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Chomsky hierarchy Closure properties Pumping lemma NFA to DFA conversion Implementation

Finite state automata

Daniël de Kok

Daniël de Kok



Today

Chomsky hierarchyClosure propertiesProving that a language is not regularNFA to DFA conversionImplementation of finite state automata

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Chomsky hierarchy

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Formal grammar

A formal grammar G consists of:

A finite set of non-terminal symbols NA finite set of terminal symbols

∑A finite set of production rules of the form(N ∪ Σ)∗N(N ∪ Σ)∗ → (N ∪ Σ)∗

A start symbol S , S ∈ N

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Example

N = {S}Σ = {a, b}Production rules:S → a S bS → ε

Defines the language anbn, n ≥ 0

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Chomsky hierarchy

A, B : non-terminalsα, β, γ: arbitrary strings of terminals and non-terminalsx is a string of terminal symbols

Grammar type Production rulesUnrestricted α→ β, α 6= εContext Sensitive αAβ → αγβ, γ 6= εContext-Free A→ γRegular A→ xB or A→ x

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Conversion of a regular grammar

Create a state si for Ni ∈ N.Designate the state corresponding to S to be the start state.Create a state sf , F = {sf }For rules of the form Ni → Nj , make a transition between siand sj with an ε transition.For rules of the form Ni → ε, make an ε transition from si tosf .For rules of the form Ni → x , make a chain of transitions andstates, recognizing x , starting at si and leading to sf .For rules of the form Ni → xNj , make a chain of transitionsand states recognizing x , starting at si and leading to sj .

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Example

Convert the following grammar to an automaton:S → AS → BA → aaSB → bbSA → cB → ε

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Closure properties

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Definition

Non-deterministic finite state automata (NFAs) are closed under:

UnionIntersectionConcatenationNegationKleene closure

(This means that operator application to the language of theoperands leads to a language that can also be described by anNFA.)

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Kleene closure

The star (*) in regular expressions.L(A∗) = {ε,A,AA,AAA, ...}Procedure:

Create a new accepting state sk .Create an epsilon transition from sk to the start state.Create an epsilon transition from all accepting states (exceptsk) to sk .Make sk the new start state.

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Example Kleene closure

0start 1 2

a

a b

0 1 2

3start

a

a b

εε

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Pumping lemma

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Intuition

Consider: anbn

An automaton has k states (it is finite).The automaton revisits some state after processing k a’s.ai and aj lead to state q

Now, suppose that the rest of the input state contains b’s.Given ai , after receiving i b’s, it must accept the input, butnot j b’s.Clearly, this is not possible, because when the b’s started, ai

and aj were in the same state.

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Pumping lemma

Citing Hopcroft, et al. 2001: Let L be a regular language. Thenthere exists a constant n (which depends on L), such for that everyw ∈ L such that |w | ≥ n, we can break w into three strings,w = xyz , such that:

1 y 6= ε

2 |xy | ≤ n

3 For all k ≥ 0, the string xykz is also in L

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NFA to DFA conversion

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Introduction

Any NFA can be converted in a DFA accepting the samelanguage.Important: it proves that an NFA is not more powerful than aDFA.The conversion is done using the so-called subset construction.

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Subset construction

Given the NFA < QN ,Σ, δN , q0,FN >, we construct the DFA< QD ,Σ, δD , {q0},FD >:

QD is the power set of QN (the set of subsets).FD contains the subsets of QD , for which hold:∀S ∈ FD : S ∩ FN 6= ε

For each subset S ∈ QD and input symbol in a in Σ:δD(S , a) =

⋃s∈S δN(s, a)

Note: we do not consider ε transitions yet.

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Example (NFA)

Automaton

0start 1 2a b

a

State table

State a b0 0,11 22*

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Power set state table

State table

State a b0 0,11 22*


State a b{0} {0,1}{1} {2}{2}*{0,1} {0,1} {2}{1,2}* {2}{0,1,2}* {0,1} {2}

Note: only states {0}, {0,1}, and {2}are reachable.

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State table

State a b0 0,11 22*


State a b

{0} {0,1}{1} {2}{2}*{0,1} {0,1} {2}{1,2}* {2}{0,1,2}* {0,1} {2}


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State table

State a b0 0,11 22*


State a b{0} {0,1}

{1} {2}{2}*{0,1} {0,1} {2}{1,2}* {2}{0,1,2}* {0,1} {2}


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State table

State a b0 0,11 22*


State a b{0} {0,1}{1} {2}

{2}*{0,1} {0,1} {2}{1,2}* {2}{0,1,2}* {0,1} {2}


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State table

State a b0 0,11 22*


State a b{0} {0,1}{1} {2}{2}*

{0,1} {0,1} {2}{1,2}* {2}{0,1,2}* {0,1} {2}


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State table

State a b0 0,11 22*


State a b{0} {0,1}{1} {2}{2}*{0,1} {0,1} {2}

{1,2}* {2}{0,1,2}* {0,1} {2}


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State table

State a b0 0,11 22*


State a b{0} {0,1}{1} {2}{2}*{0,1} {0,1} {2}{1,2}* {2}

{0,1,2}* {0,1} {2}


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State table

State a b0 0,11 22*


State a b{0} {0,1}{1} {2}{2}*{0,1} {0,1} {2}{1,2}* {2}{0,1,2}* {0,1} {2}


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Example (DFA)


