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by J. Burns and J. Pachl

Uniform Self-Stabilization. by J. Burns and J. Pachl. Prepared by: Irina Shapira Julia Mosin Modified by Mordo Shalom. Self-Stabilizing System: S = (G, R,  ,  ). G = (V,E). R - orientation of G specifies a total ordering on the neighbors for each v in V.

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by J. Burns and J. Pachl

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  1. Uniform Self-Stabilization by J. Burns and J. Pachl Prepared by: Irina Shapira Julia Mosin Modified by Mordo Shalom

  2. Self-Stabilizing System: S = (G, R, , ) • G = (V,E) • R - orientation of Gspecifies a total ordering on the neighbors for each v in V •  = 0 x1 x . . . xn-1 - set of configurationswherei is a finite set of states of processor Pi •  = 0 ,1 , . . . ,n-1 - sequence of transition relationswherei is a transition relation of processor Pi from i xv1 x . . . xvk to i ( R(i) = (v1, . . ., vk))

  3. Self-Stabilizing System: S = (G, R, , ) • For configuration  = (s0, s1, … , sn-1)   where • sj is a state of Pj (for all j). • x - possible next state of Pi , • i.e. if R(vi) = (v1, . . ., vk ) then (si,sv1, …, svk , x)  i we define i () as a set of all configurations ’ such that ’ = (s0,s1, …, si-1, x, si+1, …, sn-1) (for all possible values of x)

  4. Self-Stabilizing System: S = (G, R, , ) • Pi is enabled (privileged) at  if i () is not empty •  -> ’is a step of Pi if ’ i () i •  -> ’is a step of Pi for some i • 0, 1, ... is a computation of the system where j-1 -> j for all j Only one processor takes a step at a time - central daemon

  5. An n-processor uniform ring is a system S = (G, R, , ) where • G is a cycle • processors are denoted by 0,1, …, n-1 in order around the cycle • R(i) = ( (i-1) mod n, (i+1) mod n ) P1 P0 • 0 =1 = . . . =n-1 P2 • 0 =1 = . . . =n-1 P5 P3 P4

  6. Definition 1: A system S = (G, R, , ) is self-stabilizing iff there is a set , called the legitimate configurations of S, such that: (1) No Deadlock For every  there is a ‘ such that  -> ’ (2) Closure For every , every ’ such that  -> ’ is in  (3) No Livelock Every infinite computation of S contains a configuration in  (4) Mutual Exclusion For every , exactly one processor is enabled (5) Fairness For every processor Pi, every infinite computation consisting of configurations in  contains an infinite number of steps by Pi

  7. Theorem 1: There is no uniform n-processor self-stabilizing ring if n is composite. Proof: • [i] = {Pj | j = i mod p} • Sym()  all the Pj in [i] are in the same state in  • if Sym(0) holds, we can construct an infinite computation for which Sym(i) holds for every i s1 s0 P1 s0 s2 s0 s2 P0 P2 0: 1: 2: P5 P3 s0 s1 s0 s1 P4 s0 s2

  8. If the size of the ring is prime: • deterministic algorithm • no distinguished processor • central demon • ( n2 ) states

  9. The Protocol: • deterministic • unidirectional (the value of di(si, si-1, si+1) depends only on si and si-1) si-1si -> x means that x di(si,si-1,si+1) for any si+1

  10. A First Attempt • 0 =1 = . . . =n-1 = { 0, 1, …, n-2 } • arithmetic modulo n-1 The rule: if sisi-1+1 then si-1sisi-1+1

  11. The rule: if sisi-1+1 then si-1sisi-1+1 The cycle of the intended legitimate configurations P0 P1 P2 P3 P4 0 0 1 2 3 0 0 1 2 3 0 0 1 1 2 3 P0 P0 P0 0 1 2 2 3 0 1 2 3 3 P4 P4 P4 P1 P1 P1 1 0 0 0 3 3 0 1 2 3 0 ... P2 P3 P2 P3 P3 P2 1 2 2 1 3 2 0 0 1 2 3

