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FLP Impossibility of Consensus Yan Ji Oct 26, 2017 Slides - PowerPoint PPT Presentation

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FLP Impossibility of Consensus Yan Ji Oct 26, 2017 Slides inspired by Lorenzo Alvisi (CS5414 FA16) slides and Philip Daian (CS6410 FA16) slides I think you ought to know I'm feeling very depressed. I think you ought to know I'm feeling

  1. Model It is impossible to have a deterministic protocol that solves consensus in a message-passing asynchronous system in which at most one process may fail by crashing. p 1 0/1, b/0/1 ... Step: C’= e(C) = (s’, M’) ● (p i , m) p i M 0/1, b/0/1 ... p 1 p 1 p k 0/1, b/0/1 C = (s, M) 0/1, b/0/1 C = (s, M) 0/1, b/0/1 . . s . . . . M M p k p k 0/1, b/0/1 0/1, b/0/1 s s

  2. Model It is impossible to have a deterministic protocol that solves consensus in a message-passing asynchronous system in which at most one process may fail by crashing. p 1 0/1, b/0/1 ... Step: C’= e(C) = (s’, M’) ● (p i , m) p i M 0/1, b/0/1 ... p 1 p 1 p k 0/1, b/0/1 C = (s, M) 0/1, b/0/1 C = (s, M) 0/1, b/0/1 . ... s . (p i , m) p i 0/1, b M . M ... (p i , m) p k p k 0/1, b/0/1 0/1, b/0/1 s s

  3. Model It is impossible to have a deterministic protocol that solves consensus in a message-passing asynchronous system in which at most one process may fail by crashing. p 1 0/1, b/0/1 ... Step: C’= e(C) = (s’, M’) ● (p i , m) p i M 0/1, b/0/1 ... p 1 p 1 p k 0/1, b/0/1 C = (s, M) 0/1, b/0/1 C = (s, M) 0/1, b/0/1 . ... s . p’ i 0/1, b 0 M . M ... (p i , m) p k p k 0/1, b/0/1 0/1, b/0/1 s s’

  4. Model It is impossible to have a deterministic protocol that solves consensus in a message-passing asynchronous system in which at most one process may fail by crashing. p 1 0/1, b/0/1 ... Step: C’= e(C) = (s’, M’) ● (p i , m) p i M 0/1, b/0/1 ... p 1 p 1 p k 0/1, b/0/1 C = (s, M) 0/1, b/0/1 C = (s, M) 0/1, b/0/1 . ... s . send p’ i 0/1, b 0 M . M ... (p i , m) p k p k 0/1, b/0/1 0/1, b/0/1 s s’

  5. Model It is impossible to have a deterministic protocol that solves consensus in a message-passing asynchronous system in which at most one process may fail by crashing. p 1 0/1, b/0/1 ... Step: C’= e(C) = (s’, M’) ● (p i , m) p i M 0/1, b/0/1 ... p 1 p 1 p k 0/1, b/0/1 C = (s, M) 0/1, b/0/1 C’ = (s’, M’) 0/1, b/0/1 . ... s . p’ i 0/1, b 0 M’ . M ... (p i , m) p k p k 0/1, b/0/1 0/1, b/0/1 s s’

  6. Model It is impossible to have a deterministic protocol that solves consensus in a message-passing asynchronous system in which at most one process may fail by crashing. p 1 0/1, b/0/1 ... Step: C’= e(C) = (s’, M’) ● (p i , m) p i M 0/1, b/0/1 ... p 1 p 1 p k 0/1, b/0/1 C = (s, M) 0/1, b/0/1 C’ = (s’, M’) 0/1, b/0/1 . ... s . p’ i 0/1, b 0 M’ . M ... (p i , m) p k p k 0/1, b/0/1 0/1, b/0/1 s s’

  7. Proof It is impossible to have a deterministic protocol that solves consensus in a message-passing asynchronous system in which at most one process may fail by crashing. How to prove impossibility?

