Regular Expressions and Finite Automata

Regular Expressions and Finite Automata

Contents

Describe Regular Expression, Operators, and Identities used in RE 1

Describe Precedence of Regular Expression Operator 5

State and apply Kleene’s Theorem and Arden’s Theorem 6

Design RE for the given Sets 7

Explain the Procedure to Eliminate Epsilon Move from FA 7

Explain Algebraic Law using Arden’s Theorem to find the Regular Expression for a given FA 7

Convert Finite Automata to Regular Expression 7

Explain the rules for designing a FA for the given RE 7

Design FA for the given Regular Expression 7

Explain the Equivalence of two Finite Automata 7

Explain the Equivalence of two Regular Expressions 7

Explain Chomsky Classification of Languages 7

Describe Type0, Type1, Type2, and Type3 Grammars 7

Identify Type0, Type1, Type2, and Type3 Grammars 7

Define Grammar 7

Generate Language for a given Grammar 7

Generate Grammar for the given Language 7

Describe Regular Grammar 7

Construct Regular Grammar for the given FA 7

Construct Regular Grammar for a given Regular Expression 7

Design FA for the given Regular Grammar 7

Define Pumping Lemma for Regular Language 7

Recall Applications of Pumping Lemma 7

Apply Pumping Lemma to prove that certain Sets are not Regular 7

Describe Union and Intersection Properties of Regular Languages 7

Describe Concatenation and Difference Properties of Regular Languages 7

Complementation and Homomorphism Properties of Regular Languages 7

Describe Regular Expression, Operators, and Identities used in RE

Regular expressions are a concise and powerful notation for describing patterns in strings. They are widely used in text processing, pattern matching, and programming languages to search, manipulate, and validate strings. Regular expressions consist of various operators and identities that define the pattern matching rules.

Here’s a description of regular expressions, operators, and identities commonly used:

  • Regular Expression:

A regular expression is a sequence of characters that defines a search pattern. It consists of literal characters and metacharacters that represent classes of characters or operations.

Operators in Regular Expressions:

  1. Concatenation (AB): Concatenation is denoted by simply placing two regular expressions side by side. It matches the concatenation of strings that match the first expression followed by strings that match the second expression.
  2. Alternation (A|B): Alternation allows for matching either of the expressions separated by the “|” (pipe) symbol. It matches strings that match either the left expression or the right expression.
  3. Kleene Star (A*): The Kleene star denotes zero or more occurrences of the preceding regular expression. It matches strings that have zero or more repetitions of the expression.
  4. Kleene Plus (A+): The Kleene plus denotes one or more occurrences of the preceding regular expression. It matches strings that have one or more repetitions of the expression.
  5. Optional (A?): The optional operator “?” denotes zero or one occurrence of the preceding regular expression. It matches strings that have zero or one occurrence of the expression.
  6. Grouping (()): Grouping allows grouping subexpressions together. It helps in defining the scope of operators and enables capturing groups for further use.

Identities in Regular Expressions:

  1. Identity Element for Alternation: A|∅ = A

The identity element for alternation is the empty set (∅). When alternated with any regular expression, it yields the same regular expression.

  1. Identity Element for Concatenation: A∅ = ∅A = ∅

The identity element for concatenation is the empty set (∅). When concatenated with any regular expression, it results in the empty set.

  1. Identity Element for Kleene Star: A* = ε

The identity element for the Kleene star is the empty string (ε). The Kleene star applied to any regular expression gives the empty string.

  1. Idempotent Property of Alternation: A|A = A

The alternation operator is idempotent, meaning that if the same regular expression is alternated with itself, it remains unchanged.

  1. Identity Element for Concatenation with Empty String: Aε = εA = A

Concatenating any regular expression with the empty string (ε) results in the same regular expression.

  1. Distributive Property of Concatenation over Alternation: A(B|C) = AB|AC

The distributive property allows distributing the concatenation operation over alternation. It means that concatenating a regular expression with an alternation of two expressions is equivalent to the alternation of the concatenation of the first expression with the second expression individually.

  1. Absorption Property of Alternation with Empty String: A|ε = ε|A = A

The absorption property states that if a regular expression is alternated with the empty string (ε), the result is the same as the original regular expression.

