correzioni Gabriel piu` apici

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Francesco Mecca 2020-04-10 00:37:36 +02:00
parent 67ae30a28d
commit 1ec9393e38
2 changed files with 46 additions and 43 deletions

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@ -1,4 +1,5 @@
\begin{comment}
TODO: not all todos are explicit. Check every comment section
TODO: chiedi a Gabriel se T e S vanno bene, ma prima controlla che siano coerenti
* TODO Scaletta [1/6]
- [X] Introduction
@ -1251,10 +1252,10 @@ $(\EquivTEX S {C_S} {C_T} G)$, defined below.
* Correctness of the algorithm
Running a program tₛ or its translation 〚tₛ〛 against an input vₛ
produces as a result /r/ in the following way:
| ( 〚tₛ〛ₛ(vₛ) = Cₛ(vₛ) ) → r
| ( 〚tₛ〛ₛ(vₛ) Cₛ(vₛ) ) → r
| tₛ(vₛ) → r
Likewise
| ( 〚tₜ〛ₜ(vₜ) = Cₜ(vₜ) ) → r
| ( 〚tₜ〛ₜ(vₜ) Cₜ(vₜ) ) → r
| tₜ(vₜ) → r
where result r ::= guard list * (Match blackbox | NoMatch | Absurd)
and guard ::= blackbox.
@ -1268,10 +1269,10 @@ of decision trees
| tₛ ≃ tₜ := ∀vₛ≃vₜ, tₛ(vₛ) = tₜ(vₜ)
| Cₛ ≃ Cₜ := ∀vₛ≃vₜ, Cₛ(vₛ) = Cₜ(vₜ)
The proposed equivalence algorithm that works on a couple of
decision trees is returns either /Yes/ or /No(vₛ, vₜ)/ where vₛ and
vₜ are a couple of possible counter examples for which the constraint
trees produce a different result.
The result of the proposed equivalence algorithm is /Yes/ or /No(vₛ,
vₜ)/. In particular, in the negative case, vₛ and vₜ are a couple of
possible counter examples for which the decision trees produce a
different result.
** Statements
Theorem. We say that a translation of a source program to a decision tree
@ -1312,12 +1313,12 @@ target program we say:
*** Proof of the correctness of the translation from source programs to source decision trees
We define a source term tₛ as a collection of patterns pointing to blackboxes
We define the source term tₛ as a collection of patterns pointing to blackboxes
| tₛ ::= (p → bb)^{i∈I}
A pattern is defined as either a constructor pattern, an or pattern or
a constant pattern
| p ::= | K(pᵢ)ⁱ, i ∈ I | (p|q) | n ∈
| p ::= \vert K(pᵢ)ⁱ, i ∈ I \vert (p\vert{}q) \vert n ∈
A decision tree is defined as either a Leaf, a Failure terminal or
an intermediate node with different children sharing the same accessor /a/
@ -1327,9 +1328,9 @@ possible input values /S/. The fallback is not needed when the user
doesn't use a wildcard pattern.
%%% Give example of thing
| Cₛ ::= Leaf bb | Switch(a, (Kᵢ → Cᵢ)^{i∈S} , C?)
| a ::= Here | n.a
| vₛ ::= K(vᵢ)^{i∈I} | n ∈
| Cₛ ::= Leaf bb \vert Switch(a, (Kᵢ → Cᵢ)^{i∈S} , C?)
| a ::= Here \vert n.a
| vₛ ::= K(vᵢ)^{i∈I} \vert n ∈
\begin{comment}
Are K and Here clear here?
@ -1361,12 +1362,12 @@ We proceed to show the correctness of the invariant by a case
analysys.
Base cases:
1. [| ∅, (∅ → bbᵢ)ⁱ |] := Leaf bbᵢ where i := min(I), that is a
1. [| ∅, (∅ → bbᵢ)ⁱ |] ≡ Leaf bbᵢ where i := min(I), that is a
decision tree [|ms|] defined by an empty accessor and empty
patterns pointing to blackboxes /bbᵢ/. This respects the invariant
because a decomposition matrix in the case of empty rows returns
the first expression and we known that (Leaf bb)(v) := Match bb
2. [| (aⱼ)ʲ, ∅ |] := Failure
the first expression and (Leaf bb)(v) := Match bb
2. [| (aⱼ)ʲ, ∅ |] Failure
Regarding non base cases:
Let's first define
@ -1397,9 +1398,9 @@ that matches on all values that do not start with one of those head
constructors.
Intuitively, m is equivalent to its decomposition in the following
sense: if the first pattern of an input vector (v_i)^i starts with one
of the head constructors k, then running (v_i)^i against m is the same
as running it against the submatrix m_k; otherwise (its head
sense: if the first pattern of an input vector (vᵢ)ⁱ starts with one
of the head constructors k, then running (vᵢ)ⁱ against m is the same
as running it against the submatrix m; otherwise (its head
constructor is none of the k) it is equivalent to running it against
the wildcard submatrix.