  12. The rule: if sisi-1+1 then si-1sisi-1+1 The cycle which does not intersect the cycle of the legitimate configurations P0 P1 P2 P3 P4 0 1 20 1 0 1 20 1 0 1 2 0 0 1 2 31 21 2 31 P0 P0 P0 P0 P0 P0 P0 21 2 3 0 2 32 3 0 P1 P1 P1 P1 P1 P1 P1 P4 P4 P4 P4 1 P4 P4 3 1 2 1 0 1 132 30 13 03 0 1 203 0 P2 P2 P2 P2 P2 P2 P2 P3 P3 P3 P3 P3 P3 P3 0 1 20 10 1 3 2 2 0 0 ... 0 1 20 1 livelock

  13. A Correct Protocol

  14. The Protocol: • n  3 • states are label.tag (Li.Ti) • labels range over { 0, 1, …, n-2 } • tags range over { 0 } U { 2, 3, …, n-2 } Rule A: if Li  Li-1+1 and (Li0 or Ti-1=0 or Ti-1Li-Li-1 or Ti-1<Ti) then Li-1.Ti-1Li.Ti (Li-1+1).(Li-Li-1) Rule B: if Li = Li-1+1 and Li  0 and Ti-1 Ti then Li-1.Ti-1 Li.TiLi.Ti-1

  15. Rule A: if Li  Li-1+1 and (Li0 or Ti-1=0 or Ti-1Li-Li-1 or Ti-1<Ti) then Li-1.Ti-1Li.Ti (Li-1+1).(Li-Li-1) Rule B: if Li = Li-1+1 and Li0 and Ti-1 Tithen Li-1.Ti-1 Li.TiLi.Ti-1 The Proof of Self-Stabilization: S = (G, R, , ) LetLbe the set of all cyclic permutations of configurations of the form 0.0 1.0 … a-1.0 a.0 a.0 a+1.0 … n-2.0 for a=0, 1, …, n-2 Theorem 3: If n is prime, then the ring is self-stabilizing(and Lis a set of legitimate configurations)

  16. Definition 2: • Let gG be a configuration • if Li+1 Li+1 (mod n-1), then Pi and Pi+1 forma gap of g • the gap size (of Pi and g) g(i,g) = Li+1- Li (mod n-1) • a segment of g is a maximal cyclically contiguous sequence s=(Pi, Pi+1, …, Pj-1, Pj ) which contains no gaps • Pi is the left end of s, Pj is the right end of s • the gap size of s g(s,g) is the gap size of its right end (g(j,g)) • a segment is zero based if its left end is 0 (Li = 0) Example: P0 P1 P2 P3 P4 P5 P6 1.3 2.3 3.1 5.2 0.2 1.1 0.4 P0 P1 P2 P3 P4 P5 P6 1.3 2.3 3.1 5.2 0.2 1.1 0.4 (P6 P0 P1 P2) zero based segment gap of size 5 gap of size 2

  17. Rule A: if Li  Li-1+1 and (Li0 or Ti-1=0 or Ti-1Li-Li-1 or Ti-1<Ti) then Li-1.Ti-1Li.Ti (Li-1+1).(Li-Li-1) Rule B: if Li = Li-1+1 and Li0 and Ti-1 Tithen Li-1.Ti-1 Li.TiLi.Ti-1 P0 P1 P2 P3 P4 1.0 2.0 3.32.33.0 0.22.03.32.33.0 0.2 1.0 1.0 0.21.23.32.33.0 P0 P0 P0 P0 P0 0.21.23.32.3 3.3 0.21.23.30.33.3 P1 P1 P1 P1 P1 P4 P4 P4 P4 P4 P4 P4 0.21.23.30.31.3 P4 P4 2.0 1.2 2.0 0.2 1.0 3.0 1.3 3.3 0.21.22.20.31.3 0.21.22.23.2 1.3 0.21.22.23.2 0.2 P2 P2 P2 P2 P2 P3 P3 P3 P3 P3 P3 P3 1.01.22.23.2 0.2 0.0 3.3 2.3 3.2 3.0 0.3 2.2 1.02.02.23.2 0.2 1.02.03.03.2 0.2 1.02.03.00.0 0.2 1.02.03.00.0 1.0

