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Process Algebra Book: Chapter 8

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The Main Issue Q: When are two models equivalent? A: When they satisfy different properties. Q: Does this mean that the models have different executions?

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What is process algebra? An abstract description for nondeterministic and concurrent systems. Focuses on the transitions observed rather than on the states reached. Main correctness criterion: conformance between two models. Uses: system refinement, model checking, testing.

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Different models may have the same set of executions! a aa b bc c a-insert coin, b-press pepsi, c-press pepsi-light d-obtain pepsi, e-obtain pepsi-light d d ee

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Actions: Act={a,b,c,d} { }. Agents: E, E, F, F1, F2, G1, G2, … E E G2G1 F1F2 F a aa b bc c Agent E may evolve into agent E. Agent F may evolve into F 1 or F2. d d ee

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Events. E E G2G1 F1F2 F a aa b bc c Ea E, Fa F1, Fa F2, F1a G1, F2a G2. G1 F, G1 F.

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Actions and co-actions For each action a, except for, there is a co- action a. a and a interact (a input, a output). The coaction of a is a. G2G1 F1F2 F aa bc E E a bc

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Notation a.E – execute a, then continue according to E. E+F – execute according to E or to F. E||F – execute E and F in parallel. E GH F a bc a.(b+c) (actually, a.((b.0)+(c.0)) Ea F Fb G Fc H 0 – deadlock/termination.

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Conventions. has higher priority than +..0 or.(0||0||…||0) is omitted.

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CCS - calculus of concurrent systems [Milner]. Syntax a,b,c, … actions, A, B, C - agents. a,b,c, coactions of a,b,c. -silent action. nil - terminate. a.E - execute a, then behave like E. + - nondeterministic choice. || - parallel composition. \L - restriction: cannot use letters of L. [f] - apply mapping function f between between letters.

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Semantics (proof rule and axioms). Structural Operational Semantics SOS a.p –a p pa p |-- p+q –a p qa q |-- p+q –a q pa p |-- p|q –a p|q qa q |-- p|q –a p|q pa p, qa q |-- p|q – p|q pa p, a R |-- p\L –a p\R pa p |-- p[m]m(a) p[m]

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Action Prefixing a.Ea E (Axiom) Thus, a.(b.(c||c)+d)a (b.(c||c)+d).

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Choice Ea E Fa F (E+F)a E (E+F)a F b.(c||c)b (c||c). Thus, (b.(c||c)+e)b (c||c). If Ea E and Fa F, then E+F has a nondeterministic choice.

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Concurrent Composition Ea E Fa F E||Fa E||F Ea E, Fa F E||F E||F cc 0, cc 0, c||c 0||0, c||cc 0||c, c||cc c||0.

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Restriction Ea E, a, a R E\R –a E\R In this case: allows only internal interaction of c. c||c 0||0 c||cc 0||c c||cc c||0 (c||c) \ {c} (0||0) \{c}

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Relabeling Ea E E[m] –m(a) E[m] No axioms/rules for agent 0.

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Examples a.E||b.F a.E||F E||b.F E||F b b a a

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Derivations (0||0) a.(b.(c||c)+d) b.(c||c)+d (c||c)0 (0||c)(c||0) a b d c c c c

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Modeling binary variable C0=is_0?. C0 + set_1. C1 + set_0. C0 C1=is_1?. C1 + set_0. C0 + set_1. C1 C0C1 set_1 set_0 is_0? set_1 is_1?

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Equational Definition E=a.(b..E+c..E) Ea E, A=E F=a.b..F+a.c..F Aa E G2G1 F1F2 F aa bc E E a bc

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Trace equivalence: Systems have same finite sequences. Same traces F aa bb E a b c c E=a.(b+c)F=(a.b)+a.(b+c)

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Failures: comparing also what we cannot do after a finite sequence. F a a bb E a b c c Failure of agent E: (σ, X), where after executing σ from E, none of the events in X is enabled. Agent F has failure (a, {c}), which is not a failure of E.

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Simulation equivalence Relation over set of agents S. R S S. E R F If E R F and Ea E, then there exists F, Fa F, and E R F. E cd bb a a F c d bb a

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Simulation equivalence Relation over set of agents S. R S S. E R F If E R F and Ea E, then there exists F, Fa F, and E R F. E cd bb aa F cd bb a

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Here, simulation works only in one direction. No equivalence! Relation over set of agents S. R S S. E R F If E R F and Ea E, then there exists F, Fa F, and E R F. E cd bb aa F cd bb a want to establish symmetrically necessarily problem!!!

