correzioni gabriel #2
This commit is contained in:
Francesco Mecca
parent
14cf53ee8d
commit
97360b7f1d
4 changed files with 237 additions and 70 deletions
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@ -7,7 +7,7 @@ try:
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except:
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allsymbols = json.load(open('../unicode-latex.json'))
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mysymbols = ['≡', '≠', '≼', '→', '←', '⊀', '⋠', '≺', '∀', 'ε', '₀', '₂', '₁', '₃', '₄', 'ₐ', 'ₖ', 'ₘ', 'ₙ', 'ᵢ', 'ⁱ', '⋮', 'ₛ', 'ₜ', '≃', '⇔', '∧', '∅', 'ℕ', 'ⱼ', 'ʲ', '⊥', 'π', 'α', 'β', '∞', 'σ', '≤', '⊈', '∧', '∨', '∃', '⇒', '∩', '∉', '⋃', 'ᵏ', 'ₗ', 'ˡ', 'ₒ', 'ᵣ', 'ᴵ', '≈', '⊆' ]
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mysymbols = ['≡', '≠', '≼', '→', '←', '⊀', '⋠', '≺', '∀', 'ε', '₀', '₂', '₁', '₃', '₄', 'ₐ', 'ₖ', 'ᵥ', 'ₘ', 'ₙ', 'ᵢ', 'ⁱ', '⋮', 'ₛ', 'ₜ', '≃', '⇔', '∧', '∅', 'ℕ', 'ⱼ', 'ʲ', '⊥', 'π', 'α', 'β', '∞', 'σ', '≤', '⊈', '∧', '∨', '∃', '⇒', '∩', '∉', '⋃', 'ᵏ', 'ₗ', 'ˡ', 'ₒ', 'ᵣ', 'ᴵ', '≈', '⊆' ]
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extrasymbols = {'〚': '\llbracket', '〛': r'\rrbracket', '̸': '\neg', '¬̸': '\neg', '∈': '\in ', 'ₛ': '_S', 'ₜ': '_T'}
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symbols = {s: allsymbols[s] for s in mysymbols}
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139
tesi/drafts/partial-decomposition-proof.org
Normal file
139
tesi/drafts/partial-decomposition-proof.org
Normal file
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@ -0,0 +1,139 @@
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# done
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Auxiliary operations:
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index family of a constructor
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Idx(K) := [0; arity(K)[
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head of an ordered family (we write x for any object here, value, pattern etc.)
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head((xᵢ)^{i ∈ I}) = x_min(I)
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tail of an ordered family
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tail((xᵢ)^{i ∈ I}) := (xᵢ)^{i ≠ min(I)}
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head constructor of a value or pattern
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constr(K(xᵢ)ⁱ) = K
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constr(_) = ⊥
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constr(x) = ⊥
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first non-⊥ element of an ordered family
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First((xᵢ)ⁱ) := ⊥ if ∀i, xᵢ = ⊥
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First((xᵢ)ⁱ) := x_min{i | xᵢ ≠ ⊥} if ∃i, xᵢ ≠ ⊥
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# end
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Definition of running a pattern row against a (same-size) value vector (pᵢ)ⁱ(vᵢ)ⁱ
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∅ (∅) := []
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(_, tail(pᵢ)ⁱ) (vᵢ) := tail(pᵢ)ⁱ(tail(vᵢ)ⁱ)
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(x, tail(pᵢ)ⁱ) (vᵢ) := σ[x↦v₀] if tail(pᵢ)ⁱ(tail(vᵢ)ⁱ) = σ
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(K(qⱼ)ʲ, tail(pᵢ)ⁱ) (K(v'ⱼ)ʲ,tail(vⱼ)ʲ) := ((qⱼ)ʲ +++ tail(pᵢ)ⁱ)((v'ⱼ)ʲ +++ tail(vᵢ)ⁱ)
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(K(qⱼ)ʲ, tail(pᵢ)ⁱ) (K'(v'ₗ)ˡ,tail(vⱼ)ʲ) := ⊥ if K ≠ K'
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(q₁|q₂, tail(pᵢ)ⁱ) (vᵢ)ⁱ := First((q₁,tail(pᵢ)ⁱ)(vᵢ)ⁱ, (q₂,tail(pᵢ)ⁱ)(vᵢ)ⁱ)
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Definition of group decomposition:
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let constrs((pᵢ)^{i ∈ I}) = { K | K = constr(pᵢ), i ∈ I }
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let Groups(m) where m = ((aᵢ)ⁱ ((pᵢⱼ)ⁱ → eⱼ)ⁱʲ) =
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let (Kₖ)ᵏ = constrs(pᵢ₀)ⁱ in
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( Kₖ →
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((a₀.ₗ)ˡ +++ tail(aᵢ)ⁱ)
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(
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if pₒⱼ is Kₖ(qₗ) then
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(qₗ)ˡ +++ tail(pᵢⱼ)ⁱ → eⱼ
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if pₒⱼ is _ then