  8. Proof It is impossible to have a deterministic protocol that solves consensus in a message-passing asynchronous system in which at most one process may fail by crashing. How to prove impossibility? Assume to the contrary that there exists a consensus protocol P such that… How to define P?

  9. More terms A schedule S of P is a finite or infinite sequence of events (e 1 , e 2 , …, e k )of P, ● S(C) = e k (...(e 2 (e 1 (C)))...)

  10. More terms A schedule S of P is a finite or infinite sequence of events (e 1 , e 2 , …, e k )of P, ● S(C) = e k (...(e 2 (e 1 (C)))...) A run of P is a sequence of steps associating a schedule S, in other words, a ● run is a pair of a configuration C and a schedule S, written as (C, S)

  11. More terms A configuration C’ is reachable from a configuration C if there exist a ● schedule S such that C’ = S(C)

  12. More terms A configuration C’ is reachable from a configuration C if there exist a ● schedule S such that C’ = S(C) A configuration C ′ is accessible from an initial configuration C 0 if C’ is ● reachable from C 0

  13. More terms A configuration C’ is reachable from a configuration C if there exist a ● schedule S such that C’ = S(C) A configuration C ′ is accessible from an initial configuration C 0 if C’ is ● reachable from C 0 e 1 (p 1 , m 1 ) e 2 (p k , m 2 ) p’ 1 p 1 p’ 1 0/1, b 0/1, b C 0 = (s 0 , M 0 ) C 2 = (s 2 , M 2 ) C 1 = (s 1 , M 1 ) 0/1, b . . . . . . . . M 2 . M 0 M 1 p’ k p k p k 0/1, b 0 0/1, b 0/1, b s 2 s 0 s 1

  14. More terms A configuration C has decision value v if some process p is in a decision ● state with output=v, which is “write-once”/irreversible

  15. More terms A configuration C has decision value v if some process p is in a decision ● state with output=v, which is “write-once”/irreversible A run is a deciding run if some process reaches a decision state. ●

  16. More terms A configuration C has decision value v if some process p is in a decision ● state with output=v, which is “write-once”/irreversible A run is a deciding run if some process reaches a decision state. ● e 1 (p 1 , m 1 ) p 1 p’ 1 0/1, b C 0 = (s 0 , M 0 ) C 1 = (s 1 , M 1 ) 0/1, b . . . . . . M 0 M 1 p k p k 0/1, b 0/1, b s 0 s 1

  17. More terms A configuration C has decision value v if some process p is in a decision ● state with output=v, which is “write-once”/irreversible A run is a deciding run if some process reaches a decision state. ● e 1 (p 1 , m 1 ) e 2 (p k , m 2 ) p’ 1 p 1 p’ 1 0/1, b 0/1, b C 0 = (s 0 , M 0 ) C 2 = (s 2 , M 2 ) C 1 = (s 1 , M 1 ) 0/1, b . . . . . . . . M 2 . M 0 M 1 p’ k p k p k 0/1, b 0 0/1, b 0/1, b s 2 s 0 s 1

  18. More terms A consensus protocol P is partially correct if: ● No accessible configuration has more than one decision value ○ (agreement) For each v in {0, 1}, some accessible configuration has decision ○ value v (validity) A run is admissible if every process, except possibly one (faulty ● process), takes infinitely many steps in S

  19. Assume to the contrary that there exists P such that P is partially correct ● Agreement + Validity ○

  20. Assume to the contrary that there exists P such that P is partially correct ● Agreement + Validity ○ Every admissible run of P is a deciding run ● Termination ○

  21. Assume to the contrary that there exists P such that P is partially correct ● Agreement + Validity ○ Every admissible run of P is a deciding run ● Termination ○ What kind of contradiction should possibly be like? ●

  22. Categories of configurations Univalent, or i-valent (i in {0, 1}) ● A configuration C is univalent or i-valent if some process has ○ decided i in C, or if all configurations accessible from C are i-valent

