ITEC452Distributed Computing

Lecture 12Distributed Consensus

Hwajung Lee

Distributed Consensus

Reaching agreement is a fundamental problem in distributed computing. Some examples are

Leader election / Mutual ExclusionCommit or Abort in distributed transactions

Reaching agreement about which process has failed

Clock phase synchronizationAir traffic control system: all aircrafts must have

the same viewIf there is no failure, then reaching consensus is trivial. All-to-all broadcastFollowed by a applying a choice function … Consensus in presence of failures can however be complex.

Problem Specification

u3

u2

u1

u0 v

v

v

v

Here, v must be equal to the value at some input line.Also, all outputs must be identical.

input output

p0

p1

p2

p3

Problem Specification

Termination. Every non-faulty process must eventually decide.Agreement. The final decision of every non-faulty process

must be identical. Validity. If every non-faulty process begins with the same

initial value v, then their final decision must be v.

Asynchronous Consensus

Seven members of a busy household decided to hire a cook, since they do not have time to prepare their own food. Each member separately interviewed every applicant for the cook’s position. Depending on how it went, each member voted "yes" (means “hire”) or "no" (means “don't hire”).

These members will now have to communicate with one another to reach a

uniform final decision about whether the applicant will be hired. The process will be repeated with the next applicant, until someone is hired.

Consider various modes of communication…

Asynchronous ConsensusTheorem.In a purely asynchronous distributed system, the consensus problem is impossible to solve if even a single process crashes

Famous result due to Fischer, Lynch, Patterson(commonly known as FLP 85)

Proof

Bivalent and Univalent states

A decision state is bivalent, if starting from that state, there existtwo distinct executions leading to two distinct decision values 0 or

1.Otherwise it is univalent.

A univalent state may be either 0-valent or 1-valent.

Proof

Lemma. No execution can lead from a 0-valent to a 1-valent state or vice versa.

Proof. Follows from the definition of 0-valent and 1-valent states.

Proof

Lemma. Every consensus protocol must have a bivalent initial state.

Proof by contradiction. Suppose not. Then consider the following input patterns:

s[0] 0 0 0 0 0 0 …0 0 0 {0-valent)0 0 0 0 0 0 …0 0 1 s[j] is 0-valent0 0 0 0 0 0 …0 1 1 s[j+1] is 1-

valent… … … … (differ in jth

position)s[n-1] 1 1 1 1 1 1 …1 1 1 {1-valent}

What if process (j+1) crashes at the first step?

Proof

Lemma.

In a consensus protocol,

starting from any initial

bivalent state, there must

exist a reachable bivalent

state T, such that every

action taken by some process

p in state T leads to either a

0-valent or a 1-valent state.

Q

S R U T

T0 T1R0 R1

action 0 action 0action 1 action 1

o-valent 1-valent o-valent 1-valent

bivalent

bivalent

bivalent bivalent bivalent

Actions 0 and 1 from T must be taken by the same process p. Why?

The adversary tries to prevent The system from reaching

consensus

Proof of FLP (continued)

Lemma.

In a consensus protocol,

starting from any initial bivalent

state I, there must exist a

reachable bivalent state T,

such that every action taken by

some process p in state T leads

to either a 0-valent or a 1-valent

state.

Q

S R U T

T0 T1R0 R1

action 0 action 0action 1 action 1

o-valent 1-valent o-valent 1-valent

bivalent

bivalent

bivalent bivalent bivalent

Actions 0 and 1 from T must be taken by the same process p. Why?

Proof of FLP (continued)

T

T0

T1 Decision =1

Decision = 0p reads

q writes

e1

e0

• Starting from T, let e1 be a computation that excludes any step by p.• Let p crash after reading.Then e1 is a valid computation from T0 too.To all non-faulty processes, these two computations are identical, but the outcomes are different! This is not possible!

Case 1. 1-valent

0-valent

Assume shared memory communication. Also assume that p ≠ q. Various cases are possible

Such a computation must existsince p can crash at any time

Proof (continued)

T

T0

T1 Decision =1

Decision = 0p writes

q writes

e1

e0

Both write on the same variable, and p writes first.

• From T, let e1 be a computation that excludes any step by p.• Let p crash after writing.Then e1 is a valid computation from T0 too.

To all non-faulty processes, these two computations are identical, but the outcomes are different!

