The Byzantine Generals Problem

1. The Byzantine Generals Problem Leslie Lamport, Robert Shostak, Marshall Pease Presented by: Anna Bendersky Distributed Algorithms A

2. 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

3. 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)

4. 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 Formally: The Byzantine Generals Problem All loyal lieutenants obey the same order If the commander is loyal, then every loyal lieutenant obeys the order he sends

5. Outline • Minimal bound on the number of traitors • Oral messages algorithm to reach agreement • Written messages algorithm to reach agreement

6. Minimal Bound on the traitors number The Byzantine Generals Problem • All loyal lieutenants obey the same order • If the commander is loyal, then every loyal lieutenant obeys the order he sends 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’ Commander Commander retreat attack attack retreat attack attack L1 L2 L1 L2 retreat retreat Case 1: Source process is faulty Case 2: Process L2 is faulty

7. 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 1 0 1 1 0 0

8. 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 The Byzantine Generals Problem All loyal lieutenants obey the same order If the commanding general is loyal, then every loyal lieutenant obeys the order he sends • Assumptions: • Every message that is sent is delivered correctly • A receiver of a message knows who sent it • The absence of a message can be detected

9. Oral Messages Algorithm A recursive definition, with a base case for m=0, and a recursive step for m > 0: Algorithm OM(0) • The commander sends his value to every lieutenant. • Each lieutenant uses the value he receives from the commander. Algorithm OM(m), m > 0 • The commander sends his value to each lieutenant. • For each i, let vi be the value lieutenant i receives from the commander. Lieutenant i acts as the commander in Algorithm OM(m-1) to send the value vi to each of the n-2 other lieutenants. • For each i, and each j ≠ i, let vi be the value lieutenant i received from lieutenant j in step 2 (using Algorithm OM(m-1)). Lieutenant i uses the value Majority(v1, v2, … vn).

10. Oral Messages Example (1) G m=0 The general sends a message to all lieutenants 6 2 5 3 4 m=1 Each general sends the message he received to all other lieutenants G 0 1 0 1 … 2 0 6 3 5 4

11. Oral Messages Example (2) Messages sent in step 1: Step 2: Each lieutenant sends the message he received to all other lieutenants These are the messages being sent in the recursive algorithm – Lets see which value each lieutenant decides to use

12. Oral Messages Example (3) • We could build a “recursion tree” for the decision of each lieutenant: 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} Received message Lieutenant i: {1,1,?} {1,1,0} The messages lieutenant i receives at step OM(1) majority (v1, v2, … vn) The messages lieutenant i receives at step OM(0) All lieutenants reach the decision ‘0’

13. A more complex example {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,?} {0,13,?} {0,14,?} {0,15,?} {X,16,?} {X,17,?} {0,12,0} {0,13,0} {0,14,0} {0,15,0} {X,16,X} {X,17,X} • n=7 , m=2 G 7 2 6 3 {0,1,?} {0,1,0} 5 4

14. Oral Messages Algorithm: Proof (1) The Byzantine Generals Problem All loyal lieutenants obey the same order If the commander is loyal, then every loyal lieutenant obeys the order he sends 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 vi = ‘v’ for a majority of n-1 values i. • Majority(v1,…vn)=‘v’ in step (3).

15. Oral Messages Algorithm: Proof (2) The Byzantine Generals Problem • All loyal lieutenants obey the same order • If the commander is loyal, then every loyal lieutenant obeys the order he sends 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 vj for every loyal general j. • Majority(v1,…vn) is the same for all loyal lieutenants in step (3). QED

16. A solution with signed messages • The problem we had: traitors could lie • 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)

17. Signed messages example Commander is faulty! retreat Commander is not faulty! attack Commander Commander attack:0 retreat:0 attack:0 attack:0 retreat:0:2 retreat:2 L1 L2 L1 L2 attack:0:1 attack:0:1 Case 1: Source process is faulty Case 2: Process L2 is faulty Commander’s message cannot be forged!

18. Signed Messages Algorithm Algorithm SM(m) Initially Vi = {} • The commander signs and sends his value to every lieutenant • For each i: • If Lieutenant i receives a message of the form v:0 from the commander and he has not yet received any order, then • Vi {v} • He sends the message v:0:i to every other lieutenant • If Lieutenant i receives a message of the form v:0:j1: … :jk and v is not in the set Vi then • Vi Vi U {v} • If k<m then he sends the message v:0:j1: … : jk:i to every lieutenant other than j1,…,jk • For each i: When lieutenant i will receive no more messages, he obeys the order choice(Vi)

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

20. Signed Messages Algorithm: Proof The Byzantine Generals Problem • All loyal lieutenants obey the same order • If the commander is loyal, then every loyal lieutenant obeys the order he sends • 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 Vi. • 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: Vi = {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 Vi = Vx . • => 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. QED

21. Questions? Thank You