Theory of Computer Science C4. Regular Languages: Minimal Automata, Closure Properties and Decidability Gabriele R¨oger

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Theory of Computer Science

C4. Regular Languages: Minimal Automata, Closure Properties and Decidability

Gabriele R¨ oger

University of Basel

March 27, 2019

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Theory of Computer Science

March 27, 2019 — C4. Regular Languages: Minimal Automata, Closure Properties and Decidability

C4.1 Minimal Automata

C4.2 Closure Properties

C4.3 Decidability

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Overview

Automata &

Formal Languages

Languages

& Grammars

Regular Languages

Regular Grammars

DFAs

NFAs Regular Expressions

Pumping Lemma Minimal Automata Properties Context-free

Languages

Context-sensitive &

Type-0 Languages

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C4. Regular Languages: Minimal Automata, Closure Properties

and Decidability Minimal Automata

C4.1 Minimal Automata

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Example

The following DFAs accept the same language:

q0 q1

q2

q3 0

1

0 1 0

1

0,1

q0 q1

q2

q3

q4 0

1

0

1

0 1

0,1

Question: What is the smallest DFA that accepts this language?

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Minimal Automaton: Definition

Definition

A minimal automaton for a regular language L is a DFA M = hQ , Σ, δ, q ₀ , E i with L(M ) = L and a minimal number of states.

This means there is no DFA M ⁰ = hQ ⁰ , Σ, δ ⁰ , q ⁰ ₀ , E ⁰ i with L(M ) = L(M ⁰ ) and |Q ⁰ | < |Q |.

How to find a minimal automaton?

Idea:

I Start with any DFA that accepts the language.

I Merge states from which exactly the same words

lead to an end state.

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Minimal Automaton: Algorithm

Input: DFA M

Input:

(without states that are unreachable from the start state) Output: list of states that have to be merged

Output:

to obtain an equivalent minimal automaton

1

Create table of all pairs of states {q, q ⁰ } with q 6= q ⁰ .

2

Mark all pairs {q, q ⁰ } with q ∈ E and q ⁰ ∈ / E .

3

If there is an unmarked pair {q, q ⁰ } where {δ(q, a), δ(q ⁰ , a)}

for some a ∈ Σ is already marked, then also mark {q, q ⁰ }.

4

Repeat the last step until there are no more changes.

5

All states in pairs that are still unmarked can be merged

into one state.

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Minimal Automaton: Example

q0 q1

q2

q3

q4 0

1

0

1

0 1

0,1

q ₀ q ₁ q ₂ q ₃ q 4

q ₃ q 2

q ₁

× × × ×

×

States q ₀ , q ₂ and q ₁ , q ₃ can be merged into one state each.

Result:

q0q2 q1q3 q4 0

1 0

1

0,1

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Computation and Uniqueness of Minimal Automata

Theorem

The algorithm described on the previous slides produces a minimal automaton for the language accepted by the given input DFA.

Theorem

All minimal automata for a language L are unique up to isomorphism (i.e., renaming of states).

Without proof.

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Overview

Automata &

Formal Languages

Languages

& Grammars

Regular Languages

Regular Grammars

DFAs

NFAs Regular Expressions

Pumping Lemma Minimal Automata Properties Context-free

Languages

Context-sensitive &

Type-0 Languages

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and Decidability Closure Properties

C4.2 Closure Properties

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Closure Properties

How can you combine regular languages in a way to get

another regular language as a result?

Picture courtesy of stockimages / FreeDigitalPhotos.net

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Closure Properties: Operations

Let L and L ⁰ be regular languages over Σ and Σ ⁰ , respectively.

