The Byzantine Generals Problem

prepared by Ilya Kolchinsky

n generals, communicating through messengers some of the generals (up to m) might be traitors all loyal generals should decide on the same

plan of action a small number of traitors cannot cause the loyal

generals to adopt a bad plan every general sends v(i)

◦ every loyal general must obtain exactly same v(i)’s◦ if the i-th general is loyal, the value that he sends must

be used by all loyal generals as v(i)

Problem description

synchronous network (why?) for now: every general can directly

communicate with any other general (fully-connected network)

reduction to the problem of a single general and multiple lieutenants◦ why is it sufficient?

Problem assumptions

A commanding general must send an order to his n-1 lieutenant generals such that interactive consistency holds in presence of at most m traitors:◦ IC1: all loyal lieutenants obey the same order◦ IC2: if the commanding general is loyal, then all

loyal lieutenants obey the order he sends If the commander is loyal, IC2 is enough

Byzantine Generals Problem definition

Examine a case of three generals, one traitor

The only possible messages are “attack” and “retreat”

Sounds simple?

Is it at all possible?

Three generals, single traitor

Commander

Lieutenant

Lieutenant

attack attack

he said “retreat”

if ((X received from C) and (Y received from L)): execute X

Commander

Lieutenant

attackretreat

he said “attack”

he said “retreat”

if ((X received from C) and (Y received from L)): execute X

Retrea

t!

Attack!

No solution with fewer than 3m+1 generals can cope with more than m traitors

Proof: by contradiction and reduction to 3-generals problem◦ assume that a solution exists for 3m generals or

fewer◦ make each of the 3 generals simulate

approximately one third of 3m generals◦ since at most one general can be a traitor, there

are at most m traitors among 3m generals◦ reduction solves the problem of 3 generals, a

contradiction.

Impossibility Results

All generals execute the same algorithm◦ sending messages◦ receiving messages◦ performing local computations

Assumptions◦ every message that is sent is delivered correctly◦ the receiver of the message knows who sent it◦ the absence of the message can be detected

A solution with oral messages

majority function, capable of selecting a majority value from a set of multiple values◦

a default value to assume in case of a traitor which chooses not to send messages◦ must be consistently returned by majority if no

majority value exists

OM(m) - requirements

vvvvmajority n ,,, 21

OM(0):1. the commander sends its value to every lieutenant2. each lieutenant uses the value he receives from the

commander, or the default value if no value is received OM(m), m > 0:

1. the commander sends its value to every lieutenant2. for each i, lieutenant i sends the previously obtained

value to n-2 other lieutenants using OM(m-1)3. For each i, and each j≠i, let Vj be the value lieutenant i

received from lieutenant j in previous step (using OM(m-1)). Lieutenant i uses the value

OM(m)

nvvmajority ,,1

ExampleC

L

L

LL

L

L attack

attack

attack

attack

attack

attack

retreatattack

attackretreat

attackattackretreatattackattackretreatattack

attackretreatattackattackretreatattack

attackretreatattackattackretreatattack

attackretreatattackattackretreatattackattack

attack attack

attack

retreatretreatretreatretreatretreatretreat

retreatretreatretreatretreatretreatretreat

retreatretreatretreatretreatretreatretreat

retreatretreatretreatretreatretreatretreatretreat

retreat retreat

retreat

attackretreatattackattackretreatattack

attackretreatattackattackretreatattack

attackretreatattackattackretreatattack

attackretreatattackattackretreatattackattack

attack attack

attack

attackretreatattackattackretreatattack

attackretreatattackattackretreatattack

attackretreatattackattackretreatattack

attackretreatattackattackretreatattackattack

attack attack

attack

retreatretreatretreatretreatretreatretreat

retreatretreatretreatretreatretreatretreat

retreatretreatretreatretreatretreatretreat

retreatretreatretreatretreatretreatretreatretreat

retreat retreat

retreat

attackretreatattackattackretreatattack

attackretreatattackattackretreatattack

attackretreatattackattackretreatattack

attackretreatattackattackretreatattackattack

attack attack

attack

attack

attack attack

attack

m= 21010101010102

Why would this example fail if there were 3 traitors?

