# The Byzantine Generals Problem Leslie Lamport, Robert Shostak, Marshall Pease 10.05.2015Distributed Algorithms A1 Presented by: Anna Bendersky.

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The Byzantine Generals Problem Leslie Lamport, Robert Shostak, Marshall Pease 10.05.2015Distributed Algorithms A1 Presented by: Anna Bendersky

Motivation A reliable computer system must be able to cope with a failure of one or more of its components A failed computer behavior in this case: – Sending conflicting messages to different parts of the system – Not sending some of the messages 10.05.20152

Byzantine Generals (Abstract Expression) Several divisions of the byzantine army are camped outside a city, each division has a general. The generals can communicate with each other using a messenger The generals must decide upon a common plan of action Some of the generals might be traitors! We will assume that there is a single commanding general (Commander), and the rest of the generals are his subordinates (Lieutenants) 10.05.20153

Objective All loyal generals decide upon the same plan of action A small number of traitors will not cause the loyal generals to adopt a bad plan 10.05.20154 Formally: The Byzantine Generals Problem 1.All loyal lieutenants obey the same order 2.If the commander is loyal, then every loyal lieutenant obeys the order he sends

Outline Minimal bound on the number of traitors Oral messages algorithm to reach agreement Written messages algorithm to reach agreement 10.05.20155

Minimal Bound on the traitors number The Byzantine Generals Problem 1.All loyal lieutenants obey the same order 2.If the commander is loyal, then every loyal lieutenant obeys the order he sends 10.05.20156 Case 1: Source process is faulty Case 2: Process L2 is faulty From now forward we will assume that the generals have to decide upon a single bit: ‘1’ for ‘Attack’, or ‘0’ for ‘Retreat’. The default operation is ‘Retreat’ L2L1 Commander attackretreat L2L1 Commander attack retreat attack

Minimal Bound on the traitors number Corollary 1: In the case of 3 processes, there is no way to deal with one faulty process. Corollary 2: No solution with fewer than 3m+1 generals can cope with m traitors 10.05.20157 0 0 0 111

A solution with oral messages No solution with fewer than 3m+1 generals can cope with m traitors The paper shows an algorithm to solve the byzantine generals problem when the number of loyal generals > 3m 10.05.20158 The Byzantine Generals Problem 1.All loyal lieutenants obey the same order 2.If the commanding general is loyal, then every loyal lieutenant obeys the order he sends Assumptions: 1.Every message that is sent is delivered correctly 2.A receiver of a message knows who sent it 3.The absence of a message can be detected

Oral Messages Algorithm A recursive definition, with a base case for m=0, and a recursive step for m > 0: Algorithm OM(0) 1.The commander sends his value to every lieutenant. 2.Each lieutenant uses the value he receives from the commander. Algorithm OM(m), m > 0 1.The commander sends his value to each lieutenant. 2.For each i, let v i be the value lieutenant i receives from the commander. Lieutenant i acts as the commander in Algorithm OM(m-1) to send the value v i to each of the n-2 other lieutenants. 3.For each i, and each j ≠ i, let v i be the value lieutenant i received from lieutenant j in step 2 (using Algorithm OM(m-1)). Lieutenant i uses the value Majority(v 1, v 2, … v n ). 10.05.20159

Oral Messages Example (1) 10.05.201510 G 6 5 4 3 2 m=0 The general sends a message to all lieutenants G 6 5 4 3 2 … m=1 Each general sends the message he received to all other lieutenants Sender=P 2 Sender=P 3 Sender=P 4 Sender=P 5 Sender=P 6 DestMsgDestMsgDestMsgDestMsgDestMsg P2P2 {0,12}P2P2 {0,13}P2P2 {0,14}P2P2 {1,15}P2P2 {1,16} P3P3 {0,12}P3P3 {0,13}P3P3 {0,14}P3P3 {1,15}P3P3 {1,16} P4P4 {0,12}P4P4 {0,13}P4P4 {0,14}P4P4 {1,15}P4P4 {1,16} P5P5 {0,12}P5P5 {0,13}P5P5 {0,14}P5P5 {1,15}P5P5 {1,16} P6P6 {0,12}P6P6 {0,13}P6P6 {0,14}P6P6 {1,15}P6P6 {1,16} 0 0 0 1 1 Sender=P 2 Sender=P 3 Sender=P 4 Sender=P 5 Sender=P 6 {0,12}{0,13}{0,14}{1,15}{1,16}