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Simulation equivalent but not failure equivalent Left agent a.b+a has a failure (a,{b}). E b aa F b a

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Bisimulation: same relation simulates in both directions Not in this case: different simulation relations. E b aa F b a

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Hierarchy of equivalences Bisimulation Trace FailureSimulation

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Example: A=a.((b.nil)+(c.d.A)) B=(a.(b.nil))+(a.c.d.B) a b c d s0s0 s1s1 s2s2 s3s3 a d b a c t0t0 t1t1 t4t4 t2t2 t3t3

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Bisimulation between G 1 and G 2 Let N= N 1 U N 2 A relation R : N 1 x N 2 is a bisumulation if If (m,n) in R then 1. If ma m then n:na n and (m,n) in R 2. If na n then m:ma m and (m,n) in R. Other simulation relations are possible, I.e., m=a=> m when m …a m.

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Algorithm for bisimulation: Partition N into blocks B 1 B 2 … B n =N. Initially: one block, containing all of N. Repeat until no change: Choose a block B i and a letter a. If some of the transitions of B i move to some block B j and some not, partition B i accordingly. At the end: Structures bisimilar if initial states of two structures are in same blocks.

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Correctness of algorithm Invariant: if (m,n) in R then m and n remain in the same block throughout the algorithm. Termination: can split only a finite number of times.

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Example: a b c d s0s0 s1s1 s2s2 s3s3 a d b a c t0t0 t1t1 t4t4 t2t2 t3t3 {s 0,s 1,s 2,s 3,t 0,t 1,t 2,t 3,t 4 }

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Example: a b c d s0s0 s1s1 s2s2 s3s3 a d b a c t0t0 t1t1 t4t4 t2t2 t3t3 {s 0,s 1,s 2,s 3,t 0,t 1,t 2,t 3,t 4 } split on a. {s 0,t 0 },{s 1,s 2,s 3,t 1,t 2,t 3,t 4 }

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Example: a b c d s0s0 s1s1 s2s2 s3s3 a d b a c t0t0 t1t1 t4t4 t2t2 t3t3 {s 0,t 0 },{s 1,s 2,s 3,t 1,t 2,t 3,t 4 } split on b {s 0,t 0 },{s 1,t 1 },{s 0,s 2,s 3,t 2,t 3,t 4 }

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Example: a b c d s0s0 s1s1 s2s2 s3s3 a d b a c t0t0 t1t1 t4t4 t2t2 t3t3 {s 0,t 0 },{s 1,t 1 },{s 2,s 3,t 2,t 3,t 4 } split on c {s 0,t 0 },{s 1 },{t 1 },{s 2,s 3,t 2,t 3,t 4 }

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Example: {s 0,t 0 },{s 1 },{t 1 },{s 2,s 3,t 2,t 3,t 4 } split on c {s 0,t 0 },{s 1 },{t 1 },{t 4 },{s 2,s 3,t 2,t 3 } a b c d s0s0 s1s1 s2s2 s3s3 a d b a c t0t0 t1t1 t4t4 t2t2 t3t3

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Example: {s 0,t 0 },{s 1 },{t 1 },{t 4 },{s 2,s 3,t 2,t 3 } split on d {s 0,t 0 },{s 1 },{t 1 },{t 4 },{s 3, t 3 },{s 2,t 2 } a b c d s0s0 s1s1 s2s2 s3s3 a d b a c t0t0 t1t1 t4t4 t2t2 t3t3

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Example: {s 0,t 0 },{s 1 },{t 1 },{t 4 },{s 2,t 2 },{s 3,t 3 } split on a {s 0 },{t 0 },{s 1 },{t 1 },{t 4 },{s 3, t 3 },{s 2,t 2 } a b c d s0s0 s1s1 s2s2 s3s3 a d b a c t0t0 t1t1 t4t4 t2t2 t3t3

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Example: {s 0 },{t 0 },{s 1 },{t 1 },{t 4 },{s 2,s 3,t 2,t 3 } split on d {s 0 },{t 0 },{s 1 },{t 1 },{t 4 },{s 3 },{t 3 },{s 2,t 2 } a b c d s0s0 s1s1 s2s2 s3s3 a d b a c t0t0 t1t1 t4t4 t2t2 t3t3

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