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(_)ˡ +++ tail(pᵢⱼ)ⁱ → eⱼ
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else ⊥
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)ʲ
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), (
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tail(aᵢ)ⁱ, (tail(pᵢⱼ)ⁱ → eⱼ if p₀ⱼ is _ else ⊥)ʲ
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)
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Groups(m) is an auxiliary function that decomposes a matrix m into
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submatrices, according to the head constructor of their first pattern.
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Groups(m) returns one submatrix m_r for each head constructor K that
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occurs on the first row of m, plus one "wildcard submatrix" m_{wild}
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that matches on all values that do not start with one of those head
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constructors.
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Intuitively, m is equivalent to its decomposition in the following
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sense: if the first pattern of an input vector (v_i)^i starts with one
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of the head constructors Kₖ, then running (v_i)^i against m is the same
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as running it against the submatrix m_{Kₖ}; otherwise (its head
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constructor is none of the Kₖ) it is equivalent to running it against
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the wildcard submatrix.
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We formalize this intuition as follows:
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Lemma (Groups):
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Let $m$ be a matrix with $Groups(m) = (kᵣ → mᵣ)ᵏ, m_{wild}$.
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For any value vector $(vᵢ)ᴵ$ such that $v₀ = K(v'ₗ)ˡ$ for some
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constructor K,
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we have:
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\[
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if K = Kₖ for some k then
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m(vᵢ)ⁱ = mₖ((v'ₗ)ˡ +++ (vᵢ)^{i∈I\{0}})
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otherwise
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m(vᵢ)ⁱ = m_{wild}(vᵢ)^{i∈I\{0}}
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\]
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Proof:
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Let $m$ be a matrix ((aᵢ)ⁱ, ((pᵢⱼ)ⁱ → eⱼ)ʲ) with $Groups(m) = (Kₖ → mₖ)ᵏ, m_{wild}$.
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Below we are going to assume that m is a simplified matrix:
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the first row does not with an or-pattern or a variable
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I think it would be simpler to not consider them at all in the
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formalization. If we do want them, the discrepancy can be fixed
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later by defining a simplification step that explodes
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head-or-patterns and binds variable, and showing that it preserves
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the matrix semantics.
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Let (vᵢ)ⁱ be an input matrix with v₀ = Kᵥ(v'ₗ)ˡ for some constructor Kᵥ.
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We have to show that:
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- if Kₖ = Kᵥ for some Kₖ ∈ constrs(p₀ⱼ)ʲ, then
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m(vᵢ)ⁱ = mₖ((v'ₗ)ˡ +++ tail(vᵢ)ⁱ)
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- otherwise
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m(vᵢ)ⁱ = m_{wild}(tail(vᵢ)ⁱ)
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Let us call (rₖⱼ) the j-th row of the submatrix mₖ, and rⱼ_{wild}
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the j-th row of the wildcard submatrix m_{wild}.
