Theory of Computation Regular Expressions, Minimisation & Pumping Lemma

Theory of Computation
By Rushabh Wadkar

Topics to be covered
Regular expressions & Regular
Languages relationship
Reduction of states
Pumping Lemma
Day 5

Regular languages
● Any language that can be depicted(expressed) using a Finite State
machine is called a Regular language.
● A FSM can’t store any input variable, nor can it count.
● Hence any language that requires memory is not a regular language

Regular languages
Let us revise some examples:
● L= { 0n
1m
: m>=0,n>=1}

Regular languages
● L= { 0101n
U 0100 : n>=0}

Regular languages
● L= {an
U bn
U cn
: n>=0}

Regular languages and Regular Expressions
● A regular language can be described using regular expressions
consisting of the symbols such as alphabets in Σ.
● Additionally consisting of operators like ‘.’ | ‘+’ | ‘*’
● The symbols ‘(’ and ‘)’ can be used with regular expressions.

● + operator(union), has the least precedence.
● . operator(concatenation) has mid precedence.
● * operator(closure) has highest precedence.
However the expression enveloped by parentheses
obtains the highest precedence.
Let us try to understand the precedence of these operators:

A regular expression is recursively deﬁned as follows:
1. Φ is a regular expression denoting an empty language.
2. ε-(epsilon) is a regular expression indicates the language containing an empty string.
3. a is a regular expression which indicates the language containing only {a}
4. If R is a regular expression denoting the language LR and S is a regular expression denoting
the language Ls, then
a. R+S is a regular expression corresponding to the language LR
U LS
.
b. R.S is a regular expression corresponding to the languageLR
. LS
.
c. R* is a regular expression corresponding to the language LR
.
5. The expressions obtained by applying any of the rules from 1 to 4 are regular expressions.

a* String consisting of any number of a’s(0 or more)
a
+ String consisting of at least of a’s
a+b String consisting of either one a or one b
(a+b)* Set of strings of a’s and b’s of any length(NULL included)
(a+b)*abb Set of strings of a’s and b’s, ending with abb
ab(a+b)* Set of strings of a’s and b’s, starting with ab

(a+b)*aa(a+b)* Set of strings of a’s and b’s, having substring aa
a*b*c* String consisting of any number of a’s(0 or more) followed by any
number of b’s(0 or more) followed by any number of c’s(0 or more)
a
+
b
+
c
+ String consisting of at least 1 a, followed by string having at least 1
b, followed by string having at least 1 c
aa*bb*cc* String consisting of at least 1 a, followed by string having at least 1
b, followed by string having at least 1 c
(a+b)*(a+bb) Set of strings of a’s and b’s ending with either a or bb
(aa)*(bb)*b Set of strings of even number of a’s followed by odd number of b’s

Regular Expression Problems
Q. Obtain a regular expression having a’s and b’s having length 2.
Strings of a’s and b’s having length 2: aa, bb, ab, ba
RE is: (aa+bb+ab+ba)

Q. Obtain a regular expression having a’s and b’s having length <=2.
Strings of a’s and b’s having length <=2:
ε + a + b + aa + bb + ab + ba
This can be written as: (ε + a + b)(ε + a + b)
RE is: (ε + a + b)2

Q. Obtain a regular expression having a’s and b’s having length <=10.
From the logic of the previous problem
RE is: (ε + a + b)10

Q. Obtain a regular expression having a’s and b’s having even length.
To obtain this we will use strings aa,bb,ab,ba zero
or more times.
RE is: (aa+bb+ab+ba)*
This can also be written as:
RE is: ((a+b)(a+b))*

Q. Obtain a regular expression having a’s and b’s having odd length.
From the previous example we know that the RE
for even length is ((a+b)(a+b))*
To this we just need to add 1 more symbol.
Hence,
RE is: (a+b)((a+b)(a+b))*

Q. Obtain a regular expression having alternate a’s and b’s.
To get alternate a’s and b’s we can use multiple
concatenations of ‘ab’ or ‘ba’.
Additionally to take care of the starting and ending
variable to be either a or b, we add (ε+a) and (ε+b)
In appropriate places.
RE is: (ε+b)(ab)*(ε+a)

To obtain a FA from a RE
We have seen examples of obtaining ﬁnite automata from a regular
expression. Now let us see this conversion from the other end.
Here we have the schematic
Representation of a Finite Automata(M)
accepting a regular Expression(R).
Where q is the initial and
f is the ﬁnal state.

Let’s see the various cases for conversion.