  1. Idempotent Property of Concatenation: A.A = A

Concatenating a regular expression with itself is equivalent to the original regular expression.

  1. Identity Element for Kleene Plus: A+ = AA*

The Kleene plus operation on a regular expression is equivalent to concatenating the regular expression with its Kleene star.

  1. Identity Element for Optional Operator: A? = A|ε

The optional operator on a regular expression is equivalent to alternation with the empty string (ε).

These operators and identities provide the foundation for constructing regular expressions and enable powerful pattern matching capabilities. By combining these operators and using literals and metacharacters, complex patterns can be described concisely and efficiently.

Regular Expression:

A regular expression is a sequence of characters that defines a search pattern. It is used for matching and manipulating strings based on specific patterns.

Example of a Regular Expression:

Pattern: “abc”

This regular expression matches the string “abc” exactly.

Operators in Regular Expressions:

  1. Concatenation (AB):

Example:

Pattern: “ab”

This regular expression matches any string that has the characters “a” followed by “b”.

  1. Alternation (A|B):

Example:

Pattern: “cat|dog”

This regular expression matches either the string “cat” or the string “dog”.

  1. Kleene Star (A*):

Example:

Pattern: “a*”

This regular expression matches any string that contains zero or more occurrences of the character “a”.

  1. Kleene Plus (A+):

Example:

Pattern: “a+”

This regular expression matches any string that contains one or more occurrences of the character “a”.

  1. Optional (A?):

Example:

Pattern: “colou?r”

This regular expression matches either the string “color” or the string “colour”.

Identities in Regular Expressions:

  1. Identity Element for Alternation:

Example:

Pattern: “A|∅”

This regular expression is equivalent to “A” since alternation with an empty set (∅) yields the original expression.

  1. Identity Element for Concatenation:

Example:

Pattern: “Aε”

This regular expression is equivalent to “A” since concatenation with the empty string (ε) does not change the expression.

  1. Identity Element for Kleene Star:

Example:

Pattern: “A*”

This regular expression matches any string that contains zero or more occurrences of the regular expression “A”. An empty string is also matched since the Kleene star can represent zero occurrences.

  1. Idempotent Property of Alternation:

Example:

Pattern: “A|A”

This regular expression is equivalent to “A” since alternation with the same expression results in the original expression.

  1. Identity Element for Concatenation with Empty String:

Pattern: “Aε”

This regular expression is equivalent to “A” since concatenating any regular expression with the empty string (ε) results in the same regular expression.

  1. Distributive Property of Concatenation over Alternation:

Pattern: “A(B|C)”

This regular expression is equivalent to “AB|AC” since the concatenation of “A” with an alternation of “B” and “C” is the same as the alternation of the concatenation of “A” with “B” and the concatenation of “A” with “C”.

  1. Absorption Property of Alternation with Empty String:

Pattern: “A|ε”

This regular expression is equivalent to “A” since alternation with the empty string (ε) yields the same regular expression.

  1. Idempotent Property of Concatenation:

Pattern: “AA”

This regular expression is equivalent to “A” since concatenating a regular expression with itself is the same as the original regular expression.

  1. Identity Element for Kleene Plus:

Pattern: “A+”

This regular expression matches any string that contains one or more occurrences of the regular expression “A”. It is equivalent to “AA*”, where “A*” represents zero or more occurrences.

  1. Identity Element for Optional Operator:

Pattern: “A?”

This regular expression is equivalent to “A|ε”, as it matches either the regular expression “A” or the empty string (ε).

These examples illustrate the usage of regular expressions, operators, and identities to define patterns and match strings based on specific rules and conditions. Regular expressions provide a flexible and powerful way to manipulate and process textual data.

Describe Precedence of Regular Expression Operator

The precedence of regular expression operators determines the order in which the operators are evaluated when constructing or matching a regular expression. It helps to define the grouping and hierarchy of operations within a regular expression.

The following is a typical precedence order for the common regular expression operators, listed from highest to lowest:

  1. Grouping and Parentheses: ()

Grouping operators are evaluated first to specify the scope of operations within parentheses. They allow grouping subexpressions together and establish precedence within the expression.

  1. Kleene Operators: *, +, ?

Kleene operators have higher precedence and are evaluated after grouping. They define the repetition of the preceding element or subexpression.

