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The construction in Theorem 1.54 shows that every GNFA is equivalent to a GNFA with only two states. We can show that an opposite phenomenon occurs for DFAs. Prove that for every k>1, a language xAk{0,1}exists that is recognized by a DFA with k states but not by one with onlyk-1 states

Short Answer

Expert verified

It can be said that no DFA’s is equivalent to a DFA with lesser states.

Step by step solution

01

Introduction

The GNFA Convert G is equivalent to G. By Induction on k, the number of states of G.

02

To Prove That no DFA are equivalent to a DFA

It is being given that if there are two states GNFA then is correspondingly equal to the GNFA. But here user need to prove that in DFA subsequently opposite phenomenon occurs.

It implies user need to prove that no DFA are equivalent to a DFA with lesser states.

Assume Ak be the set of words of length at least k-1. Therefore, it can be said that role="math" localid="1663228503830" Akhas at least equivalence classes of words length0,1,2,,k-2andk-1 or more. So, it is clear from this that Akrequires a DFA with states k.

For any DFA fewer than states k, by Pigeon Hole Principle, two of thek strings cause the machine to loop in same state results in a rejection from the DFA.

03

To Consider Ak={0,1}*0k-10*

Consider Ak={0,1}*0k-10*for k>1.

Now a DFA with exactly k states can recognize the language Ak. Starting from the start state there is a state in the DFA for each 0 it had read after the last 1.

After k-10'sit arrives at an accepting state whose further transitions are self-loops. Now based on the language let say Akbe the set consisting of the strings 10,100,,10k-1. If the DFA consist of fewer thank states, then by the Pigeonhole Principle two these strings cause a loop to a single state. Hence the machine fails to accept the strings.

Hence, it can be said that no DFA’s is equivalent to a DFA with lesser number of states.

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Most popular questions from this chapter

A finite state transducer (FST) is a type of deterministic finite automaton whose output is a string and not just accept or reject. The following are state diagrams of finite state transducers T1andT2.

Each transition of an FST is labeled with two symbols, one designating the input symbol for that transition and the other designating the output symbol. The two symbols are written with a slash, I, separating them. In T1, the transition from q1toq2has input symbol 2 and output symbol 1. Some transitions may have multiple input–output pairs, such as the transition in T1from q1to itself. When an FST computes on an input string w, it takes the input symbols w1···wnone by one and, starting at the start state, follows the transitions by matching the input labels with the sequence of symbols w1···wn=w. Every time it goes along a transition, it outputs the corresponding output symbol. For example, on input 2212011, machine T1enters the sequence of states q1,q2,q2,q2,q2,q1,q1,q1and produces output 1111000. On input abbb, T2outputs 1011. Give the sequence of states entered and the output produced in each of the following parts.

a. T1on input011

b. T1on input211

c. T1on input121

d. T1on input0202

e. T2on input b

f. T2on input bbab

g. T2on input bbbbbb

h. T2on input localid="1663158267545" ε

If A is a set of natural numbers and k is a natural number greater than 1, let

Bk(A)={w|wistherepresentationinbasekofsomenumberinA}.

Here, we do not allow leading 0s in the representation of a number. For example ,B2({3,5})={11,101}and B3({3,5})={10,12}.Give an example of a set A for which B2(A)is regular butB2(A) is not regular. Prove that your example works.

Let ={0,1,#}. Let C={x#xR#x|x{0,1}*} . Show that C is a CFL.

A homomorphism is a function f:Σ-Γ*from one alphabet to strings over another alphabet. We can extend f to operate on strings by defining:f(w)=f(w1)f(w2)···f(wn),wherew=w1w2···wnandeachwiΣ.

We further extend fto operate on languages by defining f(A)={f(w)|wA},for any language A.

a. Show, by giving a formal construction, that the class of regular languages is closed under homomorphism. In other words, given a DFA Mthat recognizes Band a homomorphism f, construct a finite automaton role="math" localid="1660800566802" M0that recognizes f(B).Consider the machine role="math" localid="1660800575641" M0that you constructed. Is it a DFA in every case?

b. Show, by giving an example, that the class of non-regular languages is not closed under homomorphism.

Recall that string x is a prefix of string y if a string z exists where xz=y, and that x is a proper prefix of y if in addition x6=y. In each of the following parts, we define an operation on a language A. Show that the class of regular languages is closed under that operation.

a)NOPREFIXA={wA|noproperprefixofwisamemberofA}.b)NOEXTENDA={wA|wisnottheproperprefixofanystringinA}.

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