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Distributed Computing 8. Impossibility of consensus Shmuel Zaks zaks@cs.technion.ac.il ©

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2 Consensus Input: 1 or 0 to each processor Output: Agreement: all procssors decide 0 or 1 Termination: all processors eventually decide Validity: if all inputs x, then decide x

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The result: No completely asynchronous consensus protocol can tolerate even a single unannounced process death. e.g., the stopping of a single process at inappropriate time can cause any distributed commit protocol to fail to reach agreement.

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Motivation This problem serves a role that is similar to the one served by “the halting problem” in computability theory. Many problems equivalent to consensus (or reduce to it)

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Protocols in the industry How commit protocols in the industry deal with this outcome ? Weaken an assumption. For example: Computation model: e.g., assume bounded – delay network Computation model: e.g., assume bounded – delay network Fault model: e.g., assume faults only at start. Fault model: e.g., assume faults only at start.

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6 The Model Message System Message System Reliable Reliable Delivers all messages correctly Delivers all messages correctly Exactly once Exactly once Processing Model Completely Asynchronous No Assumptions about relative speeds No Assumptions about relative speeds Unbounded time in delivering message Unbounded time in delivering message

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Weak Consensus Every process starts with initial value in {0,1} A nonfaulty process decides on a value in {0,1} by entering an appropriate decision state All nonfaulty process that decide are required to choose the same value (note: termination not required) Both 0 and 1 are possible decision values, although perhaps for different initial configurations. (Trivial solutions – e.g., “0” - are ruled out) 7

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8 System Model Communicate by means of one global message buffer Atomic step Attempt to receive a message Perform local computation Send arbitrary but finite set of messages

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Consensus Protocol N processes (N > 1) Each process has x p – one-bit input register y p – output register with values in {b,0,1} Unbounded amount of internal storage PC – Program counter 9

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10 Consensus Protocol N processes (N > 1) process p x p 0/1 y p 0/1/b memory(unboundd)PC input register output register memory Program counter

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11 Fixed starting valued at the memory (except the input register) Output register starts with b The output register is “write once” when a value is written to the output register, the process is “in a decision state”. Process acts deterministically according to a Transition function

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12 Communication System A message is a pair (p,m) p is the name of the destination m is a “message value” message buffer Maintains messages that have been sent but not yet delivered We assume a clique topology

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13 two operations by a process : send (p,m) – place (p,m) in the message buffer ( “message (p,m) is sent to process p”) receive (p) delete a message (p,m) from the message buffer and returns m ( “message (p,m) is received”) OR returns (message buffer unchanged)

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Message system nondeterministic. For fairness we will require in the proof: if receive(p) is performed infinitely many times, then each message (p,m) in the message buffer is eventually delivered. Now: the message system can return an infinite number of times in response to receive(p), even though a message (p,m) is in the message buffer. Note: Assume a clique topology

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15 (P 1,M) Message Buffer (P 0,M’) (P 2,M’’) (P 1,M’’’) Process 0 Process 2 Process 1 receive(0) (P 0,M’)

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16 (P 1,M) Message Buffer Process 0 Process 2 Process 1 receive(1) (P 2,M’’) (P 1,M’’’) send(2,m) (P 2,m)

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17 Configurations A configurations consists of Internal state of each process Contents of the message buffer initial configuration each process p starts with x p =0 or x p =1 the message buffer is empty

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step – consists of a primitive step by a single process p. phase 1 – receive(p) is performed phase 2 – p enters a new internal state and sends a finite set of messages A step is completely determined by the pair e = (p,m), called an event. 18

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19 event e = (p,m) (“receipt of m by p”). step of a single process p: receive(p) is performed ( p receives m) p enters a new internal state p sends a finite set of messages event and step: event: syntax step: semantic

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20 Events and Schedules e(C) – denotes the resulting configuration (“e can be applied to C”) The event (p, ) can always be applied A schedule from C is a finite/infinite sequence of events that can be applied from C. The associated sequence of steps is called a run. one: event - step many: schedule - run

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21 If a schedule is finite, (C) denotes the resulting configuration C’, which is “reachable from C “. C’ is accessible if it is reachable from an initial configuration.

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22 Lemma 1 (‘commutativity’) Lemma 1 : Suppose that from some configuration C, the schedules 1, 2 lead to configurations C 1,C 2, respectively. If the sets of processes taking steps in 1 and 2, respectively, are disjoint, then 2 can be applied to C 1, and 1 can be applied to C 2, and both lead to the same configuration C 3.