Let’s see the various cases for conversion.
Case 3: R=(R1
)* we can construct a NFA that accepts L((R1
)*) as shown in ﬁg 2.6

Obtain an FSM for RE: a* + b* + c*

Adding the three graphs obtained for a*, b*, c*

Obtain a FSM for (a+b)*aa(a+b)*

We obtain FA for ‘aa’, and concatenate with (a+b)*

Finally obtained FSM is:

Pumping Lemma and Regular Languages
Pumping Lemma for Regular Languages
For any regular language L, there exists an integer n, such that for all
x ∈ L with |x| ≥ n, there exists u, v, w ∈ Σ*, such that x = uvw, and
(1) |uv| ≤ n
(2) |v| ≥ 1
(3) for all i ≥ 0: u vi
w ∈ L

Show that L = {w.wR
| w ∈(0,1)*} is not regular

Show that L = {ai
bj
| i>j } is not regular

Minimization of DFA
Suppose there is a DFA D = { Q, Σ, q0
, δ, F } which recognizes a language L.
Then the minimized DFA D’ = { Q’, Σ, q0
, δ’, F’ } can be constructed for language L as:
Step 1: We will divide Q (set of states) into two sets. One set will contain all ﬁnal states and other set
will contain non-ﬁnal states. This partition is called P0
.
Step 2: Initialize k = 1
Step 3: Find Pk
by partitioning the different sets of Pk-1
. In each set of Pk-1
, we will take all possible
pair of states. If two states of a set are distinguishable, we will split the sets into different sets in Pk
.
Step 4: Stop when Pk
= Pk-1
(No change in partition)
Step 5: All states of one set are merged into one. No. of states in minimized DFA will be equal to no.
of sets in Pk
.

Minimization of DFA
Step 1: P0
will have two sets of states.
One set will contain q1, q2, q4 which are
ﬁnal states of DFA and another set will
contain remaining states.
So P0
= { { q1, q2, q4 }, { q0, q3, q5 } }.
Step 2. To calculate P1, we will check
whether sets of partition P0 can be partitioned or not:
i) For set { q1, q2, q4 } :
δ ( q1, 0 ) = δ ( q2, 0 ) = q2 and
δ ( q1, 1 ) = δ ( q2, 1 ) = q5.
So q1 and q2 are not distinguishable.
Similarly, δ ( q1, 0 ) = δ ( q4, 0 ) = q2 and
δ ( q1, 1 ) = δ ( q4, 1 ) = q5.

Minimization of DFA
ii) For set { q0, q3, q5 } :
δ ( q0, 0 ) = q3 and δ ( q3, 0 ) = q0
δ ( q0, 1) = q1 and δ( q3, 1 ) = q4
So, q0 and q3 are not distinguishable.
δ ( q0, 0 ) = q3 and δ ( q5, 0 ) = q5 and
δ ( q0, 1 ) = q1 and δ ( q5, 1 ) = q5
Moves of q0 and q5 on input symbol 1 are q3 and q5
respectively which are in different set in partition P0. So, q0 and
q5 are distinguishable. So, set { q0, q3, q5 } will be partitioned
into { q0, q3 } and { q5 }. So,
P1 = { { q1, q2, q4 }, { q0, q3}, { q5 } }

Minimization of DFA
To calculate P2, we will check whether
sets of partition P1 can be partitioned or not:
iii)For set { q1, q2, q4 } :
δ ( q1, 0 ) = δ ( q2, 0 ) = q2 and
δ ( q1, 1 ) = δ ( q2, 1 ) = q5.
Similarly, δ ( q1, 0 ) = δ ( q4, 0 ) = q2 and
δ ( q1, 1 ) = δ ( q4, 1 ) = q5.
So, { q1, q2, q4 } set will not be partitioned in P2.

Minimization of DFA
iv)For set { q0, q3 } :
δ ( q0, 0 ) = q3 and δ ( q3, 0 ) = q0
δ ( q0, 1 ) = q1 and δ ( q3, 1 ) = q4
Moves of q0 and q3 on input symbol 0 are
q3 and q0 respectively which are in same set in
partition P1. Similarly, Moves of q0 and q3 on input
symbol 1 are q3 and q0 which are in same set in
partition P1. So, q0 and q3 are not distinguishable.

Minimization of DFA
v) For set { q5 }:
Since we have only one state in this set, it can’t be
further partitioned. So,
P2 = { { q1, q2, q4 }, { q0, q3 }, { q5 } }
Since, P1=P2. So, this is the ﬁnal partition.
Partition P2 means that q1, q2 and q4 states are
merged into one. Similarly, q0 and q3 are merged into one.

Theory of Computation Regular Expressions, Minimisation & Pumping Lemma

Theory of Computation Regular Expressions, Minimisation & Pumping Lemma

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