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Distributed Algorithms – 2g1513

Distributed Algorithms – 2g1513. Lecture 10 – by Ali Ghodsi Fault-Tolerance in Asynchronous Networks. Consensus Problems. Consensus problems very important in DS Distributed Databases All processes must agree whether to commit or abort a transaction

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Distributed Algorithms – 2g1513

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  1. Distributed Algorithms – 2g1513 Lecture 10 – by Ali Ghodsi Fault-Tolerance in Asynchronous Networks

  2. Consensus Problems • Consensus problems very important in DS • Distributed Databases • All processes must agree whether to commit or abort a transaction • If any process says abort, all processes should abort • Atomic Broadcast • All processes receive the same set of messages coming from correct processes only • Can be used to implement consensus, vice versa

  3. Fischer, Lynch, Paterson 1983/85 • Consensus cannot be solved in asynchronous model • With possibility of one process crashing • http://www.sics.se/~ali/flp85.pdf • Most influential paper award PODC 2001

  4. Modified Model • To proof the result, we will modify our model of a distributed system slightly • Processes execute local algorithms, modeled by a STS • But, given any state, a correct process can always execute a “dummy” instruction • For any state  in a process, there exists a transition  • There exists always an applicable event on every process • A crashed process, cannot make any transitions

  5. Definition: T-crash fair executions • A t-crash-robust algorithm is a consensus algorithm if it satisfies: • Termination • All correct processes eventually decides • Agreement • In every configuration, the decided processes should have decided for the same value (0 or 1) • Non-triviality • There exists at least one possible input configuration where the decision is 0 • There exists at least one possible input configuration where the decision is 1 • Example, maybe input “0,0,1”->0 while “0,1,1”->1

  6. Definitions • 0-decided configuration • A configuration with decide ”0” on some process • 1-decided configuration • A configuration with decide ”1” on some process • 0-valent configuration • A configuration in which every reachable decided configuration is a 0-decide • 1-valent configuration • A configuration in which every reachable decided configuration is a 1-decide • Bivalent configuration • A configuration which can reach a 0-decided and 1-decided configuration

  7. Definitions Illustrated 1(4) • 0-decided configuration • A configuration with decide ”0” on some process 0-decided configuration {STATE2, STATE,5 DECIDE-0, STATE7 {msg1, msg2} } P1 state2 msg2 At least of them is in state DECIDE-0 P2 state5 msg1 P3 decide0 P4 state7

  8. Definitions Illustrated 2(4) 0-valent configuration {decide-0, P2_state2, P3_state2, decide-0, { msg2} } • 0-valent configuration • No 1-decided configurations are reachable • Future determined, means ”everyone will decide 0” 0-valent configuration {decide-0, P2_state2, P3_state2, P4_state, {msg1, msg2} } 0-valent configuration {decide-0, P2_state, decide-0, P4_state, {msg1, msg2} } 0-valent configuration { P1_state, P2_state2, P3_state, P4_state, {msg1} } 0- valent configuration { P1_state, P2_state, P3_state, P4_state, {msg1} } 0-valent configuration {decide-0, P2_state, P3_state, P4_state, {msg1, msg2} } 0-valent configuration {decide-0, P2_state3, P3_state, decide-0, {} } 0-valent configuration {decide-0, P2_state, P3_state, decide-0, { msg2} }

  9. Definitions Illustrated 3(4) 0-valent configuration {decide-1, P2_state2, P3_state2, decide-1, { msg2} } • 1-valent configuration • No 0-decided configurations are reachable • Future determined, means ”everyone will decide 1” 0-valent configuration {decide-1, P2_state2, P3_state2, P4_state, {msg1, msg2} } 0-valent configuration {decide-1, P2_state, decide-1, P4_state, {msg1, msg2} } 0-valent configuration { P1_state, P2_state2, P3_state, P4_state, {msg1} } 0- valent configuration { P1_state, P2_state, P3_state, P4_state, {msg1} } 0-valent configuration {decide-1, P2_state, P3_state, P4_state, {msg1, msg2} } 0-valent configuration {decide-1, P2_state3, P3_state, decide-1, {} } 0-valent configuration {decide-1, P2_state, P3_state, decide-1, { msg2} }

