1. CSCE 668 DISTRIBUTED ALGORITHMS AND SYSTEMS
Set 10: Consensus with Byzantine Failures
CSCE 668 DISTRIBUTED ALGORITHMS AND SYSTEMS
CSCE 668
Spring 2014
Prof. Jennifer Welch

2. Consensus with Byzantine Failures
How many processors total are needed to solve consensus when f = 1 ?
Suppose n = 2. If p0 has input 0 and p1 has 1, someone has to change, but not both. What if one processor is faulty? How can the other one know?
Suppose n = 3. If p0 has input 0, p1 has input 1, and p2 is faulty, then a tie-breaker is needed, but p2 can act maliciously.
Set 10: Consensus with Byzantine Failures
CSCE 668

3. Processor Lower Bound for f = 1
Theorem (5.7): Any consensus algorithm for 1 Byzantine failure must have at least 4 processors. Proof: Suppose in contradiction there is a consensus algorithm A = (A,B,C) for 3 processors and 1 Byzantine failure. p0 p2 p1 A C B
Set 10: Consensus with Byzantine Failures
CSCE 668

4. Specifying Faulty Behavior
Consider a ring of 6 nonfaulty processors running components of A like this:
This execution  probably doesn't solve consensus (it doesn't have to).
But the processors do something -- this behavior is used to specify the behavior of faulty processors in executions of A in the triangle. p0 A p2 C p1 B p3 p4 p5 1
Set 10: Consensus with Byzantine Failures
CSCE 668

5. Getting a Contradiction
Let 0 be this execution: B p1
p1 and p2
must decide 0
act like p3
to p4 in  C p0 p2
act like p0
to p5 in 
Set 10: Consensus with Byzantine Failures
CSCE 668

6. Getting a Contradiction
Let 1 be this execution: B
p0 and p1
must decide 1 p1
act like p2
to p1 in  1 1 1 p0 p2 A
act like p5
to p0 in 
Set 10: Consensus with Byzantine Failures
CSCE 668

7. The Contradiction Contradiction! Let  be this execution: A C
What do p0 and
p2 decide?
act like p1
to p0 in 
act like p4
to p5 in  p1 ? 1 p0 p2 A C
view of p0 in  = view of p0 in  = view of p0 in 1  p0 decides 1
view of p2 in  = view of p2 in  = view of p2 in 0  p2 decides 0
Contradiction!
Set 10: Consensus with Byzantine Failures
CSCE 668

8. Views β: : 1: A C B C A B A p0 p2 p1 p3 p4 p5 1 1 1 p0 p2 p1 1 ?: p0 p2 p1 C 1 ?
act like p4
to p5 in 
act like p1
to p0 in  A p0 p2 p1 B 1
act like p2
to p1 in 
act like p5
to p0 in  A
1:
Set 10: Consensus with Byzantine Failures
CSCE 668

9. Processor Lower Bound for Any f
Theorem: Any consensus algorithm for f Byzantine failures must have at least 3f+1 processors.
Proof: Use a reduction to the 3:1 case.
Suppose in contradiction there is an algorithm A for
f > 1 failures and n = 3f total processors.
Use A as a subroutine to construct an algorithm for 1 failure and 3 processors, a contradiction to theorem just proved.
Set 10: Consensus with Byzantine Failures
CSCE 668

10. The Reduction
Partition the n ≤ 3f processors into three sets, P0, P1, and P2, each of size at most f.
In the n = 3 case, let
p0 simulate P0
p1 simulate P1
p2 simulate P2
If one processor is faulty in the n = 3 system, then at most f processors are faulty in the simulated system.
Thus the simulated system is correct.
Let the processors in the n = 3 system decide the same as the simulated processors, and their decisions will also be correct.
Set 10: Consensus with Byzantine Failures
CSCE 668

