E Language (L): {s ∈ Σ* : #b's = 1 } Alphabet (Σ) : {a, b} T is Closed (states): ∀ w∈S⋅ Σ, ∃ s ∈ S such that [w,]=[s,] If not closed, add the counter example to S T is Consistent (transitions): ∀ s1,s2∈S, if row(s1) = row(s2) then ∀ a ∈ A [s1a,]=[s2a,] If not consistent, add the counter example to E We begin with S = {ɛ} and E = {ɛ} a 0 b 1 S · Σ
E Alphabet (Σ) : {a, b} T is Closed: ∀ w∈S⋅ Σ, ∃ s ∈ S such that [w,]=[s,] If not closed, add the counter example to S T is Consistent: ∀ s1,s2∈S, if row(s1) = row(s2) then ∀ a ∈ A [s1a,]=[s2a,] If not consistent, add the counter example to E We begin with S = {ɛ} and E = {ɛ} a 0 b 1 S · Σ Language (L): {s ∈ Σ* : #b's = 1 }
Table S E a 0 ba 1 bb 0 S · Σ T is Closed: ∀ w∈S⋅ Σ, ∃ s ∈ S such that [w,]=[s,] If not closed, add the counter example to S T is Consistent: ∀ s1,s2∈S, if row(s1) = row(s2) then ∀ a ∈ A [s1a,]=[s2a,] If not consistent, add the counter example to E Language (L): {s ∈ Σ* : #b's = 1 } Alphabet (Σ) : {a, b}
Table S E a 0 ba 1 bb 0 S · Σ T is Closed: ∀ w∈S⋅ Σ, ∃ s ∈ S such that [w,]=[s,] If not closed, add the counter example to S T is Consistent: ∀ s1,s2∈S, if row(s1) = row(s2) then ∀ a ∈ A [s1a,]=[s2a,] If not consistent, add the counter example to E Language (L): {s ∈ Σ* : #b's = 1 } Alphabet (Σ) : {a, b}
Table S E a 0 ba 1 bb 0 S · Σ T is Closed: ∀ w∈S⋅ Σ, ∃ s ∈ S such that [w,]=[s,] If not closed, add the counter example to S T is Consistent: ∀ s1,s2∈S, if row(s1) = row(s2) then ∀ a ∈ A [s1a,]=[s2a,] If not consistent, add the counter example to E Alphabet (Σ) : {a, b} Language (L): {s ∈ Σ* : #b's = 1 }
Table S E a 0 ba 1 bb 0 S · Σ T is Closed: ∀ w∈S⋅ Σ, ∃ s ∈ S such that [w,]=[s,] If not closed, add the counter example to S T is Consistent: ∀ s1,s2∈S, if row(s1) = row(s2) then ∀ a ∈ A [s1a,]=[s2a,] If not consistent, add the counter example to E Alphabet (Σ) : {a, b} Language (L): {s ∈ Σ* : #b's = 1 }
∀ w∈S⋅ Σ, ∃ s ∈ S such that [w,]=[s,] If not closed, add the counter example to S T is Consistent: ∀ s1,s2∈S, if row(s1) = row(s2) then ∀ a ∈ A [s1a,]=[s2a,] If not consistent, add the counter example to E <$ɛ>:= a <$ɛ> | b <$b> <$b>:= a <$b> | b <$ɛ> | ɛ ɛ ɛ 0 b 1 S a 0 ba 1 bb 0 S · Σ Alphabet (Σ) : {a, b} Language (L): {s ∈ Σ* : #b's = 1 }
Closed: ∀ w∈S⋅ Σ, ∃ s ∈ S such that [w,]=[s,] If not closed, add the counter example to S T is Consistent: ∀ s1,s2∈S, if row(s1) = row(s2) then ∀ a ∈ A [s1a,]=[s2a,] If not consistent, add the counter example to E bbb ✘ <ɛ> := a <ɛ> | b <b> <b> := a <b> | b <ɛ> | ɛ ɛ ɛ 0 b 1 a 0 ba 1 bb 0 S · Σ Alphabet (Σ) : {a, b} Language (L): {s ∈ Σ* : #b's = 1 }
0 bbb 0 Observation Table S E T is Closed: ∀ w∈S⋅ Σ, ∃ s ∈ S such that [w,]=[s,] If not closed, add the counter example to S T is Consistent: ∀ s1,s2∈S, if row(s1) = row(s2) then ∀ a ∈ A [s1a,]=[s2a,] If not consistent, add the counter example to E Alphabet (Σ) : {a, b} Language (L): {s ∈ Σ* : #b's = 1 }
0 bbb 0 Observation Table S E T is Closed: ∀ w∈S⋅ Σ, ∃ s ∈ S such that [w,]=[s,] If not closed, add the counter example to S T is Consistent: ∀ s1,s2∈S, if row(s1) = row(s2) then ∀ a ∈ A [s1a,]=[s2a,] If not consistent, add the counter example to E Alphabet (Σ) : {a, b} Language (L): {s ∈ Σ* : #b's = 1 }
