Isothetic regions

Isothetic region $X ⊆ \vert G \vert^n$:

it admits a finite block covering

Prop: A subset $X ⊆ \vert G \vert^n$ is an isothetic region iff the collection of maximal subblocks of $X$ is finite and covers $X$.

Isothetic regions form a Booelan algebra (they’re stable under complement and intersection, and $X ∪ Y = (X^c ∩ Y^c)^c$)

Isothetic regions are stable under interior and closure too.

Factoring isothetic regions

Let

\Pi_1, …, \Pi_n

be compatible programs.

Then they are model independent iff

⟦ \Pi_1 \mid ⋯ \mid \Pi_n ⟧ = ⟦ \Pi_1 ⟧ × ⋯ × ⟦ \Pi_n ⟧

Now, the converse: assume

\Pi = G_1, …, G_n

is a family of conservative processes. Question: does there exist a permutation $σ ∈ 𝔖(n)$ such that

⟦ \Pi_σ ⟧ \; ≝ \; ⟦ G_σ(1) ⋯ G_σ(k) ⟧ × ⟦ G_σ(k+1) ⋯ G_σ(n) ⟧

? Answer: there is an (exponential) algorithm computing it.

Coming back to observational independence

Consider the program

sem 1 a
Pa Va | Pa Va


(cf. picture)

Only two execution traces: Pa, Va (first process), Pa, Va (second process) and Pa, Va (second process), Pa, Va (first process)

Swapping the first Va with the second Pa is not possible (even thought the permutation is compatible since it pertains two different processes) ⟹ not observationally equivalent

On the other hand:

x = 1 | x = 1


are observationally equivalent, since you can swap the two instructions and get the same state at the end

x = 1 | x = 0


are not observationally equivalent, since you can swap the two instructions but not get the same state at the end

Suppose

X ⊆ \vert G \vert^n

is an isothetic region.

X = \bigcup\limits_{B \text{ max block of } X} B
B = B_1 × ⋯ × B_n

where each $B_i ⊆ \vert G \vert$ is connected

Let $Σ$ be the alphabet of connected subsets of $\vert G \vert$.

The isothetic region exactly corresponds to a language over $Σ$. The alphabet may be uncountable, but the language is finite.

$D(\lbrace a \rbrace)$ (the monoid of languages over $\lbrace a \rbrace$) is not free commutative, since

\lbrace ε \rbrace \cdot a^\ast = \lbrace ε, a \rbrace \cdot a^\ast = a^\ast

so the decomposition of $a^\ast$ is not unique. But note that $\lbrace ε, a \rbrace \cdot a^\ast$ is not homogeneous (= all the words have the same length).

If

H_1 \sim H_1'\\ H_2 \sim H_2'\\

then

H_1 \cdot H_2 \sim H_1' \cdot H_2'

Since if

• $H_1’ = σ_1 H_1$
• $H_2’ = σ_2 H_2$

then

H_1' \cdot H_2' = (σ_1 ⊗ σ_2) H_1 \cdot H_2

Godement exchange law: used to show that some operation is sound with respect to some equivalence

𝔸 = \underbrace{\vert G \vert}_{\preccurlyeq = \text{ the equality}} \quad \text{ or } \quad \underbrace{\lbrace \text{finite unions of non-empty connected subsets of } \vert G \vert\rbrace}_{\text{ ≤ = inclusion }}

NB: non-empty: because $∅$ is an absorbing element

• In the first case, a word is a point $∈ \vert G \vert^n$

• In the second one, a word is a block

A subset corresponding to an isothetic region in the first corresponds to a finite language in the second one.

The order in the first one is the equality, so that

H \preccurlyeq H' \overset{déf}{⟺} ∀ ω ∈ H, ∃ ω' ∈ H'; ω \preccurlyeq^n ω'

is the inclusion order.

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