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Our goal contains same-behavior equalities between matrices, for
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a fixed input vector (vᵢ)ⁱ. It suffices to show same-behavior
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equalities between each row of the matrices for this input
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vector. We show that for any j,
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- if Kₖ = Kᵥ for some Kₖ ∈ constrs(p₀ⱼ)ʲ, then
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(pᵢⱼ)ⁱ(vᵢ)ⁱ = rₖⱼ((v'ₗ)ˡ +++ tail(vᵢ)ⁱ
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- otherwise
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(pᵢⱼ)ⁱ(vᵢ)ⁱ = rⱼ_{wild} tail(vᵢ)ⁱ
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In the first case (Kᵥ is Kₖ for some Kₖ ∈ constrs(p₀ⱼ)ʲ), we
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have to prove that
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(pᵢⱼ)ⁱ(vᵢ)ⁱ = rₖⱼ((v'ₗ)ˡ +++ tail(vᵢ)ⁱ
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By definition of mₖ we know that rₖⱼ is equal to
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if pₒⱼ is Kₖ(qₗ) then
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(qₗ)ˡ +++ tail(pᵢⱼ)ⁱ → eⱼ
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if pₒⱼ is _ then
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(_)ˡ +++ tail(pᵢⱼ)ⁱ → eⱼ
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else ⊥
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By definition of (pᵢ)ⁱ(vᵢ)ⁱ we know that (pᵢⱼ)ⁱ(vᵢ) is equal to
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(K(qⱼ)ʲ, tail(pᵢⱼ)ⁱ) (K(v'ₗ)ˡ,tail(vᵢ)ⁱ) := ((qⱼ)ʲ +++ tail(pᵢⱼ)ⁱ)((v'ₗ)ˡ +++ tail(vᵢ)ⁱ)
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(_, tail(pᵢⱼ)ⁱ) (vᵢ) := tail(pᵢⱼ)ⁱ(tail(vᵢ)ⁱ)
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(K(qⱼ)ʲ, tail(pᵢⱼ)ⁱ) (K'(v'ₗ)ˡ,tail(vⱼ)ʲ) := ⊥ if K ≠ K'
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We prove this first case by a second case analysis on p₀ⱼ.
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TODO
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In the second case (Kᵥ is distinct from Kₖ for all Kₖ ∈ constrs(pₒⱼ)ʲ),
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we have to prove that
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(pᵢⱼ)ⁱ(vᵢ)ⁱ = rⱼ_{wild} tail(vᵢ)ⁱ
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TODO
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BIN
tesi/tesi.pdf
BIN
tesi/tesi.pdf
Binary file not shown.
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@ -336,7 +336,7 @@ for content equality.
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\end{mathpar}
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If the source decision tree (left hand side) is a terminal while the
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target decistion tree (right hand side) is not, the algorithm proceeds
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target decisiorn tree (right hand side) is not, the algorithm proceeds
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by \emph{explosion} of the right hand side. Explosion means that every
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child of the right hand side is tested for equality against the left
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hand side.
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@ -1316,92 +1316,120 @@ Base cases:
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the first expression and (Leaf bb)(v) := Match bb
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2. [| (aⱼ)ʲ, ∅ |] ≡ Failure
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Regarding non base cases:
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Let's first define
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| let Idx(k) := [0; arity(k)[
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| let First(∅) := ⊥
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| let First((aⱼ)ʲ) := a_{min(j∈J≠∅)}
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\[
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m := ((a_i)^i ((p_{ij})^i \to e_j)^{ij})
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\]
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\[
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(k_k)^k := headconstructor(p_{i0})^i
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\]
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| Groups(m) := (k_{k} \to ((a)_{0.l})^{l \in Idx(k_{k})} +++ (a_{i})^{i\in{}I\DZ}),
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|(
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| \quad if p_{0j} is k(q_{l}) then
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| \quad \quad \quad (qₗ)^{l \in Idx(k_k)} +++ (p_{ij})^{i\in{}I\DZ} \to e_{j}
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| \quad if p_{0j} is "_" then
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| \quad \quad \quad ("_")^{l \in Idx(k_{k})} +++ (p_{ij})^{i\in{}I\DZ} \to e_{j}
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| \quad else \bot
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|)^{j ∈ J},
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| ((a_{i})^{i\in{}I\DZ}, ((p_{ij})^{i\in{}I\DZ} \to eⱼ if p_{0j} is \_ else \bot)^{j\in{}J})
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Groups(m) is an auxiliary function that source a matrix m into
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Let's first define some auxiliary functions
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- The index family of a constructor
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| Idx(K) := [0; arity(K)[
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- head of an ordered family (we write x for any object here, value, pattern etc.)
