CS6410 – Byzantine Agreement Kai Sun *Some slides are borrowed from Ken Birman, Andrea C. Arpaci- Dusseau, Eleanor Birrell, Zhiyuan Teo, and Indranil Gupta
So Far We’ve Talked About • State machine replication • Paxos
So Far We’ve Talked About • Assumption • Processors do not collude, lie, or otherwise attempt to subvert the protocol • But what if the assumption does not hold?
The Byzantine Generals Problem • Leslie Lamport • PhD Brandeis 1972 • LaTeX, Clocks, Paxos , … • Robert Shostak • PhD Harvard 1974 • Staff scientist for SRI International • Founder and vice president of software for Ansa Software • Founder and CTO for Portera • Founder and CTO for Vocera • Marshall Pease
The Byzantine Generals Problem “I have long felt that, because it was posed as a cute problem about philosophers seated around a table, Dijkstra's dining philosopher's problem received much more attention than it deserves. …” * Leslie Lamport *http://research.microsoft.com/en-us/um/people/lamport/pubs/pubs.html
Byzantine Agreement • General commands soldiers • If all loyal soldiers attack victory is certain • If none attack, the Empire survives • If some attack, the Empire is lost Curses! I’m surrounded! • Gong keeps time • But they don’t need to all attack at once Attack!
Byzantine Soldiers • The enemy works by corrupting the soldiers • Orders are distributed by exchange of messages • Corrupt soldiers violate protocol at will • Corrupt soldiers can’t intercept and modify messages between loyal troops • The gong sounds slowly • There is ample time for loyal soldiers to exchange messages (all to all)
More Formal • A commander must send an order to his 𝑜 − 1 lieutenants such that • IC1. All loyal lieutenants obey the same order • IC2. If the commander is loyal, then every loyal lieutenant obeys the order he sends • IC1 and IC2 are called the interactive consistency conditions.
Impossibility Results • Let 𝑢 be the maximum number of faulty processes that our protocol is supposed to tolerate • Byzantine agreement is not possible with fewer than 3𝑢 + 1 processes
Impossibility Result • With only 3 generals, no solution can work with even 1 traitor (given oral messages) commander attack L1 L2 retreat What should lieutenant 1 (L1) do? Is commander or lieutenant 2 (L2) the traitor?
Option 1: Loyal Commander commander attack attack L1 L2 retreat What must L1 do? By IC2: L1 must obey commander and attack
Option 2: Loyal L2 commander retreat attack L1 L2 retreat What must L1 do? By IC1: L1 and L2 must obey same order --> L1 must retreat
Two Options commander commander attack retreat attack attack L1 L1 L2 L2 retreat retreat Problem: L1 can’t distinguish between 2 scenarios
General Impossibility Result • No solution with fewer than 3m+1 generals can cope with m traitors • < see paper for details >
Oral Messages • Assumptions • A1) Every message sent is delivered correctly • No message loss • A2) Receiver knows who sent message • Completely connected network with reliable links • A3) Absence of message can be detected • Synchronous system only
Oral Message Algorithm • OM(0) • Commander sends his value to every lieutenant • Each lieutenant uses the value received from the commander, or uses the value RETREAT if he receives no value
Oral Message Algorithm • OM(m), m>0 • Commander sends his value to every lieutenant • For each 𝑗 , let 𝑤 𝑗 be value Lieutenant 𝑗 receives from commander (or RETREAT if he receives no value) • Act as commander for OM(m-1) and send 𝑤 𝑗 to n-2 other lieutenants • For each 𝑗 and each 𝑘 ≠ 𝑗 , let 𝑤 𝑘 be value Li. 𝑗 received from Li. 𝑘 in the above step (or RETREAT if he received no such value). • Li. 𝑗 computes majority( 𝑤 1 ,..., 𝑤 𝑜−1 )
Example: Bad Lieutenant • Scenario: m=1, n=4, traitor = L3 C Round 0 A A A L2 L3 L1 C A A A A Round 1 A A L2 L3 L1 R A R Decision L1 = majority(A, A, R); L2 = majority(A, A, R); Both attack!
Example: Bad Commander • Scenario: m=1, n=4, traitor = C C A A Round 0 R L2 L3 L1 C A A R A Round 1 R A L2 L3 L1 A R A Decision L1=majority(A, R, A); L2=majority(A, R, A); L3=majority(A,R,A); Attack!
Bigger Example: Bad Lieutenants • Scenario: m=2, n=3m+1=7, traitors=L5, L6 C A A A A A A L3 L6 L1 L2 L4 L5 Messages? L3 L6 L1 L2 L4 L5 A R A A A R majority(A,A,A,A,R,R) ==> All loyal lieutenants attack! Decision?
Bigger Example: Bad Commander+Lieutenant • Scenario: m=2, n=7, traitors=C, L6 C A x A R R A L3 L6 L1 L2 L4 L5 Messages? L3 L6 L1 L2 L4 L5 A A R R A A,R,A,R,A Decision?
