Distributed Computing On Topics of: Leader election + Byzantine algorithms)
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This document discusses algorithms for leader election and Byzantine agreement in distributed systems. It presents 3 algorithms for leader election in synchronous rings with non-anonymous processes. The most efficient uses bidirectional message passing and has O(n log n) complexity. Byzantine agreement, where some nodes may behave maliciously, is discussed for complete graphs. An algorithm is presented that works when n > 3f, where n is the number of nodes and f is the maximum faulty nodes. Additional conditions are needed for incomplete graphs.
Distributed Computing On Topics of: Leader election + Byzantine algorithms)
- 1. Distributed Computing On Topics of: Leader election + Byzantine algorithms) SLIDES BY OSAMA ASKOURA EECS 6117 Distributed Computing Set2
- 3. Slide 2 We define leader election: By the end of algorithm’s execution, only one node identifies as “leader” All other nodes identify as “follower” Leader Election in anonymous rings is impossible. Leader Election in Model Ring Topology Synchronous Non Anonymous Processes Anonymous ring Non-Anonymous ring #1 #3#9 #7 #5 #2 #3 #4
- 5. Slide 2 Leader election algorithm 1: • send id CCW; • Forward msgs until you receive own id back; • If (my id was smallest seenbefore own id comes back ) • Then I’m“leader” • Else • Not leader “follower” Total #messages =n2 Model Ring Topology Synchronous Non Anonymous Processes Non-Anonymous ring #1 #3#9 #7 #5 #2 #3 #4
- 6. Slide 2 Leader election algorithm 2: • Candidate =true; • loop • send id CCW and CW (bi-directional) • Ifreceived id< own id • Candidate =false • Exit loop (“follower”) • If got own idback • I’m“leader” • Exit loop Total #messages =2n log(n) Model Ring Topology Synchronous Non Anonymous Processes Bi-directional msgs Non-Anonymous ring #1 #3#9 #7 #5 #2 #8 #4
- 7. Leader election algorithm 2: Proof that total #messages = 2n log(n): • Every cycle each node sends 2 msgs 2n messages • Every cycle you eliminate half log(n) cycles • Total =2n log(n) Model Ring Topology Synchronous Non Anonymous Processes Bi-directional msgs Non-Anonymous ring #1 #3#9 #7 #5 #2 #8 #4 #1 #3#9 #7 #5 #2 #8 #4 #1 #9 #7 #5 #2 #8 #3 #4 #1 #9 #7 #5 #2 #8 #3 #4 #1 #9 #7 #5 #2 #8 #3 #4
- 8. Slide 2 Leader election algorithm 3: • Wait until CLOCK = n *id • Declare yourself leader • Send everybody telling you are leader Total #messages =n But what if id’s are very large numbers or IPaddresses? • Time complexity grows big • Cannot compare IPto CLOCK Model Ring Topology Synchronous Non Anonymous Processes #1 #3#9 #7 #5 #2 #8 #4
- 9. Slide 2 CLAIM: Every Leader Election algorithm in this model uses Ω( n log(n) ) messages PROOF (Contradiction): Suppose, exists algorithm that uses < n log(n) messages Let A be that L.E algorithm Transform A into A’ so that in A’ every process outputs minimum id’s So, A’ is 1. Elect leader; 2. Leader sends message around round to find smallest id 3. Leader sends smallest id to all Model Ring Topology Synchronous Non Anonymous Processes #1 #3#9 #7 #5 #2 #8 #4
- 10. Slide 2 If A uses A(n) messages, then A’ uses A(n) + 2n messages We will show that A(n) + 2n ≥ c. n log(n) i.e. A(n) ≥ c. log(n) Base n = 1: at least 0 messages m(n) : m(1) = 0 m(n) = 2 m(n) + 𝒏 𝟐 Two rings proof Model Ring Topology Synchronous Non Anonymous Processes #1 #3#9 #7 #5 #2 #8 #4
- 11. Byzantine Agreement in complete graphs
- 12. We look at the problem where a group of processes or nodes need to agree on an output. In Byzantine agreements, some nodes are allowed to be malicious. i.e. stay alive but not follow their algorithm. This problem arises in distributed systems with sensors that can be faulty; such as airplane systems Byzantine Process Correct Processes
- 13. Agreement Property All outputs of correct processes are same Validity Property If all inputs are “v”, then all outputs of correct processes are “v” Byzantine Process Correct Processes
- 14. Examples for Agreement with n=2 #failures = 0 Solution: each process send their input to each other; output AND of the inputs (ensures both output same value) Model Non Anonymous Processes ( Proc IDs) Message Passing System Complete Network Synchronous No cryptography (used in reality to avoid byzantine/malicious nodes)