  10. Definitions Illustrated 4(4) 0-valent configuration {decide-0, P2_state2, P3_state2, decide-0, { msg2} } • Bivalent configuration • Both 0 and 1-decided configurations are reachable • Future undetermined, could go either way… 0-valent configuration {decide-0, P2_state2, P3_state2, P4_state, {msg1, msg2} } 0-valent configuration {decide-0, P2_state, decide-0, P4_state, {msg1, msg2} } 0-valent configuration { P1_state, P2_state2, P3_state, P4_state, {msg1} } bivalent configuration { P1_state, P2_state, P3_state, P4_state, {msg1} } 1-valent configuration {decide-1, P2_state5, P3_state6, P4_state5, {msg1, msg3} } 1-valent configuration {decide-1, P2_state9, P3_state6, decide-1, {} } 1-valent configuration {decide-1, P2_state5, P3_state6, decide-1, { msg2} }

  11. Bivalent Initial Configuration • Theorem • For any algorithm that solves the 1-crash consensus problem there exists an initial bivalent configuration

  12. Proof 1/(10) • We know that the algorithm must be non-trivial • There should be some initial configuration that will lead to a 0-decide • There should be some initial configuration that will lead to a 1-decide • Take two such configuration i1 and i2 • E.g. 4 processes • initial values (0,1,0,1,1) lead to 1 • Initial values (0,0,1,0,0) lead to 0

  13. Proof 2/(10) • We know there exists inputs p1, p2, p3, p4, p5 • (0,1,0,1,1) leading to 1 • (0,0,1,0,0) leading to 0 Lets look at other initial configurations by flipping the inputs transforming the upper input to the lower input

  14. Proof 3/(10) • We know there exists inputs p1, p2, p3, p4, p5 • (0,1,0,1,1) leading to 1 • (0,0,0,1,1) leading to ? • (0,0,1,0,0) leading to 0 Lets look at other initial configurations by flipping the inputs transforming the upper input to the lower input

  15. Proof 4/(10) • We know there exists inputs p1, p2, p3, p4, p5 • (0,1,0,1,1) leading to 1 • (0,0,0,1,1) leading to ? • (0,0,1,1,1) leading to ? • (0,0,1,0,0) leading to 0 Lets look at other initial configurations by flipping the inputs transforming the upper input to the lower input

  16. Proof 5/(10) • We know there exists inputs p1, p2, p3, p4, p5 • (0,1,0,1,1) leading to 1 • (0,0,0,1,1) leading to ? • (0,0,1,1,1) leading to ? • (0,0,1,0,1) leading to ? • (0,0,1,0,0) leading to 0 Lets look at other initial configurations by flipping the inputs transforming the upper input to the lower input

  17. Proof 6/(10) • We know there exists inputs p1, p2, p3, p4, p5 • (0,1,0,1,1) leading to 1 • (0,0,0,1,1) leading to ? • (0,0,1,1,1) leading to ? • (0,0,1,0,1) leading to ? • (0,0,1,0,0) leading to 0 Lets look at other initial configurations by flipping the inputs transforming the upper input to the lower input There must exist two neighboring configurations here, with two different outcomes

  18. Proof 7/(10) • We know there exists inputs p1, p2, p3, p4, p5 • (0,1,0,1,1) leading to 1 • (0,0,0,1,1) leading to 1 • (0,0,1,1,1) leading to 1 • (0,0,1,0,1) leading to 0 • (0,0,1,0,0) leading to 0 Lets look at other initial configurations by flipping the inputs Assume the following two

  19. Proof 8/(10) • We know there exists inputs p1, p2, p3, p4, p5 • (0,1,0,1,1) leading to 1 • (0,0,0,1,1) leading to 1 • (0,0,1,1,1) leading to 1 • (0,0,1,0,1) leading to 0 • (0,0,1,0,0) leading to 0 Assume the following two Identical configurations except for process p4

  20. Proof 9/(10) • We know there exists inputs p1, p2, p3, p4, p5 • (0,0,1,1,1) leading to 1 • (0,0,1,0,1) leading to 0 • The consensus algorithm should tolerate if p4crashes! • (0,0,1,X,1), leads to ? (either 0 or 1) Assume the following two

  21. Proof 10/(10) • We know there exists inputs p1, p2, p3, p4, p5 • (0,0,1,1,1) leading to 1 • (0,0,1,0,1) leading to 0 • The consensus algorithm should tolerate if p4crashes! • (0,0,1,X,1), leads to ? (either 0 or 1) • If it leads to 1, then depending on whether p4 crashes or not (0,0,1,0,1) either leads to 0 or 1 (bivalent) • If it leads to 0, then depending on whether p4 crashes or not (0,0,1,1,1) either leads to 0 or 1 (bivalent) Assume the following two