Case 2. 1-valent

0-valent

Proof (continued)

T

T0

T1

Z

Decision =1

Decision = 0p writes

q writes

Let both p and q write, but on different variables.

Then regardless of the order of these writes, both computations lead

to the same intermediate global state Z, which must be univalent.

Is Z 1-valent or 0-valent? Both are absurd

Case 3

0-valent

1-valent

p writes

q writes

Proof (continued)

Similar arguments can be made for communication usingthe message passing model too (See Lynch’s book). Theselead to the fact that p, q cannot be distinct processes, and p = q. Call p the decider process.

What if p crashes in state T? No consensus is reached!

Conclusion In a purely asynchronous system, there is no solution to

the consensus problem if a single process crashes..

Note that this is true for deterministicalgorithms only. Solutions do exist for theconsensus problem using randomized algorithm, or using the synchronous model.

Byzantine Generals Problem

Describes and solves the consensus problem on the synchronous model of communication.

- Processor speeds have lower bounds and communication delays have upper bounds.

- The network is completely connected- Processes undergo byzantine failures, the worst

possible kind of failure

Byzantine Generals Problem n generals {0, 1, 2, ..., n-1} decide about whether to "attack" or

to "retreat" during a particular phase of a war. The goal is to agree upon the same plan of action.

Some generals may be "traitors" and therefore send either no

input, or send conflicting inputs to prevent the "loyal" generals from reaching an agreement.

Devise a strategy, by which every loyal general eventually agrees upon the same plan, regardless of the action of the traitors.

Byzantine Generals

0

32

1

Attack = 1Attack=1

Retreat = 0 Retreat = 0

{1, 1, 0, 0}

{1, 1, 0, 0}

Every general will broadcast his judgment to everyone else.These are inputs to the consensus protocol.

{1, 1, 0, 1}

{1, 1, 0, 0}

traitor

The traitormay send out conflicting inputs

Byzantine Generals

We need to devise a protocol so that every peer

(call it a lieutenant) receives the same value from

any given general (call it a commander). Clearly,

the lieutenants will have to use secondary information.

Note that the roles of the commander and the

lieutenants will rotate among the generals.

Interactive consistency specifications

IC1. Every loyal lieutenant receives the same order from the commander.

IC2. If the commander is loyal, then every loyal lieutenant receives the order that the commander sends.

commander

lieutenants

The Communication Model

Oral Messages

1. Messages are not corrupted in transit.2. Messages can be lost, but the absence of message

can be detected.3. When a message is received (or its absence is

detected), the receiver knows the identity of the sender (or the defaulter).

OM(m) represents an interactive consistency protocol in presence of at most m traitors.

An Impossibility Result

commander 0 commander 0

lieutenent 1 lieutenant 2 lieutenent 1 lieutenant 2

1 1

0

1 0

0

1

(a) (b)

Using oral messages, no solution to the ByzantineGenerals problem exists with three or fewer generals and one traitor. Consider the two cases:

Impossibility result

Using oral messages, no solution to the Byzantine Generals problem exists with 3m or fewer generals and m traitors (m > 0).

Hint. Divide the 3m generals into three groups of m generals each, such that

all the traitors belong to one group. This scenario is no better than the case of

three generals and one traitor.

The OM(m) algorithmRecursive algorithm

OM(m)

OM(m-1)

OM(m-2)

OM(0)

OM(0) = Direct broadcast

OM(0)

The OM(m) algorithm

1. Commander i sends out a value v (0 or 1)

2. If m > 0, then every lieutenant j ≠ i, afterreceiving v, acts as a commander and initiates OM(m-1) with everyone except i .

3. Every lieutenant, collects (n-1) values: (n-2) values sent by the lieutenants usingOM(m-1), and one direct value from the commander. Then he picks the majority of these values as the order from i

Example of OM(1)

1 1

(a)

0

1 22 3

2 3 3 1 1 2

1

1 1 1 1 0 0

1 1

0

1 22 3

2 3 3 1 1 2

1 1 0 0 1 1

0

(b)

commander commander

The OM(m) algorithm

Recall what the oral message model is.

Recall the two interactive consistency criteria IC1 & IC2.

Byzantine agreement algorithm for the Oral Message model of communication with at most m traitors

The OM(m) algorithm

Recursive algorithm

OM(m)

OM(m-1)

OM(m-2)

OM(0)

OM(0) = Direct broadcast

m= max number of traitors

The OM(m) algorithm

[1] Commander i sends out a value v (0 or 1)

[2] If m > 0, then every lieutenant j ≠ i, afterreceiving v, acts as a commander and initiates OM(m-1) with everyone except i .