State a b{0} {0,1}{1} {2}{2}*{0,1} {0,1} {2}{1,2}* {2}{0,1,2}* {0,1} {2}

DFA

{0}start {0, 1} {2}a b

a

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State creation

The number of states becomes large quickly: |QD | = 2|QN |

Usually, most states are not reachable.Solution: create states as they are needed.

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Example 2 (NFA)

Automaton

0start

1

2

3

a

a

b

c

dd

State table

State a b c d0* 1,21 32 33 0,3

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Example 2

State table

State a b c d0* 1,21 32 33 0,3


State a b c d

{0}* {1,2}{1,2} {3} {3}{3} {0,3}

{0,3}* {1,2} {0,3}

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Example 2

State table

State a b c d0* 1,21 32 33 0,3


State a b c d{0}* {1,2}

{1,2} {3} {3}{3} {0,3}

{0,3}* {1,2} {0,3}

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Example 2

State table

State a b c d0* 1,21 32 33 0,3


State a b c d{0}* {1,2}{1,2} {3} {3}

{3} {0,3}{0,3}* {1,2} {0,3}

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Example 2

State table

State a b c d0* 1,21 32 33 0,3


State a b c d{0}* {1,2}{1,2} {3} {3}{3} {0,3}

{0,3}* {1,2} {0,3}

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Example 2 (DFA)

Power set state tableState a b c d{0}* {1,2}{1,2} {3} {3}{3} {0,3}

{0,3}* {1,2} {0,3}

Automaton

{0}start {1, 2}

{3}

{0, 3}a

b,c d

a d

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Implementation

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Introduction

Broadly two ways to represent the automaton:

Table-basedAs a graph of Java objects

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Java object graph

Java has in its memory directed object graphs, e.g.:

We have a Map<String, String> reference on the stack.The stack reference points to an instance of a classimplementing Map on the heap.The map has Entry instances, where each Entry contains areference to a key object and a value object.etc.

Java objects graphs can be cyclic, e.g.: a value in a map couldrefer to that map.

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Automata in the object graph

Automata map nicely to Java object graphs.States are objects.Transitions contain references to other objects.If a state becomes unreachable, Java’s garbage collection willclaim it.

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Tricky issues

How do we represent epsilons?When are two states equal?

When they are the same object?When they have the same outgoing transitions/finalness?

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State

A state:

Has outgoing transitions (possibly multiple per character forNFAs).Is accepting or not accepting.

public class State {private final SetMultimap<Character, State> transitions;private boolean accept;

// ...}

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Constructors

public State() {this(false);

}

public State(boolean accept) {this.accept = accept;transitions = HashMultimap.create();

}

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Adding outgoing transitions

public void addTransition(char ch, State toState) {transitions.put(ch, toState);

}

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Epsilon transitions

How do we implement epsilons?

Reserve one character to represent epsilons: bad, maybesomeone wants to use that character.Do some class/interface wrapping around char, so that bothepsilons and characters can be represented.Do not explicitly represent epsilons:

Observation: an epsilon transition can be followed withoutmoving the tape.Solution: follow the epsilon transition when the transition isbeing added.Warning: this only works if the state pointed to is ‘frozen’.

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Epsilon transitions (2)

public void addEpsilonTransition(State toState) {Preconditions.checkNotNull(toState);

if (toState.isAccept()) {setAccept(true);

}

for (Map.Entry<Character, State> toTransition :toState.transitions.entries()) {

transitions.put(toTransition.getKey(),toTransition.getValue());

}}

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Automaton

The automaton should provide:

A reference to the start state.Some useful methods.

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Automaton (2)

public class FSA {private State startState;

public FSA(State startState) {Preconditions.checkNotNull(startState);this.startState = startState;

}

// ...}

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Automaton: getting all states

public Set<State> getStates() {Set<State> states = new HashSet<State>();

Queue<State> stateQueue = new LinkedList<State>();stateQueue.add(startState);while (!stateQueue.isEmpty()) {

State state = stateQueue.poll();states.add(state);

for (State toState :state.getTransitions().values()) {

if (!states.contains(toState)) {stateQueue.add(toState);

}}

}return states;

}

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Automaton: get accept states

public Set<State> getAcceptStates() {Set<State> acceptStates = new HashSet<State>();

for (State state : getStates()) {if (state.isAccept()) {

acceptStates.add(state);}

}

return acceptStates;}

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finite state automata - sfs.uni-tuebingen.deddekok/dsa3/lectures/fsa.pdffinite state automata....

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