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CSC 4170
Theory of Computation
Regular Expressions
Section 1.3
(also 1.1, 1.2)
1.3.a
Regular operations
L1  L2 = {x | xL1 or xL2}
Union:
{Good,Bad}  {Boy,Girl} =
{0,00,000,…} {1,11,111,…} =
L  =
Concatenation:
L1  L2 = {xy | xL1 and yL2}
{Good,Bad}{Boy,Girl} =
{0,00,000,…}{1,11,111,…} =
L =
Star:
L* = {x1…xk | k0 and each xiL}
{Boy,Girl}* =
{0,00,000,…}* =
 *=
1.3.b
Regular expressions
We say that R is a regular expression
(RE) iff R is one of the following:
What language is represented
by the expression:
1. a, where a is a symbol of the alphabet
{a}
2. 
{}
3. 

4. (R1)(R2), where R1 and R2 are RE
The union of the languages represented
by R1 and R2
The concatenation of the languages
represented by R1 and R2
The star of the language represented
by R1
5. (R1)  (R2), where R1 and R2 are RE
6. (R1)*, where R1 is a RE
Conventions:
 The symbol  is often omitted in RE
 Some parentheses can be omitted.
The precedence order for the operators is:
* (highest),  (medium),  (lowest)
1.3.c
Regular languages
A language is said to be regular iff it can be represented by a regular expression.
Language
{11}
{Boy, Girl, Good, Bad}
{,0,00,000,0000,…}
{0,00,000,0000,…}
{,01,0101,010101,01010101,…}
{x | x = 0k where k is a multiple of 2 or 3}
{x | x is divisible by 8}
{x | x MOD 4 = 3}
Expression
1.3.d
Exercising reading regular expressions
Expression
Language
(Good  Bad)(Boy  Girl)
(Tom  Bob)_is_(good  bad)
{Name_is_adjective | Name is an uppercase
letter followed by zero or more lowercase
letters, and adjective is a lowercase letter
followed by zero or more lowercase letters}
0*10*
(0 1)*101(0 1)*
((0 1)(0 1))*
1.3.e
Regular languages and DFA-recognizable languages are the same
Theorem 1.54* A language is regular if and only if some NFA (DFA)
recognizes it.
In other words,
a)
[The “only if” part]
For every regular expression there is an NFA that recognizes exactly
the language represented by that expression.
b)
[The “if” part]
For every NFA there is a regular expression that represents exactly
the language recognized by that NFA.
1.3.f
Constructing an NFA from a regular expression: Base cases
Case of a, where a is a symbol of the alphabet.
Case of 
Case of 
1.3.g
Constructing an NFA from a regular expression: Case of union
Case of (R1)(R2), where R1 and R2 are RE
First, construct NFAs N1
and N2 from R1 and R2:
N1
Then, combine them in the
following way:
N1
s1
s1
s2
N2
s2
N2
1.3.h
Constructing an NFA from a regular expression: Case of concatenation
Case of (R1)  (R2), where R1 and R2 are RE
First, construct NFAs N1 and N2 from R1 and R2:
N1
N2
s1
s2
Then, combine them in the following way:
N1
N2
s1
s2
1.3.i
Constructing an NFA from a regular expression: Case of star
Case of (R1)*, where R1 is a RE
First, construct an NFA N1 from R1:
N1
s1
Then, extend it in the following way:
N1
s1
1.3.j
Constructing an NFA from a regular expression: An example
#(0 1)*
(0 1)*

0 1
#
0

#

0


1
1

# 1 0 1
1.3.k
GNFA
great
mother  father
grand


grand

(great)*
greatgreatgreatgrandfather
1.3.l
About -transitions
great
mother  father
grand


grand
(great)*
Adding or removing -transitions does not change
the recognized language
1.3.m
The same GNFA simplified
great
grand


