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Use the procedure described in Lemma 1.60to convert the following finite automata to regular expressions.

Short Answer

Expert verified

Every nondeterministic finite automaton has an equivalent deterministic finite automation

Step by step solution

01

Introduction

Consider Finite Automata:

Now, inside the methods following, transform such a finite automaton to something like a regular expression:

To turn the initial allow state a non-accepting state, add the start state (S) and new accept state (F) as follows:

02

Explanation for (A)

Mostly in second stage, remove state(1); there is no need to construct a loop for state 1; instead, add a loop to state (2) directly, as well as create the expressions by transferring state (S) to state (2).

Every circuit is defined as a union step:

So, for the above finite automata, the notation isa*b(a∪ba*b)*

03

Explanation for (B)

Allow the 2nd finite automata is

Therefore, in the following stages, transform this finite automaton to a regular expression:

Add a new start state (S) and a new accepted state (A) (F). If you change the admit states to non-accepting states, the Finite Automata becomes:

Perform union on the edge from state 1 to 2 state

From the above step 2 , there are no unions or loops for the state 1 , So eliminate the state 1 as follows:

Perform unions on edges from state 3 to state 2 and from state 3 to the final state, Then the Automata becomes as below:

To minimise the automata, remove 2 then conduct a federation on 3, then transfer the state (S) statement to state(3), then apply a loop on state (3) with the state (S) expression (2

Eliminate state 3 and write the expression from state(S) to state (F), because there are no loops and unions.

Perform union on the edge from state S to state F.

So, the regular expression for the given finite automaton is in the diagram.

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Most popular questions from this chapter

Consider the problem of determining whether a Turing machine M on an input w ever attempts to move its head left when its head is on the left-most tape cell. Formulate this problem as a language and show that it is undecidable.

A queue automaton is like a push-down automaton except that the stack is replaced by a queue. A queue is a tape allowing symbols to be written only on the left-hand end and read only at the right-hand end. Each write operation (we’ll call it a push) adds a symbol to the left-hand end of the queue and each read operation (we’ll call it a pull) reads and removes a symbol at the right-hand end. As with a PDA, the input is placed on a separate read-only input tape, and the head on the input tape can move only from left to right. The input tape contains a cell with a blank symbol following the input, so that the end of the input can be detected. A queue automaton accepts its input by entering a special accept state at any time. Show that a language can be recognized by a deterministic queue automaton iff the language is Turing-recognizable.

Show that A is decidable iff A≤m0*1*.

Question: A useless state in a Turing machine is one that is never entered on any input string. Consider the problem of determining whether a Turing machine has any useless states. Formulate this problem as a language and show that it is undecidable.

LetAbe the set{x,y,z}andBbe the set{x,y}.

  1. IsAa subset ofB?
  2. IsBa subset ofA?
  3. What isA∪B?
  4. What isA∩B?
  5. What isA×B?
  6. What is the power set ofB ?
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