140 lines
No EOL
7.4 KiB
Markdown
140 lines
No EOL
7.4 KiB
Markdown
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A (finite non-deterministic) automaton is a quintuple M = (Q,Act,q0,F,T), where:
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- Q is the set of states
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- Act is the set of actions
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- q0 is the starting state
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- F is the set of final states
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- T is the transition relation (T ⊆ Q × Act × Q)
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Automata Behaviour: language equivalence
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(where L(M) is the set of all the sequences of input characters that bring the automaton M from its starting state to a final one)
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>[!note] Language equivalence
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>M1 and M2 are *language equivalent* if and only if L(M1)=L(M2)
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By considering the starting states as also final, they both generate the same language, i.e.:
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$$(20.(tea + 20.coffee))∗ = (20.tea + 20.20.coffee)∗$$
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But, do they behave the same from the point of view of an external observer??
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The essence of the difference is WHEN the decision to branch is taken
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- language equivalence gets rid of branching points
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- it is too coarse for our purposes!
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### LTSs
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In concurrency theory, we don’t use finite automata but Labeled Transition System (LTS). The main differences between the two formalisms are:
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- automata usually rely on a finite number of states, whereas states of an LTS can be infinite
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- automata fix one starting state, whereas in an LTS every state can be considered as initial (this corresponds to different possible behaviors of the process)
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- automata rely on final states for describing the language accepted, whereas in LTS this notion is not very informative
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>[!note] LTS formal definition
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>Fix a set of action names (or, simply, actions), written N.
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>
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>A Labeled Transition System (LTS) is a pair (Q, T), where Q is the set of states and T is the transition relation (T ⊆ Q × N × Q).
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We shall usually write s –a–> s′ instead of ⟨s,a,s′⟩ ∈ T.
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### Bisimulation
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Intuitively, two states are equivalent if they can perform the same actions that lead them in states where this property still holds
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P0 and Q0 are different because, after an a, the former can decide to do b or c, whereas the latter must decide this before performing a.
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Let (Q,T) be an LTS.
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A binary relation S ⊆ Q×Q is a simulation if and only if
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∀(p,q) ∈ S∀p –a–> p′∃q –a–> q′ s.t. (p′,q′) ∈ S
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We say that p is simulated by q if there exists a simulation S such that $$(p,q) ∈ S$$
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We say that S is a bisimulation if both S and S−1 are simulations (where $$S^{-1} = \{(p,q) : (q,p) ∈ S\}$$)
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Two states q and p are bisimulation equivalent (or, simply, bisimilar) if there exists a bisimulation S such that (p, q) ∈ S; we shall then write p ∼ q.
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q0 is simulated by p0; this is shown by the following simulation relation: $$S = \{(q0,p0), (q1,p1), (q2,p1), (q3,p2), (q4,p3)\}$$
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To let p0 be simulated by q0, we should have that p1 is simulated by q1 or q2.
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If S contained one among (p1,q1) or (p1,q2), then it would not be a simulation: indeed, p1 can perform both a c (whereas q1 cannot) and a b (whereas q2 cannot).
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**Remark:** for proving equivalence, it is NOT enough to find a simulation of p by q and a simulation of q by p
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p0 is simulated by q0: $$S = {(p0, q0), (p1, q2), (p2, q3)}$$
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q0 is simulated by p0: $$S′ ={(q0,p0),(q1,p1),(q2,p1),(q3,p2)}$$
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However, p0 and q0 are not bisimilar: the transition q0 -> a -> q1 is not bisimulable by any transition from p0 (ndeed, p0 –a–> p1 does not suffice, because p1 can perform a b and so cannot be bisimilar to q1).
