56 lines
No EOL
3 KiB
Markdown
56 lines
No EOL
3 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) |