Decision with Bad Commander+Lieutenant • L1: majority(A,R,A,R,A,A) ==> Attack • L2: majority(A,R,A,R,A,R) ==> Retreat • L3: majority(A,R,A,R,A,A) ==> Attack • L4: majority(A,R,A,R,A,R) ==> Retreat • L5: majority(A,R,A,R,A,A) ==> Attack • Problem: All loyal lieutenants do NOT choose the same action
Next Step of Algorithm • Verify that lieutenants tell each other the same thing • Requires 𝑛 + 1 rounds • What messages does L1 receive in this example? Round 0: A C Round 1: 2R, 3A, 4R, 5A, 6A (doesn’t know 6 is traitor) A x A R R A Round 2: L3 L6 L1 L2 L4 L5 2 { 3A, 4R, 5A, 6R} Messages? 3 {2R, 4R, 5A, 6A} L3 L6 L1 L2 L4 L5 4 {2R, 3A, 5A, 6R} A A R R A 5 {2R, 3A, 4R, 6A} A,R,A,R,A 6 { ?, ?, ?, ? } • All see same messages in round 2 from L1, L2, L3, L4, and L5 • majority(A,R,A,R,A,-) ==> All attack
Algorithm Complexity • What’s the cost? • OM(m) invokes (n-1) OM(m-1). • OM(m-1)invokes (n-2) OM(n-2). • … • OM(m-k)will be called (n-1)(n- 2)…(n -k) times. • Algorithm complexity is O(n m ). (note: m = number of failures)
Signed Messages • Problem • Traitors can lie about what others said • How can we remove that ability?
Signed Messages • New assumption (A4) -- Signed messages (Cryptography) • Loyal general’s signature cannot be forged and contents cannot be altered • Anyone can verify authenticity of signature • Simplifies problem: • When Li. 𝑗 passes on signed message from 𝑘 , receiver knows that 𝑗 didn’t lie about what j said • Lieutenants cannot do any harm alone (cannot forge loyal general’s orders) • Only have to check for traitor commander • With cryptographic primitives, can implement Byzantine Agreement with m+2 nodes, using SM(m)
Signed Messages Algorithm: SM(m) Initially 𝑊 𝑗 = ∅ 1. 2. Commander signs 𝑤 and sends to all as ( 𝑤 :0) 3. Each Li. 𝑗 : A) If receive ( 𝑤 :0) and no other order 1) 𝑊 𝑗 = {𝑤} 2) Send ( 𝑤 :0: 𝑗 ) to all B) If receive ( 𝑤 :0: 𝑘 1 :...: 𝑘 𝑙 ) and 𝑤 not in 𝑊 𝑗 1) Add 𝑤 to 𝑊 𝑗 2) If ( 𝑙 <m) send ( 𝑤 :0: 𝑘 1 :...: 𝑘 𝑙 : 𝑗 ) to all not in 𝑘 1 … 𝑘 𝑙 4. When no more messages, obey order of choice( 𝑊 𝑗 )
Signed Messages Algorithm: SM(m) • 𝐷ℎ𝑝𝑗𝑑𝑓(𝑊) • If the set 𝑊 consists of the single element 𝑤 , then 𝑑ℎ𝑝𝑗𝑑𝑓(𝑊) = 𝑤 • 𝑑ℎ𝑝𝑗𝑑𝑓 ∅ =RETREAT • One possible definition is to let 𝐷ℎ𝑝𝑗𝑑𝑓(𝑊) be the median element of 𝑊
SM(1) Example: Bad Commander • Scenario: m=1, n=m+2=3, bad commander C A:0 R:0 L2 L1 What next? A:0:L1 L2 L1 R:0:L2 𝑊 1 ={A,R} 𝑊 2 ={R,A} Both apply same decision to {A,R}
SM(2): Bad Commander+Lieutenant • Scenario: m=2, n=m+2=4, bad commander and L3 Goal? L1 and L2 C A:0 x must make same A:0 decision L2 L3 L1 A:0:L1 A:0:L3 R:0:L3:L1 L2 L3 L1 A:0:L2 L2 L1 R:0:L3 𝑊 1 = 𝑊 2 = {A,R} ==> Same decision
Other Variations • How to handle missing communication paths • < see paper for details >
Compared with Asynchronous Scenarios m = traitors Synchronous Asynchronous n = total n <= 3m m >=1 * Oral messages: fails if n >= 3m+1 no guarantee works if won’t fail unless no m >= 1 * correct processes Signed messages: fails if n >= 1 no guarantee works if *Fischer, Michael J., Nancy A. Lynch, and Michael S. Paterson. "Impossibility of distributed consensus with one faulty process." Journal of the ACM (JACM) 32.2 (1985): 374-382.
Thought?
Easy Impossibility Proofs for Distributed Consensus Problems • Michael J. Fischer • PhD from Harvard (applied mathematics) • Professor at Yale • ACM Fellow • Nancy A. Lynch • PhD from MIT • Professor at MIT • ACM Fellow, Dijkstra Prize, Knuth Prize, … • Michael Merritt • PhD from GeTech • President, Brookside Engine Company No. 1. • Vice-Chair, Advancement Committee, Patriots' Path Council
Easy Impossibility Proofs for Distributed Consensus Problems • A process is regarded as a machine processing a tape of inputs • Called an agreement device • They build a communications graph. The messages that pass over an edge from a source to a destination node are a behavior of the device on that edge • Behavior of the system is a set of node and edge behaviors • In their proofs, faulty devices often exhibit different and inconsistent behaviors with respect to different participants
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