi pjpj pipi Case 2: p i and p j read same register. Same proof.

12. CPSC 668Set 11: Asynchronous Consensus12 Critical Processors Case 3: p i writes to a register R and p j reads from R. C bival. p j reads from R p i writes to R i(C) 0-val j(C) 1-val i(j(C)) 1-val look the same to p i p i writes to R

13. CPSC 668Set 11: Asynchronous Consensus13 Critical Processors Case 4: What if p i and p j both write to the same shared variable? Can "assume away" the problem by assuming we only have single-writer shared variables. Or, can do a similar proof for this case.

14. CPSC 668Set 11: Asynchronous Consensus14 Finishing the Impossibility Proof Create an admissible execution C 0,i 1,C 1,i 2,C 2,… in which all configurations are bivalent. –contradicts termination requirement Start with bivalent initial configuration. Suppose we have bivalent C k. To get bivalent C k+1 : –Let p i k+1 be a proc. that is not critical in C k. –Let C k+1 be i k+1 (C k ).

15. CPSC 668Set 11: Asynchronous Consensus15 Impossibility of 1-Resilient Consensus: Reduction Idea Even if the ratio of nonfaulty processors becomes overwhelming, consensus still cannot be solved in asynchronous SM (with read/write registers). 1.Assume there exists an algorithm A for n processors and 1 failure. 2.Use A as a subroutine to design an algorithm A' for 2 processors and 1 failure. 3.We just showed such an A' cannot exist. 4.Thus A cannot exist.

16. CPSC 668Set 11: Asynchronous Consensus16 Impossibility of 1-Resilient Consensus: Direct Proof Idea Suppose in contradiction there is such an algorithm. Strategy: Construct an admissible execution (at most 1 fault) that never terminates: –show there is a bivalent initial configuration –show how to go from one bivalent configuration to another, forever (so can never terminate) Technically more involved because in constructing this execution, we cannot kill more than one processor.

17. CPSC 668Set 11: Asynchronous Consensus17 Impossibility of Consensus in Message Passing: Reduction Strategy: 1.Assume there exists an n-processor 1- resilient consensus algorithm A for the asynchronous message passing model. 2.Use A as a subroutine to design an n- processor 1-resilient consensus algorithm A' for asynchronous shared memory (with read/write variables). 3.Previous result shows A' cannot exist. 4.Thus A cannot exist.

18. CPSC 668Set 11: Asynchronous Consensus18 Impossibility of Consensus in MP Idea of A': Simulate message channels wth read/write registers. Then run algorithm A on top of these simulated channels. To simulate channel from p i to p j : Use one register to hold the sequence of messages sent over the channel p i "sends" a message m by writing the old value of the register with m appended p j "receives" a message by reading the register and checking for new values at the end

19. CPSC 668Set 11: Asynchronous Consensus19 Randomized Consensus To get around the negative results for asynchronous consensus, we can: –weaken the termination condition: nonfaulty processors must decide with some nonzero probability –keep the same agreement and validity conditions This version of consensus is solvable, in both shared memory and message passing!

20. CPSC 668Set 11: Asynchronous Consensus20 Motivation for Adversary Even without randomization, in an asynchronous system there are many executions of an algorithm, even when the inputs are fixed, depending on when processors take steps, when they fail, and when messages are delivered. To be able to calculate probabilities, we need to separate out variation due to causes other than the random choices Group executions of interest so that each group differs only in the random choices Perform probabilistic calculations separately for each group and then combine somehow

21. CPSC 668Set 11: Asynchronous Consensus21 Adversary Concept used to account for all variability other than the random choices is that of "adversary". Adversary is a function that takes an execution prefix and returns the next event to occur. Adversary must obey admissibility conditions of the revelant model Other conditions might be put on the adversary (e.g., what information it can observe, how much computational power it has)

