Deterministic Finite Automata

Mohammad JAWAD

Theory Of Automata And
Complier Construction

Deterministic
Finite Automata
COMPSCI 102 Lecture 2

Let me show you a
machine so simple
that you can
understand it in less
than two minutes

11 1
0
0,1
1
0111 111 1
0 0

1

The machine accepts a string if the
process ends in a double circle
Steven Rudich:
www.cs.cmu.edu/~rudich

Anatomy of a Deterministic Finite
Automaton 1
0
0,1
states accept states (F)
1
q1

0 0

q0 q1
2

The machine accepts a string if the
start state (q0) ends in 3a double circle
process q states
Steven Rudich:

Anatomy of a Deterministic Finite
Automaton

q1 1
0
0,1
1
q0 q2

0 0
The alphabet of a finite automaton is the set
where the symbols come from: 1 {0,1}
q3
The language of a finite automaton is the set
of strings that it accepts
Steven Rudich:

q0 0,1

L(M) = ∅ strings of 0s and 1s
All

The Language of Machine M

Steven Rudich:

0 0

1
q0 q1
1

L(M) = { w | w has an even number of 1s}

Steven Rudich:

Notation
An alphabet Σ is a finite set (e.g., Σ = {0,1})

A string over Σ is a finite-length sequence of
elements of Σ

For x a string, |x| is the length of x

The unique string of length 0 will be denoted
by ε and will be called the empty or null string

A language over Σ is a set of strings over Σ
Steven Rudich:

A finite automaton is a 5-tuple M = (Q, Σ, δ, q0, F)
Q is the set of states
Σ is the alphabet
δ : Q × Σ → Q is the transition function
q0 ∈ Q is the start state
F ⊆ Q is the set of accept states

L(M) = the language of machine M
= set of all strings machine M accepts

Steven Rudich:

M = (Q, Σ, δ, q0, F) Q = {q0, q1, q2, q3}
where
Σ = {0,1}
δ : Q × Σ → Q transition function *
q0 ∈ Q is start state
F = {q1, q2} ⊆ Q accept states

0
q1
0,1
1
* δ
q0
0
q0
1
q1
1
q1 q2 q2
q0
M q2
0 0 q2 q3 q2
1 q3 q0 q2
q3
Steven Rudich:

Build an automaton that accepts all and only
those strings that contain 001

0,1
1 0

0 0 1
q q0 q00 q001
1

Steven Rudich:

A language is regular if it is
recognized by a deterministic
finite automaton

L = { w | w contains 001} is regular
L = { w | w has an even number of 1s} is regular

Steven Rudich:

Union Theorem
Given two languages, L1 and L2, define
the union of L1 and L2 as
L1 ∪ L2 = { w | w ∈ L1 or w ∈ L2 }

Theorem: The union of two regular
languages is also a regular language

Steven Rudich:


Proof Sketch: Let
1
M1 = (Q1, Σ, δ1, q0, F1) be finite automaton for L1
and
2
M2 = (Q2, Σ, δ2, q0, F2) be finite automaton for L2

We want to construct a finite automaton
M = (Q, Σ, δ, q0, F) that recognizes L = L1 ∪ L2

Steven Rudich:

Idea: Run both M1 and M2 at the same time!

Q = pairs of states, one from M1 and one from M2
= { (q1, q2) | q1 ∈ Q1 and q2 ∈ Q2 }
= Q 1 × Q2

Steven Rudich:


0
0

1
q0 q1
1 1
1

0
p0 p1
Steven Rudich:
0

Automaton for Union
1
q0,p0 q1,p0
1

0 0
0 0

1
q0,p1 q1,p1
1

Steven Rudich:

Automaton for Intersection
1
q0,p0 q1,p0
1

0 0
0 0

1
q0,p1 q1,p1
1

Steven Rudich:


Corollary: Any finite language is
regular

Steven Rudich:

The Regular Operations
Union: A ∪ B = { w | w ∈ A or w ∈ B }

Intersection: A ∩ B = { w | w ∈ A and w ∈ B }

Reverse: AR = { w1 …wk | wk …w1 ∈ A }

Negation: ¬A = { w | w ∉ A }

Concatenation: A ⋅ B = { vw | v ∈ A and w ∈ B }

Star: A* = { w1 …wk | k ≥ 0 and each wi ∈ A }

Steven Rudich:

Regular Languages Are
Closed Under The
Regular Operations
We have seen part of the proof for
Union. The proof for intersection is very
similar. The proof for negation is easy.

Steven Rudich:

The “Grep” Problem
Input: Text T of length t, string S of length n
Problem: Does string S appear inside text T?
Naïve method:

a1, a2, a3, a4, a5, …, at

Cost: Roughly nt comparisons

Automata Solution
Build a machine M that accepts any string
with S as a consecutive substring

Feed the text to M

Cost: t comparisons + time to build M

As luck would have it, the Knuth, Morris,
Pratt algorithm builds M quickly

Real-life Uses of DFAs
Grep

Coke Machines

Thermostats (fridge)

Elevators

Train Track Switches

Lexical Analyzers for Parsers

Consider the language L = { anbn | n > 0 }
i.e., a bunch of a’s followed by an
equal number of b’s
No finite automaton accepts this language
Can you prove this?

anbn is not regular.
No machine has
enough states to
keep track of the
number of a’s it
might encounter

That is a fairly weak
argument

Consider the following
example…

L = strings where the # of occurrences of
the pattern ab is equal to the number of
occurrences of the pattern ba

Can’t be regular. No machine has
enough states to keep track of the
number of occurrences of ab

b
a
b
a a

b a
a
b
b

M accepts only the strings with an
equal number of ab’s and ba’s!

Let me show you a
professional strength
proof that anbn is not
regular…

Pigeonhole principle:
Given n boxes and m > n
objects, at least one box
must contain more than
one object

Letterbox principle:
If the average number of
letters per box is x, then
some box will have at
least x letters (similarly,
some box has at most x)

Theorem: L= {anbn | n > 0 } is not regular
Proof (by contradiction):
Assume that L is regular
Then there exists a machine M with k states
that accepts L
For each 0 ≤ i ≤ k, let Si be the state M is in
after reading ai
∃i,j ≤ k such that Si = Sj, but i ≠ j
M will do the same thing on aibi and ajbi
But a valid M must reject ajbi and accept aibi

Deterministic Finite
Automata
• Definition
• Testing if they accept a string
• Building automata

Regular Languages
• Definition
• Closed Under Union,
Here’s What
You Need to Intersection, Negation
Know… • Using Pigeonhole Principle to
show language ain’t regular

Deterministic Finite Automata

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