Hwajung Lee

Reaching agreement is a fundamental problem in distributed computing. Some examples are
Leader election / Mutual Exclusion
Commit or Abort in distributed transactions
R eaching agreement about which process has failed
Clock phase synchronization
Air traffic control system: all aircrafts must have the same view
If there is no failure, then reaching consensus is trivial. All-to-all broadcast
Followed by a applying a choice function … Consensus in presence of failures can however be complex.

p0 input u0 p1 p2 p3 u1 u2 u3 output v v v v
Here, v must be equal to the value at some input line.
Also, all outputs must be identical.

Termination . Every non-faulty process must eventually decide.
Agreement.
The final decision of every non-faulty process
Validity.
must be identical.
If every non-faulty process begins with the same initial value v, then their final decision must be v.

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…

Theorem .
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)

Bivalent and Univalent states
A decision state is bivalent , if starting from that state, there exist two 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.

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.

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
0 0 0 0 0 0 …0 1 1
… … … s[n-1] 1 1 1 1 1 1 …1 1 1 {1-valent}
… s[j] is 0-valent s[j+1] is 1-valent
(differ in j th position)
What if process (j+1) crashes at the first step?

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.
bivalent
S action 0
R0 o-valent
Q bivalent
R bivalent
R1
1-valent
The adversary tries to prevent
The system from reaching consensus action 1
U bivalent action 0
T0 o-valent
T bivalent action 1
T1
1-valent
Actions 0 and 1 from T must be taken by the same process p. Why?

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.
bivalent
S action 0
R0 o-valent
Q bivalent
R bivalent action 1
R1
1-valent
U bivalent action 0
T0 o-valent
T bivalent
T1
1-valent
Actions 0 and 1 from T must be action 1 taken by the same process p. Why?

Assume shared memory communication.
Also assume that p ≠ q. Various cases are possible
Case 1.
1-valent e1
T1
Decision =1 q writes
T p reads
T0
0-valent e0
Decision = 0
Such a computation must exist since p can crash at any time
• 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 2. q writes
1-valent
T1 e1
Decision =1
T p writes
T0
0-valent e0
Both write on the same variable , and p writes first.
Decision = 0
• 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 3 q writes
1-valent
T1 p writes
Decision =1
T p writes
T0
0-valent q writes
Z
Decision = 0
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

Similar arguments can be made for communication using the message passing model too (See Lynch’s book). These lead 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!


In a purely asynchronous system, there is no solution to the consensus problem if a single process crashes..

Note that this is true for deterministic algorithms only. Solutions do exist for the consensus problem using randomized algorithm , or using the synchronous model.

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


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
.

Some generals may be "traitors" and therefore send either no input, or send conflicting inputs to prevent the

Devise a strategy, by which every loyal general eventually agrees upon the same plan, regardless of the action of the traitors.

{1, 1, 0,
Attack=1
0
}
0
Attack = 1
1 {1, 1, 0,
1
} traitor
The traitor may send out conflicting inputs
{1, 1, 0,
0
}
2
Retreat = 0
3
{1, 1, 0,
Retreat = 0
0
}
Every general will broadcast his judgment to everyone else.
These are inputs to the consensus protocol.

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
.

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

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.

Using oral messages , no solution to the Byzantine
Generals problem exists with three or fewer generals and one traitor . Consider the two cases: commander 0
1 1 lieutenent 1
0
(a) lieutenant 2 commander 0
1 lieutenent 1
0
1
(b)
0 lieutenant 2

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.

