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Let \({{\bf{L}}_{\bf{n}}}\) be the set of strings with at least n bits in which the nth symbol from the end is a 0. Use Exercise 29 to show that a deterministic finite-state machine recognizing \({{\bf{L}}_{\bf{n}}}\) must have at least \({2^{\bf{n}}}\) states.

Short Answer

Expert verified

Therefore, a deterministic finite-state machine recognizing \({{\bf{L}}_{\bf{n}}}\) must have at least \({2^{\bf{n}}}\) states is proved as true.

Step by step solution

01

General form

Finite-state automaton (Definition):A finite-state automata \({\bf{M = }}\left( {{\bf{S,I,f,}}{{\bf{s}}_{\bf{0}}}{\bf{,F}}} \right)\) consists of an initial or start state \({{\bf{s}}_0}\), a finite set S of states, a finite alphabet of inputs I, a transition function f that assigns a subsequent state to each pair of states and inputs (such that \({\bf{f:S \times I}} \to {\bf{S}}\)), and a subset F of S made up of final states (or accepting states).

Regular expressions (Definition):A recursive definition of the regular expressions over a set I is as follows:

A regular expression is the symbol \(\emptyset \);

A regular expression is the symbol \({\bf{\lambda }}\);

When \({\bf{x}} \in {\bf{I}}\) occurs, the symbol x is a regular expression.

When A and B are regular expressions, the symbols \(\left( {{\bf{AB}}} \right){\bf{,}}\left( {{\bf{A}} \cup {\bf{B}}} \right){\bf{,}}\) and \({\bf{A*}}\) are also regular expressions.

Regular sets are the sets that regular expressions represent.

Kleene’s theorem: A finite-state automaton must recognize a set-in order for it to be considered regular.

Theorem 2:If and only if it is a regular set, a set is produced by a regular grammar.

Concept of distinguishable with respect to L:

Suppose that L is a subset of \({\bf{I*}}\), where \({\bf{I}}\) is a nonempty set of symbols. If \({\bf{x}} \in {\bf{I*}}\), let \(\frac{{\bf{L}}}{{\bf{x}}}{\bf{ = }}\left\{ {{\bf{z}} \in {\bf{I*}}\left| {{\bf{xz}} \in {\bf{L}}} \right.} \right\}\). If \(\frac{{\bf{L}}}{{\bf{x}}} \ne \frac{{\bf{L}}}{{\bf{y}}}\), thenthe strings \({\bf{x}} \in {\bf{I*}}\) and \({\bf{y}} \in {\bf{I*}}\) can be distinguished from one another with respect to L.

A string z is said to distinguish x and y with respect to L if \({\bf{xz}} \in {\bf{L}}\) but \({\bf{yz}} \notin {\bf{L}}\), or \({\bf{xz}} \notin {\bf{L}}\), but \({\bf{yz}} \in {\bf{L}}\).

Concept of indistinguishable with respect to L:

If\(\frac{{\bf{L}}}{{\bf{x}}}{\bf{ = }}\frac{{\bf{L}}}{{\bf{y}}}\), then x and y are said to be indistinguishable from L.

02

Step 2: Proof of the given statement

Given that,let \({{\bf{L}}_{\bf{n}}}\) be the set of strings with at least n bits in which the nth symbol from the end is a 0.

Referring to Exercise 29: every deterministic finite-state automaton recognizing L has at least n states.

To prove: a deterministic finite-state machine recognizing \({{\bf{L}}_{\bf{n}}}\) must have at least \({2^{\bf{n}}}\) states.

Proof:

Let \({\bf{M = }}\left( {{\bf{S,I,f,}}{{\bf{s}}_{\bf{0}}}{\bf{,F}}} \right)\) be a deterministic finite-state machine that recognizes \({{\bf{L}}_{\bf{n}}}\).

Let us assume that the \({2^{\bf{n}}}\) bit strings of length n.

Let x and y be two separate bit strings with lengths n and \({\bf{1}} \le {\bf{i}} \le {\bf{n}}\) as the difference in the bit i between the two strings. As a result, the \({\bf{i}}\)th bit in one of x and y is 0 and the \({\bf{i}}\)th bit in the other is 1.