  1. Concatenation: AB

Concatenation has higher precedence than alternation. It denotes the concatenation of two regular expressions or subexpressions.

  1. Alternation: |

Alternation is evaluated last among the common operators. It represents the choice between two or more alternative regular expressions or subexpressions.

It’s important to note that although this is a general precedence order, parentheses can be used to override the default precedence and explicitly specify the desired evaluation order.

Example:

Consider the regular expression (ab+c)*d | ef+g

Here is how the precedence of operators is applied:

  1. Parentheses: (ab+c)*d is evaluated as a subexpression, allowing the repetition of the sequence “ab+c” followed by the literal “d”.
  2. Kleene Operator: (ab+c)* is evaluated, indicating zero or more occurrences of the subexpression “ab+c”.
  3. Concatenation: (ab+c)d represents the concatenation of the subexpression (ab+c) with the literal “d”.
  4. Alternation: (ab+c)*d | ef+g allows a choice between the subexpression (ab+c)*d and the subexpression ef+g.

By understanding the precedence of regular expression operators, you can construct and interpret regular expressions more accurately and ensure the intended matching behavior.

State and apply Kleene’s Theorem and Arden’s Theorem

Kleene’s Theorem and Arden’s Theorem are important results in the theory of regular languages and finite automata.

  • Kleene’s Theorem:

Kleene’s Theorem states that for any regular language L, there exists a regular expression that describes L. In other words, regular languages can be characterized by regular expressions.

  • Application of Kleene’s Theorem:

To apply Kleene’s Theorem, we start with a given regular language and construct a corresponding regular expression using the following steps:

  1. Start with the regular language L and represent it using its corresponding NFA or DFA.
  2. Convert the NFA or DFA into a regular expression using methods such as the state elimination method, the Arden’s Theorem, or any other suitable technique.
  3. The resulting regular expression represents the original regular language L.
  • Arden’s Theorem:

Arden’s Theorem provides a method to find a unique solution for a system of linear equations involving regular expressions. It states that given two regular expressions A and B, the equation X = AX + B has a unique solution X, which is also a regular expression.

  • Application of Arden’s Theorem:

To apply Arden’s Theorem, we follow these steps:

  1. Given a system of linear equations involving regular expressions, set up the equations in the form X = AX + B, where A and B are regular expressions.
  2. Apply Arden’s Lemma, which states that for regular expressions A and B, the equation X = AX + B has a unique solution X = A* B.
  3. Solve the equation using the Arden’s Lemma to obtain the regular expression X, which represents the solution to the system of equations.

By applying Arden’s Theorem, we can find the solution to systems of linear equations involving regular expressions, which can be useful in various applications such as compiler design, pattern matching, and string manipulation.

Both Kleene’s Theorem and Arden’s Theorem provide important theoretical foundations for working with regular languages and regular expressions, allowing us to describe and manipulate these languages effectively.

Proof of Arden’s Theorem:

To prove Arden’s Theorem, we need to show that for any regular expressions A and B, the equation X = AX + B has a unique solution X = A* B.

Proof of Arden’s Theorem:

Step 1: Construct the Regular Expression X = A* B.

We start by assuming X = A* B and prove that it satisfies the equation X = AX + B.

Step 2: Show that X = AX + B.

We can rewrite the equation X = AX + B as X – AX = B.

Using the distributive property of concatenation over alternation, we can rewrite it as X(I – A) = B, where I is the identity element for concatenation.

Step 3: Prove that X = A* B is a solution to the equation.

To prove this, we need to show that X = A* B satisfies the equation X(I – A) = B.

Consider the expression X(I – A). Since X = A* B, we substitute it into the equation:

X(I – A) = (A* B)(I – A).

Using the identity element for concatenation, we have:

(A* B)(I – A) = A* B – A(A* B).

Now, we apply the distributive property of concatenation over alternation to the second term:

A* B – A(A* B) = A* B – A(A*)B.

Using the identity element for concatenation, A* A = A*, we simplify further:

A* B – A(A*)B = A* B – A* B.

This simplifies to:

A* B – A* B = 0.

Therefore, we have shown that X = A* B satisfies the equation X(I – A) = B.

