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Chapter 13: Problem 38

Show that there is no finite-state automaton with three states that recognizes the set of bit strings containing an even number of 1 \(\mathrm{s}\) and an even number of 0 \(\mathrm{s}\) .

Short Answer

Expert verified
A finite-state automaton with three states cannot recognize bit strings with an even number of 1s and 0s since at least four states are required.

Step by step solution

01

- Understand the Problem

We need to prove that no finite-state automaton (FSA) with three states can recognize bit strings with an even number of 1s and 0s.
02

- Define Automaton Behavior

A bit string must have an even number of both 1s and 0s. This implies maintaining a count of 1s and 0s and tracking whether these counts are even or odd.
03

- Calculate Required States

To keep track of the parities (even or odd) of the counts of 1s and 0s, we consider two binary states for each count (even or odd). This results in four possible combinations: (even, even), (even, odd), (odd, even), (odd, odd).
04

- Analyze the Combinations

The four combinations of parity states mean we need at least four states to distinguish all combinations of even and odd counts of 1s and 0s.
05

- Conclude Insufficiency of Three States

Since we need a minimum of four states to recognize all the parity combinations, a three-state automaton cannot accurately recognize the set of bit strings with an even number of both 1s and 0s. Therefore, it is impossible to have such an FSA with only three states.

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Key Concepts

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

bit strings
A bit string is a sequence of bits, where each bit is either a 0 or a 1. These sequences can vary in length and are often used in computing and digital systems. For example, '101' is a bit string of length 3. Bit strings can represent data, instructions, or be used in various algorithms. When we talk about recognizing bit strings, we often refer to processing and interpreting these sequences based on certain rules or patterns.
even and odd parity
Parity is a concept used to determine if the number of certain elements in a set is odd or even. In the context of bit strings, parity refers to the number of 1s or 0s. A bit string with even parity of 1s has an even number of 1s, e.g., '1100' (two 1s). Similarly, a bit string with odd parity of 0s has an odd number of 0s, e.g., '110' (one 0). Parity is essential for error checking and ensuring data integrity. In finite-state automata, tracking the parity of bits helps in creating states that recognize specific patterns in bit strings.
state transitions
State transitions are changes from one state to another in a finite-state automaton. These transitions depend on the input bit (0 or 1). Each state represents a condition or situation based on the input so far. For example, in an automaton designed to recognize bit strings with an even number of 1s, reading a 1 would cause a transition from an 'even' state to an 'odd' state. Conversely, reading another 1 would transition back from 'odd' to 'even'. Transitions are visualized using state diagrams with arcs labeled by the input bit.
automaton states
Automaton states are the distinct conditions or stages that an automaton can be in at any given moment. Each state represents a unique configuration of the automaton based on the inputs it has processed. For a bit string automaton recognizing even parity, states might include combinations such as (even 1s, even 0s). The automaton changes states in response to input bits, guided by defined transition rules. Understanding automaton states and their transitions is key to designing and analyzing finite-state automata.

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Most popular questions from this chapter

Construct a finite-state machine that determines whether the input string read so far ends in at least five consecutive 1s.

Construct a finite-state machine that determines whether the word computer has been read as the last eight characters in the input read so far, where the input can be any string of English letters.

Construct a Moore machine that gives an output of 1 whenever the number of symbols in the input string read so far is divisible by 4 and an output of 0 otherwise.

Let \(G\) be a grammar and let \(R\) be the relation containing the ordered pair \(\left(w_{0}, w_{1}\right)\) if and only if \(w_{1}\) is directly derivable from \(w_{0}\) in \(G .\) What is the reflexive transitive closure of \(R ?\)

Suppose that \(L\) is a subset of \(I^{*},\) where \(I\) is a nonempty set of symbols. If \(x \in I^{*},\) we let \(L / x=\left\\{z \in I^{*} | x z \in L\right\\} .\) We say that the strings \(x \in I^{*}\) and \(y \in I^{*}\) are distinguishable with respect to \(L\) if \(L / x \neq L / y .\) A string \(z\) for which \(x z \in L\) but \(y z \notin L,\) or \(x z \notin L,\) but \(y z \in L\) is said to distinguish \(x\) and \(y\) with respect to \(L .\) When \(L / x=L / y,\) we say that \(x\) and \(y\) are indistinguishable with respect to \(L .\) $$ \begin{array}{l}{\text { Let } L \text { be the set of all bit strings that end with } 01 . \text { Show }} \\ {\text { that } 11 \text { and } 10 \text { are distinguishable with respect to } L \text { and }} \\ {\text { that the strings } 1 \text { and } 11 \text { are indistinguishable with re- }} \\\ {\text { spect to } L .}\end{array} $$
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