Let z be an \({\bf{i - 1}}\)-length string.

Since this is likewise the ith bit from the beginning, \({\bf{xz}}\) and \({\bf{yz}}\)differ at the nth position from the end, and one of them ends with a 0 and the other with a 1.

One of \({\bf{xz}}\) and \({\bf{yz}}\) is in \({{\bf{L}}_{\bf{n}}}\) whereas the other is not because \({\bf{xz}}\)and \({\bf{yz}}\)differ in the nth position from the end.This means that \(\frac{{{{\bf{L}}_{\bf{n}}}}}{{\bf{x}}} \ne \frac{{{{\bf{L}}_{\bf{n}}}}}{{\bf{y}}}\)and that x and y can be separated from one another.

But then we have \({{\bf{2}}^{\bf{n}}}\) bit strings, each of length n, that can be distinguished from the others. It can conclude from the previous exercise that \({{\bf{L}}_{\bf{n}}}\)has at least \({{\bf{2}}^{\bf{n}}}\) states.

Hence, the statement is proved as true.

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Most popular questions from this chapter

Describe how productions for a grammar in extended Backus–Naur form can be translated into a set of productions for the grammar in Backus–Naur form.

This is the Backus–Naur form that describes the syntax of expressions in postfix (or reverse Polish) notation.

\(\begin{array}{c}\left\langle {{\bf{expression}}} \right\rangle {\bf{ :: = }}\left\langle {{\bf{term}}} \right\rangle {\bf{|}}\left\langle {{\bf{term}}} \right\rangle \left\langle {{\bf{term}}} \right\rangle \left\langle {{\bf{addOperator}}} \right\rangle \\{\bf{ }}\left\langle {{\bf{addOperator}}} \right\rangle {\bf{:: = + | - }}\\\left\langle {{\bf{term}}} \right\rangle {\bf{:: = }}\left\langle {{\bf{factor}}} \right\rangle {\bf{|}}\left\langle {{\bf{factor}}} \right\rangle \left\langle {{\bf{factor}}} \right\rangle \left\langle {{\bf{mulOperator}}} \right\rangle {\bf{ }}\\\left\langle {{\bf{mulOperator}}} \right\rangle {\bf{:: = *|/}}\\\left\langle {{\bf{factor}}} \right\rangle {\bf{:: = }}\left\langle {{\bf{identifier}}} \right\rangle {\bf{|}}\left\langle {{\bf{expression }}} \right\rangle \\\left\langle {{\bf{identifier}}} \right\rangle {\bf{:: = a }}\left| {{\bf{ b }}} \right|...{\bf{| z}}\end{array}\)

In Exercises 43–49 find the language recognized by the given nondeterministic finite-state automaton.

Construct phrase-structure grammars to generate each of these sets.

a) \(\left\{ {\left. {{\bf{0}}{{\bf{1}}^{{\bf{2n}}}}} \right|{\bf{n}} \ge {\bf{0}}} \right\}\)

b) \(\left\{ {\left. {{{\bf{0}}^{\bf{n}}}{{\bf{1}}^{{\bf{2n}}}}} \right|{\bf{n}} \ge {\bf{0}}} \right\}\)

c) \(\left\{ {\left. {{{\bf{0}}^{\bf{n}}}{{\bf{1}}^{\bf{m}}}{{\bf{0}}^{\bf{n}}}} \right|{\bf{m}} \ge {\bf{0}}\,{\bf{and}}\,{\bf{n}} \ge {\bf{0}}} \right\}\)

Define a Turing machine.

Find a phrase-structure grammar for each of these languages.

a) the set of all bit strings containing an even number of 0s and no 1s

b) the set of all bit strings made up of a 1 followed by an odd number of 0s

c) the set of all bit strings containing an even number of 0s and an even number of 1s

d) the set of all strings containing 10 or more 0s and no 1s

e) the set of all strings containing more 0s than 1s

f) the set of all strings containing an equal number of 0s and 1s

g) the set of all strings containing an unequal number of 0s and 1s

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