File Name: automata theory languages and computation solutions .zip
Show that a given language is or is not regular. Given a regular expression, construct an automaton that recognizes it.
It has been more than 20 years since this classic book on formal languages, automata theory, and computational complexity was first published. With this long-awaited revision, the authors continue to present the theory in a concise and straightforward manner, now with an Automata theory languages and computation solutions, as one of the most vigorous sellers here will very be along with the best options to review. Hopcroft It has been more than Introduction to automata theory, languages and computation J. Hopcroft , R.
It is the ability to do the latter that exhibits proficiency with the material. These solutions have been provided as a tool to assist students in developing their problem solving skills. The exercises in this manual were selected to complement the presentation in the text. The solutions illustrate general techniques for solving the particular type of problem or extend the the- oretical development of a topic.
Understanding the strategies employed in these solutions should provide valuable insights that can be used in solving many of the other exercises in the text.
No part of this is to be reproduced, stored in a retrieval system, or transmitted in any form without the prior written permission of the author or the Addison-Wesley Publishing Co. However, it is not onto since the range is the set of even numbers. It is not a total function from N to N, since it is not defined for odd natural numbers. This is shown using the same argument given in Example 1.
The latter set is known to be countable by Theorem 1. Diagonalization is used to prove that there are an uncountable number of monotone increasing functions. Assume that the set of monotone increasing functions is countable.
Then these functions can be listed in a sequence f0 , f1 , f2 ,. Now we will show that there is at least one in fact there are infinitely many monotone increasing function that is not in the listing. Thus the assumption that the monotone increasing functions comprise a countable set leads to a contradiction and we conclude that the set is uncountable.
To show that the set of real numbers in the interval [0, a] is uncountable, we first observe that every real number in 0, 1] can be expressed by an infinite decimal.
With such a representation, the number 12 is represented by both. Assume the set of real numbers in 0, 1] is countable. This implies that there is a sequence r0 , r 1 , r 2 ,. Let the decimal expansion of r n be denoted. The enumeration given above is used to construct an infinite two-dimensional array, the ith row of which consists of the expansion of ri.
With the unique decimal representation, two numbers are distinct if they differ at any position in the decimal expansion. Therefore the assumption that the enumeration contains all real numbers in 0, 1] fails, and we conclude that the set is uncountable. The function h demonstrates that A and B have the same cardinality. The function f is defined for all elements in A which includes all elements of B and the values of f , indicated by the heads of the arrows, must all be in B.
There are four cases to consider. Thus h x cannot equal h y in this case. Same argument as case 2. Let f i denote the composition of f with itself i times, and f 0 denote the identity function.
The proof uses the fact that the composition of one-to-one functions is one- to-one. Although you will be familiar with functional composition from previous mathematical studies, a description and formal definition are given in Section 9. We now show that h maps A onto B. The product of two natural numbers can be defined recursively using addition and the successor operator s. Closure: A set X is in F only if it can be obtained from the basis elements by a finite number of applications of the recursive step.
The first rule in the recursive step generates all sets containing a single natural number. The second rule combines previously generated sets to obtain sets of larger cardinality. We explicitly show that the equality holds for this case.
The set R of nodes reachable from a given node x in a directed graph is defined recursively using the adjacency relation A. Recall that by our definition is node is an ancestor of itself, which is certainly not the case in family trees. Let nodes T and leaves T denote the number of nodes and leaves in a tree T.
Basis: The basis consists of trees of depth zero; that is, trees consisting solely of the root. T is obtained by adding two children to each leaf of a complete binary tree T0 of depth n.
Basis: The basis consists of the null string. Note that every string in L has length divisible by three. Thus, one additional application of the recursive step produces u from xyzw. The proof is by induction on the number of applications of the recursive step in the definition of palindrome required to generate the string.
Basis: The basis consists of strings of P that are generated with no applications of the recursive step. Inductive hypothesis: Assume that every string generated by n or fewer applications of the recursive step is in W. The proof is by induction on the length of the strings in W. Similarly, strings of length one in W are also in P. L3 , obtained by applying the Kleene star operation to L2 , contains all strings with length divisible by four.
The null string, with length zero, is in L3. Since the null string is not allowed in the language, each string must contain at least one a or one b or one c. However, it need not contain one of every symbol. Clearly every string in this language has substrings ab and ba.
Every string described by the preceding expression has length four or more. Have we missed some strings with the desired property? Consider aba, bab, and aaaba. However, every string in this expression ends with b while strings in the language may end in a or aa as well as b.
This expression is obtained by combining an expression for each of the component subsets with the union operation. Exercises 37 and 38 illustrate the significant difference between union and intersection in describing patterns with regular expressions.