  23. Categories of configurations Univalent, or i-valent (i in {0, 1}) ● A configuration C is univalent or i-valent if some process has ○ decided i in C, or if all configurations accessible from C are i-valent ... S 1 p 1 0, b C = (s, M) . . S 2 ... Decide on 0 . M p k 0, b ... S 3 s

  24. Categories of configurations Univalent, or i-valent (i in {0, 1}) ● A configuration C is univalent or i-valent if some process has ○ decided i in C, or if all configurations accessible from C are i-valent ... S 1 p 1 1, b C = (s, M) . . S 2 ... Decide on 1 . M p k 1, b ... S 3 s

  25. Categories of configurations Bivalent ● A configuration C is bivalent if some of the configurations ○ accessible from it are 0-valent while others are 1-valent ... S 1 Decide on 0 p 1 0, b C = (s, M) . . S 2 ... Decide on 1 . M p k 1, b ... S 3 Decide on 0 s

  26. Categories of configurations Bivalent (see Bivalent, read Undeciding) ● A configuration C is bivalent if some of the configurations ○ accessible from it are 0-valent while others are 1-valent ... S 1 Decide on 0 p 1 0, b C = (s, M) . . S 2 ... Decide on 1 . M p k 1, b ... S 3 Decide on 0 s

  27. What kind of contradiction should possibly be like? INDISTINGUISHABILITY between processes: Crashed ● Simply slow in processing or having a terrible network condition ●

  28. What kind of contradiction should possibly be like? INDISTINGUISHABILITY between processes: Crashed ● Simply slow in processing or having a terrible network condition ● For any protocol, there exists a configuration that is always bivalent.

  29. What kind of contradiction should possibly be like? INDISTINGUISHABILITY between processes: Crashed ● Simply slow in processing or having a terrible network condition ● For any protocol, there exists a configuration that is always bivalent. Remaining UNDECIDED in the value

  30. What kind of contradiction should possibly be like? INDISTINGUISHABILITY between processes: Crashed ● Simply slow in processing or having a terrible network condition ● For any protocol, there exists a configuration that is always bivalent. Remaining UNDECIDED in the value

  31. Proof Outline For any protocol, there is an initial configuration that is bivalent ● Then there is another bivalent configuration reachable from it after ● applying some event And another reachable bivalent configuration ● … ● An infinite undeciding run ●

  32. Most exciting part!!! Lemma 1 (commutativity of schedules) ● Suppose that from some C, the schedules S 1 , S 2 lead to C 1 , C 2 ○ respectively. If the steps in S 1 and in S 2 are disjoint, then S 2 can be applied to C 1 and S 1 can be applied to C 2 and both lead to the same C 3 .

  33. Most exciting part!!! Lemma 1 (commutativity of schedules) ● Suppose that from some C, the schedules S 1 , S 2 lead to C 1 , C 2 ○ respectively. If the steps in S 1 and in S 2 are disjoint, then S 2 can be applied to C 1 and S 1 can be applied to C 2 and both lead to the same C 3 . C S 1 S 2 C 1 C 2 S 2 S 1 C 3

  34. Proof Outline For any protocol, there is an initial configuration that is bivalent ● Then there is another bivalent configuration reachable from it after ● applying some event And another reachable bivalent configuration ● … ● An infinite undeciding run ●

  35. Most exciting part!!! Lemma 2 ● P has a bivalent initial configuration. ○

  36. Most exciting part!!! Lemma 2 ● P has a bivalent initial configuration. ○ Assume all initial configurations are either 0-valent or 1-valent. p 1 0, b 0, b 0, b 0, b 1, b 1, b 1, b 1, b p 2 0, b 0, b 1, b 1, b 1, b 1, b 0, b 0, b 1, b 0, b 0, b 1, b 1, b 0, b 0, b 1, b p 3