  18. Definition 1: A system S = (G, R, , ) is self-stabilizing iff there is a set , called the legitimate configurations of S, such that: (1) No Deadlock (2) Closure (3) No Livelock (4) Mutual Exclusion (5) Fairness

  19. Rule A: if Li  Li-1+1 and (Li0 or Ti-1=0 or Ti-1Li-Li-1 or Ti-1<Ti) then Li-1.Ti-1Li.Ti (Li-1+1).(Li-Li-1) Rule B: if Li = Li-1+1 and Li0 and Ti-1 Tithen Li-1.Ti-1 Li.TiLi.Ti-1 Lemma 1: Conditions 2 (Closure), 4 (Mutual Exclusion), and 5 (Fairness) of Definition 1 hold for L and S. Proof: • all configurations in Lare cyclic permutations of the forml = 0.0 1.0 … a-1.0 a.0 a.0 a+1.0 … n-2.0 • Ti = Tj for all i and j, hence rule B does not apply • rule A applies to exactly one processor - Mutual Exclusion • l -> l’by rule A l’ = 0.0 1.0 … a-1.0 a.0 a+1.0 a+1.0 … n-2.0 • l’  L- Closure • if Pi is enabled in l , then Pi+1(mod n) will be enabled in l ’- Fairness

  20. Rule A: if Li  Li-1+1 and (Li0 or Ti-1=0 or Ti-1Li-Li-1 or Ti-1<Ti) then Li-1.Ti-1Li.Ti (Li-1+1).(Li-Li-1) Rule B: if Li = Li-1+1 and Li0 and Ti-1 Tithen Li-1.Ti-1 Li.TiLi.Ti-1 Lemma 2: If n is prime, then some processor is enabled for every configuration in G, and therefore Condition 1 (No Deadlock) holds Proof: • if rule A doesn’t apply, then all the segments are zero-based … Li-1.Ti-1 0.Ti . … • Case 1: one zero-based segment of length n … 0.Ti-1 0.Ti ... if Ti-1 = 0 rule A applies else Ti-1  0 and so Ti-1  Li - Li-1 and again rule A applies

  21. Rule A: if Li  Li-1+1 and (Li0 or Ti-1=0 or Ti-1Li-Li-1 or Ti-1<Ti) then Li-1.Ti-1Li.Ti (Li-1+1).(Li-Li-1) Rule B: if Li = Li-1+1 and Li0 and Ti-1 Tithen Li-1.Ti-1 Li.TiLi.Ti-1 • Case 2: there is more than one segment rule B is not applicable - all tags within each segment are equal ... 0.t2 1.t2 … m2.t2 0.tl 1.tl … ml.tl 0.t1 1.t1 … m1.t1 rule A doesn’t apply - Ti-1= Li- Li-1for each gap tj = 0 - mj rule A doesn’t apply - tj  tj+1for each gap all the gaps and all the segments have the same size Since n is prime, all the segment lengths must be 1 ... 0.0 0.0 0.0 0.0 and rule A applies for every gap

  22. Rule A: if Li  Li-1+1 and (Li0 or Ti-1=0 or Ti-1Li-Li-1 or Ti-1<Ti) then Li-1.Ti-1Li.Ti (Li-1+1).(Li-Li-1) Rule B: if Li = Li-1+1 and Li0 and Ti-1 Tithen Li-1.Ti-1 Li.TiLi.Ti-1 Lemma 3: In every infinite computation of S, every processor takes a step by rule A infinitely many times Proof: • if somePi takes an infinite number of steps by rule A, so do all • suppose that there is no such Pi • all labels are fixed and so are all segments • the left end of the segment cannot take a step by rule B • Pi which is not the left end can take at most one step more than Pi-1 Pi … ... Pi-1