@ -1466,21 +1467,23 @@ TODO: put ^i∈I where needed
\end{comment}
\subsubsection{The trimming lemma}
The trimming lemma allows to reduce the size of a decision tree given
an accessor → π relation (TODO: expand)
| ∀vₜ ∈ (a→π), Cₜ(vₜ) = C_{t/a→π(kᵢ)}(vₜ)
an accessor /a/ → π relation (TODO: expand)
| ∀vₜ ∈ (a→π), Cₜ(vₜ) = C_{t/a→π}(vₜ)
We prove this by induction on Cₜ:
a. Cₜ = Leaf_{bb}: when the decision tree is a leaf terminal, we
know that
| Leaf_{bb/a→π}(v) = Leaf_{bb}(v)
That means that the result of trimming on a Leaf is the Leaf itself
b. The same applies to Failure terminal
| Failure_{/a→π}(v) = Failure(v)
c. When Cₜ = Switch(b, (π→Cᵢ)ⁱ)_{/a→π} then
we look at the accessor /a/ of the subtree Cᵢ and
we define πᵢ' = πᵢ if a≠b else πᵢ∩π
Trimming a switch node yields the following result:
| Switch(b, (π→Cᵢ)ⁱ)_{/a→π} := Switch(b, (π'ᵢ→C_{i/a→π})ⁱ)
- Cₜ = Leaf_{bb}: when the decision tree is a leaf terminal, the result of trimming on a Leaf is the Leaf itself
| Leaf_{bb/a→π}(v) = Leaf_{bb}(v)
- The same applies to Failure terminal
| Failure_{/a→π}(v) = Failure(v)
- When Cₜ = Switch(b, (π→Cᵢ)ⁱ)_{/a→π} then we look at the accessor
/a/ of the subtree Cᵢ and we define πᵢ' = πᵢ if a≠b else πᵢ∩π Trimming
a switch node yields the following result:
| Switch(b, (π→Cᵢ)^{i∈I})_{/a→π} := Switch(b, (π'ᵢ→C_{i/a→π})^{i∈I})
\begin{comment}
TODO: understand how to properly align lists
check that every list is aligned
\end{comment}
\begin{comment}
Actually in the proof.org file I transcribed:
e. Unreachabe → ⊥
@ -1488,14 +1491,14 @@ This is not correct because you don't have Unreachable nodes in target decision
\end{comment}
For the trimming lemma we have to prove that running the value vₜ against
the decistion tree Cₜ is the same as running vₜ against the tree
the decision tree Cₜ is the same as running vₜ against the tree
C_{trim} that is the result of the trimming operation on Cₜ
| Cₜ(vₜ) = C_{trim}(vₜ) = Switch(b, (πᵢ'→C_{i/a→π}))(vₜ)
| Cₜ(vₜ) = C_{trim}(vₜ) = Switch(b, (πᵢ'→C_{i/a→π})^{i∈I})(vₜ)
We can reason by first noting that when vₜ∉(b→πᵢ)ⁱ the node must be a Failure node.
In the case where ∃k| vₜ∈(b→πₖ) then we can prove that
| C_{k/a→π}(vₜ) = Switch(b, (πᵢ'→C_{i/a→π}))(vₜ)
In the case where ∃k \vert{} vₜ∈(b→πₖ) then we can prove that
| C_{k/a→π}(vₜ) = Switch(b, (πᵢ'→C_{i/a→π})^{i∈I})(vₜ)
because when a ≠ b then πₖ'= πₖ and this means that vₜ∈πₖ'
while when a = b then πₖ'=(πₖ∩π) and vt∈πₖ' because:
while when a = b then πₖ'=(πₖ∩π) and v∈πₖ' because:
- by the hypothesis, vₜ∈π
- we are in the case where vₜ∈πₖ
So vₜ ∈ πₖ' and by induction
@ -1517,7 +1520,7 @@ Covering lemma:
\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)
| equiv(S, Cₛ, Cₜ) → Yes \vert{} No(vₛ, vₜ)
When the algorithm returns Yes and the input space is covered by /Cₛ/
we can say that the couple of decision trees are the same for
@ -1563,8 +1566,8 @@ I think the unreachable case should go at the end.
|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, Switch(a, (πᵢ → Cₜᵢ)ⁱ)) := Forall(equiv( S∩a→π(kᵢ)), Cₛ, Cₜᵢ)ⁱ)
| equiv(S, Leaf bbₛ as Cₛ, Switch(a, (πᵢ → Cₜᵢ)ⁱ)) := Forall(equiv( S∩a→π(kᵢ)), Cₛ, Cₜᵢ)ⁱ)
| equiv(S, Failure as Cₛ, Switch(a, (πᵢ → Cₜᵢ)^{i∈I})) := Forall(equiv( S∩a→π(kᵢ)), Cₛ, Cₜᵢ)^{i∈I})
| equiv(S, Leaf bbₛ as Cₛ, Switch(a, (πᵢ → Cₜᵢ)^{i∈I})) := Forall(equiv( S∩a→π(kᵢ)), Cₛ, Cₜᵢ)^{i∈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
@ -1577,13 +1580,13 @@ I think the unreachable case should go at the end.
| equiv(Sᵢ, Cₛ, Cₜᵢ) = No(vₛ, vₜ) for some minimal k∈I
4. When we have a Switch on the right we define πₙ as the domain of
values not covered but the union of the constructors kᵢ
| πₙ = ¬(⋃π(kᵢ))
| πₙ = ¬(⋃π(kᵢ)^{i∈I})
The algorithm proceeds by trimming
| equiv(S, Switch(a, (kᵢ → Cₛᵢ), 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}))
| equiv(S, Switch(a, (kᵢ → Cₛᵢ)^{i∈I}, C_{sf}), Cₜ) :=
| Forall(equiv( S∩(a→π(kᵢ)^{i∈I}), Cₛᵢ, C_{t/a→π(kᵢ)})^{i∈I} +++ equiv(S∩(a→πₙ), 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
| Forall(equiv( S∩(a→π(kᵢ)^{i∈I}), Cₛᵢ, C_{t/a→π(kᵢ)})^{i∈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