Malicious Motes and Suspicoius Sensors: Byzantine Interference in Wireless Networks

1. Malicious Motes and Suspicoius Sensors:Byzantine Interference in Wireless Networks Seth Gilbert February 13, 2006 Part II TexPoint fonts used in EMF. Read the TexPoint manual before you delete this box.: AAAAAAAAAAAA

2. Part I Errata • Clarified “bounded collision” assumption: • Adversary can cause  collisions. • Adversary can broadcast out of turn  times? • Adversary can spoof ’ messages. • Adversary can broadcast arbitrary numbers of correct messages.

3. Part II: Overview • Alice, Bob, and Collin • Basic 3-player game. • Algorithm • Synchronized • Unsynchronized • Extensions: Reliable broadcast, consensus, etc. • Lower Bound • Construct bad execution. • Extensions: Reliable broadcast, consensus, etc. • Discussion, Conclusions, and Open Questions.

4. Alice, Bob, and Collin The Game: Validity: Bob outputs only va, Alice outputs only vb. Termination: Eventually Alice or Bob output a value. Alice Bob

5. Alice, Bob, and Collin Key assumption: Alice and Bob can detect when a collision occurs. The Game: Alice Bob Collin

6. Alice, Bob, and Collin The Game: Alice Bob Collin

7. Alice, Bob, and Collin • What can Collin do? • Spoofing: Bob cannot distinguish a message sent by Alice from a message sent by Collin. • Jamming: Collin can obliterate any message by causing a collision. • Collin has at most  broadcasts. • Then he runs out of energy. •  is unknown to Alice and Bob. • Alice and Bob are deterministic. Can Alice and Bob communicate? How long can Collin delay Alice and Bob from communicating?

8. Alice, Bob, and Collin • Trivial Result: • Collin can delay Alice and Bob for at least  rounds. • Can Collin do any better? • “Jamming Gain” : How much better can Collin do than just jamming? • Does the protocol have any semantic vulnerability?

9. Alice, Bob, and Collin The Game: • Gilbert, Guerraoui, Newport, On the efficiency of malicious interference in wireless networks. OPODIS, 2006. Theorem: Assume Alice and Bob are transmitting values with logV bits. Then Collin can delay Alice and Bob for exactly rounds.

10. Basic Algorithm • Binary Value: • Alice and Bob start at the same time. • Synchronized start. • Alice sends her value to Bob. • Alice uses silence to communication information to Bob. • Collin can not forge a silent round!

11. Basic Algorithm • Binary value: • Round 1: (broadcast) • If (value=0), Alice silent. • If (value=1), Alice broadcasts. • Round 2: (veto) • If (value=0) and (Collin broadcasts in Round 1), Then Alice broadcasts. • Repeat…

12. Basic Algorithm • Binary Value: • Bob receives:

13. Rnd 1 (bcast) Rnd 2 (veto) Result --- --- Out: 0 1 --- Out: 1 --- 1 1 1 Basic Algorithm • Binary Value: • What can Collin do? • If (value=1): Alice Bob Round 2: Round 1: Collin

14. Rnd 1 (bcast) Rnd 2 (veto) Result --- --- Out: 0 1 --- Out: 1 --- 1 1 1 Basic Algorithm • Binary Value: • What can Collin do? • If (value=0): Alice Bob Round 2: Round 1: Collin

15. Basic Algorithm • Binary Value: • What can Collin do? • Broadcast in Round 1 • If (value=0): • Alice “veto”s in Round 2 • The protocol continues. • If (value=1): • Bob receives 1. • Broadcast in Round 2 • The protocol continues.

16. Basic Algorithm • Binary Value: Theorem 4: Bob receives Alice’s value within 2 +2 rounds.

17. Basic Algorithm • Binary Value:Asynchronous wake-up. • Adversary wakes up Bob

18. 0 1 Basic Algorithm • Binary Value:Asynchronous wake-up. • Adversary wakes up Bob Round 1 2 3 4 5 6 7 8 9 Alice 1 0 1 0 1 0 1 0 1 Bob W 1 0 Bob W 0 1

19. Basic Algorithm • Binary Value:Asynchronous wake-up. • Synchronization sequence: Bob W x x x 1 1 1 - P P

20. Basic Algorithm • Binary Value:Asynchronous wake-up. • Synchronization sequence: - P P Bob - 1 1 1 1 P P

21. Basic Algorithm • Binary Value:Asynchronous wake-up. • Collin: • During synch: • 1 broadcast ) Alice and Bob delay 1 rounds • During broadcast: • 1 broadcast ) Alice and Bob delay 2 rounds • Initial Cost: • 9 rounds: 3 init + 2 protocol + 4 init

22. Basic Algorithm • Binary Value: Theorem 5: Bob receives Alice’s value within 2 + 11 rounds after Bob wakes up.

23. Full Algorithm • Multivalued Messages: • Two changes: • Alice sends bits in sequence: • bit 0, bit 1, bit 2, … • Each bit is encoded as a two-bit sequence: • 0 = (00) • 1 = (10) • Synchronization sequence: • 3 log|V| sequence of 1’s, followed by a ---

24. 1 0 Full Algorithm • Multivalued Messages: 1 1 1 1 1 … … … … -- ----1 --

25. Full Algorithm • Multivalued Messages: • Why is the initialization string so long? • Collin can “fake” the initialization string by broadcasting in a long sequence of rounds. • Then the protocol restarts. • How many rounds are wasted when Collin fakes initialization?