  22. Initial Bivalence • Intuition • Given any algorithm, we can find some start state, that depending on the failure of one process, will either lead to a 0-decide or a 1-decide 1-valent configuration {decide-1, P2_state2, P3_state2, P4_state, {msg1, msg2} } 1-valent configuration { P1_state, P2_state, decide-1, P4_state, {msg1, msg2} } 1-valent configuration { P1_state, P2_state2, P3_state, P4_state, {msg1} } Bivalent Initial Config { P1_state, P2_state, P3_state, P4_state, {msg1} } 0-valent configuration { P1_state, P2_state, P3_state, P4_state, {msg1, msg2} } 0-valent configuration {decide-0, decide-0, P3_state, decide-0, {} } 0-valent configuration {decide-0, P2_state, P3_state, P4_state, { msg2} }

  23. Coarse-grained Model of Distributed Systems • In our model, we will now let each event be the receipt of a message • After the receipt of a message m, a process deterministically makes all internal and send events it can do • In other words, we make our course-grained model a bit more fine-grained • An event represents the receipt of a message, some internal transitions and the sending of some messages • A receipt of message m at process p is always applicable if a message m with destination p is in the network

  24. Intuition behind model Initial state of p receive <tok, y> from q for x:=1 to 3 do begin y:=y+1; send <tok, y> neighp[x]; end receive <tok, z> from q; print z+y Receipt event e Deterministic transitions State of p after receipt of e Receipt event f Deterministic transitions State of p after receipt of f

  25. Order of events • Intuition • The order in which two applicable events are executed is not important! • Order Theorem • Let ep and eq be two events on two different processors p and q which are both applicable in configuration . Then ep can be applied to eq(), and eq can be applied to ep(). • Moreover, ep(eq()) = eq(ep() ).

  26. Definitions • A sequence of events =( e1, e2,…,ek) is applicable in configuration  if • e1 is applicable in , • e2 applicable in e1() • ... • If the resulting configuration is  we write ()= or  • If  only contains events of a subset of the processes P, we write P

  27. Order of sequences • Diamond Theorem • Let sequences 1 and 2 be applicable in configuration , and let no process participate in both 1 and 2. Then 2 is applicable in 1(), 2 is applicable in 2(), and 1(2())=2(1()) • Proof • By induction using the order theorem

  28. Illustration of the Diamond Theorem  1 2 2() 1() 2 1   =2(1())=1(2())

  29. Bivalent Configuration • Any configuration of the 1-robust consensus algorithm is exactly one of these three • Bivalent • 0-valent • 1-valent • Why? • Any configuration leads to a decide because of termination • We know bivalent configurations exist • If it is not bivalent, it must lead to either 0-decide or 1-decide, so it is either 0-valent or 1-valent

  30. Bivalent Configurations • In any bivalent config , either • one applicable event goes to a bivalent config, or • there exists two applicable events, leading to a 0-valent and 1-valent configurations (respectively) Case 1 Case 2 0-valent Bivalent Bivalent Bivalent 1-valent

  31. Staying Bivalent • Theorem • Given any bivalent config  and an event e applicable in  • There exists another reachable config  where e is applicable, and e() is bivalent Theorem Illustration    e e Bivalent … … e Bivalent Bivalent … …

  32. Proof definitions • Assume e involves process p • Call the set of all possible configs reachable from  without applyinge the set C • Apply eventeto all configs inCand call the resulting configsD Theorem Illustration C … … e … … e e … … … Bivalent e … e … … D e … …

  33. Proof intuition • We will proof that D contains a bivalent config by contradiction • I.e., assume there exists no bivalent config in D, show that this will lead to a contradiction or absurdity Theorem Illustration C … … e … … e e … … … Bivalent e … e … … D e … …

  34. Proof • Assume D contains no bivalent configs • I.e. all configs in D are either 0-valent or 1-valent • Then it follows that there exists a 0-valent and a 1-valent config in D (next slides)

  35. Proof • We know we can reach a 0-valent and 1-valent config from , call them 1and 2 (non-triviality) • Either 1 and 2 are in C or they are not in C • If inside C, then e(1) and e(2) is in D and they are 0-valent/1-valent 1 and 2 are in C 1 and 2 are not in C C C 1 … … … e e … … … … e e e e 1 … … … … Bivalent Bivalent e e … … e e 2 … e e … … 2 … …

  36. Proof • If not inside C, then some1 and 2 exists on the path to 1and 2, such that e(1) ande(2) are inDand they are 0-valent/1-valent • [Remember we assumed no bivalent config available inD] 1 and 2 are in C 1 and 2 are not in C C C 1 … … … e e … … … … e e e e 1 1 … … … … Bivalent Bivalent e e … … 2 e e 2 … e e … … 2 … …

  37. Reflection • We now know that D must always contain a 0-valent and 1-valent config, assuming no bivalent config exists in D • Lets call the two 0-valent and 1-valent configs in D, d0 and d1 • We will now show that this situation is a contradiction itself. Hence, D must contain a bivalent config

  38. Deriving the contradiction • There must exist two configs c0 and c1 in C such that c1=f(c0), and d0=e(c0) and d1=e(c1) f C c0 c1 e e d0 d1 D • Lets see why!