- 15. Examples for Agreement with n=2 #failures = 1 (upto1failure:wedon’tknowifanywillfail) Solution Attempt: output whatever your input is; No communication in between. - This satisfies validity property (if all input was v; all output is v) - But doesn’t satisfy agreement property. How?: So if n=2 & f =1, it’s impossible to solve Model Non Anonymous Processes ( Proc IDs) Message Passing System Complete Network Synchronous No cryptography (used in reality to avoid byzantine/malicious nodes) v1 v1 v1 v1 1 1 0 0
- 16. Examples for Agreement with n=3 #failures = 1 (upto1failure:wedon’tknowifanywillfail) Solution Attempt: Round 1: send to your neighbors your input Round 2: ask other nodes what others sent them? Problem: Bad guy can send inconsistent answers to bias their decisions and violate agreement or validity property. n = 3 and f = 1 is impossible Model Non Anonymous Processes ( Proc IDs) Message Passing System Complete Network Synchronous No cryptography (used in reality to avoid byzantine/malicious nodes) 1 1 01 0 0 0 1 Byzantine: sends inconsistent values to each 0 1 01 0 1 0 1 Byzantine: sends inconsistent values to each Round 1 Round 2
- 17. Byzantine Agreement in n = 4, f =1 is possible. Byzantine Agreement is possible if n > 3f Proving impossibility for n ≤ 3f Claim: it’s impossible for n ≤ 3f Proof (by contradiction): assume it’s possible for n ≤ 3f i.e. exists algorithm A, that solve it, then we can show that n=3, f=1 is solvable by A
- 18. If there is an algorithm for n=3, f=1 take it and run it on 6 nodes as follows: Then processes a & b can’t tell difference if P3’ and P3 send you same as P3’ P3 a b
- 19. Algorithm for Byzantine Agreement if n > 4f: fori=1tof+1 round1:sendpreferencetoeveryone round2:if i=my_idthen sendmajorityvaluereceivedinprevround Lemma 1: if all correct processes have same preference v at start of phase P, they have preference v at end of phase P. Proof: In round 1, all correct processes send v (≥ n-f v’s sent) n-f > f + 𝑛 2 ≡ 𝑛 2 > 2f ≡ n > 4f Therefore, satisfying validity property follows from Lemma 1
- 20. Algorithm for Byzantine Agreement if n > 4f: Lemma 2: if process i is correct, then all correct processes have same preference at end of phase P. Proof: Any 2 correct proc that update to leader’s value choose same preference (since leader is correct) Any 2 that chooses value because they get > f+ 𝑛 2 copies must choose same value. (since > 𝑛 2 copies caused from correct processes + cannot be each of 2 different values )
- 21. Proof (continued): if any proc gets > f + 𝑛 2 copies of v then: • > 𝑛 2 v’s came from correct processes • Leader gets > 𝑛 2 copies of v • Leader sends v • all correct processes choose v Thus satisfying agreement property follows from Lemma 2 Since we proved that our algorithm satisfies validity and agreement, we prove that our algorithm is correct.
- 22. Byzantine Agreement in incomplete graphs
- 23. Byzantine agreement works in complete graphs (model below) if n > 3f But for incomplete graphs, an extra condition holds. Why? Model Non Anonymous Processes ( Proc IDs) Message Passing System Complete Network Synchronous No cryptography (used in reality to avoid byzantine/malicious nodes) A C B D Nodes B or C can fail, but NOT both If this happens, the agreement is not solvable.
- 24. Byzantine agreement works in complete graphs (model below) if n > 3f But for incomplete graphs, an extra condition holds. Why? Model Non Anonymous Processes ( Proc IDs) Message Passing System Complete Network Synchronous No cryptography (used in reality to avoid byzantine/malicious nodes) A C B D Nodes B or C can fail, but NOT both (so n > 3f) BUT If this happens, the agreement is not solvable: as A and D can only communicate reliably one non-byzantine process. However, the byzantine process can reach all other 3.
- 25. Generality: Byzantine Agreement is solvable in any arbitrary graph if and only if: n > 3f AND connectivity > 2f
- 26. When we say model A is as strong as model B, we need to prove any of: 1. All executions in A are also executions in B 2. Problems solvable in B are subset of problems solvable in A • i.e. any problem the can be solved in B; can also be solved in A. (Corollary: Ifa problem is impossible in A; it’s also impossible in B) 3. Show that A simulates B ( ᴟ transformation that converts any algorithm for B intoalgorithm forA) • Strong simulates weak B executions A executions