Recursive algorithm
OM(m)
OM(m-1)
OM(m-2)
OM(0)
OM(0) = Direct broadcast
OM(0)

1. Commander i sends out a value v (0 or 1)
2. If m > 0, then every lieutenant j ≠ i , after receiving 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 using
OM(m-1) , and one direct value from the commander. Then he picks the majority of these values as the order from i

1 commander
0
1 1
2
1
1
1 1
3
3
2 2
1
1 0
3
0
1 2
(a)
2
1
1
1 commander
0
0
1
2 2
0 1
3
1
1 0
3
3
1 1 2
(b)

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

Recursive algorithm
OM( m )
OM(m-1
)
OM(m-2)
OM(0)
m= max number of traitors

[1] Commander i sends out a value v (0 or 1)
[2] If m > 0, then every lieutenant j ≠ i , after receiving v, acts as a commander and initiates OM(m-1) with everyone except i .
[3] Every lieutenant, collects
values:
(n-2) values sent by the lieutenants
and
from the commander. Then he picks the majority of these values as the
from i

1 commander
0
1 1
2
1
1
1 1
3
3
2 2
1
1 0
3
0
1 2
(a)
2
1
1
1 commander
0
0
1
2 2
0 1
3
1
1 0
3
3
1 1 2
(b)

OM(2)
Commander
0 v v v v v v
OM(2)
OM(1)
4 5 6
1 v v v
2
2 3 v v v
4 5 v v v
6 v v v
OM(1)
5
6 2 4 6 2 4 5
OM(0) 5 v v
6 v
2 v
6 v
2 v
5
OM(0)

Lemma.
Let the commander be loyal , and n > 2m + k, where m = maximum number of traitors.
Then OM(k) satisfies IC2 loyal commander values received via OM(r) n-m-1 loyal lieutenants m traitors

Proof
If 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 that it holds for k = r + 1 too.
loyal commander
By definition n > 2m+ r+1, so n-1 > 2m+ r
So OM(r) holds for the lieutenants in the bottom row. Each loyal lieutenant will collect n-m-1 identical good values and m bad values. So bad values are voted out ( n-m-1 > m + r implies n-m-1 > m ) values received via OM(r) n-m-1 loyal lieutenants m traitors

Theorem. 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.

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

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.

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.

commander 0 commander 0
1{0} 1{0} lieutenent 1
0{0,2} lieutenant 2 lieutenent 1
1{0}
0{0,2}
1{0,1}
0{0} lieutenant 2 discard
(a) (b)
Using signed messages, byzantine consensus is feasible with 3 generals and 1 traitor

0
V{0}
1
V{0,1}
V{0,1,7}
7
V{0,1,7,4}
4
2

1. 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.

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 of
Each process will contain exactly one message, that was sent by the commander.

Case 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
.




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?

In asynchronous distributed systems, the detection of crash failures is imperfect. This is why, consensus cannot be solved. But how close can we get towards a perfect failure detector ? Two properties are relevant:
Completeness
Every crashed process is eventually suspected .
Accuracy
No correct process is suspected .

crashed
0
1
3
6
5
7
4
2
0 suspects {1,2,3,7} to have failed. Does this satisfy completeness ? Does this satisfy accuracy ?

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)

(These are safety properties)

Transforming 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

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 .

strong completeness strong accuracy weak accuracy ◊ strong accuracy ◊ weak accuracy
Perfect P Strong S ◊P ◊S weak completeness
Weak W
◊W
Perfect P.
(Strongly) Complete and strongly accurate
Strong S . (Strongly) Complete and weakly accurate
Eventually perfect ◊P .
(Strongly) Complete and eventually strongly accurate
Eventually strong ◊S
(Strongly) Complete and eventually weakly accurate

Question 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)

3
4
1
2 input output
Agreed value

{program for process p, t = max number of faulty processes}
initially V p
:= (  ,  ,  , …,  ); D p
:= V p;
{V p
[q] ≠  means, process p thinks q is a suspect}
{Phase 1} for round r p
= 1 to t +1 send (r p
, D p, p) to all; wait to receive (r p
, D q
, q) from all q, {or else q becomes a suspect};
for k = 1 to n V p
[k] =    (r p
, D q
, q): D q
[k] ≠   V p
[k] := D q
[k] end for end for
{at the end of Phase 1, V p for each correct process is identical}
{Phase 2} Final decision value is the first element V p
[j]: V p
[j] ≠ 
Q. What is there is no suspect?

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 , V p 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.

Sends (1, D i then crashes
) and i
Sends (2, D j then crashes
) and ij
Sends (t+1,D l
) l
Completely connected topology l
Sends (t, D k then crashes
) and k l l