Oral Messages Example (2) 10.05.201511 Step 2: Each lieutenant sends the message he received to all other lieutenants Sender=P 2 Sender=P 3 Sender=P 4 Sender=P 5 Sender=P 6 {0,132} {0,142} {1,152} {1,162} {0,123} {0,143} {1,153} {1,163} {0,124} {0,134} {1,154} {1,164} {0,125} {0,135} {0,145} {1,165} {0,126} {0,136} {0,146} {1,156} Sender=P 2 Sender=P 3 Sender=P 4 Sender=P 5 Sender=P 6 {0,12}{0,13}{0,14}{1,15}{1,16} Messages sent in step 1: These are the messages being sent in the recursive algorithm – Lets see which value each lieutenant decides to use

Oral Messages Example (3) We could build a “recursion tree” for the decision of each lieutenant: 10.05.201512 Lieutenant i: {1,1,?} Received message Message source Decision {0,12,?} {0,13,?} {0,14,?} {1,15,?} {1,16,?} {0,12,0} {0,13,0} {0,14,0} {1,15,1} {1,16,1} {1,1,0} majority (v 1, v 2, … v n ) All lieutenants reach the decision ‘0’ The messages lieutenant i receives at step OM(1) The messages lieutenant i receives at step OM(0)

A more complex example n=7, m=2 13 G 6 54 3 27 {0,1,?} {0,12,?} {0,13,?} {0,14,?} {0,15,?} {X,16,?} {X,17,?} {0,123,?} {0,124,?} {0,125,?} {X,126,?} {X,127,?} {0,132,?} {0,134,?} {0,135,?} {X,136,?} {X,137,?} {0,142,?} {0,143,?} {0,145,?} {X,146,?} {X,147,?} {0,152,?} {0,153,?} {0,154,?} {X,156,?} {X,157,?} {X,162,?} {X,163,?} {X,164,?} {X,165,?} {X,167,?} {X,172,?} {X,173,?} {X,174,?} {X,175,?} {X,176,?} {0,123,0} {0,124,0} {0,125,0} {X,126,X} {X,127,X} {0,132,0} {0,134,0} {0,135,0} {X,136,X} {X,137,X} {0,142,0} {0,143,0} {0,145,0} {X,146,X} {X,147,X} {0,152,0} {0,153,0} {0,154,0} {X,156,X} {X,157,X} {X,162,X} {X,163,X} {X,164,X} {X,165,X} {X,167,X} {X,172,X} {X,173,X} {X,174,X} {X,175,X} {X,176,X} {0,12,0} {0,13,0} {0,14,0} {0,15,0} {X,16,X} {X,17,X} {0,1,0}

Oral Messages Algorithm: Proof (1) Lemma 1: For any m and k, Algorithm OM(m) satisfies (2) if there are more than 2k+m generals and at most k traitors Proof: (by induction on m) Base: Algorithm OM(0) satisfies (2) when the commander is loyal. Assumption: the algorithm OM(m-1) satisfies (2) if there are more than 2k+m-1 generals and at most k traitors Step: In step (1) every loyal commander sends the value ‘v’ to all n-1 lieutenants. In step (2) each loyal lieutenant applies OM(m-1) with n-1 lieutenants By hypothesis,  A majority of the n-1 lieutenants are loyal  By assumption, each loyal lieutenant has v i = ‘v’ for a majority of n-1 values i.  Majority(v 1,…v n )=‘v’ in step (3). 10.05.201514 The Byzantine Generals Problem 1.All loyal lieutenants obey the same order 2.If the commander is loyal, then every loyal lieutenant obeys the order he sends

