1. 1 The Byzantine Generals Problem Leslie Lamport, Robert Shostak, Marshall Pease Presented by Radu Handorean

2. 2 Byzantine Generals Problem (metaphor)

3. 3 GBP – the Generals  Loyal Generals Behave according to THE algorithm which should ensure that  They decide upon the same plan (A)  A small number of traitors shouldn’t be able to force a bad decision (B)  Traitorous Generals Try to mess the final decision Send any info they want

4. 4 GBP – the Generals  (A) => Every loyal general must obtain the same v(1)…v(n)  (B) => If the i th general is loyal => v(i) must be used by all (loyal) generals

5. 5 Byzantine Generals Problem (formal)  0.. N-1 processes in a complete graph  Process 0 needs to send a value v to all others such that (IC1) If process 0 is non faulty then any non faulty process i receives v (IC2) If processes i and j are non faulty, they receive the same value  Note: 0 is non faulty, then IC1=>IC2

6. 6 Impossibility Results – Oral Msg  Oral message – the content is entirely under the control of the sender  No solution if more than 1/3 of the generals are traitorous

7. 7 Traitorous Lieutenant attack he said “retreat”

8. 8 Traitorous General retreat attack he said “retreat”

9. 9 Impossibility Results – Generalization  No solution with fewer than 3m+1 generals for m traitors  Proof by contradiction: reduce the problem to the 3 generals problem Assume 3m (let’s call them Albanians) or fewer generals can cope with m traitors Build the solution with Byzantine generals

10. 10 Proof  1 Byzantine simulates ~1/3 Albanians 1 Byzantine simulates 1 Albanian general & m-1 Albanian lieutenants (m, m, respectively) Max m traitor Albanians IC1 & IC2 hold for Albanians (assumed) IC1 & IC2 hold for Byzantine (implied)  IMPOSSIBLE SOLUTION

11. 11 Solution with Oral Messages  A1. Every msg. is delivered correctly  A2. The receiver knows where the msg. comes from  A3.The absence of a msg. can be detected A1&A2 – a traitor cannot interfere with a msg. between others A3 – a traitor cannot drop msg.

12. 12 Oral Messages – Cont.  No order from a traitorous commander => RETREAT by default  OM(m) – alg. for 3m+1 generals with at most m traitors  Use the majority function for decision Majority value if exists or RETREAT Median value if they are an ordered set

13. 13 OM(0)  (1) The commander sends his value to each lieutenant  (2) Each lieutenant uses the value from the commander or RETREAT if the commander is silent

14. 14 OM(m)  (1) The commander sends his value to each lieutenant (v i )  (2) Each L acts as commander for OM(m-1) and sends V i to the other n-2 (or RETREAT)  (3) For each i and j!=i, Li receives v j from Lj in (2) (or RETREAT); Li uses majority(v 1..v n-1 )

15. 15 Example m=1, n=4, L traitor v v v v x

16. 16 Example m=1, n=4, L traitor x y z x z y y x z

17. 17 OM(m) - Proof of Correctness  Lemma1: for any m, k, OM(m) has IC2 for more than 2k+m generals and at most k traitors IC2: if the commander is loyal, every loyal general obeys commander’s order  Proof: induction on m OM(0) – trivial m>0  Commander sends v to n-1 lieutenants

18. 18 OM(m) – Proof - Cont.  Each loyal general applies OM(m-1) with n-1 generals  (*) n>2k+m => n-1>2k+(m-1)  >each loyal Li gets v j =v from each loyal Lj  At most k traitors and (*) =>a majotiry of n-1 lieutenants are loyal

19. 19 OM(m) – Proof – Cont.  Theorem: OM(m) satisfies IC1 and IC2 if there are more than 3m generals and at most m traitors  Proof: induction on m OM(0) satisfies IC1 and IC2 (no traitors) Commander = loyal & k=m in Lemma => IC2 => IC1 Commander = traitor => at most m-1 traitorous lieutenants

20. 20 OM(m) – Proof – Cont. There are more than 3m generals => more than 3m-1 lieutenants 3m-1>3(m-1) & apply induction (OM(m-1) satisfies IC1 & IC2) => for each j, any 2 loyal Ls get the same value for v j in step 3 => any 2 loyal Ls get the same array (v 1...v n-1 ) in step 3 => the same majority(…) => IC1

21. 21 Solution with Written Messages  Generals send unforgeable signed messages  Add A4 to A1-A3: A loyal G’s signature cannot be forged and any alteration can be detected Anyone can verify the auth of a G’s signature  NO assumptions about a traitorous G’s signature

22. 22 New Solution  C sends signed orders to Ls  Each L adds its signature and forwards the message, etc…  Use a function choice(…) to obtain a single order choice(V) = v if v if the only elem. in V choice(V) = RETREAT if V is empty Any choice() function must have these properties

23. 23 Notations  x:i = msg. x signed by G i  v:j:i = msg. v signed by Gs j and I  G0 = commander (C)  Vi = set of properly signed orders received by Li Loyal C => Vi has only 1 element Do NOT confuse with the set of msg. !!! (many different msg can carry the same order)

24. 24 SM(m)  Initially Vi = empty for each I  (1) C signs and send v to each L  (2) For each i: (A) if Li receives v:0 and Vi=empty  (i) Vi={v}  (ii) Send v:0:i to all other Ls (B) if Li receives v:j 1 …:j k and v not in Vi  (i) Add v to Vi  (ii) if k<m send v:j 1 …:j k :I to all other agents  (3) When Li receives no more msg., he obeys choice(Vi)