22. CPSC 668Set 11: Asynchronous Consensus22 Probabilistic Definitions An execution of a specific algorithm, exec( A,C 0,R), is uniquely determined by –an adversary A –an initial configuration C 0 –a collection of random numbers R Given a predicate P on executions and a fixed adversary A and initial config C 0, Pr[P] is the probability of {R : exec( A,C 0,R) satisfies P} Let T be a random variable (e.g., running time). For a fixed A and C 0, the expected value of T is x Pr[T = x] ∑ x is a value of T

23. CPSC 668Set 11: Asynchronous Consensus23 Probabilistic Definitions We define the expected value of a complexity measure to be the maximum over all admissible adversaries A and initial configurations C 0, of the expected value for that particular A and C 0. So this is a "worst-case" average: worst possible adversary (pattern of asynchrony and failures) and initial configuration, averaging over the random choices.

24. CPSC 668Set 11: Asynchronous Consensus24 A Randomized Consensus Algorithm Works in message passing model Tolerates f crash failures –more complicated version handles Byzantine failures Works in asynchronous case –circumvents asynchronous impossibility result Requires n > 2f –this is optimal

25. CPSC 668Set 11: Asynchronous Consensus25 Consensus Algorithm Code for processor p i : Initially r = 1 and prefer = p i 's input 1.while true do 2. votes := get-core( ) 3. let v be majority of phase r votes 4. if all phase r votes are v then decide v 5. outcomes := get-core( ) 6. if all phase r outcome values are w 7. then prefer := w 8. else prefer := common-coin() 9. r := r + 1 ensures a high level of consistency b/w what different procs get uses randomization to imitate tossing a coin

26. CPSC 668Set 11: Asynchronous Consensus26 Properties of Get-Core Executed by n processors, at most f of which can crash. Input parameter is a value supplied by the calling processor. Return parameter is an n-array, one entry per processor Every nonfaulty processor's call to get-core returns. There exists a set C of more than n/2 processors such that every array returned by a call to get-core contains the input parameter supplied by every processor in C.

27. CPSC 668Set 11: Asynchronous Consensus27 Properties of Common-Coin Subroutine implements an f-resilient common coin with bias . Executed by n processors, at most f of which can crash. No input parameter Return parameter is a 0 or 1. Every nonfaulty processor's call to common- coin returns. Probability that a return value is 0 is at least . Probability that a return value is 1 is at least .

28. CPSC 668Set 11: Asynchronous Consensus28 Correctness of Consensus Algorithm For now, don't worry about how to implement get-core and common-coin. Assuming we have subroutines with the desired properties, we'll show –validity –agreement –probabilistic termination (and expected running time)

29. CPSC 668Set 11: Asynchronous Consensus29 Unanimity Lemma Lemma (14.6): if all procs. that reach phase r prefer v, then all nonfaulty procs decide v by phase r. Proof: Since all prefer v, all call get-core with v Thus get-core returns a majority of votes for v Thus all nonfaulty procs. decide v

30. CPSC 668Set 11: Asynchronous Consensus30 Validity If all processors have input v, then all prefer v in phase 1. By unanimity lemma, all nonfaulty processors decide v by phase 1.

31. CPSC 668Set 11: Asynchronous Consensus31 Agreement Claim: If p i decides v in phase r, then all nonfaulty procs. decide v by phase r + 1. Proof: Suppose r is earliest phase in which any proc. decides. p i decides v in phase r all its phase r votes are v p i 's call to get-core( ) returns more than n/2 non-nil entries and all are all entries for procs. in C are

32. CPSC 668Set 11: Asynchronous Consensus32 Agreement Thus every p j receives more than n/2 entries p j does not decide a value other than v in phase r Also if p j calls get-core a second time in phase r, it uses input Every p k gets only as a result of its second call to get-core in phase r p k sets preference to v at end of phase r in round r + 1, all prefer v and Unanimity Lemma implies they all decide v in that round.