Step 4: Show uniqueness of the solution.

To prove uniqueness, assume there exists another regular expression Y that satisfies the equation X(I – A) = B.

Let’s assume Y is a different solution such that Y = AY + B.

Substituting X = A* B and Y = AY + B into the equation, we have:

A* B(I – A) = B.

By simplifying the equation, we get:

A* B – A* BA = B.

Using the distributive property and the identity element for concatenation, we have:

A* B – A* BA = A* B – A* B = 0.

Therefore, Y = A* B is the unique solution to the equation X = AX + B.

Hence, we have proved Arden’s Theorem, which states that for any regular expressions A and B, the equation X = AX + B has a unique solution X = A* B.

Design RE for the given Sets

Here are 10 examples of regular expressions for different sets, listed from lower to higher difficulty:

  1. Set: {a}

Regular Expression: a

Description: Matches the single character ‘a’.

  1. Set: {a, b}

Regular Expression: a|b

Description: Matches either the character ‘a’ or ‘b’.

  1. Set: {ab, ac}

Regular Expression: a[b|c]

Description: Matches the string ‘ab’ or ‘ac’.

  1. Set: {a, aa, aaa}

Regular Expression: a{1,3}

Description: Matches the strings ‘a’, ‘aa’, or ‘aaa’.

  1. Set: {0, 10, 100, 1000}

Regular Expression: 10{0,3}

Description: Matches the strings ‘0’, ’10’, ‘100’, or ‘1000’.

  1. Set: {a, ab, abb, abbb}

Regular Expression: ab{1,3}

Description: Matches the strings ‘a’, ‘ab’, ‘abb’, or ‘abbb’.

  1. Set: {aa, aba, abba, abbba}

Regular Expression: (a(ba){0,3})

Description: Matches the strings ‘aa’, ‘aba’, ‘abba’, or ‘abbba’.

  1. Set: {abc, acb, adc}

Regular Expression: a(b|d)c

Description: Matches the strings ‘abc’, ‘acb’, or ‘adc’.

  1. Set: {ab, abb, abbb, abbbb}

Regular Expression: ab{1,}

Description: Matches the strings ‘ab’, ‘abb’, ‘abbb’, or ‘abbbb’. Allows one or more occurrences of ‘b’.

  1. Set: {ab, aabb, aaabbb, aaaabbbb}

Regular Expression: a{1,}b{2,}

Description: Matches the strings ‘ab’, ‘aabb’, ‘aaabbb’, or ‘aaaabbbb’. Allows one or more occurrences of ‘a’ followed by two or more occurrences of ‘b’.

These examples gradually increase in difficulty, introducing more complex patterns and quantifiers to match different sets of strings.

Here aresome examples of finding regular expressions for different languages or strings:

  1. Language: Set of all strings over {a, b} that start with ‘aa’:

Regular Expression: aa(a|b)*

Description: Matches any string that starts with ‘aa’, allowing any number of ‘a’ or ‘b’ characters after ‘aa’.

  1. Language: Set of all strings over {0, 1} that have an odd number of ‘0’ characters:

Regular Expression: (1(01)*0)*1

Description: Matches any string that has an odd number of ‘0’ characters, allowing any number of alternating ‘1’ and ’01’ sequences before the last ‘1’.

  1. Language: Set of all strings over {a, b} that contain at least one ‘ab’ or ‘ba’ substring:

Regular Expression: (a|b)((ab)|(ba))(a|b)

Description: Matches any string that contains at least one occurrence of ‘ab’ or ‘ba’, allowing any number of ‘a’ or ‘b’ characters before, in between, and after.

  1. Language: Set of all strings over {a, b} that end with ‘aa’:

Regular Expression: (a|b)*(aa)

Description: Matches any string that ends with ‘aa’, allowing any number of ‘a’ or ‘b’ characters before ‘aa’.