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#### Bisimulation is NOT isomorfism
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$S = \{(p0,q0), (p1,q1), (p2,q1), (p0,q2)\}$
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$S^{−1} = \{(q0,p0), (q1,p1), (q1,p2), (q2,p0)\}$
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>[!def] Theorem
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>Bisimilarity is an equivalence relation
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*Proof:*
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**Reflexivity:** we have to show that q ∼ q, for every q. Consider the following relation $S ={(p,p):p∈Q}$ and observe that it is a bisimulation (the inverse is S itself).
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**Symmetry:** we have to show that p ~ q implies q ~ p, for every p, q.
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By hypothesis, there exists a bisimulation S that contains the pair (p, q). By definition of bisimulation, $S^{-1}$ is a simulation; hence, $(q, p) \in S^{-1}$ and, consequently, q ~ p.
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**Transitivity:** we have to show that `p ~ q` and `q ~ r` imply `p ~ r` for all p, r, q. Let's consider the following relation: $S = \{(x, y): \exists y : (x, y) \in S_{1} \land(y,z)\in S_{2}\}$
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where S1 and S2 are bisimulations, then we have to show that S is a bisimulation.
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- let $(x, y) \in S$ and x -a-> x'
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- if $(x, z)$ belongs to S, then, by definition, there exists y such that $(x, y) \in S_{1}$ and $(y, z) \in S_{2}$
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- since S1 is a bisimulation, there exists y -a-> y' such that $(x', y') \in S_{1}$
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- since S2 is a bisimulation, there exists z -a-> z' such that $(y', z') \in S_{2}$
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- hence, from x -a-> x', we found z -a-> z' such that $(x', z') \in S$, because there exists a y' such that $(x', y') \in S_{1}$ and $(y',z') \in S_{2}$.
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**Theorem:** ∼ is a bisimulation.
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*Proof:*
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The proof is done by showing that ∼ is a simulation.
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By definition of similarity, we have to show that ∀(p,q)∈∼ ∀p –a–> p′ ∃q –a–> q′ s.t.(p′,q′)∈∼
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Let us fix a pair (p,q) ∈∼
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Bisimilarity of p and q implies the existence of a bisimulation S such that (p,q) ∈ S.
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Hence, for every transition p –a–> p′, there exists a transition q –a–> q′ such that (p′, q′) ∈ S.
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So, (p', q') ∈ ∼
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**Theorem:** For every bisimulation S, it holds that S ⊆ ∼.
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*Proof:*
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Let (p,q) ∈ S. Then, there exists a bisimulation that contains the pair (p, q); thus, (p, q) ∈ ∼.
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## La parte difficile
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The only ingredients we used to write down an LTS are:
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- sequential compsition (of an action and a process)
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- non-deterministic choice (between a finite set of prefixed processes)
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- recursion
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To simplify process writing, we shall assume to have a finite set Id of processes identifiers and that the definitions we shall give will be parametric
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For every identifier (denoted with capital letters A,B,..), we shall assume a unique definition of the form $A(x1,x2,..,xn) := P$, where names x1,x2,..,xn are all distinct and all included in the names of P.
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Let us denote with P{b1/x1 . . . bn/xn} the process obtained from P by replacing name xi with name bi, for every i = 1,..., n.
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>[!def]
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>The set of non-deterministic processes is given by the following grammar:
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$$P::= \sum \alpha_{i}.P_{i} | A(a_{1}\dots a_{n})$$
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where I is a finite set of indices and $\alpha_{i} \in A$ for every $i \in I$.
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**Remark:** we now fuse together in a unique operator sequential composition and nondeterministic choice.
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If the index set $I$ is empty, then $\sum_{i\in I} \alpha_{i}.P_{i}$ is the terminated process, that cannot perform any action; this process will be represented with the symbol **0**.
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We shall usually omit tail occurrences of ‘.0’ and, for example, simply write a.b instead of a.b.0
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### From the syntax to the LTS
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From the syntax to the LTS We have shown how it is possible, starting from an LTS, to generate a corresponding process
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It is also possible the inverse translation and then the two formalisms do coincide; the rules that have to be used in this translation are:
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examples |