What is the closure property of regular languages under concatenation? How are finite state machines combined to represent the union of languages recognized by two machines?

by Baldo Gagliano / Tuesday, 11 June 2024 / Published in Cybersecurity, EITC/IS/CCTF Computational Complexity Theory Fundamentals, Finite State Machines, Operations on Regular Languages

The closure properties of regular languages and the methods for combining finite state machines (FSMs) to represent operations such as union and concatenation are fundamental concepts in the theory of computation and have significant implications in the domain of cybersecurity, particularly in the analysis and design of algorithms for pattern matching, intrusion detection systems, and other applications.

Closure Property of Regular Languages under Concatenation

A set of languages is said to be closed under an operation if applying that operation to languages within the set results in a language that is also within the set. Regular languages, which can be recognized by finite state machines, are closed under several operations, including union, intersection, complementation, and concatenation.

For the concatenation operation, if $L_1$ and $L_2$ are regular languages, then the concatenation $L_1L_2$ is also a regular language. The concatenation of two languages $L_1$ and $L_2$ is defined as:

$L_1L_2 = \{ xy \mid x \in L_1 \text{ and } y \in L_2 \}$

To demonstrate the closure property under concatenation, consider that $L_1$ and $L_2$ are recognized by finite state machines $M_1$ and $M_2$ , respectively. The goal is to construct a new finite state machine $M$ that recognizes the language $L_1L_2$ .

Construction of the Finite State Machine for Concatenation

Let $M_1 = (Q_1, \Sigma, \delta_1, q_1, F_1)$ and $M_2 = (Q_2, \Sigma, \delta_2, q_2, F_2)$ be the finite state machines recognizing $L_1$ and $L_2$ , respectively. Here, $Q_1$ and $Q_2$ are the sets of states, $\Sigma$ is the common input alphabet, $\delta_1$ and $\delta_2$ are the transition functions, $q_1$ and $q_2$ are the initial states, and $F_1$ and $F_2$ are the sets of accepting states.

To construct the finite state machine $M$ that recognizes $L_1L_2$ :

1. States: The set of states for $M$ is $Q = Q_1 \cup Q_2$ . This means $M$ will have all the states from both $M_1$ and $M_2$ .

2. Alphabet: The input alphabet $\Sigma$ remains the same.

3. Transition Function: The transition function $\delta$ of $M$ is defined as follows:
– For states in $Q_1$ and input $a \in \Sigma$ , $\delta$ behaves as $\delta_1$ . That is, $\delta(q, a) = \delta_1(q, a)$ for $q \in Q_1$ .
– For states in $Q_2$ and input $a \in \Sigma$ , $\delta$ behaves as $\delta_2$ . That is, $\delta(q, a) = \delta_2(q, a)$ for $q \in Q_2$ .
– Additionally, for any accepting state $f \in F_1$ and the empty string $\epsilon$ , $\delta(f, \epsilon) = q_2$ . This transition essentially allows the machine to move from an accepting state of $M_1$ to the initial state of $M_2$ without consuming any input.

4. Initial State: The initial state of $M$ is the initial state of $M_1$ , which is $q_1$ .

5. Accepting States: The set of accepting states of $M$ is $F = F_2$ . This means that $M$ accepts a string if it reaches an accepting state of $M_2$ after processing the entire string.

By following this construction, $M$ will recognize the concatenation of $L_1$ and $L_2$ .

Example

Consider two regular languages $L_1 = \{ a, ab \}$ and $L_2 = \{ b, ba \}$ . Let $M_1$ and $M_2$ be finite state machines recognizing $L_1$ and $L_2$ , respectively.