Explain the Procedure to Eliminate Epsilon Move from FA

The procedure to eliminate epsilon (ε) moves from a finite automaton (FA) involves several steps. Here is a general procedure to follow:

  1. Identify epsilon closure: Compute the epsilon closure for each state in the FA. The epsilon closure of a state is the set of all states that can be reached from it by following epsilon transitions.
  2. Create a new DFA: Create a new deterministic finite automaton (DFA) that will replace the original FA without epsilon moves.
  3. Initialize the new DFA: Start by adding the epsilon closure of the initial state of the original FA as the initial state of the new DFA.
  4. Explore transitions: For each state in the new DFA, explore transitions on input symbols other than epsilon. Determine the set of states that can be reached from the current state by following transitions on each input symbol.
  5. Add transitions: Add transitions to the new DFA based on the sets of states reached in the previous step. Each set of states becomes a new state in the DFA, and transitions are added accordingly.
  6. Finalize states: Identify the final states of the new DFA. A state in the new DFA is a final state if it contains any final state of the original FA.
  7. Repeat steps 4-6: Repeat steps 4 to 6 for each new state added to the DFA until no more new states can be added.
  8. The resulting DFA: The resulting DFA obtained after eliminating epsilon moves will have no epsilon transitions, and it recognizes the same language as the original FA.

It is important to note that the procedure to eliminate epsilon moves can be complex for more intricate FAs. In some cases, it may involve constructing and analyzing an epsilon-NFA and then converting it to a DFA.

Explain Algebraic Law using Arden’s Theorem to find the Regular Expression for a given FA

The two commonly used methods for converting a given DFA to its regular expression are:

  1. Arden’s Method
  2. State Elimination Method

In this article, we will focus on Arden’s Theorem.

Arden’s Theorem is a popular technique employed to convert a given DFA to its corresponding regular expression. The theorem states the following:

Let P and Q be two regular expressions over ∑.

If P does not contain the empty string ε, then the equation R = Q + RP has a unique solution, which is R = QP*.

To use Arden’s Theorem, certain conditions must be satisfied:

  1. The transition diagram should not include any ε transitions.
  2. There should be only a single initial state.

The process of converting a DFA to its regular expression using Arden’s Theorem involves the following steps:

Step 1:

Formulate an equation for each state, considering the transitions leading to that state. Include the empty string ε in the equation of the initial state.

Step 2:

Manipulate the equations to bring the final state into the form R = Q + RP, which yields the desired regular expression.

Important notes to consider:

  • Arden’s Theorem can be applied to both DFAs and NFAs.
  • If there are multiple final states, create a separate regular expression for each one and combine them by adding all the regular expressions to obtain the final regular expression.

Convert Finite Automata to Regular Expression

Example 1: Find regular expression for the following DFA using Arden’s Theorem-

PyZMbSOjJBrcAAAAAElFTkSuQmCC

Form a equation for each state-

  • A = ∈ + B.1 ……(1)
  • B = A.0 ……(2)

Bring final state in the form R = Q + RP.

Using (1) in (2), we get-

B = (∈ + B.1).0

B = ∈.0 + B.1.0

B = 0 + B.(1.0) ……(3)

Using Arden’s Theorem in (3), we get-

B = 0.(1.0)*

Thus, Regular Expression for the given DFA = 0(10)*

Example 2: Find regular expression for the following DFA using Arden’s Theorem-

2PoegBDw6j2fWUYjJwmSkg378skQwrEnIiGEEEIIo0QkhBBCCGGUiIQQQgghjBKREEIIIYRRIhJCCCGEMEpEQgghhBBGiUgIIYQQwigRCSGEEEIYJSIhhBBCCKNEJIQQQghhlIiEEEIIIYwSkRBCCCGEUSISQgghhDBKREIIIYQQRolICCGEEMIoEQkhhBBCGCUiIYQQQgijRCSEEEIIYZSIhBBCCCGMEpEQQgghhFEiEkIIIYQwSkRCCCGEEEb5f5rI7zllJspfAAAAAElFTkSuQmCC

Form a equation for each state-

  • q1 = ∈ ……(1)
  • q2 = q1.a ……(2)
  • q3 = q1.b + q2.a + q3.a …….(3)

Bring final state in the form R = Q + RP.