We consider the following operations:

I union L ∪ L ⁰ = {w | w ∈ L or w ∈ L ⁰ } over Σ ∪ Σ ⁰

I intersection L ∩ L ⁰ = {w | w ∈ L and w ∈ L ⁰ } over Σ ∩ Σ ⁰

I complement L ¯ = {w ∈ Σ ^∗ | w ∈ / L} over Σ

I product LL ⁰ = {uv | u ∈ L and v ∈ L ⁰ } over Σ ∪ Σ ⁰

I

special case: L

ⁿ

= L

ⁿ⁻¹

L, where L

⁰

= {ε}

I star L ^∗ = S

k≥0 L ^k over Σ

German: Abschlusseigenschaften, Vereinigung, Schnitt, Komplement,

German: Produkt, Stern

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Closure Properties

Definition (Closure)

Let K be a class of languages.

Then K is closed. . .

I . . . under union if L, L ⁰ ∈ K implies L ∪ L ⁰ ∈ K

I . . . under intersection if L, L ⁰ ∈ K implies L ∩ L ⁰ ∈ K

I . . . under complement if L ∈ K implies ¯ L ∈ K

I . . . under product if L, L ⁰ ∈ K implies LL ⁰ ∈ K

I . . . under star if L ∈ K implies L ^∗ ∈ K

German: Abgeschlossenheit, K ist abgeschlossen unter Vereinigung

German: (Schnitt, Komplement, Produkt, Stern)

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Clourse Properties of Regular Languages

Theorem

The regular languages are closed under:

I union

I intersection

I complement

I product

I star

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Closure Properties

Proof.

Closure under union, product, and star follows because for regular expressions α and β, the expressions

(α|β), (αβ) and (α ^∗ ) describe the corresponding languages.

Complement: Let M = hQ, Σ, δ, q 0 , E i be a DFA with L(M ) = L.

Then M ⁰ = hQ , Σ, δ, q ₀ , Q \ E i is a DFA with L(M ⁰ ) = ¯ L.

Intersection: Let M 1 = hQ ₁ , Σ 1 , δ 1 , q 01 , E 1 i and

M 2 = hQ ₂ , Σ 2 , δ 2 , q 02 , E 2 i be DFAs. The product automaton M = hQ ₁ × Q 2 , Σ 1 ∩ Σ 2 , δ, hq ₀₁ , q 02 i, E 1 × E 2 i

with δ(hq ₁ , q ₂ i, a) = hδ ₁ (q ₁ , a), δ ₂ (q ₂ , a)i accepts L(M ) = L(M ₁ ) ∩ L(M ₂ ).

German: Kreuzproduktautomat

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and Decidability Decidability

C4.3 Decidability

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Decision Problems and Decidability (1)

“Intuitive Definition:” Decision Problem, Decidability A decision problem is an algorithmic problem where

I for a given input

I an algorithm determines if the input has a given property

I and then produces the output “yes” or “no” accordingly.

A decision problem is decidable if an algorithm for it (that always gives the correct answer) exists.

German: Entscheidungsproblem, Eingabe, Eigenschaft, Ausgabe, German: entscheidbar

Note: “exists” 6= “is known”

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Decision Problems and Decidability (2)

Notes:

I not a formal definition: we did not formally define

“algorithm”, “input”, “output” etc. (which is not trivial)

I lack of a formal definition makes it difficult to prove that something is not decidable

studied thoroughly in the next part of the course

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Decision Problems: Example

For now we describe decision problems in a semi-formal

“given”/“question” way:

Example (Emptiness Problem for Regular Languages) The emptiness problem P ∅ for regular languages is the following problem:

Given: regular grammar G Question: Is L(G ) = ∅?

German: Leerheitsproblem

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Word Problem

Definition (Word Problem for Regular Languages) The word problem P ∈ for regular languages is:

Given: regular grammar G with alphabet Σ and word w ∈ Σ ^∗

Question: Is w ∈ L(G )?

German: Wortproblem (f¨ ur regul¨ are Sprachen)

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Decidability: Word Problem

Theorem

The word problem for regular languages is decidable.

Proof.

Construct a DFA M with L(M) = L(G ).

(The proofs in Chapter C2 describe a possible method.) Simulate M on input w . The simulation ends after |w | steps.

The DFA M is an end state after this iff w ∈ L(G ).

Print “yes” or “no” accordingly.

