latex symbols and python
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@ -3,10 +3,12 @@ import re
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from sys import argv
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allsymbols = json.load(open('./unicode-latex.json'))
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mysymbols = ['≡', '≠', '≼', '→', '←', '⊀', '⋠', '≺', '∀', '∈', 'ε','₀', '₂', '₁', '₃', 'ₐ', 'ₖ', 'ₘ', 'ₙ', 'ᵢ', 'ⁱ', '⋮']
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mysymbols = ['≡', '≠', '≼', '→', '←', '⊀', '⋠', '≺', '∀', '∈', 'ε','₀', '₂', '₁', '₃', 'ₐ', 'ₖ', 'ₘ', 'ₙ', 'ᵢ', 'ⁱ', '⋮', 'ₛ', 'ₜ', '≃', '⇔', '∧', '∅', 'ℕ', 'ⱼ', 'ʲ', '⊥', 'π']
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extrasymbols = {'〚': '\llbracket', '〛': '\rrbracket'}
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symbols = {s: allsymbols[s] for s in mysymbols}
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mathsymbols = {s: '$'+allsymbols[s]+'$' for s in symbols}
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symbols.update(extrasymbols)
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mathsymbols = {s: '$'+v+'$' for s, v in symbols.items()}
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def read_by_char(fname):
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# Yield character and True/False if inside mathmode block
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BIN
tesi/tesi.pdf
BIN
tesi/tesi.pdf
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@ -921,17 +921,17 @@ Theorem: Given an input space /S/ and a couple of decision trees, where
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the target decision tree Cₜ covers the input space /S/, we say that
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the two decision trees are equivalent when:
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| (equiv S Cₛ Cₜ gs = Yes) ∧ covers(Cₜ, S) ⇒ ∀vₛ≃vₜ ∈ S, Cₛ(vₛ) ≃gs Cₜ(vₜ)
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| equiv(S, Cₛ, Cₜ, gs) = Yes ∧ covers(Cₜ, S) → ∀vₛ≃vₜ ∈ S, Cₛ(vₛ) ≃gs Cₜ(vₜ)
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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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| 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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| (equiv S [|tₛ|]ₛ [|tₜ|]ₜ ∅ = Yes) ⇔ tₛ ≃ tₜ
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| equiv(S, 〚tₛ〛ₛ, 〚tₜ〛ₜ, ∅) = Yes ⇔ tₛ ≃ tₜ
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*** Proof of the correctness of the translation from source programs to source decision trees
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@ -1011,13 +1011,13 @@ we can say that the couple of decision trees are the same for
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every couple of source value /vₛ/ and target value /vₜ/ that are equivalent.
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\begin{comment}
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Define "covered"
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Is it correct to say the same? How to correctly distinguish in words ≊ and = ?
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Is it correct to say the same? How to correctly distinguish in words ≃ and = ?
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\end{comment}
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| equiv(S, Cₛ, Cₜ) = Yes and cover(Cₜ, S) → ∀ vₛ ≊ vₜ∈S ∧ Cₛ(vₛ) = Cₜ(vₜ)
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| equiv(S, Cₛ, Cₜ) = Yes and cover(Cₜ, S) → ∀ vₛ ≃ vₜ∈S ∧ Cₛ(vₛ) = Cₜ(vₜ)
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In the case where the algorithm returns No we have at least a couple
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of counter example values vₛ and vₜ for which the two decision trees
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outputs a different result.
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| equiv(S, Cₛ, Cₜ) = No(vₛ,vₜ) and cover(Cₜ, S) → ∀ vₛ ≊ vₜ∈S ∧ Cₛ(vₛ) ≠ Cₜ(vₜ)
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| equiv(S, Cₛ, Cₜ) = No(vₛ,vₜ) and cover(Cₜ, S) → ∀ vₛ ≃ vₜ∈S ∧ Cₛ(vₛ) ≠ Cₜ(vₜ)
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We define the following
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| Forall(Yes) = Yes
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@ -1054,10 +1054,10 @@ In the other subcases S is always non-empty.
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subtree Cₜᵢ
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| equiv(Sᵢ, Cₛ, Cₜᵢ) = Yes ∀i
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or we have a counter example vₛ, vₜ for which
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| vₛ≊vₜ∈Sₖ ∧ cₛ(vₛ) ≠ Cₜₖ(vₜ)
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| vₛ≃vₜ∈Sₖ ∧ cₛ(vₛ) ≠ Cₜₖ(vₜ)
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then because
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| vₜ∈(a→πₖ) ⇒ Cₜ(vₜ) = Cₜₖ(vₜ)
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then
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| vₛ≊vₜ∈S ∧ Cₛ(vₛ)≠Cₜ(vₜ)
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| vₛ≃vₜ∈S ∧ Cₛ(vₛ)≠Cₜ(vₜ)
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and the result of the algorithm is
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| equiv(Sᵢ, Cₛ, Cₜᵢ) = No(vₛ, vₜ) for some minimal k∈I
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