26. Full Algorithm • Multivalued Messages: • What can Collin do? • Disrupt some bit: • Alice resends bit: 2 bit sequence. • All the bits requires 2 + log|V| rounds. • Disrupt synchronization: • Alice resends entire value. • Cost: 2log|V| rounds. • Collin uses 3log|V|-1 rounds.

27. Full Algorithm • Multivalued Messages: Theorem 5: Bob receives Alice’s value within 2 + (log|V|) rounds after Bob wakes up.

28. Extensions • Reliable Broadcast:n players • All receivers play Bob. • Result: • Binary Consensus:n players, t crashes • Each of 2t+1 players transmits their initial value. • A crashed player can be faked by Collin. • Decide majority value. • Result:

29. Extensions • Leader Election: k competitors • Tournament tree… • k-Selection: k competitors • Combined leader election and reliable broadcast…

30. Lower Bound • Key claim: For any algorithm A: Collin can delay Bob for 2 + log|V| - 1 rounds using only  broadcasts. ** Note: Focus on Bob; same argument holds for Alice.

31. Lower Bound • Main Idea: • Build two indistinguishable executions. • Show that Bob cannot decide in either execution for 2 + log|V| - 1 rounds. • Show that Collin only has to broadcast  times in one of the executions. 4 : two pairs of 2 executions

32. Lower Bound • Step One: Initial values • Find two initial values, v and v’, such that: • Algorithm A has the same pattern of broadcast and silence for the first log|V|-1 rounds.

33. Lower Bound • Step One: Initial values • Find two initial values, v and v’, such that: • Algorithm A has the same pattern of broadcast and silence for the first log|V|-1 rounds.

34. Lower Bound • Step One: Initial values • Counting: • There are |V |/2 broadcast/silence patterns of length log|V |-1. • There are only |V | initial values. • Thus at least two values must have the same broadcast/silence pattern (when Collin is silent).

35. Lower Bound • Step One: • Two initial values, v and v’, such that: • Algorithm A has the same pattern of broadcast and silence for the first log|V|-1 rounds.

36. Lower Bound • Step Two:Building executions. • Consider two pairs of executions • In each pair, one begins with v, one with v’. • We show how Collin behaves. • In each round, 4 possible patterns: • Alice broadcasts in both • Alice broadcasts in neither • Alice broadcasts in one and not the other A vA  v’A  vA  v’

37. Lower Bound • Step Two:Collin’s Rules. • Fill in blanks. x x x x A vA  v’A  vA  v’

38. Lower Bound • Step Two:Collin’s Rules. • Fill in blanks. x x x x x x x x A vA  v’A  vA  v’

39. Lower Bound • Step Two:Collin’s Rules. • Fill in blanks. x x x x x x x x A vA  v’A  vA  v’

40. Lower Bound • Step Two:Collin’s Rules. • Fill in blanks. • Overwrite where necessary. y y y x x x x x x x x x x x A vA  v’A  vA  v’

41. Lower Bound • Step Two:Constructing executions. • Continue for 2 + log|V| - 1 rounds. x x x x x x x x y x x x x x x x x x x x x y y y x x x y y y x x x x y y A vA  v’A  vA  v’

42. Lower Bound Bob cannot distinguish: x x x x x x x x y x x x x x x x x x x x x y y y x x x y y y x x x x y y A vA  v’A  vA  v’

43. Lower Bound Bob cannot output v or v’. x x x x x x x x x x x x x x x x x x x x y y y x x x y y y x x x x y y A vA  v’A  vA  v’

44. Lower Bound Alice cannot distinguish: x x x x x x x x x x x x x x x x x x x x y y y x x x y y y x x x x y y A vA  v’A  vA  v’

45. Lower Bound • Key Claim:Collin uses only  broadcasts. x x x x x x x x x x x x x x x x x x x x y y y x x x y y y x x x x y y A vA  v’A  vA  v’

46. Lower Bound Since v and v’ have the same broadcast pattern, Collin does 0 broadcasts in the initial rounds: • Key Claim:Collin uses only  broadcasts. x x x x x x x x x x x x x x x x x x x x y y y x x x y y y x x x x y y A vA  v’A  vA  v’

47. Lower Bound One of these 2 executions has · broadcasts. Which one? • Key Claim:Collin uses only  broadcasts. x x x x x x x x y x x x x x x x x x x x x y y y x x x y y y x x x x y y A vA  v’A  vA  v’

48. Lower Bound • Conclusions: • There exists an execution in which: • Bob does not output a value for: 2 + log|V | - 1 rounds. • Collin uses · broadcasts.

49. Lower Bound Theorem 6: For any algorithm A, there exists an execution of A in which Bob does not output Alice’s value for at least 2 + log|V | - 1 rounds. ** Note: Same argument shows that Alice does not receive Bob’s value.

50. Extensions • Reliable Broadcast: • By reduction: • Assume there exists an algorithm A’ that can solve reliable broadcast faster. • Alice simulates source. • Bob simulates all the receivers. • Therefore, reliable broadcast requires: 2 + log|V | - 1 rounds.