  37. Most exciting part!!! Lemma 2 ● P has a bivalent initial configuration. ○ Assume all initial configurations are either 0-valent or 1-valent. p 1 0, b 0, b 0, b 0, b 1, b 1, b 1, b 1, b p 2 0, b 0, b 1, b 1, b 1, b 1, b 0, b 0, b 1, b 0, b 0, b 1, b 1, b 0, b 0, b 1, b p 3 1 1 0 0 0 0 1 1

  38. Most exciting part!!! Lemma 2 ● P has a bivalent initial configuration. ○ Assume all initial configurations are either 0-valent or 1-valent. p 1 0, b 0, b 0, b 0, b 1, b 1, b 1, b 1, b p 2 0, b 0, b 1, b 1, b 1, b 1, b 0, b 0, b 1, b 0, b 0, b 1, b 1, b 0, b 0, b 1, b p 3 1 1 0 0 0 0 1 1

  39. Most exciting part!!! Lemma 2 ● P has a bivalent initial configuration. ○ Assume all initial configurations are either 0-valent or 1-valent. p 1 0, b 0, b 0, b 0, b 1, b 1, b 1, b 1, b p 2 0, b 0, b 1, b 1, b 1, b 1, b 0, b 0, b 1, b 0, b 0, b 1, b 1, b 0, b 0, b 1, b p 3 1 0 0 0 1 1 1 0

  40. Most exciting part!!! Lemma 2 ● P has a bivalent initial configuration. ○ Assume all initial configurations are either 0-valent or 1-valent. p 1 0, b 0, b 0, b 0, b 1, b 1, b 1, b 1, b p 2 0, b 0, b 1, b 1, b 1, b 1, b 0, b 0, b 1, b 0, b 0, b 1, b 1, b 0, b 0, b 1, b p 3 1 0 0 0 1 1 1 0

  41. Most exciting part!!! Lemma 2 ● P has a bivalent initial configuration. ○ Assume all initial configurations are either 0-valent or 1-valent. Adjacent: differ in the C 1 C 2 initial state of a single process

  42. Most exciting part!!! Lemma 2 ● P has a bivalent initial configuration. ○ Assume all initial configurations are either 0-valent or 1-valent. Adjacent: ... ... differ in the C 1 C 2 C i C i+1 C k initial state of a single process

  43. Most exciting part!!! Lemma 2 ● P has a bivalent initial configuration. ○ Assume all initial configurations are either 0-valent or 1-valent. Adjacent: ... ... differ in the C 1 C 2 C i C i+1 C k initial state of a single process 0 0 0 1 1

  44. Most exciting part!!! Lemma 2 ● P has a bivalent initial configuration. ○ Assume all initial configurations are either 0-valent or 1-valent. Adjacent: ... ... differ in the C 1 C 2 C i C i+1 C k initial state of a single process 0 0 0 1 1

  45. Most exciting part!!! differ in the initial state C i C i+1 of a single process p Lemma 2 ● P has a bivalent initial configuration. ○ Assume all initial configurations are either 0-valent or 1-valent. Adjacent: ... ... differ in the C 1 C 2 C i C i+1 C k initial state of a single process 0 0 0 1 1

  46. Most exciting part!!! differ in the initial state C i C i+1 of a single process p S in Lemma 2 ● which p takes no P has a bivalent initial configuration. ○ step C/p Assume all initial configurations are either 0-valent or 1-valent. Adjacent: ... ... differ in the C 1 C 2 C i C i+1 C k initial state of a single process 0 0 0 1 1

  47. Most exciting part!!! differ in the initial state C i C i+1 of a single process p S in S in Lemma 2 ● which p which p takes no takes no P has a bivalent initial configuration. ○ step step C/p Assume all initial configurations are either 0-valent or 1-valent. Adjacent: ... ... differ in the C 1 C 2 C i C i+1 C k initial state of a single process 0 0 0 1 1