mother  father
1.3.n
Ripping a state out
(great)* grand

mother  father
1.3.o
Eliminating parallel transitions
  (great)*grand
mother  father
1.3.p
Again ripping out
(  (great)*grand) (mother  father)
1.3.q1
How, exactly, to do ripping out
L
R
T
S
Assume, we are ripping out the state r from a GNFA that has no parallel transitions.
Let L be the label of the loop from r to r (if there is no loop, then L=).
1.3.q2
How, exactly, to do ripping out
RL*T
L
R
T
S
SL*T
Assume, we are ripping out the state r from a GNFA that has no parallel transitions.
Let L be the label of the loop from r to r (if there is no loop, then L=).
1. For every pair s1,s2 of states such that there is an E1-labeled transition from
s1 to r and an E2-labeled transition from r to s2, add an R1L*R2-labeled transition
from s1 to s2;
1.3.q3
How, exactly, to do ripping out
RL*T
SL*T
Assume, we are ripping out the state r from a GNFA that has no parallel transitions.
Let L be the label of the loop from r to r (if there is no loop, then L=).
1. For every pair s1,s2 of states such that there is an E1-labeled transition from
s1 to r and an E2-labeled transition from r to s2, add an R1L*R2-labeled transition
from s1 to s2;
2.
Delete r together with all its incoming and outgoing transitions.
1.3.r
How, exactly, to eliminate parallel transitions
Whenever you see parallel transitions labeled with R1 and R2,
R1
R2
Replace them by a transition labeled with R1R2.
R1R2
Repeat until there are no parallel transitions remaining.
1.3.s
From NFA to RE
b
a
b
b
1.3.t
From NFA to RE: Step 1
Step 1: If there are incoming arrows to the start state,
or the start state is an accept state, then
add a new start state and connect it with an -arrow to the old
start state.
b
a
a

b
b
1.3.u
From NFA to RE: Step 2
Step 2: If there are more than one, or no, accept states,
or there is an accept state that has outgoing arrows,
then add a new accept state,
make all the old accept states non-accept states and connect
each of them with an -arrow to the new accept state.
b

a
a

b
b

1.3.v
From NFA to RE: Step 3
Step 3: Eliminate all parallel transitions.
b

a
a

b
b

1.3.w1
From NFA to RE: Step 4
Step 4: While there are internal states (states that are neither the start
nor the accept state), do the following:
Step 4.1: Select an internal state and rip it out;
Step 4.2: Eliminate all parallel transitions.
b
aa
a

b
ab

b
1.3.w2
From NFA to RE: Step 4
Step 4: While there are internal states (states that are neither the start
nor the accept state), do the following:
Step 4.1: Select an internal state and rip it out;
Step 4.2: Eliminate all parallel transitions.
baa
a

b
ab

b
1.3.w3
From NFA to RE: Step 4
Step 4: While there are internal states (states that are neither the start
nor the accept state), do the following:
Step 4.1: Select an internal state and rip it out;
Step 4.2: Eliminate all parallel transitions.
a(baa)*
a(baa)*ab
b

b(baa)*
b(baa)*ab
1.3.w4
From NFA to RE: Step 4
Step 4: While there are internal states (states that are neither the start
nor the accept state), do the following:
Step 4.1: Select an internal state and rip it out;
Step 4.2: Eliminate all parallel transitions.
a(baa)*
ba(baa)*ab
b(baa)*ab
  b(baa)*
1.3.w5
From NFA to RE: Step 4
Step 4: While there are internal states (states that are neither the start
nor the accept state), do the following:
Step 4.1: Select an internal state and rip it out;
Step 4.2: Eliminate all parallel transitions.
(b a(baa)*ab) (b(baa)*ab)* (  b(baa)*)
a(baa)*
1.3.w6
From NFA to RE: Step 4
Step 4: While there are internal states (states that are neither the start
nor the accept state), do the following:
Step 4.1: Select an internal state and rip it out;
Step 4.2: Eliminate all parallel transitions.
((b a(baa)*ab) (b(baa)*ab)* (  b(baa)*)(a(baa)*)
1.3.x
From NFA to RE: Step 5
Step 5: Return the label of the only remaining arrow
(if there is no arrow, return ).
((b a(baa)*ab) (b(baa)*ab)* (  b(baa)*)(a(baa)*)
((b a(baa)*ab) (b(baa)*ab)* (  b(baa)*)(a(baa)*)
Claim: The resulting RE represents exactly the language
recognized by the original NFA.
This completes the proof of Theorem 1.54.