  23. Rule A: if Li  Li-1+1 and (Li0 or Ti-1=0 or Ti-1Li-Li-1 or Ti-1<Ti) then Li-1.Ti-1Li.Ti (Li-1+1).(Li-Li-1) Rule B: if Li = Li-1+1 and Li0 and Ti-1 Tithen Li-1.Ti-1 Li.TiLi.Ti-1 Lemma 4: For any computation of S, the number of segments is nonincreasing. Proof: • rule B does not change the number of segments • rule A cannot increase the number of segments … Pi-1 Pi Pi+1 … … Pi-1 Pi Pi+1 … Pi … Pi-1 Pi … Pi-1 Pi+1 … Pi+1 …

  24. Rule A: if Li  Li-1+1 and (Li0 or Ti-1=0 or Ti-1Li-Li-1 or Ti-1<Ti) then Li-1.Ti-1Li.Ti (Li-1+1).(Li-Li-1) Rule B: if Li = Li-1+1 and Li0 and Ti-1 Tithen Li-1.Ti-1 Li.TiLi.Ti-1 Definition: A computation in which the number of segments is constant is called quiet.

  25. Rule A: if Li  Li-1+1 and (Li0 or Ti-1=0 or Ti-1Li-Li-1 or Ti-1<Ti) then Li-1.Ti-1Li.Ti (Li-1+1).(Li-Li-1) Rule B: if Li = Li-1+1 and Li0 and Ti-1 Tithen Li-1.Ti-1 Li.TiLi.Ti-1 Definition : Let C = g0g1 … be a quiet computation • a dynamic gap is a function z from indices of computation into indices of processors, satisfying the following: (1) processor z(0) is the right end of the segment in g0 (2) if gj-1-> gj is a step of Pi by rule A and z(j-1) = i-1 then z(j) = i, else z(j)=z(j-1) z(j-1) gj-1: Pi Pi+1 ... ... Pi-1 z(j) gj: Pi+1 ... ... Pi-1 Pi

  26. Rule A: if Li  Li-1+1 and (Li0 or Ti-1=0 or Ti-1Li-Li-1 or Ti-1<Ti) then Li-1.Ti-1Li.Ti (Li-1+1).(Li-Li-1) Rule B: if Li = Li-1+1 and Li0 and Ti-1 Tithen Li-1.Ti-1 Li.TiLi.Ti-1 Definition (cont): • a dynamic segmentis a function sz from indices of computation into segments, such that sz(j) is a segment ingj whose right end is z(j). sz(j) gj: ... Pi z(j)

  27. Rule A: if Li  Li-1+1 and (Li0 or Ti-1=0 or Ti-1Li-Li-1 or Ti-1<Ti) then Li-1.Ti-1Li.Ti (Li-1+1).(Li-Li-1) Rule B: if Li = Li-1+1 and Li0 and Ti-1 Tithen Li-1.Ti-1 Li.TiLi.Ti-1 Lemma 5: Let C = g0g1 … be a quiet computation and z a dynamic gap of C. Then g(z(j-1), gj-1) = g(z(j), gj) for 0 < j < length(C). Proof: • Pi= z(j-1)takes a step - the segment disappears-contradiction z(j-1) Pi • Piz(j-1)+1takes a step - does not affect the gap • Pi=z(j-1)+1 takes a step length > 1 z(j-1) gj-1: Pi Pi+1 ... ... Pi-1 g =Li-Li-1 z(j) L’i+1=Li +1 L’i=Li-1+1 gj: Pi+1 ... ... Pi-1 Pi

  28. Rule A: if Li  Li-1+1 and (Li0 or Ti-1=0 or Ti-1Li-Li-1 or Ti-1<Ti) then Li-1.Ti-1Li.Ti (Li-1+1).(Li-Li-1) Rule B: if Li = Li-1+1 and Li0 and Ti-1 Tithen Li-1.Ti-1 Li.TiLi.Ti-1 Lemma 6: • Let C = g0g1 … be an infinite quiet computation and z a dynamic gap of C. Then • there is j0, such that the state of processor z(j) in gj is 0.g(z) and • there is k0, such that the state of processor z(k) +1 in gk is 0.u for some u Proof: sz gj1: ... a.* ... z(j1) gj2: ... ... a.* (a+1).g(z) z(j2) ... at most n-1 times gj: … 0.g(z) ... z(j)