25. 25 SM(1) - Example Attack:0 Retreat:0 Attack:0:1 Retreat:0:2 0 1 2

26. 26 SM(1) – Proof  Theorem2: SM(m) solves GBP for at most m traitors C = loyal => sends v:0 to all Ls  Every loyal L receives v in (2)  No loyal L can receive v’:0 in (2B)  Vi = {v} for all i  Loyal Ls obey choice() in (3) => IC2 => IC1 C = traitorous

27. 27 SM(m) – Proof – Cont. C = traitorous  Loyal Li and Lj obey the same order in (3) if Vi = Vj from (2)  If Li receives v in (2A), it sends it to Lj in (2Aii)  If Li adds v to Vi in (2B) => must receive a first message v:j 1 …:j k

28. 28 SM(m) – Proof – Cont. If j is one of the j r, v must have already been added to Vi If not  (1) k<m : i sends v:j 1 …j k :i to j  (2) k=m : since C=traitor= > max m-1 traitor Ls => at least 1 of j 1 …j m is loyal  This loyal L must have sent v to j so j has that order

29. 29 Missing Communication Paths  The Generals’ graph is no longer complete 3-regular graph not 3-regular

30. 30 Definitions  (a) {i 1,…,i p } is a regular set of neighbors of I if Each i j is a neighbor of I For any k!=i there are paths g j,k from i j to k not passing through i s.t. any 2 such path only have k in common  A graph G is p-regular if any node has a set of p regular neighbors  Note: a 3m-regular graph has min 3m+1 nodes

31. 31 OM(m,p)  G must be p-regular  (0) N = p-regular set of C’s neighbors  C sends the order to every L in N  For each i in N, Li receives v i from C or RETREAT; L i sends v i to every other L k as follows:

32. 32 OM(m,p) – Cont. (A) if m=1, it sends along g j,k (B) if m>1, it acts as commander for OM(m-1, p-1), after removing C  For each k and i in N, k!=i, Lk receives v i from Li, or v i =RETREAT; Lk uses majority(v i1,…, v ip ), where N = {i1,…ip}

33. 33 OM(m, 3m) – GBP  O(m,3m) solves GBP for at most m traitors (proof below)  Lemma1: for any m>0 and any p>=2k+m, OM(m,p) satisfies IC2 for at most m traitors m=1  L obtains majority(v 1..v p )  At most k traitors and p>=2k+1 => more than half of the p paths –> loyal Ls -> if C is loyal then the majority() if his command m>1

34. 34 Lemma2 – Cont. m>1  Assume for m-1  If C = loyal, each of the p Ls in N has the correct order  p>2k -> a majority are loyal & each sends the correct order  Each loyal L gets a majority of correct orders

35. 35 GBP – Cont.  Theorem 3: for any m>0 and any p>=3m, OM(m,p) solves GBP for max. m traitors Lemma 2 & k=m => IC2 C = loyal then IC2 implies IC1 C = traitorous  m=1 => all Ls = loyal and g j,k do not pass through C  m>1: induction since p>=3m implies p-1>=3(m-1)

36. 36 Comments  For 3m+1 generals, 3m-regularity = complete connectivity  IC2 cannot be satisfied if a message C->L is “routed” by traitors  IC1 cannot be satisfied if L1 and L2 can only communicate via traitors  These assumptions are too strong

37. 37 SM(m)  If the subgraph of loyal Ls is connected =>SM(n-2) is a solution (n=# of Gs) regardless of # of traitors  Definition: the diameter of a graph is the smallest # of edges to connect any 2 nodes

38. 38 GBP - SM  Theorem 4: If there are at most m traitors, and d=the diameter of loyal Ls subgraph, SM(m-d+1) solves GBP  Proof: similar to Theorem 2

39. 39 SO WHAT ???  Use of redundancy and voting to achieve reliability  Majority voting All non faulty processes produce the same result (from the same input - e.g. 2 non faulty processors read a clock) If the input unit (G) is non faulty, all non faulty (loyal) processes (Ls) use the provided value

40. 40 SO WHAT – Cont. A1..A3(A4)  A1 – every msg. sent by a non faulty proc. Is delivered correctly The failure of a communication line cannot be distinguished from the failure of a component => max m failures Real life effect: lowers connectivity, does not forge information

41. 41 SO WHAT – Cont. A1..A3(A4)  A2 – a processor can determine the origin of a msg. Most important is that a faulty proc. cannot impersonate a non faulty one In practice we should use IPC over fixed lines rather than fancy network switching A4 obsoletes A2, is satisfied

42. 42 SO WHAT – Cont. A1..A3(A4)  A3 – the absence of a message can be detected Use of time-outs:  Fixed maximum time to produce and deliver a message  Sender’s and receiver’s clock’s are reasonably synchronized

43. 43 SO WHAT – Cont. A1..A3(A4)  A4 – processors sign messages s.t. a non faulty processor cannot forged Signature = redundant info. Message signed by i = (M, S i (M))  S i must satisfy If I is non faulty, no other processor can generate S i (M) – cannot be guaranteed  Random multiplication  Malicious intelligence Given M and X, any processor can verify X=S i (M)

44. 44 DO YOU STILL HAVE QUESTIONS? raduh@cse.wustl.edu