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| head((xᵢ)^{i ∈ I}) = x_min(I)
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- tail of an ordered family
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| tail((xᵢ)^{i ∈ I}) := (xᵢ)^{i ≠ min(I)}
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- head constructor of a value or pattern
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| constr(K(xᵢ)ⁱ) = K
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| constr(_) = ⊥
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| constr(x) = ⊥
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- first non-⊥ element of an ordered family
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| First((xᵢ)ⁱ) := ⊥ \quad \quad \quad if ∀i, xᵢ = ⊥
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| First((xᵢ)ⁱ) := x_min{i \vert{} xᵢ ≠ ⊥} \quad if ∃i, xᵢ ≠ ⊥
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- definition of group decomposition:
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| let constrs((pᵢ)^{i ∈ I}) = { K \vert{} K = constr(pᵢ), i ∈ I }
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| let Groups(m) where m = ((aᵢ)ⁱ ((pᵢⱼ)ⁱ → eⱼ)ⁱʲ) =
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| \quad let (Kₖ)ᵏ = constrs(pᵢ₀)ⁱ in
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| \quad ( Kₖ →
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| \quad \quad ((a₀.ₗ)ˡ +++ tail(aᵢ)ⁱ)
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| \quad \quad (
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| \quad \quad if pₒⱼ is Kₖ(qₗ) then
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| \quad \quad \quad (qₗ)ˡ +++ tail(pᵢⱼ)ⁱ → eⱼ
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| \quad \quad if pₒⱼ is _ then
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| \quad \quad \quad (_)ˡ +++ tail(pᵢⱼ)ⁱ → eⱼ
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| \quad \quad else ⊥
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| \quad \quad )ʲ
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| \quad ), (
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| \quad \quad tail(aᵢ)ⁱ, (tail(pᵢⱼ)ⁱ → eⱼ if p₀ⱼ is _ else ⊥)ʲ
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| \quad )
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Groups(m) is an auxiliary function that decomposes a matrix m into
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submatrices, according to the head constructor of their first pattern.
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Groups(m) returns one submatrix m_r for each head constructor k that
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Groups(m) returns one submatrix m_r for each head constructor K that
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occurs on the first row of m, plus one "wildcard submatrix" m_{wild}
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that matches on all values that do not start with one of those head
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constructors.
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Intuitively, m is equivalent to its decomposition in the following
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sense: if the first pattern of an input vector (vᵢ)ⁱ starts with one
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of the head constructors k, then running (vᵢ)ⁱ against m is the same
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as running it against the submatrix mₖ; otherwise (its head
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constructor is none of the k) it is equivalent to running it against
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sense: if the first pattern of an input vector (v_i)^i starts with one
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of the head constructors Kₖ, then running (v_i)^i against m is the same
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as running it against the submatrix m_{Kₖ}; otherwise (its head
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constructor is none of the Kₖ) it is equivalent to running it against
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the wildcard submatrix.
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We formalize this intuition as follows:
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Lemma (Groups):
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Let /m/ be a matrix with \[Groups(m) = (k_r \to m_r)^k, m_{wild}\].