  29. Rule A: if Li  Li-1+1 and (Li0 or Ti-1=0 or Ti-1Li-Li-1 or Ti-1<Ti) then Li-1.Ti-1Li.Ti (Li-1+1).(Li-Li-1) Rule B: if Li = Li-1+1 and Li0 and Ti-1 Tithen Li-1.Ti-1 Li.TiLi.Ti-1 Lemma 6 (b): • Let C = g0g1 … be an infinite quiet computation and z a dynamic gap of C. Then there is j0, such that the state of processor z(j) in gj is 0.g(z) and • there is k0, such that the state of processor z(k) +1 in gk is 0.u for some u Proof: sz sz’ gj: ... … Pi 0.g(z’) z’(j)=Pi+1 ... sz’ sz gk: 0.g(z’) ... ... Pi z(k) Pi+1

  30. Rule A: if Li  Li-1+1 and (Li0 or Ti-1=0 or Ti-1Li-Li-1 or Ti-1<Ti) then Li-1.Ti-1Li.Ti (Li-1+1).(Li-Li-1) Rule B: if Li = Li-1+1 and Li0 and Ti-1 Tithen Li-1.Ti-1 Li.TiLi.Ti-1 Definition: A segment is well formed if all its tags are equal to its gap size.

  31. Rule A: if Li  Li-1+1 and (Li0 or Ti-1=0 or Ti-1Li-Li-1 or Ti-1<Ti) then Li-1.Ti-1Li.Ti (Li-1+1).(Li-Li-1) Rule B: if Li = Li-1+1 and Li0 and Ti-1 Tithen Li-1.Ti-1 Li.TiLi.Ti-1 Lemma 7: Let C = g0g1 … be an infinite quiet computation. Then there is j0  0, such that for all j  j0 all segments in gj are well formed Proof: sz gj1: ... 0.g(z) z(j1)=Pi gj’1: ... 0.g(z) 1.g(z) z(j’1)=Pi+1 ... gj’m: … 0.g(z) 1.g(z) … m.g(z) z(j’m)=Pi+m ... gj2: 0.g(z) 1.g(z) … *.g(z) z(j2)

  32. Rule A: if Li  Li-1+1 and (Li0 or Ti-1=0 or Ti-1Li-Li-1 or Ti-1<Ti) then Li-1.Ti-1Li.Ti (Li-1+1).(Li-Li-1) Rule B: if Li = Li-1+1 and Li0 and Ti-1 Tithen Li-1.Ti-1 Li.TiLi.Ti-1 Lemma 8: There is a point in every infinite computation of S after which every configuration has exactly one segment. Proof: C’ - quiet computation with well formed segments - rule B is void If for all dynamic gaps of C’ g(z) were 0, then there would be only one segment … a.0 a.0... If there is some dynamic gap of C’ such that its size is greater than 0: sz sz’ a.g(z) 0.g(z’) g(z)  g(z’) and g(z)  0 ... ... z(j1)=Pi Pi+1 Pi+1 can make a step only after Pi, but such a step would destroy sz contradiction

  33. Rule A: if Li  Li-1+1 and (Li0 or Ti-1=0 or Ti-1Li-Li-1 or Ti-1<Ti) then Li-1.Ti-1Li.Ti (Li-1+1).(Li-Li-1) Rule B: if Li = Li-1+1 and Li0 and Ti-1 Tithen Li-1.Ti-1 Li.TiLi.Ti-1 Conclusion: The system never livelocks. Proof: After finitely many step the system reaches the following configuration : : … a-1.0 a.0 a.0a+1.0 ... one well formed segment, g(z) = 0, all tags are 0 Recall:the set of legitimate configurationsLis the set of all cyclic permutations of configurations of the form 0.0 1.0 … a-1.0 a.0 a.0 a+1.0 … n-2.0 for a=0, 1, …, n-2 Hence,  is a legitimate configuration

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