  48. Most exciting part!!! differ in the initial state C i C i+1 of a single process p S in S in Lemma 2 ● which p which p takes no takes no P has a bivalent initial configuration. ○ step step C/p Assume all initial configurations are either 0-valent or 1-valent. Adjacent: ... ... 0 differ in the C 1 C 2 C i C i+1 C k initial state of a single process 0 0 0 1 1

  49. 0/1 Most exciting part!!! differ in the initial state C i C i+1 of a single process p S in S in Lemma 2 ● which p which p takes no takes no P has a bivalent initial configuration. ○ step step C/p Assume all initial configurations are either 0-valent or 1-valent. Adjacent: ... ... 0 differ in the C 1 C 2 C i C i+1 C k initial state of a single process 0 0 0 1 1

  50. Most exciting part!!! differ in the initial state C i C i+1 of a single process p S in S in Lemma 2 ● which p which p takes no takes no P has a bivalent initial configuration. ○ step step C/p Assume all initial configurations are either 0-valent or 1-valent. Adjacent: ... ... 1 differ in the C 1 C 2 C i C i+1 C k initial state of a single process 0 0 0 1 1

  51. 0/1 Most exciting part!!! differ in the initial state C i C i+1 of a single process p S in S in Lemma 2 ● which p which p takes no takes no P has a bivalent initial configuration. ○ step step C/p Assume all initial configurations are either 0-valent or 1-valent. Adjacent: ... ... 1 differ in the C 1 C 2 C i C i+1 C k initial state of a single process 0 0 0 1 1

  52. Proof Outline For any protocol, there is an initial configuration that is bivalent ● Then there is another bivalent configuration reachable from it after ● applying some event And another reachable bivalent configuration ● … ● An infinite undeciding run ●

  53. Most exciting part!!! Lemma 3 ● Let C be a bivalent configuration of P, and e=(p, m) be an event that is ○ applicable to C. Let E be the set of configurations reachable from C without applying e, and let D=e(E), the set of configurations after applying e to all those in E. Then, D contains a bivalent configuration.

  54. Most exciting part!!! Lemma 3 ● Let C be a bivalent configuration of P, and e=(p, m) be an event that is ○ applicable to C. Let E be the set of configurations reachable from C without applying e, and let D=e(E), the set of configurations after applying e to all those in E. Then, D contains a bivalent configuration. C 0/1

  55. Most exciting part!!! Lemma 3 ● Let C be a bivalent configuration of P, and e=(p, m) be an event that is ○ applicable to C. Let E be the set of configurations reachable from C without applying e, and let D=e(E), the set of configurations after applying e to all those in E. Then, D contains a bivalent configuration. Any schedule E 1 without C applying e E 2 E 3 ... 0/1 E

  56. Most exciting part!!! Lemma 3 ● Let C be a bivalent configuration of P, and e=(p, m) be an event that is ○ applicable to C. Let E be the set of configurations reachable from C without applying e, and let D=e(E), the set of configurations after applying e to all those in E. Then, D contains a bivalent configuration. Any schedule E 1 D 1 Apply e without C applying e E 2 D 2 E 3 D 3 ... ... 0/1 E D

  57. Most exciting part!!! Lemma 3 ● Let C be a bivalent configuration of P, and e=(p, m) be an event that is ○ applicable to C. Let E be the set of configurations reachable from C without applying e, and let D=e(E), the set of configurations after applying e to all those in E. Then, D contains a bivalent configuration. Any schedule E 1 D 1 Apply e without C applying e E 2 D 2 E 3 D 3 0/1 ... ... 0/1 E D

  58. Most exciting part!!! Assume all configurations in D are univalent. ●

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Disjoint-Access Parallelism: Impossibility, Possibility, and Cost of Transactional Memory Implementations Sebastiano Peluso 1 , Roberto Palmieri 1 , Paolo Romano 2 , Binoy Ravindran 1 and Francesco Quaglia 3 3 1 2

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