33. CPSC 668Set 11: Asynchronous Consensus33 Termination Lemma (4.10): Probability that all nonfaulty procs decide by any particular phase is at least . Proof: Case 1: All nonfaulty procs set preference in that phase using common-coin. –Prob. that all get the same value is at least 2  (  for 0 and  for 1), by property of common-coin –Then apply Unanimity Lemma (14.6)

34. CPSC 668Set 11: Asynchronous Consensus34 Termination Case 2: Some processor does not set its preference using common-coin. –All procs. that don't use common-coin to set their preference for that round have the same preference, v (convince yourself) –Probability that the common-coin subroutine returns v for all procs. that use it is at least . –Then apply the Unanimity Lemma (14.6).

35. CPSC 668Set 11: Asynchronous Consensus35 Expected Number of Phases What is the expected number of phases until all nonfaulty processors have decided? Probability of all deciding in any given phase is at least . Probability of terminating after i phases is (1–  ) i-1 . Geometric random variable whose expected value is 1/ .

36. CPSC 668Set 11: Asynchronous Consensus36 Implementing Get-Core Difficulty in achieving consistency of messages is due to combination of asynchrony and crash possibility: –a processor can only wait to receive n - f messages –the first n - f messages that p i gets might not be from the same set of processors as p j 's first n - f messages Overcome this by exchanging messages three times

37. CPSC 668Set 11: Asynchronous Consensus37 Get-Core First exchange ("round"): –send argument value to all –wait for n - f first round msgs Second exchange ("round"): –send values received in first round to all –wait for n - f second round msgs –merge data from second round msgs Third exchange ("round"): –send values received in second round to all –wait for n - f third round msgs –merge data from third round msgs –return result

38. CPSC 668Set 11: Asynchronous Consensus38 Analysis of Get-Core Lemmas 14.4 and 14.5 show that it satisfies the desired properties (termination and consistency). Time is O(1) (using standard way of measuring time in an asynchronous system)

39. CPSC 668Set 11: Asynchronous Consensus39 Implementing Common-Coin A simple algorithm: Each processor independently outputs 0 with probability 1/2 and 1 with probability 1/2. Bias  = 1/2 n Advantage: simple, no communication Disadvantage: Expected number of phases until termination is 2 n

40. CPSC 668Set 11: Asynchronous Consensus40 A Common Coin with Constant Bias 0 with probability 1/n 1 with probability 1 – 1/n coins := get-core( ) if there exists j s.t. coins[j] = 0 then return 0 else return 1 c :=

41. CPSC 668Set 11: Asynchronous Consensus41 Correctness of Common Coin Lemma (14.12): Common-coin implements a (  n/2  – 1)-resilient coin with bias 1/4. Proof: Fix any admissible adversary that is weak (cannot see the contents of messages) and any initial configuration. All probabilities are calculated with respect to them.

42. CPSC 668Set 11: Asynchronous Consensus42 Probability of Flipping 1 Probability that all nonfaulty processors get 1 for the common coin is at least the probability that they all set c to 1. This probability is at least (1 – 1/n) n When n = 2, this function is 1/4 This function increases up to its limit of 1/e. Thus the probability that all nonfaulty processors get 1 is at least 1/4.

43. CPSC 668Set 11: Asynchronous Consensus43 Probability of Flipping 0 Let C be the set of core processors (whose existence is guaranteed by properties of get- core). If any processor in C sets c to 0, then all the nonfaulty processors will observe this 0 after executing get-core, and thus return 0. Probability at least one processor in C sets c to 0 is 1 – (1 – 1/n) |C|. This expression is at least 1/4 (by arithmetic).

44. CPSC 668Set 11: Asynchronous Consensus44 Summary of Randomized Consensus Algorithm Using the given implementations for get-core and common-coin, we get an asynchronous randomized consensus algorithm for f crash failures with –n > 2f –O(1) expected time complexity expected number of phases is 4 time per phase is O(1)