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