ba 1 bba 0 bbba 0 bbbb 0 S · Σ T is Closed: ∀ w∈S⋅ Σ, ∃ s ∈ S such that [w,]=[s,] If not closed, add the counter example to S T is Consistent: ∀ s1,s2∈S, if row(s1) = row(s2) then ∀ a ∈ A [s1a,]=[s2a,] If not consistent, add the counter example to E ɛ ɛ 0 b 1 bb 0 bbb 0 Alphabet (Σ) : {a, b} Language (L): {s ∈ Σ* : #b's = 1 }
Closed: ∀ w∈S⋅ Σ, ∃ s ∈ S such that [w,]=[s,] If not closed, add the counter example to S T is Consistent: ∀ s1,s2∈S, if row(s1) = row(s2) then ∀ a ∈ A [s1a,]=[s2a,] If not consistent, add the counter example to E s1 = [ɛ,] 0 s2 = [bb,] 0 s1.a = [ɛb,] 1 s2.a = [bbb,] 0 ɛ ɛ 0 b 1 bb 0 bbb 0 Alphabet (Σ) : {a, b} Language (L): {s ∈ Σ* : #b's = 1 } a 0 ba 1 bba 0 bbba 0 bbbb 0 S · Σ b b
Closed: ∀ w∈S⋅ Σ, ∃ s ∈ S such that [w,]=[s,] If not closed, add the counter example to S T is Consistent: ∀ s1,s2∈S, if row(s1) = row(s2) then ∀ a ∈ A [s1a,]=[s2a,] If not consistent, add the counter example to E ɛ b ɛ 0 1 b 1 0 bb 0 0 bbb 0 0 Alphabet (Σ) : {a, b} Language (L): {s ∈ Σ* : #b's = 1 } a 0 1 ba 1 0 bba 0 0 bbba 0 0 bbbb 0 0 S · Σ
an observation table -Minimise the counter examples through exponential backo ff and binary search of the su ffi x -Intuition LG(w) != BB(w) -So, there should be a short pre fi x u such that w = uv where they di ff er originally. Find it through binary search. v is then the distinguishing su ffi x. Minimise that su ff i x. -You need to only add all pre fi xes of u to S. -Remove su ff i xes that do not add any value. Improvements (TTT)
Equivalences Queries are not possible in software engineering scenarios <$ɛ> := a <$ɛ> | b <$b> <$b> := a <$b> | ɛ Alphabet (Σ) : {a, b} Language (L): {s ∈ Σ* : #b's = 1 } ✓
Pr(L(A)≢X ≤ ϵ) ≥ 1−δ 1-∈: accuracy 1-δ: confidence Equivalence Query = Multiple Membership Checks Checks come from some sampling distribution D over A* We only get a PAC guarantee based on D qi = [1/ϵ (ln(1/δ) + i ln(2))] Checks made in place of ith equivalence query:
between D,ϵ,δ and F1 score On Arithmetic (depth limited) L(*) Eq = Pre fi x Sampler Eq = Pre fi x Sampler) (p=0.05) (p=0.5) Eq = Pre fi x Sampler) (p=1.0) Red is good, Blue is bad PL(*) PL(*) PL(*) 1-δ: confidence 1-∈: accuracy
between D,ϵ,δ and F1 score On JSON (depth limited) L(*) Eq = Pre fi x Sampler (p=0.05) Eq = Pre fi x Sampler) (p=0.5) Eq = Pre fi x Sampler) (p=1.0) Red is good, Blue is bad 1-δ: confidence 1-∈: accuracy PL(*) PL(*) PL(*)
(class {}){}; Issue 2937 from Closure const [y,y] = []; var {baz:{} = baz => {}} = baz => {}; Issue 385 from Rhino {while ((l_0)){ if ((l_0)) {break;;var l_0; continue }0}} Issue 2842 from Closure Delta Minimization is useful but not su ff i cient
<string> : null & def jsoncheck(json): if no_key_is_empty_string(json): fail(’one key must be empty’) if any_key_has_null_value(json): fail(’key value must not be null’) process(json) {"": 124} ✔ Start Symbol 101
is "": <elt> <item N> is <string>:null <calc not(D or F)> where <factor F> is ((<expr>)) <term D> is <factor> / 0 <ipv4addr O and H> where <quad O> is "0" <num> <quad H> is "0x" <num> <C not(F) not(EW or ED or F)> where <forCondition F> is ";;" <iterationStatement EW> is <WHILE> "()" <statement> <iterationStatement ED> is <DO> <statement> <WHILE> "()" <eos> Alternative to Regular Expressions 104
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