There is nothing intuitive about a regular ex- pression for this language. The exercise is given, along with the hint, to indicate the need for an algorithmic approach for producing regular expressions. A technique accomplish this will follow from the ability to reduce finite state machines to regular expressions developed in Chapter 6. There is no trick to obtaining this answer, you simply must count the possible derivations represented by the tree.
The basis of the recursive definition consists of the null string. The first S rule produces this string. The recursive step consists of inserting an a and b in a previously generated string or by concatenating two previously generated strings. These operations are captured by the second set of S rules. If the derivation begins with an S rule that begins with an a, the A rules ensure that an a occurs in the middle of the string. Similarly, the S and B rules combine to produce strings with a b in the first and middle positions.
While the grammar described above generates the desired language, it is not regular. A regular grammar can be obtained by explicitly replacing S1 and S2 in the S rules with the right-hand sides of the S1 and S2 rules. We will refer to this condition as the prefix property.
The proof is by induction of the length of derivations of G. The prefix property is seen to hold for these strings by inspection. Inductive hypothesis: Assume that every sentential form that can be obtained by a deriva- tion of length n or less satisfies the prefix property. By the inductive hypothesis uSv satisfies the prefix property. Thus w also satisfies the prefix property.
To construct an unambiguous grammar that generates L G , it is necessary to determine precisely which strings are in L G. Each of these strings can be generated by two distinct leftmost derivations. To produce an unambiguous grammar, it is necessary to ensure that these strings are generated by only one of A and B. Consequently, there is only one leftmost derivation of ai bj and the grammar is unambiguous. Thus every rule of G1 is either in G2 or derivable in G2. Now if w is derivable in G1 , it is also derivable in G2 by replacing the application of a rule of G1 by the sequence of rules of G2 that produce the same transformation.
Since every regular grammar is also right-linear, all regular languages are generated by right-linear grammars. We now must show that every language generated by a right-linear grammar is regular.
Let G be a right-linear grammar.
The authors present the theory in a concise and straightforward manner, with an eye out for the practical applications. This book is an introduction to the theory of computation. Rajeev Motwani contributed to the , and later, edition. Solutions for Section 6. Solutions for Section 3. Introduction to Automata Theory, Languages, and. Exercises at the end of each chapter, including some that have been solved, help readers confirm and enhance their understanding of the material.
Hopcroft, Rajeev Motwani, Jeffrey D. Please note that you should not expect these notes to be a complete record of all that is said and discussed during the lectures. Lecture attendance is compulsory, and reading the main references strongly encouraged. That said, the typeset lecture notes are fairly comprehensive. Ullman is the main reference for the course. Note that this book is quite different from the classic first edition see below.
Solutions for Section 2. 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, r. Here is the transition table: A.
Theory and Formal Languages taught at Clarkson University. The course is also Look at the solutions only to check your answer once you think you know how to do an compute the carries as it reads the numbers. So it will do the op-.
It was inevitable that he should come to know of, the problem they faced, but he could be trusted to maintain secrecy. Then he pushed himself back from the reading table at which he had been seated and rubbed his eyes wearily. He rose, stretched, and walked slowly to the curtained windows, peering through the folds into the darkness beyond.
Она выглядела как первокурсница, попавшая под дождь, а он был похож на студента последнего курса, одолжившего ей свою куртку. Впервые за многие годы коммандер почувствовал себя молодым. Его мечта была близка к осуществлению. Однако, сделав еще несколько шагов, Стратмор почувствовалчто смотрит в глаза совершенно незнакомой ему женщины. Ее глаза были холодны как лед, а ее обычная мягкость исчезла без следа. Сьюзан стояла прямо и неподвижно, как статуя. Глаза ее были полны слез.
- Это Мидж. - Королева информации! - приветствовал ее толстяк. Он всегда питал слабость к Мидж Милкен. Умница, да к тому же единственная женщина, не упускавшая случая с ним пококетничать. - Как твои дела.
Чепуха. Ты никогда не смог бы проникнуть в почту коммандера. - Ты ничего не понимаешь! - кричал Хейл.
Сдерживая подступившую к горлу тошноту, Беккер успел заметить, что все пассажиры повернулись и смотрят на. Все как один были панки. И, наверное, у половины из них - красно-бело-синие волосы. - Sientate! - услышал он крик водителя.
Нет, вообще-то я… - Из туристического бюро. - Нет, я… - Слушайте, я знаю, зачем вы пришли! - Старик попытался сесть в кровати. - Меня не удастся запугать. Я уже говорил это и могу повторить тысячу раз - Пьер Клушар описывает мир таким, каким его видит. Некоторые ваши туристические путеводители старательно скрывают правду, обещая бесплатный ночлег в городе, но Монреаль тайме не продается.
Где она изучала математику. Как она попала в АНБ. Как ей удалось стать столь привлекательной.
- Он замолчал, словно подбирая нужные слова.
Your email address will not be published. Required fields are marked *