11. Exponential Tree Algorithm
This algorithm uses
f + 1 rounds (optimal)
n = 3f + 1 processors (optimal)
exponential size messages (sub-optimal)
Each processor keeps a tree data structure in its local state
Values are filled in the tree during the f + 1 rounds
At the end, the values in the tree are used to calculate the decision.
Set 10: Consensus with Byzantine Failures
CSCE 668

12. Local Tree Data Structure
Each tree node is labeled with a sequence of unique processor indices.
Root's label is empty sequence ; root has level 0
Root has n children, labeled 0 through n - 1
Child node labeled i has n - 1 children, labeled i : 0 through i : n-1 (skipping i : i)
Node at level d labeled v has n - d children, labeled v : 0 through v : n-1 (skipping any index appearing in v)
Nodes at level f + 1 are leaves.
Set 10: Consensus with Byzantine Failures
CSCE 668

13. Example of Local Tree The tree when n = 4 and f = 1 :
Set 10: Consensus with Byzantine Failures
CSCE 668

14. Filling in the Tree Nodes
Initially store your input in the root (level 0)
Round 1:
send level 0 of your tree to all
store value x received from each pj in tree node labeled j (level 1); use a default if necessary
"pj told me that pj 's input was x"
Round 2:
send level 1 of your tree to all
store value x received from each pj for each tree node k in tree node labeled k : j (level 2); use a default if necessary
"pj told me that pk told pj that pk's input was x"
Continue for f + 1 rounds
Set 10: Consensus with Byzantine Failures
CSCE 668

15. Calculating the Decision
In round f + 1, each processor uses the values in its tree to compute its decision.
Recursively compute the "resolved" value for the root of the tree, resolve(), based on the "resolved" values for the other tree nodes:
resolve() =
value in tree node labeled  if it is a leaf
majority{resolve( ') : ' is a child of } otherwise (use a default if tied)
Set 10: Consensus with Byzantine Failures
CSCE 668

16. Example of Resolving Values
The tree when n = 4 and f = 1 :
(assuming 0 is the default) 1 1 1 1 1 1 1 1
Set 10: Consensus with Byzantine Failures
CSCE 668

17. Resolved Values are Consistent
Lemma (5.9): If pi and pj are nonfaulty, then pi 's resolved value for its tree node labeled ' j (what pj tells pi for node ') equals what pj stores in its node '. ' '
original
value = v
'j
resolved
value = v
part of pi's tree
part of pj's tree
Set 10: Consensus with Byzantine Failures
CSCE 668

18. Resolved Values are Consistent
Proof Ideas:
By induction on the height of the tree node.
Uses inductive hypothesis to know that resolved values for children of the tree node corresponding to nonfaulty procs are consistent.
Uses fact that n > 3f and fact that each tree node has at least n - f children to know that majority of children are nonfaulty.
Set 10: Consensus with Byzantine Failures
CSCE 668

19. Validity Suppose all inputs are v.
Nonfaulty proc. pi decides resolve(), which is the majority among resolve(j), 0 ≤ j ≤ n-1, based on
pi's tree.
Since resolved values are consistent, resolve(j) (at pi) is value stored at the root of pj's tree, which is pj's input value if pj is nonfaulty.
Since there are a majority of nonfaulty processors, pi decides v.
Set 10: Consensus with Byzantine Failures
CSCE 668

20. Common Nodes
A tree node  is common if all nonfaulty procs. compute the same value of resolve().  
same resolved value
part of pi's tree
part of pj's tree
Set 10: Consensus with Byzantine Failures
CSCE 668

21. Common Frontiers
A tree node  has a common frontier if every path from  to a leaf contains a common node. 
pink means
common
Set 10: Consensus with Byzantine Failures
CSCE 668

22. Common Nodes and Frontiers
Lemma (5.10): If  has a common frontier, then  is common. Proof Ideas: By induction on height of . Uses fact that resolve is defined using majority.
Set 10: Consensus with Byzantine Failures
CSCE 668