[3] Every lieutenant, collects (n-1) values: (n-2) values sent by the lieutenants usingOM(m-1), and one direct value from the commander. Then he picks the majority of these values as the order from i

Example of OM(1)

1 1

(a)

0

1 22 3

2 3 3 1 1 2

1

1 1 1 1 0 0

1 1

0

1 22 3

2 3 3 1 1 2

1 1 0 0 1 1

0

(b)

commander commander

Example of OM(2)

0

1 2

2

3 4 5 6

4 5 6 5 6 2 4 6 2 4 5

Commander

OM(2)

OM(1)

v v vvv v

v v v v v v v v v v v v

5 6 2 6 2 5OM(0)

v v vv v v

OM(2)

OM(1)

OM(0)

Proof of OM(m)

Lemma.Let the commander beloyal, and n > 2m + k,where m = maximumnumber of traitors.

Then OM(k) satisfies IC2m traitors

n-m-1 loyal lieutenants

values received via OM(r)

loyal commander

Proof of OM(m)ProofIf k=0, then the result trivially holds.

Let it hold for k = r (r > 0) i.e. OM(r)satisfies IC2. We have to show thatit holds for k = r + 1 too.

By definition n > 2m+ r+1, so n-1 > 2m+ rSo OM(r) holds for the lieutenants in the bottom row. Each loyal lieutenant

willcollect n-m-1 identical good values andm bad values. So bad values are votedout (n-m-1 > m + r implies n-m-1 > m)

m traitors

n-m-1 loyal lieutenants

values received via OM(r)

loyal commander

The final theoremTheorem. If n > 3m where m is the maximum number of traitors, then OM(m) satisfies both IC1 and IC2.

Proof. Consider two cases:Case 1. Commander is loyal. The theorem follows from the previous lemma (substitute k = m).Case 2. Commander is a traitor. We prove it by induction. Base case. m=0 trivial. (Induction hypothesis) Let the theorem hold for m = r. We have to show that it holds for m = r+1 too.

Proof (continued)

There are n > 3(r + 1) generals and r + 1 traitors. Excluding the commander, there are > 3r+2 generals of which there are r traitors. So > 2r+2 lieutenants are loyal. Since 3r+ 2 > 3.r, OM(r) satisfies IC1 and IC2

> 2r+2 r traitors

Proof (continued)

In OM(r+1), a loyal lieutenant chooses the

majority from (1) > 2r+1 values obtained

from the loyal lieutenants via OM(r),

(2) the r values from the traitors, and

(3) the value directly from the commander. > 2r+2 r traitors

The set of values collected in part (1) & (3) are the same for all loyal lieutenants –

it is the same set of values that these lieutenants received from the commander.

Also, by the induction hypothesis, in part (2) each loyal lieutenant receives

identical values from each traitor. So every loyal lieutenant collects the same set of values.

Solution using signed messages

A signed message satisfies all the conditions of oral message, plus two extra conditions

Signature cannot be forged. Forged message are detected and discarded.

Anyone can verify its authenticity of a signature.

Signed messages improve resilience.

Example

commander 0 commander 0

lieutenent 1 lieutenant 2 lieutenent 1 lieutenant 2

(a) (b)

1{0} 1{0}

0{0,2}

1{0}0{0}

0{0,2}

1{0,1}

discard Using signed messages, byzantine consensus is feasible with 3 generals and 1 traitor

Signature list

0

1

7 2

4

V{0} V{0,1}

V{0,1,7}

V{0,1,7,4}

The SM(m) algorithm1. Commander i sends out a signed message v{i} to each lieutenant j ≠

i

2. Lieutenant j, after receiving v{S}, appends it to a set V.j, only if (i) it is not forged, and (ii) it has not been received before.

3. If the length of S is less than m+1, then lieutenant j (i) appends his own signature to S, and (ii) sends out the signed message to every other lieutenant

whose signature does not appear in S. 4. Lieutenant j applies a choice function on V.j to make the final

decision.

Theorem of signed messages

If n ≥ m + 2, where m is the maximum number of traitors, then SM(m) satisfies both IC1 and IC2.