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23 C2C2C2C2 C0C0C0C0 C1C1C1C1 C3C3C3C3 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 when 1 and 2 contain a single event (p,m) event when 1 and 2 contain a single event (p,m) event

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24 (P 1,M 1 ) (P 2,M 2 ) (P 1,M 1 ) 1111 2222 1111 2222 The message buffer of C 3 The message buffer of C 1 The message buffer of C 2 The message buffer of C 0 Message buffer

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25 P 1 Internal state - A P 2 Internal state - X P 1 Internal state - B P 2 Internal state - Y P 1 Internal state - B P 2 Internal state - X P 1 Internal state - A P 2 Internal state - Y 1111 2222 1111 2222 All other processors – change unchanged states

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26 C2C2C2C2 C0C0C0C0 C1C1C1C1 C3C3C3C3 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 when 1 and 2 contain a single event (p,m) event - ok when 1 and 2 contain a single event (p,m) event - ok when 1 and 2 contain any run – use induction when 1 and 2 contain any run – use induction

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27 A configuration C has a decision value v if some process p is in a decision state with y p = v (v =0 or v=1). if some process p is in a decision state with y p = v (v =0 or v=1). A consensus protocol is partially correct if it satisfies two conditions: 1. No accessible configuration has more than one decision value. 2. For each v {0,1}, some accessible configuration has decision value v. good news - it is non trivial - sometimes it decides - it never decides incorrectly bad news - termination not guaranteed - what about delivering all messages? -what about failures?

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28 A process p is nonfaulty in a run if it takes steps. It is faulty otherwise. A process p is nonfaulty in a run if it takes steps. It is faulty otherwise. bad news: a process can be declared faulty only at !! A run is admissible if A run is admissible if - at most one process is faulty, and - at most one process is faulty, and - all messages sent to non-faulty - all messages sent to non-faulty processes are eventually received. processes are eventually received.

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29 A run is deciding if some process reaches a decision state. A run is deciding if some process reaches a decision state. A consensus protocol is totally correct in spite of one fault if it is: A consensus protocol is totally correct in spite of one fault if it is: partially correct, and partially correct, and every admissible run is a deciding run. every admissible run is a deciding run.

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30 Theorem: No consensus protocol is totally correct in spite of one fault. No consensus protocol is totally correct in spite of one fault.

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31 Sketch of Proof: Assume that P is totally correct in spite of one fault. sssshow an initial configuration from which each decision is still possible ( Lemma 2 ) sssshow that from such a configuration one can always reach another similar configuration ( Lemma 3 ) cccconclude – by induction – with an admissible run that never decides – a contradiction.

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32 Let C be a configuration and let V be the set of decision values of configurations reachable from C. C is bivalent if |V| = 2 C is bivalent if |V| = 2 C is univalent if |V| = 1 C is univalent if |V| = 1 if V = {0} then C is 0-valent if V = {0} then C is 0-valent if V = {1} then C is 1-valent if V = {1} then C is 1-valent (Note: |V|≠0, since P is totally correct) Theorem: No consensus protocol is totally correct in spite of one fault. totally correct in spite of one fault. Proof: Assume that P is totally correct in spite of one fault. We will reach a contradiction.

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33 0-valent configuration From now on: 1-valent configuration 2-valent configuration Unknown

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34 Proof: Assume there is no bivalent initial configuration. But P is partially correct. So, there are both 0-valent and 1-valent initial configurations. initial configurations. Lemma 2: P has a bivalent initial configuration.

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35...... bivalentconfiguration initial configurations C

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36 C0C0C0C0...... 0-valentconfiguration C1C1C1C1 initial configurations 1-valentconfiguration

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37 Two initial configurations are called adjacent if they differ only in the initial value of a single process. 0 1 0 1 1 0 1 0 1 0 x 0 x 1 x 2 x 3 x 4

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38 Claim: Claim: There exist a 0-valent initial configuration C 0 adjacents to a 1-valent initial configuration C 1.

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39 0 1 0 1 1 1 1 0 1 1 1 1 0 1 0 1 1 0 0 0 1 0 0 0 0 x 0 x 1 x 2 x 3 x 4 C0C0 C1C1 Proof by example: 0-valent 1-valent

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40 So: So: There exist a 0-valent initial configuration C 0 adjacents to a 1-valent initial configuration C 1. p Let p be the process in whose initial value they differ

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41 P is a consensus protocol that is totally correct in spite of one fault. P is a consensus protocol that is totally correct in spite of one fault. Consider an admissible deciding run (with schedule ) from C 0 in which process p takes no steps. Consider an admissible deciding run (with schedule ) from C 0 in which process p takes no steps. can be applied to C 1 can be applied to C 1 The two corresponding configurations are identical, except for the internal state in p The two corresponding configurations are identical, except for the internal state in p Both runs reach the same decision x Both runs reach the same decision x