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Emptiness Problem

Definition (Emptiness Problem for Regular Languages) The emptiness problem P ∅ for regular languages is:

Given: regular grammar G Question: Is L(G ) = ∅?

German: Leerheitsproblem

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Decidability: Emptiness Problem

Theorem

The emptiness problem for regular languages is decidable.

Proof.

Construct a DFA M with L(M) = L(G ).

We have L(G ) = ∅ iff in the transition diagram of M there is no path from the start state to any end state.

This can be checked with standard graph algorithms

(e.g., breadth-first search).

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Finiteness Problem

Definition (Finiteness Problem for Regular Languages) The finiteness problem P ∞ for regular languages is:

Given: regular grammar G Question: Is |L(G )| < ∞?

German: Endlichkeitsproblem

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Decidability: Finiteness Problem

Theorem

The finiteness problem for regular languages is decidable.

Proof.

Construct a DFA M with L(M) = L(G ).

We have |L(G )| = ∞ iff in the transition diagram of M there is a cycle that is reachable from the start state and from which an end state can be reached.

This can be checked with standard graph algorithms.

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Intersection Problem

Definition (Intersection Problem for Regular Languages) The intersection problem P ∩ for regular languages is:

Given: regular grammars G and G ⁰ Question: Is L(G ) ∩ L(G ⁰ ) = ∅?

German: Schnittproblem

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Decidability: Intersection Problem

Theorem

The intersection problem for regular languages is decidable.

Proof.

Using the closure of regular languages under intersection, we can construct (e.g., by converting to DFAs, constructing the product automaton, then converting back to a grammar) a grammar G ⁰⁰ with L(G ⁰⁰ ) = L(G ) ∩ L(G ⁰ )

and use the algorithm for the emptiness problem P _∅ .

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Equivalence Problem

Definition (Equivalence Problem for Regular Languages) The equivalence problem P = for regular languages is:

Given: regular grammars G and G ⁰ Question: Is L(G ) = L(G ⁰ )?

German: Aquivalenzproblem ¨

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Decidability: Equivalence Problem

Theorem

The equivalence problem for regular languages is decidable.

Proof.

In general for languages L and L ⁰ , we have

L = L ⁰ iff (L ∩ L ¯ ⁰ ) ∪ (¯ L ∩ L ⁰ ) = ∅.

The regular languages are closed under intersection, union and complement, and we know algorithms for these operations.

We can therefore construct a grammar for (L ∩ L ¯ ⁰ ) ∪ (¯ L ∩ L ⁰ )

and use the algorithm for the emptiness problem P _∅ .

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and Decidability Summary

Theory of Computer Science C4. Regular Languages: Minimal Automata, Closure Properties and Decidability Gabriele R¨oger

Theory of Computer Science

C4. Regular Languages: Minimal Automata, Closure Properties and Decidability

Gabriele R¨ oger

University of Basel

March 27, 2019

Theory of Computer Science

March 27, 2019 — C4. Regular Languages: Minimal Automata, Closure Properties and Decidability

C4.1 Minimal Automata

C4.2 Closure Properties

C4.3 Decidability

Overview

Automata &

Formal Languages

Languages

& Grammars

Regular Languages

Regular Grammars

DFAs

NFAs Regular Expressions

Pumping Lemma Minimal Automata Properties Context-free

Languages

Context-sensitive &

Type-0 Languages

C4.1 Minimal Automata

Example

The following DFAs accept the same language:

Question: What is the smallest DFA that accepts this language?

Minimal Automaton: Definition

Definition

A minimal automaton for a regular language L is a DFA M = hQ , Σ, δ, q 0 , E i with L(M ) = L and a minimal number of states.

This means there is no DFA M 0 = hQ 0 , Σ, δ 0 , q 0 0 , E 0 i with L(M ) = L(M 0 ) and |Q 0 | < |Q |.

How to find a minimal automaton?

Idea:

I Start with any DFA that accepts the language.

I Merge states from which exactly the same words

lead to an end state.

