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gabriel draft

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Francesco Mecca 2020-04-09 01:00:31 +02:00
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@ -1,4 +1,6 @@
\begin{comment}
TODO: neg is parsed incorrectly
TODO: chiedi a Gabriel se T e S vanno bene, ma prima controlla che siano coerenti
* TODO Scaletta [1/6]
- [X] Introduction
- [-] Background [80%]
@ -1396,7 +1398,6 @@ m := ((a_i)^i ((p_{ij})^i \to e_j)^{ij})
(k_k)^k := headconstructor(p_{i0})^i
\]
\begin{equation}
\begin{align}
Groups(m) := ( k_k \to ((a)_{0.l})^{l \in Idx(k_k)} +++ (a_i)^{i \in I\backslash \{0\} }), \\
( if p_{0j} is k(q_l) then \\
(qₗ)^{l \in Idx(k_k)} +++ (p_{ij})^{i \in I\backslash \{0\}} \to e_j \\
@ -1404,7 +1405,6 @@ m := ((a_i)^i ((p_{ij})^i \to e_j)^{ij})
(\_)^{l \in Idx(k_k)} +++ (p_{ij})^{i \in I\backslash \{0\}} \to e_j \\
else \bot )^j ), \\
((a_i)^{i \in I\backslash \{0\}}, ((p_{ij})^{i \in I\backslash \{0\}} \to eⱼ if p_{0j} is \_ else \bot)^{j \in J})
\end{align}
\end{equation}
Groups(m) is an auxiliary function that decomposes a matrix m into
@ -1516,7 +1516,7 @@ an accessor → π relation (In other words???)
Should I swap π and π'
\end{comment}
\subsubsection Proof of equivalence checking
\subsubsection{Equivalence checking}
The equivalence checking algorithm takes as parameters an input space
/S/, a source decision tree /Cₛ/ and a target decision tree /Cₜ/:
| equiv(S, Cₛ, Cₜ) → Yes | No(vₛ, vₜ)
@ -1539,9 +1539,9 @@ We define the following
| Forall(Yes::l) = Forall(l)
| Forall(No(vₛ,vₜ)::_) = No(vₛ,vₜ)
There exists and are injective:
| int(k)∈ℕ (ar(k) = 0)
| tag(k)∈ℕ (ar(k) > 0)
| π(k) = {n|int(k) = n} x {n|tag{k} = n}
| int(k) ∈ ℕ (arity(k) = 0)
| tag(k) ∈ ℕ (arity(k) > 0)
| π(k) = {n\vert int(k) = n} x {n\vert tag(k) = n}
where k is a constructor.
\begin{comment}
@ -1554,51 +1554,52 @@ We proceed by case analysis:
I start numbering from zero to leave the numbers as they were on the blackboard, were we skipped some things
I think the unreachable case should go at the end.
\end{comment}
0. in case of unreachable: Cₛ(vₛ) = Absurd(Unreachable) ≠ Cₜ(vₜ) ∀vₛ,vₜ
0. in case of unreachable:
| Cₛ(vₛ) = Absurd(Unreachable) ≠ Cₜ(vₜ) ∀vₛ,vₜ
1. In the case of an empty input space
| equiv(∅, Cₛ, Cₜ) := Yes
and that is trivial to prove because there is no pair of values (vₛ, vₜ) that could be
tested against the decision trees.
In the other subcases S is always non-empty.
2. equiv(S, Failure, Failure) := Yes
the statement holds because of equality between Failure nodes in
the case of every possible value /v/.
3. The result of the subcase where we have a source decision tree
/Cₛ/ that is either a Leaf terminal or a Failure terminal and a
target decision tree defined by an accessor /a/ and a positive
number of couples constraint πᵢ and children nodes Cₜᵢ. The output
the output of the algorithm is:
| equiv(S, (Leaf bbₛ|Failure) as Cₛ, Node(a, (πᵢ → Cₜᵢ)ⁱ)) := Forall(equiv( S∩a→π(kᵢ)), Cₛ, Cₜᵢ)ⁱ)
The statement holds because defined let Sᵢ := S∩(a→πᵢ)
either the algorithm is true for every sub-input space Sᵢ and
| equiv(∅, Cₛ, Cₜ) := Yes
and that is trivial to prove because there is no pair of values (vₛ, vₜ) that could be
tested against the decision trees.
In the other subcases S is always non-empty.
2. When there are /Failure/ nodes at both sides the result is /Yes/:
|equiv(S, Failure, Failure) := Yes
Given that ∀v, Failure(v) = Failure, the statement holds.
