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Concurrency 6

This article explores the specification and verification of distributed schedulers and counters in the CCS framework. It provides examples and proofs of correctness for the implementation of these systems.

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Concurrency 6

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  1. Concurrency 6 Specification and Verification in CCS Catuscia Palamidessi catuscia@lix.polytechnique.fr

  2. Example: A distributed scheduler • 1,...,n are tasks identifiers. Tasks have to be executed repeatedly,in a cyclic order. There can be more than one task executed at the same time, but the next instance of Task i cannot start before previous instance has finished. • Specification: We use: • ak as the signal start to Taks k and • bk as the signal that Task k has terminated Assume: • X  {1,...,n } are the tasks in progress • Task i is next ScSpec(i,X)  { bk. ScSpec(i,X-{k}) | k  X} if i  X ScSpec(i,X)  ai.ScSpec(i+1,XU{i}) +  { bk. ScSpec(i,X-{k}) | k  X} if i  X Concurrency 6

  3. Example: A distributed scheduler • Implementation: We build the scheduler, Sched, as a ring of n cells each linked to one task • Cell: A  a.C C  c.E E  b.D + d.B B  b.A D  d.A Note: A stands for A(a,b,c,d), B stands for B(a,b,c,d), etc. We will also use Ak for A(ak,bk,ck,ck-1), Bk for B(ak,bk,ck,ck-1), etc. • Definition Sched  (n c1)...(n cn) ( A1 |  { Dk | k  1}) • Theorem 1 (Correctness of the implementation wrt the specification): Sched = ScSpec(1,) Concurrency 6

  4. Scheduler: Proof of correctness • The meaning of the various cells: • Ai : Task i is next, and it is ready to initiate • Bi: Task i is next, but it is not ready to initiate • Di: Task i is not next, but it is ready to initiate • Ei: Task i is not next, and it is not ready to initiate • Definition: Sched(i,X) (n c)(Bi |  {Dk | k  X} |  {Em | m  X-{i} } ) if i  X Sched(i,X) (n c)(Ai |  {Dk | k  X U {i} } |  { Em | m  X } ) if i  X • Proposition 2: Sched(i,X) = ScSpec(i,X) • Theorem 1 is a particular case of Proposition 2 Concurrency 6

  5. Implementation of the scheduler: how it works D cn b1 next A a1 c1 b2 D a2 c2 D b3 a3 Concurrency 6

  6. Implementation of the scheduler: how it works D cn b1 C a1 T1 c1 b2 D a2 c2 D b3 a3 Concurrency 6

  7. Implementation of the scheduler: how it works D cn b1 E a1 T1 c1 b2 A next a2 c2 D b3 a3 Concurrency 6

  8. Implementation of the scheduler: a possible future configuration E cn b1 B Tn next a1 T1 c1 b2 D a2 c2 T3 E b3 a3 Concurrency 6

  9. Scheduler: Proof of Correctness Proposition 2: Sched(i,X) = ScSpec(i,X) Proof • Lemma 3 • (1) (n ci) ( Ci | Di+1 ) = t.(n ci) ( Ei | Ai+1 ) • (2) (n ci) ( Ci | Ei+1 ) = t.(n ci) ( Ei | Bi+1 ) Proof: By exapansion law • Lemma 4 • Sched(i,X) =  { bk. Sched(i,X-{k}) | k  X } if i  X • Sched(i,X) = ai.Sched(i+1,XU{i}) +  { bk. Sched(i,X-{k}) | k  X } if i  X Proof: By Expansion law and Lemma 3 From Lemma 4 and the Definition law we obtain that Sched(i,X) = ScSpec(i,X)□ Concurrency 6

  10. Example: Counter • It is possible in CCS to create structures which grow and shrink dynamically. Examples include unbounded queues and stacks, and counters. • Specification of a Counter A counter is an object that can be • tested for zero zero • incremented inc • decremented dec Count0 inc.Count1 + zero. Count0 Countn inc.Countn+1 + dec. Countn-1 n > 0 Concurrency 6

  11. Example: Counter • Implementation: A structure obtained by linking together a process B and n copies of a process C specified as follows: B inc.(B^C) + zero.B C inc.(C^C) + dec.D D  d.C + z.B P^Q  (n i’)(n z’)(n d’)(P(z,d,i’,z’,d’) | Q(z’,d’, inc,zero,dec)) Note: B, C and D stand for B(z,d,inc,zero,dec), C(z,d,inc,zero,dec), and D(z,d,inc,zero,dec) respectively. (P^Q) stands for (P^Q)(z,d,inc,zero,dec). Proposition: ^ is associative, i.e. P^(Q^R) = (P^Q)^R inc B inc C zero z zero dec dec d Concurrency 6

  12. Example: Counter • Implementation: Definition: C(n) B^C^C^...^C (n times) • Theorem (Correctness): C(n) = Countn Proof Lemma: (1) C^D  D^C (2) B^D  B^B (3) B^B = B We can now prove that • C(0) = inc. C(1) + zero. C(0) and • C(n) = C(n-1) ^C for n > 0 by definition = inc. (C(n-1) ^C^C) + dec. (C(n-1) ^D) by expansion law = inc. C(n+1) + dec. C(n-1) by the lemma above Hence C(n) satisfies the same equations as Countn. By the unique solution law we can conclude C(n) = Countn □ Concurrency 6

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