Separating regular languages by piecewise testable and ......Thomas Place, Lorijn van Rooijen, Marc...

Separating regular languages by piecewise testableand unambiguous languages

Thomas Place, Lorijn van Rooijen, Marc Zeitoun

LaBRI · Univ. Bordeaux · CNRS

September, 2013 · Highlights of Logic, Games and Automata

Thomas Place, Lorijn van Rooijen, Marc Zeitoun Separation by PTL and UL 1/14

Separation problem

S is a fixed class of languages.

In this talk, S = BΣ1(<), FO2(<).

L1, L2 ∈ Reg are S-separable iff there exists

K ∈ S such that

K ∈ S

A∗ \K

Separation problem

K ∈ S such that

K ∈ S

A∗ \K

Separation problem

K ∈ S such that

K ∈ S

A∗ \K

Separation problem

Decision problem: Given L1, L2 ∈ Reg, are they S-separable?

Can we compute a separator?

What is the complexity of the decision problem?And of computing a separator?

Separation problem

Motivation

Captures discriminative power of logics

Generalization of membership problem

L1 L2 = A∗ \ L1

K ∈ S

Motivation

L1 L2 = A∗ \ L1

K ∈ S

Motivation

L1 L2 = A∗ \ L1

K ∈ S

Motivation

L1 L2 = A∗ \ L1

K ∈ S

Classes of separators

Piecewise testable languages: definable by BΣ1(<) formulas(recall previous talk).

Unambiguous languages: definable by FO2(<) formulas.

Membership is decidable for both BΣ1(<) and FO2(<)languages.

Simon ’75Schutzenberger ’76, Therien, Wilke ’98

Classes of separators

Piecewise testable languages: definable by BΣ1(<) formulas(recall previous talk).

Unambiguous languages: definable by FO2(<) formulas.

Membership is decidable for both BΣ1(<) and FO2(<)languages.

Simon ’75Schutzenberger ’76, Therien, Wilke ’98

Stratification of S

In general, there is no smallest separator in S.

→ Restriction of S to quantifier depth k: S[k].

If two languages are S[k]-separable, there is a smallestseparator in S[k].

Eg. (a2)∗ and b(b2)∗ have no smallest BΣ1(<)-separator,

while a∗ \ {a, a3, a5} is the smallest BΣ1(<)[6]-separator.

Stratification of S

Computing smallest separator for S[k]

Quantifier depth provides indexed equivalence relations for BΣ1(<)resp. FO2(<) languages:

w1 ∼k w2 ⇔ w1 and w2 satisfy the same BΣ1(<)resp. FO2(<) formulas up to depth k.

L is a BΣ1(<) resp. FO2(<) language

L is a union of ∼k-classes for some k ∈ N.

Increasing k refines the smallest potential separator:

[L]∼1[L]∼2[L]∼3

Refining [L]∼kgives a semi-algorithm for separability.

From which level of refinement can we conclude non-separability?

[L]∼1[L]∼2[L]∼3

[L]∼1

[L]∼2[L]∼3

[L]∼1

[L]∼2[L]∼3

[L]∼1

[L]∼2

[L]∼3

[L]∼1[L]∼2

[L]∼3

[L]∼1[L]∼2

[L]∼3

[L]∼1[L]∼2

[L]∼3

Testing non-separability

A pair of words w1 ∈ L1, w2 ∈ L2 is a k-witness of non-separabilityprovided

w1 ∼k w2.

Abstraction: tuples (i1, f1, i2, f2) A1

L1, L2 are k-separable ⇔ {k-witnesses} = ∅.

{k + 1-witnesses} ⊆ {k-witnesses}.Limit behaviour?

We can compute K such that

limit = ∅ ⇔ {K-witnesses} = ∅

w1 ∼k w2.

{k + 1-witnesses} ⊆ {k-witnesses}.

Limit behaviour?

w1 ∼k w2.

1 Verify that the languages are not separable by any language ofa sufficient, computable, index K.

2 Find a pattern in the automata producing k-witnesses forarbitrarily large k. (For BΣ1(<): same as in previous talk)

Patterns in the automata, independently of the bound K, provide

- Better complexity- A yes/no answer for separability- But no separator

1 Verify that the languages are not separable by any language ofa sufficient, computable, index K.

2 Find a pattern in the automata producing k-witnesses forarbitrarily large k. (For BΣ1(<): same as in previous talk)

Patterns in the automata, independently of the bound K, provide

- Better complexity- A yes/no answer for separability- But no separator

Main result

Theorem

For S = BΣ1(<), FO2(<), we can compute K ∈ N such thatTFAE

i. L1, L2 are S-separable.

ii. L1, L2 are S[K]-separable.

iii. [L1]∼K separates L1 from L2.

iv. A1,A2 do not contain a pattern witnessing non-separability.

Main result

Condition iv. yields the following complexity results:

Theorem

Given NFAs A1,A2, one can determine whether the languagesL(A1) and L(A2) are

BΣ1(<)-separable in PTIME,

FO2(<)-separable in EXPTIME,

with respect to |Q1|,|Q2|,|A|.

Future work

Obtain tight bounds on the size of BΣ1(<) resp. FO2(<) -separators

Efficient computation of the separators

Consider other classes S

Future work

Thank you

BΣ1(<): Witnesses in automata

L(A1) and L(A2) are not BΣ1(<)-separableiff

both A1 and A2 have a (~u, ~B)-path:

u0 u1 up−1 upB1 Bp

This can be determined in PTIME(|Q1|, |Q2|, |A|).

Detecting forbidden patterns

One can determine in PTIME(|Q1|, |Q2|, |A|) whether∃ (~u, ~B) such that both A1 and A2 have a (~u, ~B)-path.

By adding meta-transitions in Ai, finding a (~u, ~B)-witnessreduces to:

Given states pi, qi, ri of Ai, is there B ⊆ A st. the paths

p1 q1 r1 p2 q2 r2⊆ B ⊆ B ⊆ B ⊆ B

= B = B

occur in A1 resp. A2?

This is in Ptime: iteratively use Tarjan’s algorithm.

p1 q1 r1 p2 q2 r2⊆ B ⊆ B ⊆ B ⊆ B

= B = B

p1 q1 r1 p2 q2 r2⊆ B ⊆ B ⊆ B ⊆ B

= B = B

If B exists, B ⊆ C1def= alph scc(q1,A1) ∩ alph scc(q2,A2)

(linear)

Restrict the automata to alphabet C1, and repeat the process:

Ci+1def= alph scc(q1,A1 �Ci) ∩ alph scc(q2,A2 �Ci).

After n 6 |A| iterations, Cn = Cn+1.

If Cn = ∅, the answer is no.If Cn 6= ∅, it is the maximal possible B with (= B)-loopsaround q1, q2.

Then, determine the remaining paths. (linear)

Overall linear algorithm.

Separating regular languages by piecewise testable and ......Thomas Place, Lorijn van Rooijen, Marc...

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