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42 x = 1 C 0 is bivalent x = 0 C 1 is bivalent Contradiction. C1C1C1C1 C0C0C0C0 C’ C’’ Decision: x x 0-valent 1-valent Lemma 2: P has a bivalent initial configuration. So, we proved:

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43 Lemma 3: Let: C be a bivalent configuration of P, C be a bivalent configuration of P, e = (p,m) be an event that is applicable to C. e = (p,m) be an event that is applicable to C. S be the set of configurations reachable from S be the set of configurations reachable from C without applying e, and C without applying e, and D = e(S) = {e(E)| E S and e is applicable to E}. D = e(S) = {e(E)| E S and e is applicable to E}. Then, D contains a bivalent configuration.

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44 E e2e2e2e2 e1e1e1e1 e4e4e4e4 e i ≠ e bivalent configuration e e e e S e D=e(S) e e5e5e5e5 e6e6e6e6 e7e7e7e7 C Need to prove: D contains a bivalent configuration

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45 Note: e =(p,m) is applicable to C so: message (p,m) is in the message buffer, so: e is applicable to every E S.

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46 Prove by contradiction Assume that D contains no D=e(S) e i ≠ e e e e e S e e C 0-valent 1-valent

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47 Step 1: Claim: D contains both and 0-valent 1-valent So: every configuration d D is or The proof has three steps.

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48 S e D=e(S) D0D0D0D0 D1D1D1D1 e e=(p,m) Step 1

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49 C is bivalent There exist E i,, i=0,1, i-valent configurations reachable from C. e i ≠ e e e e e S e D=e(S) e C

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50 L et F 0 = e (E 0 ). E0E0E0E0 0 e2e2e2e2 e1e1e1e1 e4e4e4e4 e i ≠ e bivalent configurati on F0F0F0F0 e e e e S e D=e(S) e e5e5e5e5 e6e6e6e6 e7e7e7e7 C 0-valent 1-valent so: D contains

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51 e was applied in reaching E 0 so, either E 0 is in D, or there exists F 0 D from which E 0 is reachable. e2e2e2e2 e1e1e1e1 e4e4e4e4 e i ≠ e bivalent configurati on e e e e S e D=e(S) e e5e5e5e5 e6e6e6e6 e7e7e7e7 F0F0F0F0 E0E0E0E0 C 0-valent 1-valent so: D contains

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52 So: F i is i-valent (not bivalent) One of E i and F i is reachable from the other. both and So, we know that D contains 0-valent 1-valent End of step 1 Start of step 2

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53 Step 2 Claim: There exist C 0, C 1 S such that: C 0 and C 1 are neighbors ( C 1 = e’(C 0 ), e’=(p’,m’) ) D 0 = e(C 0 ) is D 1 = e(C 1 ) is (two configurations neighbors if one results from the other in a single step.) 0-valent 1-valent

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54 S e D=e(S) D0D0D0D0 D1D1D1D1 e’ C1C1C1C1 C0C0C0C0 e e=(p,m) e’=(p’,m’) Step 2

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55 e(C) is or. Suppose it is. There is in D. It has predecessor in S. e(C) S D=e(S) e(C) C e e 0-valent 1-valent

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56 Consider the path in S from C to the predecessor of e(C) S e D=e(S) e e(C) C 0-valent 1-valent

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57 Applying e to each configuration on this path, we get a configuration in D, which is or. bivalent configurati on S e D=e(S) e e(C) e e e C e

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58 So we get two configurations C 0 and C 1, that are neighbors in S; i.e., there is e’ s.t. S e D=e(S) e(C) D0D0D0D0 D1D1D1D1 e’ C1C1C1C1 C0C0C0C0 C e

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59 So, we proved the claim: There exist C 0, C 1 S such that: C 0 and C 1 are neighbors ( C 1 = e’(C 0 ), e’=(p’,m’) ) D 0 = e(C 0 ) is D 1 = e(C 1 ) is End of step 2 Start of step 3

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60 D 1 = e’(D 0 ) by Lemma 1 Case 1 : Case 1 : p’ ≠ p contradiction S e D=e(S) e(C) D0D0D0D0 D1D1D1D1 e’ C1C1C1C1 C0C0C0C0 C e Step 3: get to a contradiction Recall: e=(p,m)

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61 S e D=e(S) D0D0D0D0 D1D1D1D1 e’ C1C1C1C1 C0C0C0C0 e e=(p,m) e’=(p’,m’) p’ p Case 2 : Case 2 : p’ = p recall:

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62 C1C1C1C1 C0C0C0C0 D0D0D0D0 D1D1D1D1 A Case 2 : Case 2 : p’ = p e - deciding run from C 0 in which p takes no steps A = (C 0 ) deciding run 1-valent 0-valent e e’ e e E0E0E0E0 E1E1E1E1 A is a deciding run. But it cannot be and it cannot be. a contradiction !!!

