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BYZANTINE AGREEMENT

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Byzantine generals have to agree on attack or retreaval ... Messengers are sent to each other camps ... Impossibility Results ... – PowerPoint PPT presentation

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Title: BYZANTINE AGREEMENT


1
BYZANTINE AGREEMENT
2
The Byzantine generals problem
  • Turkish invasion into Byzantium
  • Byzantine generals have to agree on attack or
    retreaval
  • The enemy works by corrupting the soldiers
  • Byzantine generals are notoriously treacherous
    ...
  • The loyal generals have to prevent traitors from
    spoiling a coordinated attack
  • Messengers are sent to each other camps
  • Orders are distributed by exchange of messages,
    corrupt soldiers violate protocol at will
  • But corrupt soldiers cant intercept and modify
    messages between loyal troops
  • The gong sounds slowly there is ample time for
    loyal soldiers to exchange messages (all to all)

3
Byzantine Agreement
  • Commanding General commands other generals
  • If all loyal generals attack victory is certain
  • If none attack, the Empire survives
  • If some attack, Empire is lost
  • Gong keeps time
  • but they dont need toall attack at once

4
Consensus problem
  • With a leader leader gives an order, like
    attack, and non-faulty participants either
    attack or do nothing, despite some limited number
    of failures
  • Byzantine Agreement (agreement about a
    value)
  • atomic broadcast
  • Without a leader participants have an initial
    vote protocol runs and eventually all non-faulty
    participants chose the same outcome, and it is
    one of the initial votes (typically, 0 or 1)
  • Fault-tolerant Consensus (distributed
    consensus)
  • maintaining replicated data
  • monitoring a distributed computation
  • detecting failed processor (sensors)

5
The formal setting of the consensus problem
  • There are M processors P p1, ..., pM that are
    trying to reach consensus.
  • A subset F of the processors are faulty, and the
    remaining processors are nonfaulty.
    Each processor pi ? P stores a
    value Vi.
  • During the consensus protocol, the processors
    calculate an consensus value Ai.
  • After the protocol ends, the following two
    conditions should hold

6
  • 1. For every pair pi and pj of nonfaulty
    processors,
  • Ai Aj. This value is the consensus
    value.
  • (Every correct node chooses the same
    value)
  • 2. The consensus value is a function of the
    initial values
  • Vi of the nonfaulty processors (P -
    F).
  • ( If all the correct nodes have the
    same input, that
  • input must be the value chosen)
  • Conditions
  • synchronous system
  • receivers can reliably identify the sender

7
Byzantine Agreement ( reliable broadcast )
  • each general broadcasts reliably its opinion Vj
    to all other generals
  • all generals will learn other generals opinions
    and decide for the same action
  • BA is equivalent to the consensus
  • each loyal general can agree on the opinion Vj
    of every other general pj
  • it means, they can agree on the consensus value

8
Single broadcast
  • we will consider only a single broadcast in the
    following
  • commanding general is broadcasting (can be
    treacherous)
  • all others are lieutenant generals
  • broadcast is reliable if
    interactive consistency conditions hold
  • If sender pS is loyal, the loyal generals will
    agree on VS
  • If the sender pS is treacherous, the loyal
    processors will agree on the same value for VS

9
Impossibility Results
  • Let t be the maximum number of faulty processes
    that our protocol is supposed to tolerate
  • To reach agreement
  • At least 3t1 generals must be present.
  • At least t1 rounds of communication
  • Same result holds for fault-tolerant consensus in
    the Byzantine model
  • Byzantine agreement is not possible with fewer
    than 3t1 processes
  • No solution exists for 3 generals in the presence
    of a single traitor.

10
Consensus of 3 processes
  • Example 3 processes, 1 is faulty (A, B, C)
  • Non-faulty processes start with input 0 and 1,
    respectively
  • They exchange messages
    each now has a set of
    inputs 0, 1, x, where x comes from C
  • C sends 0 to A and 1 to B
  • A has 0, 1, 0 and wants to pick 0. B has 0,
    1, 1 and wants to pick 1. What to do?
  • Can we prove that consensus is impossible with
    just 3 processes?

11
Scenario A - 3 generals with a single traitor
  • 1 commanding general and 2 lieutenants
  • one of them is treacherous

A B 1, 0 gt 0 C 0, 1 gt 0
A gt 1 B 1, 0 gt 0 C 1, 1
12
Theorem
If M lt 3t, the system cannot reach
agreement. ---------------------------------------
--------------------------- M generals t
traitors 3 groups of processes one of the groups
contains all treacherous processes

S
U
T
13
scenario 0
the
general broadcasts 0
processes in U send out
the same
messages as in the
1. scenario, to processes in T
gt
assume that the agreement
works and correct processes
in T
decide correctly on 0
S
0
0
0
0
1
U
T
0
0
14
scenario 1
the
general broadcasts 1
processes in T send out
the same messages as
in the
0. scenario, to processes in U
gt processes in U must
decide on
1 to achieve the correct result