  39. Proofing two neighbors exist 1(4) • We know  is bivalent, and e() is in D and is either 0-valent or 1-valent, assume 0-valent C  e 0-valent D

  40. Proofing two neighbors exist 2(4) • We know  is bivalent, and e() is in D and is either 0-valent or 1-valent, assume 0-valent • There is a reachable 1-valent config in D f0 2 … m C  1 e e 1-valent 0-valent D

  41. Proofing two neighbors exist 3(4) • We know  is bivalent, and e() is in D and is either 0-valent or 1-valent, assume 0-valent • There is a reachable 1-valent config in D • e is applicable in each i, and must be 0-valent or 1-valent f0 2 … m C  1 e e e e e x-valent z-valent 1-valent y-valent 0-valent D

  42. Proofing two neighbors exist 4(4) • We know  is bivalent, and e() is in D and is either 0-valent or 1-valent, assume 0-valent • There is a reachable 1-valent config in D • e is applicable in each i, and must be 0-valent or 1-valent f0 f1 f3 f2 2 … m C  1 e e e e e There exists two neighbors, one 1-valent and one 0-valent 0-valent z-valent 1-valent 1-valent 0-valent D

  43. Proofing two neighbors exist 4(4) • We know  is bivalent, and e() is in D and is either 0-valent or 1-valent, assume 0-valent • There is a reachable 1-valent config in D • e is applicable in each i, and is 0/1-valent f 2 C 1 e e There exists two neighbors, one 1-valent and one 0-valent 0-valent 1-valent D

  44. Neighbors lead to contradiction 1(3) • We now know there exist two configs c0 and c1 in C such that c1=f(c0), and d0=e(c0) and d1=e(c1) • Either the events e and f happen on the same processor or on different processors, both cases will lead to contradictions f 2 C 1 e e There exists two neighbors, one 1-valent and one 0-valent 0-valent 1-valent D

  45. f Neighbors lead to contradiction 2(3) • We now know there exist two configs c0 and c1 in C such that c1=f(c0), and d0=e(c0) and d1=e(c1) • Assume e and f happen on two different processes p and q • Then, the order of their execution can be exchanged f C c0 c1 e e d0 d1 0-valent 1-valent D Contradiction as d0 is 0-valent, but it can lead to a 1-valent config, hence d0 must be bivalent, but we assumed no bivalent configs exist in D

  46. If p is silent, the algorithm should continue and terminate with a decision in some config A If p is silent, some execution leading to 1should exist If p is silent, some execution leading to 0should exist f e e A 2 0 1 0-valent 1-valent Neighbors lead to contradiction 3(3) • We now know there exist two configs c0 and c1 in C such that c1=f(c0), and d0=e(c0) and d1=e(c1) • Assume e and f happen on the same process p, the algorithm should still work if p is silent C f e e c0 c1 d0 d1 Contradiction as A should be a 0/1-valent configuration, but we have shown that A can lead to both 0and 1

  47. Proof Map Assume there is no bivalent config in D • We know all configs in D are 0-valent or 1-valent • Show that we can find a 0-valent and 1-valent config in D • Show that two neighboring configs c0─e→c1 exist, where c0 ─f→”0-valent config”, c1 ─f→”1-valent config” • Show this is a contradiction Assumption must be incorrect D must contain a bivalent configuration

  48. Final Theorem • No deterministic 1-crash-robust consensus algorithm exists for the asynchronous model • Proof • Start in a initial bivalent config • Given the bivalent config, pick the event e that has been applicable longest • Pick the execution taking us to another config where e is applicable • Apply e, and get a bivalent config • Repeat 2.

  49. Consensus not Impossible! • Lets do deterministic consensus algorithm for the a different failure model • Initially dead processes • Assume t failures can happen initially • Where t=4 for N=10, t=5 for N=11 • Let Ldenote L=6 for N=10, L=6 for N=11 N=t+L

  50. Intuition • Assume N processes are connected in a underlying graph, and at most t fail • We know L processes are alive after the start • Broadcast your identity, and receive/collect L identities • For any two correct processes, their set of collected identities will overlap • Quorom concept • There are N nodes, any two processes have L identities each, i.e. total • N+1 identities, total N nodes, at least two must be same (PHP)

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