Why is a single iteration insufficient and two are required?

Example - questions

Lemma: For any m and k, OM(m) satisfies IC2 if there are more than 2k+m generals and at most k traitors

Proof: by induction on m◦ m=0: OM(0) works if the commander is loyal◦ assume the claim holds for m-1,m>0◦ in step (2), each lieutenant i receives value v and applies

OM(m-1) with n-1 generals n > 2k + m ⇒ n-1 > (2k + m) – 1 = 2k + (m-1) by induction hypothesis, each loyal lieutenant gets the value v

◦ there are at most k traitors and n-1 > 2k + (m-1)≥ 2k, majority of the lieutenants are loyal

OM(m) correctness

vvvmajority n ,,1

Theorem: For any m, Algorithm OM(m) satisfies conditions IC1 and IC2 if there are more than 3m generals and at most m traitors.

Proof: by induction on m◦ m=0: no traitors, hence OM(0) satisfies IC1 and IC2◦ assume the claim holds for m-1,m>0◦ if the commander is loyal, conditions hold by the

Lemma◦ if the commander is a traitor

there are more than 3m-1 lieutenants, at most m-1 are traitors

by induction hypothesis, OM(m-1) satisfies IC1 and IC2 any two loyal lieutenants get the same vector of values by determinism of majority function, any two loyal

lieutenants obtain the same value v

OM(m) correctness

Number of messages?◦ (n-1)(n-2)…(n-m-1)

Time required?◦ m+1

OM(m) complexity

Introducing new assumptions:◦ a loyal general's signature cannot be forged, and

any alteration of the contents of his signed messages can be detected

◦ anyone can verify the authenticity of a general's signature

A solution with signed messages

The algorithm requires a choice function which receives a set of values and satisfies the following properties:◦ if the set consists of a single element v, choice

returns v◦ if the set is empty, choice returns the default

value

SM(m) requirements

1. For each i: 2. The commander signs and sends his value to

every lieutenant3. For each lieutenant i:

A. If a message of the form v:0 received from the commander and no order has been received yet, then1. 2. Send v:0:i to every other lieutenant

B. If a message of the form and , then1. 2. if k < m, send the message to every

lieutenant other than

4. When lieutenant i will receive no more messages, he obeys the order

SM(m)

iV

vVi

kjjv :::0: 1 iVv

vVV ii ijjv k ::::0: 1

kjj ,,1

iVchoice

Example

C

L1

L2

L3

attack : 0

attack : 0

retreat : 0

attack : 0 : 1

attack : 0 : 2attack : 0 : 2

retreat : 0 : 3

retreat : 0 : 3

retreat : 0 : 3 : 2

attack : 0 : 2 : 3

V2}{=

V3}{=

V2={attack}

V3={retreat}V3={retreat,attac

k}

V3={attack,retreat}

Order = CHOICE(attack,retreat)

Theorem: For any m, Algorithm SM(m) solves the Byzantine Generals problem if there are at most m traitors.

IC2:◦ each loyal lieutenant receives commander’s order◦ no additional order can be received by any loyal

lieutenant◦ hence, for each lieutenant the set will consist

of a single order, which he will obey by property of choice function

SM(m) correctness

iV

IC1 follows from IC2 if the commander is loyal

Assume the commander is a traitor Lemma (see proof below): if lieutenant i

puts an order v into , then lieutenant j must put an order v into .

