FInite Automata

Introduction to Formal Languages
Finite Automata
Mobeen Mustafa1
Chapter 5: Finite Automata
We introduce the simplest deterministic theoretical
machines: Finite Automata.

Mobeen Mustafa2
 A finite automaton (FA) is the following 3 things:
1. a finite set of states, one of which is designated as the
start state, and some (maybe none) of which are
designated the final states (or accepting states)
2. an alphabet Σ of input letters
3. a finite set of transitions that indicate, for each state
and letter of the input alphabet, the state to go to next
state state
letter

Mobeen Mustafa3
The language defined or accepted by a finite
automaton is the set of words that end in a final
state.
If w is in the language defined by a finite
automaton, then we also say that the finite
automaton accepts w.

Mobeen Mustafa4
Example: ∑= {a,b}
states = {x,y,z}
start state : x
final states: {z}
a b
x y z
y x z
z z z
aaa: a
x ya ax y y is not final;
aaa is not accepted
aaba: a
x ya bx z z is final;
aaba is accepted
za
transitions:

5
x
–
y
z
+
a
a
a
b b
b
a b
x y z
y x z
z z z
Regular expression: (a+b)*b(a+b)*
aaaabba?
bbaabbbb?
Transition Diagram

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x
–
y
z
+
a
a
a
b b
b
x
–
y
z
+
a
a
a
b
b
a
a
b
x
–
y
z
+a
b
a
b
b
bb
aaaabba?
bbaabbbb?

Mobeen Mustafa7
+ a,b
(a+b)(a+b)* = (a+b)+
– +
a
b
a,b
(a+b)*

8
No final state
The middle state is not a final state and all transitions that go
into this state do not exit.
–
a
b
a,b
–
a,b
a,b
+a,b
Finite Automata that Accept No Words

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Two Ways to Study Finite Automata
1. Starting with a finite automaton (FA), analyze it to determine
the language it accepts.
2. Starting from a language, build an FA.

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Example of 2: The language of all words with an even
number of letters over the alphabet {a,b}:
● Two states: 1 – even number, 2 – odd number
● Start state: 1
● Final state: 1
● The transitions:
a b
1 2 2
2 1 1
1+ 2
a,b
a,b

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x
–
y
z
+
a,b
a,b
b
a
a(a+b)*
Not necessarily unique:
–
+
a,b
a,b
b
a
+
a,b
Remark: 2 final states

Mobeen Mustafa12
From Languages to Finite Automata
 There is not necessarily a unique FA that accepts a given
language.
 Is there always at least one FA:
 that accepts each possible language?
 that defines a language associated with a given regular
expression?

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Example: Build an FA that accepts all words containing a
triple letter (either aaa or bbb).
1.Build an FA that accepts aaa
2.Add a path that accepts bbb.
3.Add paths for words that contain a’s and b’s before or
after the aaa or bbb.

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 Does this FA accept the word ababa?
 The word babbb?
 2 ways to get to state 4
 The only way to get to state 2 is by reading an input a.
 The only way to get to state 3 is by reading an input b.
 What language?
1–
2
4
+
a,ba
b 3
a
a
b
b
This example shows that it is
possible to characterize
states by the purposes they
serve.
From Finite Automata to Languages

Mobeen Mustafa15
The third letter is b.
(aab + abb + bab + bbb)(a+b)*
(a+b)(a+b)b(a+b)*
The regular expression is not unique.
Is there always at least one regular expression defining the
language accepted by an FA?
+
a,b
a,b
a
b
– a,ba,b

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Regular expression: baa
A “collecting bucket” state for all other words.
–
a,b
b +a
a
a
bb a,b

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Regular expression: baa + ab
–
a,b
b +aa
bb a,b
+ b a
a,b a

Mobeen Mustafa
18
+
a,b
a,b
+
b
a
ba
– +
a
a
bb
– +
a
b
ab
Λ (a+b)*a + Λ
Words that do not end in b.
(a+b)*a
Words that end
in a.
?

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The only letter that can change state is a.
(b*ab*)(ab*ab*)*
an odd number of a’s
– +
a
a
bb

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Words that contain a double a
(a+b)*aa(a+b)*
Start state: the previous letter (if there is one) was not an a.
Middle state: we have just seen an a that was not preceded
by an a.
Final state: we have seen a double a.
– +
a
b
a,bb
a

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+
b
a
ba
+
a
b
a
b
–
b
a
+
ba
–
a
a
a
b
b
b
aabbaabb

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1
+
2
b
b
a
3 4
b
b
a a a
The Language EVEN-EVEN

FInite Automata

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