Case 1. Commander is loyal. The bag ofEach process will contain exactly

one message, that was sent by the

commander.

Theorem of signed messagesCase 2. Commander is traitor.

The signature list has a size (m+1), and there are m traitors, so

at least one lieutenant signing the message must be loyal.

Every loyal lieutenant i will receive every other loyal lieutenant’s

message. So, every message accepted by j is also accepted by

i and vice versa. So V.i = V.j.

Concluding remarks

The signed message version tolerates a larger number (n-2) of faults.

Message complexity however is the same in both cases

Failure detector

Recall FLP’85 impossibility result. If crash failure can be detected, then consensus is trivial!

In synchronous systems with bounded delay channels, crash failures can definitely be detected using timeouts.

But what about asynchronous systems? Can we design a failure detector for purely asynchronous distributed systems?

Failure detectors for asynchronous systems

In asynchronous distributed systems, the detection ofcrash failures is imperfect. This is why, consensus cannotbe solved. But how close can we get towards a perfectfailure detector? Two properties are relevant:

Completeness Every crashed process is eventually suspected.

AccuracyNo correct process is suspected.

Example

0

6

1 3

5

247

0 suspects {1,2,3,7} to have failed. Does this satisfy

completeness? Does this satisfy accuracy?

crashed

Classification of completeness

Strong completeness. Every crashed process is eventually suspected by every correct process, and remains a suspect thereafter.

Weak completeness. Every crashed process is eventually suspected by at least one correct process, and remains a suspect thereafter.

(These are liveness properties)

Classification of accuracy

Strong accuracy. No correct process is ever suspected.

Weak accuracy. There is at least one correct process that is never suspected.

(These are safety properties)

Transforming completenessTransforming Weak completeness into strong completeness

Program strong completeness (program for process i};define D: set of process ids (representing the suspects);initially D is generated by the weakly complete failure detector; do true

send D(i) to every process j ≠ i;receive D(j) from every process j ≠ i;D(i) := D(i) - D(j);if j - D(i) D(i) := D(i) \ j fi

od

Eventual accuracy

Accuracy is a safety property. A failure detector is eventually strongly accurate, if there exists a time T after which no correct process is suspected. (Before that time, a correct process be added to and removed from the list of suspects any number of times)

A failure detector is eventually weakly accurate, if there exists a time T after which at least one process is no more suspected.

Classifying failure detectors

Perfect P. (Strongly) Complete and strongly accurateStrong S. (Strongly) Complete and weakly accurateEventually perfect ◊P. (Strongly) Complete and eventually strongly accurateEventually strong ◊S (Strongly) Complete and eventually weakly accurate

strong completeness

weak completeness

strong accuracy weak accuracy ◊ strong accuracy ◊ weak accuracy

Perfect P Strong S ◊P ◊S

Weak W ◊W

MotivationQuestion 1. Given a failure detector of a certain type,

how can we solve the consensus problem?

Question 2. How can we implement these classes of failure detectors in asynchronous distributed systems?

Question 3. What is the weakest class of failure detectors that can solve the consensus problem?

(Weakest class of failure detectors is closer to reality)

Consensus using failure detector

input output

1 2 3 4

Agreed value

Consensus using P{program for process p, t = max number of faulty processes}

initially Vp := (-, -, -, …, -); Dp := Vp;

{Vp[q] ≠ - means, process p thinks q is a suspect}

{Phase 1} for round rp= 1 to t +1

send (rp, Dp, p) to all;

wait to receive (rp, Dq, q) from all q, {or else q becomes a suspect};

for k = 1 to n Vp[k] = - - - (rp, Dq, q): Dq[k] ≠ - Vp[k] := Dq[k] end for

end for

{at the end of Phase 1, Vp for each correct process is identical}

{Phase 2} Final decision value is the first element Vp[j]: Vp[j] ≠ -

Q. What is there is no suspect?

Understanding consensus using P

It is possible that a process p sends out the first message

to q and then crashes. If there are n processes and t of

them crashed, then after at most (t + 1) asynchronous

rounds, Vp for each correct process p becomes identical,

and contains all inputs from processes that may have

transmitted at least once. The choice function leads to

a unique decision.

Understanding consensus using P

i ij k

l l

ll

Sends (1, Di) and then crashes

Sends (2, Dj) and then crashes

Sends (t, Dk) and then crashes

Sends (t+1,Dl)

Completely connected topology