Using (1) in (2), we get-

q2 = ∈.a

q2 = a …….(4)

Using (1) and (4) in (3), we get-

q3 = q1.b + q2.a + q3.a

q3 = ∈.b + a.a + q3.a

q3 = (b + a.a) + q3.a …….(5)

Using Arden’s Theorem in (5), we get-

q3 = (b + a.a)a*

Thus, Regular Expression for the given DFA = (b + aa)a*

Example 3: Find regular expression for the following DFA using Arden’s Theorem-

4fNRiIgkJ+R3AAAAAASUVORK5CYII=

Form a equation for each state-

  • q1 = ∈ + q1.b + q2.a ……(1)
  • q2 = q1.a + q2.b ……(2)

Bring final state in the form R = Q + RP.

Using Arden’s Theorem in (2), we get-

q2 = q1.a.b* …….(3)

Using (3) in (1), we get-

q1 = ∈ + q1.b + q1.a.b*.a

q1 = ∈ + q1.(b + a.b*.a) …….(4)

Using Arden’s Theorem in (4), we get-

q1 = ∈.(b + a.b*.a)*

q1 = (b + a.b*.a)*

Thus, Regular Expression for the given DFA = (b + a.b*.a)*

Explain the rules for designing a FA for the given RE

When designing a finite automaton (FA) for a given regular expression, there are certain rules and guidelines to follow. The purpose of designing an FA is to construct a machine that recognizes the language defined by the regular expression.

Here are the rules for designing an FA based on a regular expression:

  1. Rule 1: Start with the simplest cases:
    • If the regular expression is ε (empty string), design an FA with a single state that is also a final state.
    • If the regular expression is a single symbol, design an FA with two states connected by a transition on that symbol. The initial state is not a final state, and the destination state is a final state.
  2. Rule 2: Handle concatenation (concatenation operator):
    • For a regular expression R1R2 (concatenation of R1 and R2), design an FA for R1 and an FA for R2.
    • Make the final state(s) of the FA for R1 non-final.
    • Connect the final state(s) of the FA for R1 to the initial state(s) of the FA for R2 with epsilon transitions.
    • The initial state of the FA for R1 becomes the initial state of the resulting FA, and the final state(s) of the FA for R2 become the final state(s) of the resulting FA.
  3. Rule 3: Handle union (OR operator):
    • For a regular expression R1 + R2 (union of R1 and R2), design an FA for R1 and an FA for R2.
    • Create a new initial state and connect it to the initial states of the FAs for R1 and R2 with epsilon transitions.
    • Create a new final state and make the final states of the FAs for R1 and R2 non-final, connecting them to the new final state with epsilon transitions.
    • The new initial state becomes the initial state of the resulting FA, and the new final state becomes the final state of the resulting FA.
  4. Rule 4: Handle Kleene star (closure operator):
    • For a regular expression R*, design an FA for R.
    • Create a new initial state and a new final state.
    • Connect the new initial state to the initial state of the FA for R and to the new final state with epsilon transitions.
    • Make the final states of the FA for R non-final, connecting them to the initial state of the FA for R and to the new final state with epsilon transitions.
    • The new initial state becomes the initial state of the resulting FA, and the new final state becomes the final state of the resulting FA.

By applying these rules recursively, you can design an FA for any given regular expression. The resulting FA will recognize the language defined by the regular expression.

Design FA for the given Regular Expression

Example 1: Design a FA from given regular expression 10 + (0 + 11)0* 1.

Solution: First we will construct the transition diagram for a given regular expression.

Conversion of RE to FA

Conversion of RE to FA

Conversion of RE to FA

Conversion of RE to FA

Conversion of RE to FA

Now we have got NFA without ε. Now we will convert it into required DFA for that, we will first write a transition table for this NFA.

State 0 1
→q0 q3 {q1, q2}
q1 qf ϕ
q2 ϕ q3
q3 q3 qf
*qf ϕ ϕ

The equivalent DFA will be:

State 0 1
→[q0] [q3] [q1, q2]
[q1] [qf] ϕ
[q2] ϕ [q3]
[q3] [q3] [qf]
[q1, q2] [qf] [qf]
*[qf] ϕ ϕ

Example 2: Design a NFA from given regular expression 1 (1* 01* 01*)*.

Solution: The NFA for the given regular expression is as follows:

Conversion of RE to FA

Conversion of RE to FA

Conversion of RE to FA

Example 3: Construct the FA for regular expression 0*1 + 10.

Solution:

We will first construct FA for R = 0*1 + 10 as follows:

Conversion of RE to FA

Conversion of RE to FA