Oral Messages Algorithm: Proof (2) Theorem 1: For any m, algorithm OM(m) satisfies conditions 1 and 2 if there are more than 3m generals, and at most m traitors. Proof: (By induction on m) Base: if there are no traitors, OM(0) satisfies conditions 1 and 2 Assumption: OM(m-1) satisfies conditions 1 and 2 if there are more than 3(m-1) generals, and at most m-1 traitors Step: We can use lemma1 with k=m, and get that condition 2 holds. Condition 1 follows from condition 2 when the commander is loyal. Else, there are at most m traitors and the commander is one of them  At most m-1 of the lieutenants are traitors  At step (2) of the algorithm there are 3m-1 > 3(m-1) generals, and at most m-1 traitors  From the assumption, OM(m-1) satisfies conditions 1 and 2.  All loyal generals get the same values v j for every loyal general j.  Majority(v 1,…v n ) is the same for all loyal lieutenants in step (3). 10.05.201515 The Byzantine Generals Problem 1.All loyal lieutenants obey the same order 2.If the commander is loyal, then every loyal lieutenant obeys the order he sends QED

A solution with signed messages The problem we had: traitors could lie 10.05.201516 Assumption 4: A loyal general’s signature cannot be forged, and any alteration of the contents of his signed message can be detected Anyone can verify the authenticity of a general’s signature The previous lower bound no longer holds! There exists an algorithm that copes with m traitors for any number of generals (n≥m+2)

Signed messages example 10.05.201517 Case 1: Source process is faulty L2L1 Commander attack:0retreat:0 retreat:0:2 Commander is faulty! retreat Commander is faulty! retreat attack:0:1 Case 2: Process L2 is faulty L2L1 Commander attack:0 retreat:2 Commander is not faulty! attack Commander is not faulty! attack attack:0:1 Commander’s message cannot be forged!

Signed Messages Algorithm Algorithm SM(m) Initially V i = {} 1.The commander signs and sends his value to every lieutenant 2.For each i: 1.If Lieutenant i receives a message of the form v:0 from the commander and he has not yet received any order, then 1.V i  {v} 2.He sends the message v:0:i to every other lieutenant 2.If Lieutenant i receives a message of the form v:0:j 1 : … :j k and v is not in the set V i then 1.V i  V i U {v} 2.If k<m then he sends the message v:0:j 1 : … : j k :i to every lieutenant other than j 1,…,j k 3.For each i: When lieutenant i will receive no more messages, he obeys the order choice(V i ) 10.05.201518

Signed messages example 10.05.201519 Case 1: Source process is faulty L2L1 Commander attack:0retreat:0 retreat:0:2 Commander is faulty! retreat Commander is faulty! retreat attack:0:1 Case 2: Process L2 is faulty L2L1 Commander attack:0 retreat:2 Commander is not faulty! attack Commander is not faulty! attack attack:0:1 Commander’s message cannot be forged! V(L1) = {attack, retreat} V(L1) = {attack}

Signed Messages Algorithm: Proof If the commander is loyal, then he sends his signed order v:0 to every lieutenant in step (1), and every loyal lieutenant will add v to V i. Since no traitorous lieutenant can forge a message of the form v’:0, a loyal lieutenant can receive no other order in step (2.2). For all loyal lieutenants: V i = {v} => every loyal lieutenant obeys the order the general sends. (condition 2 OK) It remains to prove condition 1 for the case where the commander is not loyal. Two loyal lieutenants i and x obey the same order in step (3) if the sets V i = V x. => i received the message v1: – If it was received from the general – It was sent to x in step 1. – It was received by v1:0:{list}. If x is in the list, then x has. – It was received by v1:0:{list}, and x is not in the list: If one of the lieutenants in the list is loyal, then x received it when the loyal lieutenant sent it There are at most m-1 traitorous lieutenants, so in step m lieutenant i will send the message to x. 10.05.201520 The Byzantine Generals Problem 1.All loyal lieutenants obey the same order 2.If the commander is loyal, then every loyal lieutenant obeys the order he sends QED

Thank You 10.05.201521 Questions?

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