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For any value vector $(v_i)^l$ such that $v_0 = k(v'_l)^l$ for some
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constructor k,
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we have:
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| if k = kₖ \text{ for some k then}
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| \quad m(vᵢ)ⁱ = mₖ((v_{l}')ˡ +++ (v_{i})^{i∈I\DZ})
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| \text{else}
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| \quad m(vᵢ)ⁱ = m_{wild}(vᵢ)^{i∈I\DZ}
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We formalize this intuition as follows
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*** Lemma (Groups):
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Let /m/ be a matrix with
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| Groups(m) = (k_r \to m_r)^k, m_{wild}
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For any value vector $(v_i)^l$ such that $v_0 = k(v'_l)^l$ for some
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constructor k,
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we have:
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| if k = kₖ \text{ for some k then}
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| \quad m(vᵢ)ⁱ = mₖ((v_{l}')ˡ +++ (v_{i})^{i∈I\DZ})
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| \text{else}
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| \quad m(vᵢ)ⁱ = m_{wild}(vᵢ)^{i∈I\DZ}
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\begin{comment}
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TODO: fix \{0}
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\end{comment}
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*** Proof:
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Let $m$ be a matrix with \[Group(m) = (k_r \to m_r)^k, m_{wild}\]
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Let $(v_i)^i$ be an input matrix with $v_0 = k(v'_l)^l$ for some k.
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We proceed by case analysis:
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- either k is one of the kₖ for some k
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- or k is none of the (kₖ)ᵏ
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Let /m/ be a matrix ((aᵢ)ⁱ, ((pᵢⱼ)ⁱ → eⱼ)ʲ) with
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| Groups(m) = (Kₖ → mₖ)ᵏ, m_{wild}
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Below we are going to assume that m is a simplified matrix such that
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the first row does not contain an or-pattern or a binding to a
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variable.
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Both m(vᵢ)ⁱ and mₖ(vₖ)ᵏ are defined as the first matching result of
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a family over each row rⱼ of a matrix
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Let (vᵢ)ⁱ be an input matrix with v₀ = Kᵥ(v'_{l})ˡ for some constructor Kᵥ.
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We have to show that:
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- if Kₖ = Kᵥ for some Kₖ ∈ constrs(p₀ⱼ)ʲ, then
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m(vᵢ)ⁱ = mₖ((v'ₗ)ˡ +++ tail(vᵢ)ⁱ)
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- otherwise
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m(vᵢ)ⁱ = m_{wild}(tail(vᵢ)ⁱ)
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Let us call (rₖⱼ) the j-th row of the submatrix mₖ, and rⱼ_{wild}
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the j-th row of the wildcard submatrix m_{wild}.
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We know, from the definition of
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Groups(m), that mₖ is
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| ((a){0.l})^{l∈Idx(kₖ)} +++ (aᵢ)^{i∈I\DZ}),
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| (
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| \quad if p_{0j} is k(qₗ) then
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| \quad \quad (qₗ)ˡ +++ (p_{ij})^{i∈I\DZ } → eⱼ
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| \quad if p_{0j} is _ then
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| \quad \quad (_)ˡ +++ (p_{ij})^{i∈I\DZ} → eⱼ
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| \quad else ⊥
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| )^{j∈J}
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Our goal contains same-behavior equalities between matrices, for
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a fixed input vector (vᵢ)ⁱ. It suffices to show same-behavior
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equalities between each row of the matrices for this input
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vector. We show that for any j,
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- if Kₖ = Kᵥ for some Kₖ ∈ constrs(p₀ⱼ)ʲ, then
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| (pᵢⱼ)ⁱ(vᵢ)ⁱ = rₖⱼ((v'ₗ)ˡ +++ tail(vᵢ)ⁱ
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- otherwise
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| (pᵢⱼ)ⁱ(vᵢ)ⁱ = rⱼ_{wild} tail(vᵢ)ⁱ
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In the first case (Kᵥ is Kₖ for some Kₖ ∈ constrs(p₀ⱼ)ʲ), we
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have to prove that
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| (pᵢⱼ)ⁱ(vᵢ)ⁱ = rₖⱼ((v'ₗ)ˡ +++ tail(vᵢ)ⁱ
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By definition of mₖ we know that rₖⱼ is equal to
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| if pₒⱼ is Kₖ(qₗ) then
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| \quad (qₗ)ˡ +++ tail(pᵢⱼ)ⁱ → eⱼ
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| if pₒⱼ is _ then
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| \quad (_)ˡ +++ tail(pᵢⱼ)ⁱ → eⱼ
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| else ⊥
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\begin{comment}
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Maybe better as a table?