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63 Lemma 3: Let: C be a bivalent configuration of P, C be a bivalent configuration of P, e = (p,m) be an event that is applicable to C. e = (p,m) be an event that is applicable to C. S be the set of configurations reachable from S be the set of configurations reachable from C without applying e, and C without applying e, and D = e(S) = {e(E)| E S and e is applicable to E}. D = e(S) = {e(E)| E S and e is applicable to E}. Then, D contains a bivalent configuration. Lemma 2: P has a bivalent initial configuration. So, we proved:

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64 Any deciding run from a bivalent initial configuration goes to univalent configuration, so there must be some single step that goes from a bivalent to univalent configuration. We construct a run that avoids such a step: bivalent configuration deciding run bivalent configuration … univalent configuration end of proof:

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65 we construct an infinite non-deciding run bivalent configuration non-deciding run bivalent configuration … …

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66 Start with a bivalent initial configuration ( Lemma 2) The run constructed in stages. Every stage starts with a bivalent configuration and ends with a bivalent configuration A queue of processes, initially in arbitrary order Message buffer is ordered according to the time messages were sent

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67 In each stage: C is a bivalent configuration that the stage starts with. Suppose that process p heads the queue Suppose that m is the earliest message to p in the message buffer if any (or otherwise) e = (p,m)

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68 By Lemma 3 there is a bivalent configuration C’ reachable from C by a schedule in which e is the last event. After applying this schedule: move p to the back of the queue

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Message Buffer Message Buffer eeee P3P3P3P3 P2P2P2P2 P1P1P1P1 P0P0P0P0 (P 1,M) (P 0,M) (P 2,M) (P 3,M) P0P0P0P0 P3P3P3P3 P2P2P2P2 P1P1P1P1 (P 1,M) (P 2,M) (P 3,M) P1P1P1P1 P0P0P0P0 P3P3P3P3 P2P2P2P2 (P 2,M) (P 3,M) P2P2P2P2 P1P1P1P1 P0P0P0P0 P3P3P3P3 P3P3P3P3 P2P2P2P2 P1P1P1P1 P0P0P0P0

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70 in any infinite sequence of stages every process takes infinitely many steps every process receives every message sent to it Therefore, the constructed run is admissible never reaches a univalent configuration The protocol never reaches a decision The protocol is not totally correct in spite of one fault. contradiction

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Conclusion Theorem: No consensus protocol is totally correct in spite of one fault. Proof Construct the run that was shown before which is an admissible run which never reaches a univalent configuration The protocol never reaches a decision The protocol is not totally correct in spite of one fault.

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72 Conclusion Theorem: No consensus protocol is totally correct in spite of one fault. hw: which process fails in the infinite run that was constructed for the proof?

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Main lesson: In an asynchronous system, there is no way to distinguish between a faulty process and a slow process.

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74 One importance lesson: In an asynchronous system, there is no way to distinguish between a faulty process and a slow process. Other tasks not solvable with one faulty processor: Input graph – connected Output graph - disconnected Many extensions and uses

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Other tasks not solvable with one faulty processor: Input graph – connected Output graph - disconnected

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Extensions 1 fault t faults

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Non-Asynchronous Models Synchronous f+1 rounds if f failures Asynchronous plus eventual sychrony eventual synchronized clocks eventual message delivery bound d Some communication links good Consensus terminates: O((f+4)*d) after stabilization Asynchronous Consensus 1 stop failure - impossible Initially crash failures - possible

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Other Consensus Problems Weak Consensus k-set Consensus Approximate Consensus Byzantine failures What if some nodes lie? (Non-Asynchronous Models) Synchronous model f stopping failures, n nodes 2f+1 ≤ n

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Failure Detectors Assume total asynchrony Assume failure detector service Notifies node i when node j fails Eventually… Allow solving consensus Weakest failure-detector? Leader-election failure-detector (Non-Asynchronous Models)

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References M. Fischer, N. Lynch, M. Paterson, Impossibility of distributed consensus with one faulty processor, 1985 O. Biran, S. Moran and S. Zaks, A Combinatorial Characterization of the Distributed Tasks Which Are Solvable in the Presence of One Faulty Processor, J. of Algorithms, 1990.

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