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CPSC 668 Distributed Algorithms and Systems

CPSC 668 Distributed Algorithms and Systems. Fall 2006 Prof. Jennifer Welch. Consensus with Byzantine Failures. How many processors total are needed to solve consensus when f = 1 ?

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CPSC 668 Distributed Algorithms and Systems

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  1. CPSC 668Distributed Algorithms and Systems Fall 2006 Prof. Jennifer Welch Set 10: Consensus with Byzantine Failures

  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

  3. B p1 A C p2 p0 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. Set 10: Consensus with Byzantine Failures

  4. 1 1 1 0 B C 0 0 p1 p2 A A p0 p3 p5 p4 C B 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. Set 10: Consensus with Byzantine Failures

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

  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 p2 p0 A act like p5 to p0 in  Set 10: Consensus with Byzantine Failures

  7. The Contradiction • Let  be this execution: What do p0 and p2 decide? act like p1 to p0 in  act like p4 to p5 in  p1 ? 1 0 p2 p0 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

  8. 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 to construct an algorithm for 1 failure and 3 processors, a contradiction. Set 10: Consensus with Byzantine Failures

  9. 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 • p1simulate P1 • p2simulate 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 do the same as the simulated processors, and it will also be correct. Set 10: Consensus with Byzantine Failures

  10. 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

  11. 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

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

  13. 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

  14. 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: value in tree node labeled  if it is a leaf resolve() = majority{resolve( ') : ' is a child of } otherwise (use a default if tied) Set 10: Consensus with Byzantine Failures

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

  16. Resolved Values are Consistent Lemma (5.9): If pi and pj are nonfaulty, then pi 's resolved value for tree node labeled ' j (what pj tells pifor node ') equals what pj stores in its node '. 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

  17. 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

  18. Common Nodes and Frontiers • A tree node  is common if all nonfaulty procs. compute the same value of resolve(). • A tree node  has a common frontier if every path from  to a leaf contains a common node. 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

  19. Agreement • The nodes on each path from a child of the root to a leaf correspond to f + 1 different processors. • Since there are at most f faulty processors, at least on 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

  20. 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

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