– $M_1$ might have states $\{ q_1, q_2, q_3 \}$ with transitions:
– $\delta_1(q_1, a) = q_2$
– $\delta_1(q_2, b) = q_3$
– $F_1 = \{ q_2, q_3 \}$

– $M_2$ might have states $\{ p_1, p_2, p_3 \}$ with transitions:
– $\delta_2(p_1, b) = p_2$
– $\delta_2(p_2, a) = p_3$
– $F_2 = \{ p_2, p_3 \}$

To construct $M$ for $L_1L_2$ :

– States: $\{ q_1, q_2, q_3, p_1, p_2, p_3 \}$
– Transitions include those of $M_1$ and $M_2$ , plus $\delta(q_2, \epsilon) = p_1$ and $\delta(q_3, \epsilon) = p_1$
– Initial state: $q_1$
– Accepting states: $\{ p_2, p_3 \}$

Combining Finite State Machines for Union

The union of two languages $L_1$ and $L_2$ is defined as:

$L_1 \cup L_2 = \{ x \mid x \in L_1 \text{ or } x \in L_2 \}$

To construct a finite state machine $M$ that recognizes the union of $L_1$ and $L_2$ , we utilize the non-deterministic finite automaton (NFA) construction. If $M_1 = (Q_1, \Sigma, \delta_1, q_1, F_1)$ and $M_2 = (Q_2, \Sigma, \delta_2, q_2, F_2)$ are the finite state machines for $L_1$ and $L_2$ , respectively, the construction of $M$ is as follows:

1. States: The set of states for $M$ is $Q = Q_1 \cup Q_2 \cup \{ q_0 \}$ , where $q_0$ is a new initial state.

2. Alphabet: The input alphabet $\Sigma$ remains the same.

3. Transition Function: The transition function $\delta$ is defined as:
– $\delta(q, a) = \delta_1(q, a)$ for $q \in Q_1$
– $\delta(q, a) = \delta_2(q, a)$ for $q \in Q_2$
– $\delta(q_0, \epsilon) = \{ q_1, q_2 \}$ . This transition allows the machine to non-deterministically choose to start in either $M_1$ or $M_2$ .

4. Initial State: The initial state of $M$ is the new state $q_0$ .

5. Accepting States: The set of accepting states of $M$ is $F = F_1 \cup F_2$ .

This construction ensures that $M$ accepts a string if it is accepted by either $M_1$ or $M_2$ .

Example

Consider two regular languages $L_1 = \{ a, ab \}$ and $L_2 = \{ b, ba \}$ . Let $M_1$ and $M_2$ be finite state machines recognizing $L_1$ and $L_2$ , respectively.

– $M_1$ might have states $\{ q_1, q_2, q_3 \}$ with transitions:
– $\delta_1(q_1, a) = q_2$
– $\delta_1(q_2, b) = q_3$
– $F_1 = \{ q_2, q_3 \}$

– $M_2$ might have states $\{ p_1, p_2, p_3 \}$ with transitions:
– $\delta_2(p_1, b) = p_2$
– $\delta_2(p_2, a) = p_3$
– $F_2 = \{ p_2, p_3 \}$

To construct $M$ for $L_1 \cup L_2$ :

– States: $\{ q_0, q_1, q_2, q_3, p_1, p_2, p_3 \}$
– Transitions include those of $M_1$ and $M_2$ , plus $\delta(q_0, \epsilon) = \{ q_1, p_1 \}$
– Initial state: $q_0$
– Accepting states: $\{ q_2, q_3, p_2, p_3 \}$

This construction demonstrates how finite state machines can be combined to represent the union of languages they recognize, illustrating the closure properties of regular languages under union and concatenation.

EITCA Academy

What is the closure property of regular languages under concatenation? How are finite state machines combined to represent the union of languages recognized by two machines?

Other recent questions and answers regarding EITC/IS/CCTF Computational Complexity Theory Fundamentals:

More questions and answers:

EITCA Academy is a part of the European IT Certification framework

EITCA Academy

SIGN IN YOUR ACCOUNT TO HAVE ACCESS TO DIFFERENT FEATURES

FORGOT YOUR DETAILS?

CREATE ACCOUNT

What is the closure property of regular languages under concatenation? How are finite state machines combined to represent the union of languages recognized by two machines?

Other recent questions and answers regarding EITC/IS/CCTF Computational Complexity Theory Fundamentals:

More questions and answers: