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Introduction to automata theory languages and computation solutions for chapter 2

Introduction to Automata Theory, Languages, and Computation Solutions for Chapter 2 Solutions for Chapter 3 Solutions for Chapter 4 Solutions for Chapter 5 Solutions for Chapter 6 Solutions for Chapter 7 Solutions for Chapter 8 Solutions for Chapter 9 Solutions for Chapter 10 Solutions for Chapter 11 Introduction to
Automata Theory, Languages, and Computation Solutions for Section 2.2 Solutions for Section 2.3 Solutions for Section 2.4 Solutions for Section 2.5 Revised 9/6/01. Solutions for Section 2.2 Exercise 2.2.1(a) States correspond to the eight combinations of switch positions, and also must indicate whether the previous roll
came out at D, i.e., whether the previous input was accepted. Let 0 represent a position to the left (as in the diagram) and 1 a position to the right. Each state can be represented by a sequence of three 0's or 1's, representing the directions of the three switches, in order from left to right. We follow these three bits by
either a indicating it is an accepting state or r, indicating rejection. Of the 16 possible states, it turns out that only 13 are accessible from the initial state, 000r. Here is the transition table: AB ->000r100r011r *000a100r011r *001a101r000a 010r110r001a *010a110r001a 011r111r010a 100r010r111r *100a010r111r
101r011r100a *101a011r100a 110r000a101a *110a000a101a 111r001a110a Exercise 2.2.2 The statement to be proved is δ-hat(q,xy) = δ-hat(δ-hat(q,x),y), and we proceed by induction on the length of y. Basis: If y = ε, then the statement is δ-hat(q,x) = δ-hat(δ-hat(q,x),ε). This statement follows from the basis in the
definition of δ-hat. Note that in applying this definition, we must treat δ-hat(q,x) as if it were just a state, say p. Then, the statement to be proved is p = δ-hat(p,ε), which is easy to recognize as the basis in the definition of δ-hat. Induction: Assume the statement for strings shorter than y, and break y = za, where a is the last
symbol of y. The steps converting δ-hat(δ-hat(q,x),y) to δ-hat(q,xy) are summarized in the following table: ExpressionReason δ-hat(δ-hat(q,x),y)Start δ-hat(δ-hat(q,x),za)y=za by assumption δ(δ-hat(δ-hat(q,x),z),a)Definition of δ-hat, treating δ-hat(q,x) as a state δ(δ-hat(q,xz),a)Inductive hypothesis δ-hat(q,xza)Definition of
δ-hat δ-hat(q,xy)y=za Exercise 2.2.4(a) The intuitive meanings of states A, B, and C are that the string seen so far ends in 0, 1, or at least 2 zeros. Exercise 2.2.6(a) The trick is to realize that reading another bit either multiplies the number seen so far by 2 (if it is a 0), or multiplies by 2 and then adds 1 (if it is a 1). We
don't need to remember the entire number seen --- just its remainder when divided by 5. That is, if we have any number of the form 5a+b, where b is the remainder, between 0 and 4, then 2(5a+b) = 10a+2b. Since 10a is surely divisible by 5, the remainder of 10a+2b is the same as the remainder of 2b when divided by 5.
Since b, is 0, 1, 2, 3, or 4, we can tabulate the answers easily. The same idea holds if we want to consider what happens to 5a+b if we multiply by 2 and add 1. The table below shows this automaton. State qi means that the input seen so far has remainder i when divided by 5. 01 ->*q0q0q1 q1q2q3 q2q4q0 q3q1q2
q4q3q4 There is a small matter, however, that this automaton accepts strings with leading 0's. Since the problem calls for accepting only those strings that begin with 1, we need an additional state s, the start state, and an additional ``dead state'' d. If, in state s, we see a 1 first, we act like q0; i.e., we go to state q1.