For each two loyal lieutenants i,j, the sets are the same

Each two loyal lieutenants i,j obey the same order, which completes the proof

SM(m) correctness (contd.)

iV

jV

ji VV ,

Case 1: i receives the order v:0 in step 3a◦ in this case, i sends it to j in step 3a(2), hence j receives it

• Case 2: i receives the order in step 3b, and j is one of the◦ in this case, j must already have received the order v

• Case 3: i receives the order in step 3b, j is not one of the and k<m◦ in this case, i sends the message to j, so j must

receive the order v• Case 4: i receives the order in step 3b, j is not one

of the and k=m◦ since the commander is a traitor, at most m-1 of the

lieutenants are traitors◦ hence, at least one of the lieutenants is loyal◦ this loyal lieutenant must have sent j the value v when he first

received it, so j must therefore receive that value

SM(m) correctness – proof of the lemma

kjjv :::0: 1 rj

kjjv :::0: 1 rj

ijjv k ::::0: 1

kjjv :::0: 1 rj

kjj ,,1

removing an assumption regarding the fully-connected network

what is the new upper bound on number of traitors?

General Network Topology

every processor knows the topology of the network every message includes the route through which it

is supposed to be delivered a reliable processor validates the route

◦ drops a message if received from a wrong neighbor◦ forwards a message to a neighbor only if it appears next

to itself in the route Definition: t-connected graph is a graph in which

there exist at least t disjoint paths between every pair of nodes

Claim: if t >= 2m+1 and there are more than 3m processors, the problem is solvable

Assumptions for BGP in general networks

Definition: Let be the set of copies of the commander’s order received by the lieutenant i. Let be a set of lieutenants. A set is called a set of suspicious lieutenants determined by i if every message that did not pass through lieutenants in carries the same value.

Algorithm 1. If a set of up to m suspicious lieutenants exists, then

the purified value is the value of the messages that did not pass through . If no message is left, the default value is returned.

2. If there is no set of cardinality up to m, then the default value is returned.

The Algorithm raa ,1

iU

iU

ia

iU

iaaPurifying r ;,1 iU

iU

iU

BG(0):1. the commander sends its value to every lieutenant via

2m+1 disjoint paths2. each lieutenant applies the purifying algorithm on the

received values to find the value the commander intended to deliver

BG(k), k > 0:1. the commander sends its value to every lieutenant via

2m+1 disjoint paths2. each lieutenant applies the purifying algorithm on the

received values to find the value the commander intended to deliver

3. for each i, lieutenant i sends the previously obtained value to n-2 other lieutenants using BG(k-1)

4. For each i, and each j≠i, let Vj be the value lieutenant i received from lieutenant j in previous step (using BG(k-1)). Lieutenant i uses the value

The Algorithm (contd.)

nvvmajority ,,1

Example

CL1

L2

L3

L4

attack

attack

attack

attack

retreat

retreat

attackattack

attackattack

attackattack

attack

attack

attack

CL1

attack

attack

attack

attack

retreat

retreat

attack

L2

attack

retreat

attack attack

attack

attack attack

attack

attack

attack

attack

L4retreat

retreat

retreatL3

attack

attack

attackattack

Lemma: by use of Purifying algorithm every lieutenant can obtain the order sent by a loyal commander in presence of m traitors, if the connectivity of the network is at least 2m+1

Proof:◦ let be the set of all the copies of the commander’s

order that lieutenant i receives◦ there are at most m faulty lieutenants, therefore at most m

copies might be lost. This implies that the number of received copies, r, is at least m+1, a majority

◦ at least m+1 of the messages are relayed through routes which contain only reliable lieutenants

◦ hence, at least m+1 of the received copies carry the original order

◦ by definition of the Purifying Algorithm, only up to m independent copies can be eliminated, there are m+1 of independent copies of the correct value, hence they cannot be eliminated, which completes the proof.

Proof of correctness

raa ,,1

The rest of the proof is identical to the proof of correctness for OM(m), with an additional usage of the above lemma

Proof of correctness (contd.)

Solutions exist for Byzantine Generals Problem under various hypotheses

These solutions are expensive◦ in amount of time◦ in amount of messages required

The main reason is the requirement of achieving reliability in presence of arbitrary malfunctioning

Assumptions can be changed and agreements can be relaxed

Conclusions

L. Lamport, R. Shostak, M. Pease, “The Byzantine Generals Problem”, TPLS 4(3), 1982.

Dolev, D. “The Byzantine generals strike again” J. Algorithms 3, 1 (Jan. 1982).

References