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Basics of automata theory

Nondeterministic Finite Automata (NFA)

Nondeterministic Finite Automata on infinite words (NFW)

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Nondeterministic finite automaton (NFA)

Transitions: (S0,A,S0), (S0,B,S0),(S0,A,S1),(S1,A,S1).

What is the language of this automaton?

All words that have a path to an accepting state.

A,B A AS0 S1

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Equivalent deterministic automaton

Every NFAcan be transformed to DFA. How ? The price may be exponential.

A,B A AS0 S1

B

A

AS0 S0,S1

B

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Determinization

Let M = (S, , , I, F) be an NFA.

Define a DFA Md = (Sd, d, d, Id, Fd), where

Sd= P(S)// power-set of S

d=

Id = I d(q, a) = { (r,a) | r 2 q} for all q2 Sd, a 2

Fd = {q| q 2 Sd q F ;}

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Example

Sd = {}{S0}{S1}{S0,S1}

d

= = {A,B}

Id = I = {S0}

d= ...

FD = {S1}{S0,S1}

A,B A AS0 S1

S0 S1

S0,S1

B B

A

A

AB

A,B

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Example 2

B A,B0 2

1 AA

0 1 2

01 02 12

012

Complete it yourself

A B

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Example 2

B A,B0 2

1 AA

0 2

12

012

A B B

A

A

B

1

01 02

A

B

A

BA B

A

B

A,B

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Few important questions (1)

Given two automataA1,A

2...

How do we build an automatonA3such that

L(A3) = L(A1) L(A2)

A method to buildA3: compute the product

A1 A2

We already saw how to compute a product...

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Product of two NFA-s (finite words)

A1=h, S1, 1, I1, F1i and A2= h, S2, 2, I2, F2i

A1

A2

=

Each state is a pair (s,t): s 2 S1 and t 2 S2.

Initial states: pairs (s,t) such that s 2 I1 and t 2 I2.

Accepting states: pairs (s,t) such that s 2 F1 and t 2 F2

((s,t) a (s,t)) is a transition if(s,a,s) 21, and (t,a,t) 22.

Reminder

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Example product of two automata

L(A1) = (a+b)

*

a +

(words ending with a+ empty word)

What should be the language ofA1 A2 ?

L(A2) = (ba)* + (ba)*b

A1:b

a

ab

s0 s1

A2:

a

bt0 t1

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1. States: (s0,t0), (s0,t1), (s1,t0), (s1,t1).

2. Initial state: (s0,t0).

3. Accepting states: (s0,t0), (s0,t1).A1 A2:

A1:

A2:

b

a

ab

s0 s1

a

bt0 t1


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s0,t0

s0,t1

s1,t1

s1,t0b

b

a

aA1 A2:

L(A1 A2) = (ba)*

A1:

A2:

b

a

ab

s0 s1

a

bt0 t1


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Example 2 product of two automata

A1:

A2:

L(A1) = (ab)* + (ab)*a(words that alternate

between a and b )

What should be the language ofA1 A2 ?

L(A2) = (ba)* + (ba)*b

(words that alternatebetween b and a)

a

bs0 s1

a

bt0 t1

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1. States: (s0,t0), (s0,t1), (s1,t0), (s1,t1).

2. Initial state: (s1,t0).

3. Accepting states: (s0,t0), (s0,t1), (s1,t0), (s1,t1).A1 A2:

A1:

A2:

ab

s0 s1

a

bt0 t1


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s0,t0

s0,t1

s1,t1

s1,t0b aA1 A2:

L(A

1

A2) =

A1:

A2:

ab

s0 s1

a

bt0 t1


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Given a DFA A:

How do we construct A such thatL(A) = * - L(A)

In other words, how do we build an automaton thataccepts exactly those words that are rejected byA?

Answer: compute the complement automaton.

How ? Let F = S F.(i.e., substitute accepting and non-accepting states.)

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Example: complementation

The complementation of A is denoted by

b

a

a

bs0 s1

b

a

a

bs0 s1

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Given an automaton A:

Universality: is L(A) = * ?

Emptiness: is L(A) = ;?

Emptiness: is an accepting state reachable?

Universality: check whether

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And now....

... Automata on infinite words These are called !-automata

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Automata over infinite words (DFW / NFW)

Similar definition.

Runs on infinite words over .

Accepts when an accepting state occurs

infinitely often in the computation. This is called a Buchi automaton

a

a

bb

S0 S1

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Formally, let F be the set of accepting states.