3. When we have a Leaf or a Failure at the left side:
| equiv(S, Failure as Cₛ, Node(a, (πᵢ → Cₜᵢ)ⁱ)) := Forall(equiv( S∩a→π(kᵢ)), Cₛ, Cₜᵢ)ⁱ)
| equiv(S, Leaf bbₛ as Cₛ, Node(a, (πᵢ → Cₜᵢ)ⁱ)) := Forall(equiv( S∩a→π(kᵢ)), Cₛ, Cₜᵢ)ⁱ)
The algorithm either returns Yes for every sub-input space Sᵢ := S∩(a→π(kᵢ)) and
subtree Cₜᵢ
| equiv(Sᵢ, Cₛ, Cₜᵢ) = Yes ∀i
or we have a counter example vₛ, vₜ for which
| vₛ≃vₜ∈Sₖ ∧ cₛ(vₛ) ≠ Cₜₖ(vₜ)
then because
| vₜ∈(a→πₖ) ⇒ Cₜ(vₜ) = Cₜₖ(vₜ)
then
| vₜ∈(a→πₖ) → Cₜ(vₜ) = Cₜₖ(vₜ) ,
| vₛ≃vₜ∈S ∧ Cₛ(vₛ)≠Cₜ(vₜ)
and the result of the algorithm is
we can say that
| equiv(Sᵢ, Cₛ, Cₜᵢ) = No(vₛ, vₜ) for some minimal k∈I
4. equiv(S, Node(a, (kᵢ → Cₛᵢ)ⁱ, C_{sf}), Cₜ) :=
let π' = ⋃π(kᵢ) ∀i in
Forall(equiv( S∩(a→π(kᵢ)ⁱ), Cₛᵢ, C_{t/a→π(kᵢ)})ⁱ +++ equiv(S∩(a→π(kᵢ)), Cₛ, C_{/a̸¬̸π'}))
The statement holds because:
a. Forall(equiv( S∩(a→π(kᵢ)ⁱ), Cₛᵢ, C_{t/a→π(kᵢ)})ⁱ = Yes
In the yes case let's reason by case analysis:
i. When k∈(kᵢ)ⁱ
there is a k=kₖ for some k and this means that Cₛ(vₛ) = Cₛᵢ(vₛ)
By induction we know that Cₛᵢ(vₛ) = c_{t/a→πᵢ}(vₜ)
and because of the trimming lemma:
Cₜ(vₜ) = C_{t/a→πᵢ}(vₜ)
Putting all together:
Cₛ(vₛ) = Cₛᵢ(vₛ) = C_{t/a→πᵢ}(vₜ) = Cₜ(vₜ)
4. When we have a Node on the right we define πₙ as the domain of
values not covered but the union of the constructors kᵢ
| πₙ = ¬(⋃π(kᵢ)ⁱ)
The algorithm proceeds by trimming
| equiv(S, Node(a, (kᵢ → Cₛᵢ)ⁱ, C_{sf}), Cₜ) :=
| Forall(equiv( S∩(a→π(kᵢ)ⁱ), Cₛᵢ, C_{t/a→π(kᵢ)})ⁱ +++ equiv(S∩(a→π(kᵢ)), Cₛ, C_{a→πₙ}))
The statement still holds and we show this by first analyzing the
/Yes/ case:
| Forall(equiv( S∩(a→π(kᵢ)ⁱ), Cₛᵢ, C_{t/a→π(kᵢ)})ⁱ = Yes
The constructor k is either included in the set of constructors kᵢ:
| k \vert k∈(kᵢ)ⁱ ∧ Cₛ(vₛ) = Cₛᵢ(vₛ)
We also know that
| (1) Cₛᵢ(vₛ) = C_{t/a→πᵢ}(vₜ)
| (2) C_{T/a→πᵢ}(vₜ) = Cₜ(vₜ)
(1) is true by induction and (2) is a consequence of the trimming lemma.
Putting everything together:
| Cₛ(vₛ) = Cₛᵢ(vₛ) = C_{T/a→πᵢ}(vₜ) = Cₜ(vₜ)
ii. when k∉(kᵢ)ⁱ ???
When the k∉(kᵢ)ⁱ [TODO]
b. Forall(...) = No(vₛ, vₜ)
for a minimum k, equiv(Sₖ, Cₛₖ, C_{t/a→πₖ} = No(vₛ, vₜ)
then Cₛₖ(vₛ) ≠ C_{t/a→πₖ}(vₜ) and C_{t/a→πₖ}(vₜ) = Cₜ(vt)
=> (Cₛₖ(vₛ) = Cₛ(vₛ)) ≠ Cₜ(vₜ)
# Same for fallback?
The auxiliary Forall function returns /No(vₛ, vₜ)/ when, for a minimum k,
| equiv(Sₖ, Cₛₖ, C_{T/a→πₖ} = No(vₛ, vₜ)
Then we can say that
| Cₛₖ(vₛ) ≠ C_{t/a→πₖ}(vₜ)
that is enough for proving that
| Cₛₖ(vₛ) ≠ (C_{t/a→πₖ}(vₜ) = Cₜ(vₜ))