master-degree-notes/Concurrent Systems/notes/12b - CCS cose varie.md

An n-ary semaphore S(n)(p,v) is a process used to ensure that there are no more than n istances of the same activity concurrently in execution. An activity is started by action p and is terminated by action v.

The specification of a unary semaphore is the following:

S^{(1)} \triangleq p \cdot S_{1}^{(1)}

S_{1}^{(1)} \triangleq p \cdot S_{}^{(1)}

The specification of a binary semaphore is the following:

S_{}^{(2)} \triangleq p \cdot S_{1}^{(2)}

S_{1}^{(2)} \triangleq p \cdot S_{2}^{(2)}+v\cdot S^{(2)}

S_{2}^{(2)} \triangleq v \cdot S_{1}^{(2)}

If we consider S(2) as the specification of the expected behavior of a binary semaphore and S(1) | S(1) as its concrete implementation, we can show that S^{(1)}|S^{(1)} \space \textasciitilde \space S^{2} This means that the implementation and the specification do coincide. To show this equivalence, it suffices to show that following relation is a bisimulation:

Restrictions

Proposition: a.P \textbackslash a ∼ 0 Proof:

• S = {(a.P\a , 0)} is a bisimulation

Which challenges can (a.P)\a have?

• a.P can only perform a (and become P)
• however, because of restriction, a.P\a is stuck

No challenge from a.P\a, nor from 0 -> bisimilar!

Proposition: \bar{a}.P \textbackslash a ∼ 0 Proof is similar.

Idempotency of Sum

Proposition: α.P+α.P+M ∼ α.P+M Proof:

S = \{ (α.P+α.P+M , α.P+M) \}

Is it a bisimulation? NO: the problem is that, for example:

• α.P+α.P+M –α–> P
• α.P+M –α–> P
• BUT (P,P) in general does NOT belong to S!

So we can try with S = \{ (α.P+α.P+M , α.P+M) \} ∪ \{(P,P)\} But it is not yet a bisimulation. P –β–> P’ (challenge and reply), but we don't have (P', P') in S. (Questo per garantire che funzioni in qualsiasi caso, con qualsiasi P che eventualmente evolve in P')

So let's try with: S = \{ (α.P+α.P+M , α.P+M) \} ∪ Id Let's go!

Congruence

One of the main aims of an equivalence notion between processes is to make equational reasonings of the kind: “if P and Q are equivalent, then they can be interchangeably used in any execution context”.

This feature on an equivalence makes it a congruence Not all equivalences are necessarily congruences (even though most of them are). To properly define a congruence, we first need to define an execution context, and then what it means to run a process in a context. Intuitively:

where C is a context (i.e., a process with a hole ☐), P is a process, and C[P] denotes filling the hole with P

Example: if \space C = (☐ | Q) \textbackslash a, \space then \space C[P] = (P | Q) \textbackslash a The set C of CCS contexts is given by the following grammar:

C ::= ☐ \space | \space C|P \space | \space C \textbackslash a \space | \space a.C + M

where M denotes a sum.

An equivalence relation R is a congruence if and only if

\forall (P, Q) \in R, \forall C.(C[P], C[Q]) \in R

Is bisimilarity a congruence? Yes. Theorem:

if \space P ∼ Q \space then \space \forall C.C[P] ∼ C[Q]

Proof on the slides.

So today we learned that bisimulation is a good equivalence.