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Chapter 11: Problem 34

Create a NDFSA that accepts the regular language over \(\\{a, b\\}\) of strings that: Contain \(b a^{2} b\) as a substring.

Short Answer

Expert verified
The Non-Deterministic Finite Automaton (NDFA) that accepts a regular language over the alphabet {a, b} of strings containing the substring "bab" can be represented as: States: {q0, q1, q2, accept_state} Alphabet: {a, b} Start state: q0 Accept state: accept_state Transitions: - q0 --a--> q0 - q0 --b--> q1 - q1 --a--> q2 - q1 --b--> q1 - q2 --a--> q2 - q2 --b--> accept_state - accept_state --a--> accept_state - accept_state --b--> accept_state

Step by step solution

01

1. Define the alphabet and start state

The alphabet of this language is {a, b}. Let's define the initial state of the NDFA as q0, which is the start state.
02

2. Create states for recognizing substring 'bab'

In order to recognize the required substring "bab", we need to create multiple states. - q1: State reached when symbol 'b' is read. - q2: State reached when symbol sequence 'ba' is read. - accept_state: State reached when the sequence 'bab' is recognized.
03

3. Define the state transitions for recognizing the substring

Now, we need to define transitions between the states that will help in recognizing the required substring "bab". 1. From q0 (start state), upon reading symbol 'a', stay at q0. 2. From q0, upon reading symbol 'b', move to q1. 3. From q1, upon reading symbol 'a', move to q2. 4. From q1, upon reading symbol 'b', stay at q1. 5. From q2, upon reading symbol 'a', stay at q2. 6. From q2, upon reading symbol 'b', move to the 'accept_state'. 7. Lastly, from the 'accept_state', on reading either 'a' or 'b', stay at the 'accept_state'.
04

4. Create the final NDFA

Based on the steps above, the NDFA can be represented as: States: {q0, q1, q2, accept_state} Alphabet: {a, b} Start state: q0 Accept state: accept_state Transitions: - q0 --a--> q0 - q0 --b--> q1 - q1 --a--> q2 - q1 --b--> q1 - q2 --a--> q2 - q2 --b--> accept_state - accept_state --a--> accept_state - accept_state --b--> accept_state This NDFA accepts the given language over {a, b} of strings containing the substring "bab".

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Regular Languages
Regular languages form a foundational concept in computer science and automata theory. They are a class of languages that can be expressed using regular expressions. Each regular language can be recognized by some finite automaton, be it deterministic or non-deterministic.
They are known for their simplicity and efficiency when it comes to algorithmic processing. Regular languages are crucial in computer science applications like text processing, compiler construction, and network protocols.
When we refer to the regular language of strings that contain "bab" as a substring, we describe a specific pattern that can be expressed and recognized within these languages. This means any string within this language will necessarily contain the sequence "bab" somewhere within it.
State Transitions
State transitions are the cornerstone of finite automata. They describe how an automaton moves from one state to another based on input symbols.
In our NFA example, the transitions are clearly defined by the input alphabet of {a, b}. The initial state is q0, and as the automaton reads the input, it transitions between various states according to predefined rules.
  • From q0, reading 'a' leads back to q0, and reading 'b' transitions it to q1.
  • At state q1, reading 'a' goes to q2, while reading 'b' stays at q1.
  • From q2, reading 'a' remains in q2, but 'b' takes it to the accept state.
  • Once at the accept state, any further input keeps it in that state.

These transitions ensure that the automaton can "keep track" of its progress in recognizing the sequence "bab".
Substring Recognition
Substring recognition in automata involves checking whether a specific sequence of symbols appears in the input string.
The NFA designed to recognize substring "bab" is defined by a series of transitions that guide the automaton to accept a string containing the target sequence. As it processes the input, each character moves the automaton closer or further from its accepting state, depending on whether it matches the expected symbol in "bab."
The key to substring recognition is that the automaton must have a clear path of states reflecting each character of the substring. In our case, states q0, q1, q2, and the accept state collectively ensure that when the substring "bab" is encountered, the automaton reaches its accept state.
Finite State Machines
Finite state machines (FSMs) are one of the simplest models of computation. They consist of a finite number of states and operate based on state transitions governed by input symbols.
FSMs come in two main types: deterministic (DFAs) and non-deterministic (NFAs). In deterministic machines, each state has precisely one transition per input symbol. Non-deterministic machines, however, may have multiple transitions for a particular input or none at all, allowing more flexibility.
The NFA created for the substring "bab" is an example of a non-deterministic finite automaton. It effectively uses its state transitions to ensure the substring "bab" is found in any given string. Its structure is simpler than a DFA for the same language, as it leverages its non-deterministic nature to manage different possibilities concurrently. Thus, NFAs, while potentially more efficient in design, additionally require conversion to a DFA for some computational purposes.

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Most popular questions from this chapter

Consider the grammar \(G=(N, T, P, \sigma),\) where \(N=\\{\sigma\\}, T=\\{a, b\\},\) and \(P=\\{\sigma \rightarrow a \sigma b, \sigma \rightarrow a b\\} .\) Determine if each word belongs to \(L(G)\) . $$\mathrm{abab}$$

For Exercises \(68-73,\) use the following definition of a simple algebraic expression: $$\langle\text {expression}\rangle : :=\langle\text { term }\rangle |\langle\text { sign }\rangle\langle\text { term }\rangle |$$ $$\langle\text { expression }\rangle\langle\text { adding operator }\rangle\langle\text { term }\rangle$$ $$\langle\operatorname{sign}\rangle \therefore=+ 1-$$ $$\langle\text { adding operator}\rangle: :=+1-$$ $$\langle\text { term }\rangle : :=\langle\text { factor }\rangle |$$ $$\langle\text { term }\rangle\langle\text { multiplying operator }\rangle\langle\text { factor }\rangle$$ $$\langle\text { multiplying operator }\rangle := *| /$$ $$\langle\text { factor }\rangle : :=\langle\text { letter }|\rangle (\langle\text { expression }\rangle |\langle\text { expression }\rangle$$ $$\langle\text { letter }\rangle : := a|b| c | \ldots : z$$ Construct a derivation tree for each expression. $$(a * b)+c / d$$

For Exercises \(68-73,\) use the following definition of a simple algebraic expression: $$\langle\text {expression}\rangle : :=\langle\text { term }\rangle |\langle\text { sign }\rangle\langle\text { term }\rangle |$$ $$\langle\text { expression }\rangle\langle\text { adding operator }\rangle\langle\text { term }\rangle$$ $$\langle\operatorname{sign}\rangle \therefore=+ 1-$$ $$\langle\text { adding operator}\rangle: :=+1-$$ $$\langle\text { term }\rangle : :=\langle\text { factor }\rangle |$$ $$\langle\text { term }\rangle\langle\text { multiplying operator }\rangle\langle\text { factor }\rangle$$ $$\langle\text { multiplying operator }\rangle := *| /$$ $$\langle\text { factor }\rangle : :=\langle\text { letter }|\rangle (\langle\text { expression }\rangle |\langle\text { expression }\rangle$$ $$\langle\text { letter }\rangle : := a|b| c | \ldots : z$$ Determine if each is a legal expression. $$a+b *(c / d)$$

Prove each, where \(A, B,\) and \(C\) are arbitrary languages over \(\Sigma\) and \(x \in \Sigma\). \(\left(A^{*}\right)^{*}=A^{*}\)

Develop a grammar that generates each language over {0,1}. $$\\{1,11,1111,11111111, \ldots\\}$$
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