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| pₒⱼ | rₖⱼ |
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|--------+---------------------------|
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| Kₖ(qₗ) | (qₗ)ˡ +++ tail(pᵢⱼ)ⁱ → eⱼ |
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| _ | (_)ˡ +++ tail(pᵢⱼ)ⁱ → eⱼ |
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| else | ⊥ |
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\end{comment}
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By definition of (pᵢ)ⁱ(vᵢ)ⁱ we know that (pᵢⱼ)ⁱ(vᵢ) is equal to
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| (K(qⱼ)ʲ, tail(pᵢⱼ)ⁱ) (K(v'ₗ)ˡ,tail(vᵢ)ⁱ) := ((qⱼ)ʲ +++ tail(pᵢⱼ)ⁱ)((v'ₗ)ˡ +++ tail(vᵢ)ⁱ)
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| (_, tail(pᵢⱼ)ⁱ) (vᵢ) := tail(pᵢⱼ)ⁱ(tail(vᵢ)ⁱ)
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| (K(qⱼ)ʲ, tail(pᵢⱼ)ⁱ) (K'(v'ₗ)ˡ,tail(vⱼ)ʲ) := ⊥ if K ≠ K'
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By definition, m(vᵢ)ⁱ is
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| m(vᵢ)ⁱ = First(rⱼ(vᵢ)ⁱ)ʲ for m = ((aᵢ)ⁱ, (rⱼ)ʲ)
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| (pᵢ)ⁱ (vᵢ)ⁱ = {
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| \quad if p₀ = k(qₗ)ˡ, v₀ = k'(v'ₖ)ᵏ, k=Idx(k') and l=Idx(k)
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| \quad \quad if k ≠ k' then ⊥
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| if k = k' then ((qₗ)ˡ +++ (pᵢ)^{i∈I\DZ}) ((v'ₖ)ᵏ +++ (vᵢ)^{i∈I\DZ} )
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| if p₀ = (q₁\vert{}q₂) then
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| First( (q₁pᵢ^{i∈I\DZ}) vᵢ^{i∈I\DZ}, (q₂pᵢ^{i∈I \DZ}) vᵢ^{i∈I\DZ})}
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We prove this first case by a second case analysis on p₀ⱼ.
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For this reason, if we can prove that
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| ∀j, rⱼ(vᵢ)ⁱ = r'ⱼ((v'ₖ)ᵏ ++ (vᵢ)ⁱ)
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it follows that
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| m(vᵢ)ⁱ = mₖ((v'ₖ)ᵏ ++ (vᵢ)ⁱ)
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from the above definition.
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TODO
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We can also show that aᵢ = (a_{0.l})ˡ +++ a_{i∈I\DZ} because v(a₀) = K(v(a){0.l})ˡ)
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In the second case (Kᵥ is distinct from Kₖ for all Kₖ ∈ constrs(pₒⱼ)ʲ),
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we have to prove that
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| (pᵢⱼ)ⁱ(vᵢ)ⁱ = rⱼ_{wild} tail(vᵢ)ⁱ
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TODO
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** Target translation
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@ -1597,7 +1625,7 @@ Similarly we say that a couple of decision trees in the presence of
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an input space /S/ are /not/ equivalent when:
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| equiv(S, Cₛ, Cₜ, gs) = No(vₛ,vₜ) ∧ covers(Cₜ, S) → vₛ≃vₜ ∈ S ∧ Cₛ(vₛ) ≠gs Cₜ(vₜ)
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Corollary: For a full input space /S/, that is the universe of the
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target program we say:
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target program:
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| equiv(S, 〚tₛ〛ₛ, 〚tₜ〛ₜ, ∅) = Yes ⇔ tₛ ≃ tₜ
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Loading…
Reference in a new issue