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Chapter 11 - 2

- Finite Automata

Section 11.2 Finite Automata

- Can a machine (i.e., algorithm) recognize a

regular language Yes! - Deterministic Finite Automata
- A deterministic finite automaton (DFA) over an

alphabet A is a finite digraph (where vertices or

nodes are called states) for which each state

emits one labeled edge for each letter of A. One

state is designated as the start state and a set

of states may be final states.

Example

- Either of the following alternatives is

acceptable for representing a DFA, where final

states are indicated by double circles. - The Execution of DFA for input string w A

begins at the start state and follows a path

whose edges concatenate to w. The DFA accepts w

if the path ends in a final state. - Otherwise the DFA rejects w. The language of a

DFA is the set of accepted strings.

Example

- The example DFA accepts the strings
- a, b, ab, bb, abb, bbb, , abn, bbn,
- So the language of the DFA is given by the

regular expression (a b)b.

Theorem (Kleene)

- The class of regular languages is exactly the

same as the class of languages accepted by DFAs. - Quiz. Find an DFA for each of the following

languages over the alphabet a, b. - (a) (b) . (c) (ab)n n N, which has

regular expression (ab).

Quiz

- Find a DFA for the language of a aab.

Table Representation of a DFA

- A DFA over A can be represented by a transition

function T States A States, where T(i, a)

is the state reached from state i along the edge

labeled a, and we mark the start and final

states. For example, the following figures show a

DFA and its transition table. - Note T can be extended to T States A

States by - T(i, ) i and T(i, aw) T(T(i, a), w) for a

A and w A.

Quiz

- Quiz Calculate T(0, bba).
- Solution T(0, bba) T(1, ba) T(1, a) T(2,

) 2. - Example/Quiz. Back to the problem of describing

input strings over a, b that contain exactly

one substring bb. We observed that the strings

could be described by the regular expression - (a ba)bb(a ab).
- Find a DFA to recognize the language.
- A solution

Nondeterministic Finite Automata

- A nondeterministic finite automaton (NFA) over an

alphabet A is similar to a DFA except that

-edges are allowed, there is no requirement to

emit edges from a state, and multiple edges with

the same letter can be emitted from a state.

Example

- The following NFA recognizes the language of a

aab ab.

Table representation of NFA

- An NFA over A can be represented by a function T

States A power(States), where T(i, a)

is the set of states reached from state i along

the edge labeled a, and we mark the start and

final states. The following figure shows the

table for the preceding NFA.

Theorem (Rabin and Scott)

- The class of regular languages is exactly the

same as the class of languages accepted by NFAs. - Quizzes. Find an NFA for each of the following

languages over a, b. - (a) (b) . (c) (ab)n n N, which has

regular expression (ab).

Example/Quiz

- Back to the problem of describing input strings

over a, b that contain exactly one substring

bb. We observed that the strings could be

described by the regular expression (a ba)bb(a

ab). Find an NFA to recognize the language.

Algorithm

- Transform a Regular Expression into a Finite

Automaton - Start by placing the regular expression on the

edge between a start and final state - Apply the following rules to obtain a finite

automaton after erasing any -edges.

Quiz

- Use the algorithm to construct a finite automaton

for (ab) ba. - Answer

Algorithm

- Algorithm Transform a Finite Automaton into a

Regular Expression - Connect a new start state s to the start state of

the FA and connect each final state of the FA to

a new final state f as shown in the figure. - If needed, combine all multiple edges between the

same two nodes into one edge with label the sum

of the labels on the multiple edges. If there is

no edge between two states, assume there is an

-edge. - Now eliminate each state k of the FA by

constructing a new edge (i, j) for each pair of

edges (i, k) and (k, j) where i k and j k.

The new label new(i, j) is defined in terms of

the old labels by the formula - new(i, j) old(i, j) old(i, k)old(k,

k)old(k, j)

Example

Quiz

- Use the algorithm to transform the following NFA

into a regular expression. (Half the class

eliminate state 0 then state 1 and half the class

eliminate state 1 then state 0.)

Solutions

- Solution Connect the NFA to new a start state s

and a new final state f as pictured. - First Solution Eliminate state 0 to obtain
- new(s, 1) aa aa.
- new(1, 1) baa baa.
- Eliminate state 1 to obtain
- new(s, f) aa(baa) aa(baa).
- Second Solution Eliminate state 1 to obtain
- new(0, f) a a.
- new(0, 0) a ab a ab.
- Eliminate state 0 to obtain
- new(s, f) (a ab)a (a ab)a.

Quiz

- Use regular algebra to show the following

equality from the previous quiz. - aa(baa) (a ab)a.

Automata with Output (Mealy and Moore)

- Mealy machine Associate an output action with

each transition between states. - Moore machine Associate an output action with

each state.

Theorem

- Theorem Mealy and Moore machines are equivalent.

- Challenge Describe a two-floor elevator system

with a Mealy or Moore machine.

The End of Chapter 11 - 2

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