Regular expressions
Regular expressions
Regular expressions
Regular expressions
Regular expressions
Regular expressions
Regular expressions
Regular expressions
Regular expressions
Regular expressions
Regular expressions
Regular expressions
Regular expressions
Describe the following languages by regular expressions
(a) L1=set of all strings of 0s and 1s ending in 00.
(b) L2=set of all strings of 0s and 1s beginning with 0 and ending
with 1.
(c) L3={ε , 11, 1111,111111,…}
Solution:
(a) (0+1)* 00.
(b) 0(0+1)*1
(c) (11)*
Regular expressions
Two regular expressions P and Q are equivalent if P and Q
represent same set of strings. We write P = Q.
The following identities( I ) are useful for simplifying regular
expressions:
7) RR* = R*R
1) Ø + R = R
8) (R*)* = R*
2) ØR = RØ = Ø
9) ε + RR* = R*
3) εR = Rε = R
10) (PQ)*P = P(QP)*
4) ε* = ε
11) (P+Q)* = (P*Q*)* = (P*+Q*)*
5) R + R = R
12) (P+Q)R = PR + QR
6) R*R* = R*
And R(P+Q) = RP + RQ
Regular expressions
a) Find a regular expression for representing the set L of strings
in which every 0 is immediately followed by at least two 1s.
b) Prove that the regular expression R = ε + 1*(011)*(1*(011)*)*
also describes the same set of strings.
Solution:
a) (1 + 011)*
b) R = ε + PP* , where P = 1*(011)*
= P* using I9
= (Q*S*)* where Q=1, S=011
= (Q + S)* using I11
= (1 + 011)*
Regular expressions
Prove (1+00*1) + (1 + 00*1)(0 +10*1)*(0 + 10*1) = 0*1(0 + 10*1)*
Solution:
L.H.S = (1 + 00*1)(εε + (0 + 10*1)*(0 + 10*1)) using I12
= (1 + 00*1)(0 + 10*1)* using I9
= (εε + 00*)1(0 + 10*1)* using I12
= 0*1(0+10*1)*
= R.H.S
using I9
Arden’s Theorem
Let P and Q be two regular expressions over Σ. If P does not
contain ε, then the following equation in R,
R = Q + RP
has a unique solution (one and only one) given by R = QP*
Finite Automata and Regular Expressions
Every regular expression R can be recognized by a transition
diagram. For every string there exists a path from the initial
state to a final state.
Finite Automata and Regular Expressions
A regular expression can be of the following forms:
1) R = P + Q
2) R = PQ
3) R = P*
Where P and Q are also regular
expressions.
Assume that P and Q can be
recognized by transition systems(FA)
G and H respectively.
Finite Automata and Regular Expressions
It is important to note that some authors use Λ (Lambda) symbol
instead of ε.
Regular Expression of a FA
Regular Expression of a FA
Regular Expression of a FA
Regular Expression of a FA
Find the regular expression recognized by the following transition
diagram.
Regular Expression of a FA
We get the following equations:
As q1 is the only final state and the equation for q1 involves only
q2 and q3, we use only q2 and q3 equations. Substituting for q2
and q3 in the equation for q1 , we get
Regular Expression of a FA
Describe in English the set of strings accepted by the following
finite automata
Regular Expression of a FA
As q1, q2 are final states
Answer: Strings containing any number of 0 s followed by any
number of 1 s.
Regular Expression of a FA
Construct the regular expression corresponding to the following
FA
Regular Expression of a FA
Regular Expression of a FA
Find the regular expression corresponding to the following FA
Regular Expression of a FA
q1 = q10 + q30 + q40 + Λ ,
q3 = q20, q4 = q31
q2= q11 + q21 + q41
FA of a Regular Expression
Construct an FA corresponding to the following RE.
(0+1)*(00+11)(0+1)*
First of all we need to construct a transition graph with Λ
moves. Then all the Λ moves need to be removed after the
construction of the graph.
FA of a Regular Expression
FA of a Regular Expression
After removing
the Λ moves
What is the DFA corresponding to the above diagram?
FA of a Regular Expression
FA of a Regular Expression
As the last two rows of the transition table are identical, the two
states can be combined into a single state as shown below.
FA of a Regular Expression
Construct a DFA with reduced states equivalent to the regular
expression 10 + (0 + 11)0*1