However, if the first input is 0, we should never accept, so we go to state d, which we never leave. The complete automaton is: 01 ->sdq1 *q0q0q1 q1q2q3 q2q4q0 q3q1q2 q4q3q4 ddd Exercise 2.2.9 Part (a) is an easy induction on the length of w, starting at length 1. Basis: |w| = 1. Then δ-hat(q0,w) = δ-hat(qf,w),
because w is a single symbol, and δ-hat agrees with δ on single symbols. Induction: Let w = za, so the inductive hypothesis applies to z. Then δ-hat(q0,w) = δ-hat(q0,za) = δ(δ-hat(q0,z),a) = δ(δ-hat(qf,z),a) [by the inductive hypothesis] = δ-hat(qf,za) = δ-hat(qf,w). For part (b), we know that δ-hat(q0,x) = qf. Since xε, we
know by part (a) that δ-hat(qf,x) = qf. It is then a simple induction on k to show that δ-hat(q0,xk) = qf. Basis: For k=1 the statement is given. Induction: Assume the statement for k-1; i.e., δ-hat(q0,xSUP>k-1) = qf. Using Exercise 2.2.2, δ-hat(q0,xk) = δ-hat(δ-hat(q0,xk-1),x) = δ-hat(qf,x) [by the inductive hypothesis] = qf [by
(a)]. Exercise 2.2.10 The automaton tells whether the number of 1's seen is even (state A) or odd (state B), accepting in the latter case. It is an easy induction on |w| to show that dh(A,w) = A if and only if w has an even number of 1's. Basis: |w| = 0. Then w, the empty string surely has an even number of 1's, namely zero
1's, and δ-hat(A,w) = A. Induction: Assume the statement for strings shorter than w. Then w = za, where a is either 0 or 1. Case 1: a = 0. If w has an even number of 1's, so does z. By the inductive hypothesis, δ-hat(A,z) = A. The transitions of the DFA tell us δ-hat(A,w) = A. If w has an odd number of 1's, then so does z.
By the inductive hypothesis, δ-hat(A,z) = B, and the transitions of the DFA tell us δ-hat(A,w) = B. Thus, in this case, δ-hat(A,w) = A if and only if w has an even number of 1's. Case 2: a = 1. If w has an even number of 1's, then z has an odd number of 1's. By the inductive hypothesis, δ-hat(A,z) = B. The transitions of the
DFA tell us δ-hat(A,w) = A. If w has an odd number of 1's, then z has an even number of 1's. By the inductive hypothesis, δ-hat(A,z) = A, and the transitions of the DFA tell us δ-hat(A,w) = B. Thus, in this case as well, δ-hat(A,w) = A if and only if w has an even number of 1's. Return to Top Solutions for Section 2.3
Exercise 2.3.1 Here are the sets of NFA states represented by each of the DFA states A through H: A = {p}; B = {p,q}; C = {p,r}; D = {p,q,r}; E = {p,q,s}; F = {p,q,r,s}; G = {p,r,s}; H = {p,s}. 01 ->ABA BDC CEA DFC *EFG *FFG *GEH *HEH Exercise 2.3.4(a) The idea is to use a state qi, for i = 0,1,...,9 to represent the idea
that we have seen an input i and guessed that this is the repeated digit at the end. We also have state qs, the initial state, and qf, the final state. We stay in state qs all the time; it represents no guess having been made. The transition table: 01...9 ->qs{qs,q0}{qs,q1}...{qs,q9} q0{qf}{q0}...{q0} q1{q1}{qf}...{q1} ...............