23. Agreement
The nodes on each path from a child of the root to a leaf correspond to f + 1 different procs.
Since there are at most f faulty processors, at least one such node corresponds to a nonfaulty processor.
This node is common (by the lemma about the consistency of resolved values).
Thus the root has a common frontier.
Thus the root is common (by preceding lemma).
Set 10: Consensus with Byzantine Failures
CSCE 668

24. Complexity Exponential tree algorithm uses n > 3f processors
f + 1 rounds
exponential size messages:
each msg in round r contains
n(n-1)(n-2)…(n-(r-2)) values
When r = f + 1, this is exponential if f is more than constant relative to n
Set 10: Consensus with Byzantine Failures
CSCE 668

25. A Polynomial Algorithm
We can reduce the message size to polynomial with a simple algorithm
The number of processors increases to
n > 4f
The number of rounds increases to
2(f + 1)
Uses f + 1 phases, each taking two rounds
Set 10: Consensus with Byzantine Failures
CSCE 668

26. Phase King Algorithm Code for each processor pi: pref := my input
first round of phase k, 1 ≤ k ≤ f+1:
send pref to all
receive prefs of others
let maj be value that occurs > n/2 times (0 if none)
let mult be number of times maj occurs
second round of phase k:
if i = k then send maj to all // I am the phase king
receive tie-breaker from pk (0 if none)
if mult > n/2 + f
then pref := maj
else pref := tie-breaker
if k = f + 1 then decide pref
Set 10: Consensus with Byzantine Failures
CSCE 668

27. Unanimous Phase Lemma
Lemma (5.12): If all nonfaulty processors prefer v at start of phase k, then all do at end of phase k.
Proof:
Each nonfaulty proc. receives at least n - f preferences for v in first round of phase k
Since n > 4f, it follows that n - f > n/2 + f
So each nonfaulty proc. still prefers v.
Set 10: Consensus with Byzantine Failures
CSCE 668

28. Phase King Validity Unanimous phase lemma implies validity:
Suppose all procs have input v.
Then at start of phase 1, all nf procs prefer v.
So at end of phase 1, all nf procs prefer v.
So at start of phase 2, all nf procs prefer v.
So at end of phase 2, all nf procs prefer v. …
At end of phase f + 1, all nf procs prefer v and decide v.
Set 10: Consensus with Byzantine Failures
CSCE 668

29. Nonfaulty King Lemma
Lemma (5.13): If king of phase k is nonfaulty, then all nonfaulty procs have same preference at end of phase k. Proof: Let pi and pj be nonfaulty. Case 1: pi and pj both use pk's tie-breaker. Since pk is nonfaulty, they both have same preference.
Set 10: Consensus with Byzantine Failures
CSCE 668

30. Nonfaulty King Lemma
Case 2: pi uses its majority value v and pj uses king's tie-breaker.
Then pi receives more than n/2 + f preferences
for v
So pk receives more than n/2 preferences for v
So pk's tie-breaker is v
Set 10: Consensus with Byzantine Failures
CSCE 668

31. Nonfaulty King Lemma
Case 3: pi and pj both use their own majority values.
Suppose pi's majority value is v
Then pi receives more than n/2 + f preferences
for v
So pj receives more than n/2 preferences for v
So pj 's majority value is also v
Set 10: Consensus with Byzantine Failures
CSCE 668

32. Phase King Agreement Use previous two lemmas to prove agreement:
Since there are f + 1 phases, at least one has a nonfaulty king.
Nonfaulty King Lemma implies at the end of that phase, all nonfaulty processors have same preference
Unanimous Phase Lemma implies that from that phase onward, all nonfaulty processors have same preference
Thus all nonfaulty decisions are same.
Set 10: Consensus with Byzantine Failures
CSCE 668

33. Complexities of Phase King
number of processors n > 4f
2(f + 1) rounds
O(n2f) messages, each of size log|V|
Set 10: Consensus with Byzantine Failures
CSCE 668