1. Distributed Systems CS 15-440
Fault Tolerance – Part II
Lecture 22, November 29, 2017
Mohammad Hammoud

2. Today… Last Session: Fault-tolerance - Part I Today’s Session:
Fault-tolerance – Part II
Handling byzantine failures
Overview
Final Announcement:
The final exam is tomorrow, Nov 30 from 4:00 to 7:00PM in the classroom. It is open book, open notes.

3. Handling Byzantine Failures
Byzantine failures are hard to detect and mask
A server may produce an output it should have never produced (due to an erroneous computation)
Worse yet, a server may be maliciously working (either singlehandedly or collaboratively with other servers) to produce intentionally wrong answers
Byzantine failures can be typically handled using agreement (or consensus) protocols

4. Consensus in Faulty Systems
Reaching a consensus in a distributed system is only possible in the following circumstances:
Lamport’s assumption
Message Ordering
Unordered
Ordered
Process Behavior
Synchronous
Bounded
Communication Delay
Unbounded
Asynchronous
Unicast
Multicast
Message Transmission

5. Byzantine Generals Problem
Lamport’s Assumptions:
There are n processes (or generals)
Each process i will send a value vi (i.e., “attack” or “wait” or something else) to every other process
There are at most m faulty processes (or traitors)
Each process will construct a vector V of length n
If process i is non-faulty, V[i] = vi
Else, V[i] is undefined

6. Byzantine Generals Problem
Case I: n = 4 and m = 1
Step3: Every process
passes its vector to every
other process
Step1: Each process sends
its value to all others
Step2: Each process
collects values received
in a vector
1 Got
(1, 2, y, 4)
(a, b, c, d)
(1, 2, z, 4)
2 Got
(1, 2, x, 4)
(e, f, g, h)
(1, 2, z, 4) 1 1 2
1 Got(1, 2, x, 4)
2 Got(1, 2, y, 4)
3 Got(1, 2, 3, 4)
4 Got(1, 2, z, 4) 2 1 2 4 2 1 x
4 Got
(1, 2, x, 4)
(1, 2, y, 4)
(i, j, k, l) y 4 3 4 4 z
Faulty
process

7. Byzantine Generals Problem
Step 4:
Each process examines the ith element of each of the newly received
vectors
If any value has a majority, that value is put into the result vector
If a value has no majority, the corresponding element of the result vector is
marked UNKNOWN
1 Got
(1, 2, y, 4)
(a, b, c, d)
(1, 2, z, 4)
The algorithm detected the traitor!
2 Got
(1, 2, x, 4)
(e, f, g, h)
(1, 2, z, 4)
4 Got
(1, 2, x, 4)
(1, 2, y, 4)
( i, j, k, l)
Result Vector:
(1, 2, UNKNOWN, 4)
Result Vector:
(1, 2, UNKNOWN, 4)
Result Vector:
(1, 2, UNKNOWN, 4)

8. Byzantine Generals Problem
Case II: n = 3 and m = 1
Step3: Every process
passes its vector to every
other process
Step1: Each process sends
its value to all others
Step2: Each process
collects values received
in a vector 1 1 2
1 Got
(1, 2, y)
(a, b, c)
2 Got
(1, 2, x)
(d, e, f)
1 Got(1, 2, x)
2 Got(1, 2, y)
3 Got(1, 2, 3) 2 1 2 1 x y 3
Faulty
process

9. Byzantine Generals Problem
Step 4:
Each process examines the ith element of each of the newly received
vectors
If any value has a majority, that value is put into the result vector
If a value has no majority, the corresponding element of the result vector is
marked UNKNOWN
The algorithm failed to detect the traitor!
1 Got
(1, 2, y)
(a, b, c)
2 Got
(1, 2, x)
(d, e, f)
Result Vector:
(UNKOWN, UNKNOWN, UNKNOWN)
Result Vector:
(UNKOWN, UNKNOWN, UNKNOWN)

10. Concluding Remarks on the Byzantine Generals Problem
Lamport et al. (1982) proved that in a system with m faulty processes, an agreement can be achieved only if 2m+1 correctly functioning processes are present, for a total of 3m+1
BGP (n, m) is solvable iff n ≥ (3m + 1)
In other words, a consensus is only possible if more than two- thirds of the processes are working properly