q9{q9}{q9}...{qf} *qf{}{}...{} Return to Top Solutions for Section 2.4 Exercise 2.4.1(a) We'll use q0 as the start state. q1, q2, and q3 will recognize abc; q4, q5, and q6 will recognize abd, and q7 through q10 will recognize aacd. The transition table is: abcd ->q0{q0,q1,q4,q7}{q0}{q0}{q0} q1{}{q2}{}{} q2{}{}{q3}{} *q3{}{}{}{} q4{}
{q5}{}{} q5{}{}{}{q6} *q6{}{}{}{} q7{q8}{}{}{} q8{}{}{q9}{} q9{}{}{}{q10} *q10{}{}{}{} Exercise 2.4.2(a) The subset construction gives us the following states, each representing the subset of the NFA states indicated: A = {q0}; B = {q0,q1,q4,q7}; C = {q0,q1,q4,q7,q8}; D = {q0,q2,q5}; E = {q0,q9}; F = {q0,q3}; G = {q0,q6}; H =
{q0,q10}. Note that F, G and H can be combined into one accepting state, or we can use these three state to signal the recognition of abc, abd, and aacd, respectively. abcd ->ABAAA BCDAA CCDEA DBAFG EBAAH *FBAAA *GBAAA *HBAAA Return to Top Solutions for Section 2.5 Exercise 2.5.1 For part (a): the
closure of p is just {p}; for q it is {p,q}, and for r it is {p,q,r}. For (b), begin by noticing that a always leaves the state unchanged. Thus, we can think of the effect of strings of b's and c's only. To begin, notice that the only ways to get from p to r for the first time, using only b, c, and ε-transitions are bb, bc, and c. After getting
to r, we can return to r reading either b or c. Thus, every string of length 3 or less, consisting of b's and c's only, is accepted, with the exception of the string b. However, we have to allow a's as well. When we try to insert a's in these strings, yet keeping the length to 3 or less, we find that every string of a's b's, and c's
with at most one a is accepted. Also, the strings consisting of one c and up to 2 a's are accepted; other strings are rejected. There are three DFA states accessible from the initial state, which is the ε closure of p, or {p}. Let A = {p}, B = {p,q}, and C = {p,q,r}. Then the transition table is: Return to Top Engineering
Introduction to Automata Theory, Languages, and Computation 3rd Edition | ISBN: 9780321455369 / 0321455363 Buy on Amazon.com 3rd Edition | ISBN: 9780321455369 / 0321455363 0 expert-verified solutions in this book Buy on Amazon.com Automata: The Methods And The Madness Finite Automata 2-2
Deterministic Finite Automata Exercises p.52 2-3 Nondeterministic Finite Automata Exercises p.65 2-4 An Application: Text Search Exercises p.71 2-5 Finite Automata With Epsilon-Transitions Exercises p.79 Problems p.81 Regular Expressions And Languages 3-1 Regular Expressions Exercises p.91 3-2 Finite
Automata and Regular Expressions Exercises p.107 3-3 Applications of Regular Expressions Exercises p.114 3-4 Algebraic Laws for Regular Expressions Exercises p.121 Problems p.124 Properties Of Regular Languages 4-1 Proving Languages Not to Be Regular Exercises p.131 4-2 Closure Properties of Regular
Languages Exercises p.147 4-3 Decision Properties of Regular Languages Exercises p.155 4-4 Equivalence and Minimization of Automata Exercises p.165 Problems p.167 Context-Free Grammars And Languages 5-1 Context-Free Grammars Exercises p.181 5-2 Parse Trees Exercises p.193 5-3 Applications of
Context-Free Grammars Exercises p.206 5-4 Ambiguity in Grammars and Languages Exercises p.215 Problems p.218 Pushdown Automata 6-1 Definition of the Pushdown Automaton Exercises p.233 6-2 The Languages of a PDA Exercises p.241 6-3 Equivalence of PDA's and CFG's Exercises p.251 6-4 Determinate
Pushdown Automata Exercises p.256 Problems p.258 Properties Of Context-Free Languages 7-1 Normal Forms for Context-Free Grammars Exercises p.275 7-2 The Pumping Lemma for Context-Free Languages Exercises p.286 7-3 Closure Properties of Context-Free Languages Exercises p.297 7-4 Decision
Properties of CFL's Exercises p.307 Problems p.309 Introduction To Turning Machines 8-1 Problems that Computer Cannot Solve Exercises p.324 8-2 The Turning Machine Exercises p.335 8-3 Programming Techniques for Turning Machines Exercises p.343 8-4 Extensions to the Basic Turning Machine Exercises
p.349 8-5 Restricted Turning Machines Exercises p.361 Problems p.372 Undecidability 9-1 A Language That is Not Recursively Enumerable Exercises p.382 9-2 An Undecidable Problem that Is RE Exercises p.390 9-3 Undecidable Problems About Turning Machines Exercises p.400 9-4 Post's Correspondence Problem
Exercises p.412 9-5 Other Undecidable Problems Exercises p.418 Problems p.420 Intractable Problems 10-1 The Classes P and NP Exercises p.435 10-2 An NP-Complete Problem Exercises p.447 10-3 A Restricted Satisfiability Problem Exercises p.458 10-4 Additional NP-Complete Problems Exercises p.473
Problems p.478 Additional Classes Of Problems 11-1 Complements of Languages in NP Exercises p.486 11-3 A Problem That is Complete for PS Exercises p.501 11-5 The Complexity of Primality Testing Exercises p.521 Problems p.523 YES! Now is the time to redefine your true self using Slader’s Introduction to
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