Minimal Automaton: Algorithm

Input: DFA M

Input:

(without states that are unreachable from the start state) Output: list of states that have to be merged

Output:

to obtain an equivalent minimal automaton

Create table of all pairs of states {q, q 0 } with q 6= q 0 .

Mark all pairs {q, q 0 } with q ∈ E and q 0 ∈ / E .

If there is an unmarked pair {q, q 0 } where {δ(q, a), δ(q 0 , a)}

for some a ∈ Σ is already marked, then also mark {q, q 0 }.

Repeat the last step until there are no more changes.

All states in pairs that are still unmarked can be merged

into one state.

Minimal Automaton: Example

q 0 q 1 q 2 q 3 q 4

q 3 q 2

q 1

× × × ×

×

×

×

×

States q 0 , q 2 and q 1 , q 3 can be merged into one state each.

Result:

Computation and Uniqueness of Minimal Automata

Theorem

The algorithm described on the previous slides produces a minimal automaton for the language accepted by the given input DFA.

Theorem

All minimal automata for a language L are unique up to isomorphism (i.e., renaming of states).

Without proof.

Overview

Automata &

Formal Languages

Languages

& Grammars

Regular Languages

Regular Grammars

DFAs

NFAs Regular Expressions

Pumping Lemma Minimal Automata Properties Context-free

Languages

Context-sensitive &

Type-0 Languages

A minimal automaton for a regular language L is a DFA M = hQ , Σ, δ, q ₀ , E i with L(M ) = L and a minimal number of states.

This means there is no DFA M ⁰ = hQ ⁰ , Σ, δ ⁰ , q ⁰ ₀ , E ⁰ i with L(M ) = L(M ⁰ ) and |Q ⁰ | < |Q |.

Create table of all pairs of states {q, q ⁰ } with q 6= q ⁰ .

Mark all pairs {q, q ⁰ } with q ∈ E and q ⁰ ∈ / E .

If there is an unmarked pair {q, q ⁰ } where {δ(q, a), δ(q ⁰ , a)}

for some a ∈ Σ is already marked, then also mark {q, q ⁰ }.

q ₀ q ₁ q ₂ q ₃ q 4

q ₃ q 2

q ₁

States q ₀ , q ₂ and q ₁ , q ₃ can be merged into one state each.

Let L and L ⁰ be regular languages over Σ and Σ ⁰ , respectively.

I union L ∪ L ⁰ = {w | w ∈ L or w ∈ L ⁰ } over Σ ∪ Σ ⁰

I intersection L ∩ L ⁰ = {w | w ∈ L and w ∈ L ⁰ } over Σ ∩ Σ ⁰

I complement L ¯ = {w ∈ Σ ^∗ | w ∈ / L} over Σ

I product LL ⁰ = {uv | u ∈ L and v ∈ L ⁰ } over Σ ∪ Σ ⁰

I star L ^∗ = S

k≥0 L ^k over Σ

I . . . under union if L, L ⁰ ∈ K implies L ∪ L ⁰ ∈ K

I . . . under intersection if L, L ⁰ ∈ K implies L ∩ L ⁰ ∈ K

I . . . under product if L, L ⁰ ∈ K implies LL ⁰ ∈ K

I . . . under star if L ∈ K implies L ^∗ ∈ K

(α|β), (αβ) and (α ^∗ ) describe the corresponding languages.

Then M ⁰ = hQ , Σ, δ, q ₀ , Q \ E i is a DFA with L(M ⁰ ) = ¯ L.

Intersection: Let M 1 = hQ ₁ , Σ 1 , δ 1 , q 01 , E 1 i and

M 2 = hQ ₂ , Σ 2 , δ 2 , q 02 , E 2 i be DFAs. The product automaton M = hQ ₁ × Q 2 , Σ 1 ∩ Σ 2 , δ, hq ₀₁ , q 02 i, E 1 × E 2 i

with δ(hq ₁ , q ₂ i, a) = hδ ₁ (q ₁ , a), δ ₂ (q ₂ , a)i accepts L(M ) = L(M ₁ ) ∩ L(M ₂ ).