Let inf() S be the set of states appearing infinitenumber of times in a computation .

is accepted by the automaton ifinf() F ;.

a

a

bb

S0 S1

Automata over infinite words (DFW)

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Consider the word a b a b a b a b The computation is S0 S0 S1 S0 S1 S0 S1

This computation is accepting, since S0 appears

infinitely many times.

a

a

bb

S0 S1

Automata over infinite words (DFW)

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Other computations

For the word b b b b b the computation isS0 S1 S1 S1 S1 and is not accepting.

For the word a a a b b b b b, thecomputation is S0 S0 S0 S0 S1 S1 S1 S1and ... (?)

What is the computation for a b a b b a b b b? Is itaccepting ?

a

a

bb

S0 S1

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The language of a Buchi automaton

The language of a Buchi automaton is the set ofinfinite strings that are accepted by it.

Such languages are called ! languages.

Buchi automaton defines ! regular languages.

L(B) = (ab*)!

a

a

b

bS0 S1B =

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As before, a string is accepted if there exists anaccepting run.

L(B) = aa*b!

Surprise: there is no determinization procedure

We will focus on DFW

NonDeterministic Buchi automata (NFW)

a

a bS0 S1B =

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Why do we need Buchi automata ?

The compilation theorem: every LTL formula can be translated to a Buchi automaton B thataccepts the same language as .

-a

a

a

-a

a

-a

a

Fa

Ga

GFa

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What about the other direction ?

Can every Buchi automata be translated to an LTLformula ?

No LTL formula can express this property.

a

a holds on every even step

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About the alphabet...

So far a 2 meant that a is some atom (a,b...)from a given setAP.

We will also use a 22AP

This is useful for representing states

(contd on next slide)

b

a

a

b

a,b

a,:b

:a,:b

:a,b

a,b

a

b

(same thing, only write positive literals)

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... or even a 2 22AP

This can give us a more compact representaiton

a b

a b

:a :b

:ab a:b

:a b a:b

a,:b

:a, b

:a, :b

a,b:a, b

:a, b

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A Buchi automaton can also be represented withlabels on states.

There is 1-1 translation to a Buchi automaton withlabels on transitions:

Move labels to outgoing edges.

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From labels on states to labels on transitions:

Example.

Let 2AP

p,q

p

p

p,q

p

Recall that this

is {p,q}

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For Buchi automata B1, B2:

How to compute L(B1) L(B2) ?

How to complement ? Find Bs.t.

How to check for emptiness ? Is L(B) = ; ?

Again, important questions...

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Intersecting two Buchi automata (infinite words)

Previous method doesnt work:

ab

s0 s1ab Infinite as

b

a

t0 t1ba

Infinite bs

s0, t0

s0, t1 s1, t0

a b

s1, t1

a

b

a

b

Empty language !

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The reason: a path should be accepted if it fulfills

two separate acceptance conditions: passes infinitely many times through s0

passes infinitely many times through t0

An automaton has such multiple acceptance

conditions is called a generalized Buchi automata. We will learn about this later on.

For now, we will see a reduction of this condition to astandard Buchi automaton.

s0, t0

s0, t1 s1, t0

a b

s1, t1

a

b

a

b

Empty language !

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Strategy:

Multiply the product automaton by 3(S = S1 S2 {0,1,2} )

Start from the 0 copy.

Transition to the 1 copy when entering a state fromF1

Transition to the 2 copy if in a 1 state and enteringa state from F2, and in the next state back to a 0

state.

Make the 2 copy an accepting set.

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s0, t0

s0, t1 s1, t0

a b

s1, t1

a

b

a b

s0, t0

s0, t1 s1, t0

a b

s1, t1

a

b

a b

s0, t0

s0, t1 s1, t0

a b

s1, t1

a

b

a b

0

1

2

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s0, t0

s0, t1 s1, t0

a ba

b

a b

s0, t0

s0, t1 s1, t0

a b

a

b

a b

s0, t0

s0, t1 s1, t0

a b

a

b

a b

a

a

simplify by removing unreachable states

bb

0

1

2

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s0, t0

s0, t1 s1, t0

b

bb

s0, t1 s1, t0

aa

s0, t1 s1, t0

a

b

a b

a

a

simplify by removing unreachable states

bba

a b

b

a

a0

1

2

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a

bs0 s1a

b

b

at0 t1b

a

a

a

b

b

aa

a

a b

h s1,t0,2 i

h s0,t1,1 i

b b

h s0,t0,0 i

h s1,t0,0 i

h s0,t1,0 i

There are total of 12 states in theproduct automaton.

The reachable part ofA1 A2 is:

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c4.1 Nfa+Buchi