S
1
1
1
1
1
U
T
0
1
15
scenario 0, 1
The interaction between T and U is the same in
both 0 and 1 scenario. The difference makes
only S, that changes (defines) an input and
therefore the output
S
0, 1
0, 1
0, 1
0, 1
1
U
T
0
1
0
16
scenario 2
the
commanding general
is treacherous
T
receives the same messages
as in 0. case and decides on
0 U
receives the same messages
as in 1. case and decides on
1 Now U and T see the same messages as in the
previous scenarios gt they make the same
decisions as before, but now the decisions are
contradictory
S
0
1
0
1
1
U
T
1
0
0
17
Scenario B 4 Generals and a single traitor
G
  • A B C
  • ------------------
  • 1 1 1
  • 1,0 1,0 1,1
  • Output(1,1,1)

1
1
1
1
1
C
B
A
0
1
1
0
  • there are enough loyal generals to reach a
    consensus opinion
  • a loyal general gets a correct value from the
    commanding general and the other loyal lieutenant
    general one possibly wrong value

18
Scenario B 4 Generals and a single traitor
G
  • A B C
  • ------------------
  • 1 1 0
  • 1,0 1,0 1,1
  • Output(1,1,1)

0
1
1
1
1
C
B
A
0
1
1
0
  • all loyal generals get the same set of values
    from each other and they decide for the same
    output (majority)
  • Agreement Protocol works for N 3t 1.

19
BG(k) protocol
  • k ... the maximum number of traitors that can be
    tolerated (t . . . the real number of traitors)
  • M gt 3k and t lt k
  • majority consensus (default value)
  • processors communicate with each other to
    determine the majority decision
  • all lieutenants become commanding generals after
    receiving the commanders order
  • their broadcast does not have to be reliable as
    opinions of disloyal lieutenants may be discarded

20
  • BG(k) protocol is recursive
  • lieutenants perform BG(k - 1) that tolerates k
    traitors
  • time-out -gt default value
  • senders of messages are known
  • protocol works synchronously in rounds

Majority and default decisions.
Majority(v1,v2, ..., vn) Return the majority
v among v1,v2, ..., vn , or Retreat if no
majority exists
21
Byzantine Generals broadcast
Base case for a Byzantine Generals broadcast
BG_Send(0,v,l) The commander broadcasts v to
every lieutenant in l. BG_Receive(0) Return
the value they sent to you or Retreat if no
message is received.
22
Byzantine Generals broadcast
General case for Byzantine Generals broadcast
BG_Send(k,v,l) send v to every lieutenant
on l BG_Receive(k) let v be the value sent
to you or Retreat if no value is sent let
l be the set of lieutenants who have never
broadcast v BG_Send(k-1,v,l-Self) Use
BG_Receive(k-1) to receive v(i) for every i in
l-Self Return majority(v, v(1), ..., v(l
- 1))
23
An execution of BG(2)
C
The commander broadcasts his order using BG(2)
L1
L2
L3
L4
L5
L6
L1
BG(1)
L1v
L2
L3
L4
L5
L6
L2
BG(0)
L2L1v
L3
L4
L5
L6
24
Lemma 1
  • For any t and k, if the commanding general is
    loyal, the BG(k) protocol is correct if there are
    no more that t traitors and at least 2t k
    generals.
  • Proof (by induction)
  • BG(0) works
  • assume BG(k-1) works for M gt 2t k - 1, k
    gt 0, t lt k
  • consider BG(k) for M 2t k

25
  • commander broadcasts his order
  • lieutenants use BG(k-1) for M 2t k -1
    to rebroadcast
  • 1. lieutenant is loyal gt BG(k-1) works
    according to the I.H.
  • 2. lieutenant is treacherous gt sends
    different orders to different lieutenants
  • loyal lieutenants compute majority of orders of
    all other lieutenants
  • t k - 1 values are correct t values may be
    incorrect
  • t k - 1 gt t as k gt 1

26
Lemma 2
  • For any k, the BG(k) protocol is correct if there
    are more than 3k generals and no more that k
    traitors.
  • Proof (by induction)
  • BG(0) works
  • assume BG(k-1) works for M gt 3(k - 1) ,
    t lt k - 1
  • consider BG(k) for M 3t 1 t k

27
  • 1. commander is loyal and M 3k 2t k gt
    BG(k) works according to Lemma 1
  • 2. commander is treacherous
  • sends different orders
  • lieutenants rebroadcast the orders using BG(k
    -1)
  • there is t lt k - 1 traitors and M gt 3k
    generals gt BG(k - 1) works according to I.H.
  • even if rebroadcasting lieutenant is treacherous
    all loyal l. will compute the same value for his
    order
  • gt they will have the same input for the majority
    function
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