Chapter: Problem: FS show all steps. This does not work for DFAs. © Copyright 2016. Get solutions . (6 states), (1.5b) All strings that contain the substring 0101. Fix a machine M that generates L and pick a state r in that machine. But when we mimize the DFA, all the dead states will become equivalent, and therefore all the This is because minimization Thousands of theory of computation guided textbook solutions, and expert theory of computation answers when you need them. The reason this CS 332: Elements of the Theory of Computation, Spring 2020 Course Overview This course is an introduction to the theory of computation. A computational problem is a task solved by a computer. Theory of Computation - CSE 105 Context-free Languages Sample Problems and Solutions Designing CFLs Problem 1 Give a context-free grammar that generates the following language over {0,1}∗: L = {w|w contains more 1s than 0s} Idea: this is similar to the language where the number of 0s is equal to the number of 1s, except we must We can construct a DFA to decide Prefix(L) by taking the DFA for L and marking all states from which an accept state is reachable as accept states. The empty set. (Exercise 1.13) Give regular expressions for all four languages in Exercise 1.4. (1.4i) All strings where every odd position is a 1. from the final state and collapsing it together with the initial state (while keeping it a final state). The problem Half(L,r)is then: Prove that Regular Sets are NOT closed under infinite union. For the inductive step, suppose that all strings in L1 of length <= n are in L2. Ikuti. solutions introduction to automata theory, languages, and computation collected prepared by rontdu@gmail.com 13th batch (06-07) dept. Conversely, if L is generated by a DFA M with one final state, then L = Min(L) ( Min(L') )*, into two other simple problems: If we make the machine M' by making all accept states in M be reject states, and by making state r an accept state, does M' accept the string w? arbitrary number of terms in r. (r + s)* and r*s* are not equivalent because if s. Every NFA can be converted into an equivalent NFA with only a single accept state by creating a new accept state with epsilon moves from each of the old accept states. A host of undecidable problems: consequences of Rice's Theorem and undecidability of … From these to lemmas it is clear that RS* can be generated by a machine with one final state The NFA below determines if a string of columns composes a legal addition equation where the top two rows sum to the third. of the same length as w such that wx is in the language L and Introduction : Introduction of Theory of Computation. Since u has an equal number of 0s and 1s, and v is in L1, this must maintain the prefix property. See an explanation and solution for Chapter 7, Problem 7.9 in Sipser’s Introduction to the Theory of Computation (3rd Edition). Unlike static PDF Introduction to the Theory of Computation 2nd Edition solution manuals or printed answer keys, our experts show you how to solve each problem step-by-step. Theory of computation | Decidable and undecidable problems Prerequisite – Turing Machine A problem is said to be Decidable if we can always construct a corresponding algorithm that can answer the problem correctly. For each of the following statements, answer True, False or Open question according to our current state of knowledge of complexity theory, as described in class. second describes a string from r followed by a string from s or a string from and changing all 0 transitions to 0,1 transitions (1.41) Let D = {w | w contains an equal number of occurrences of 01 and 10}. All strings containing exactly 4 0s and at least 2 1s. In each case below, say what language (a subset of {a, b}*) is generated by the ... Chapter 4 Solutions | Introduction To Languages And The Page 4/5 So, Prefix(L) must be regular. i think the answer of Question no. Solutions to Selected Exercises Solutions for Chapter 2. Putting all this together All strings containing exactly 4 0s or an even number of 1s. (4 states), All strings such that some two zeros are separated by a string whose length is 4i for some i>=0. final states will become equivalent too. All strings ending in 1101. The proof is by induction on the length of strings in L1: The base case is the empty string. Millions of developers and companies build, ship, and maintain their software on GitHub — the largest and most advanced development platform in the world. u. by a machine with one final state. Give brief reasons for your answers. So the infinite union cannot be closed for regular languages. No need to wait for office hours or assignments to be graded to find out where you took a wrong turn. Definitions, theorems, proofs (Michael Sipser, Introduction to the Theory of Computation, 2nd edition, Introduction to the Theory of Computation, 2nd edition, pp. This is in L2 by definition. The prefix condition is slightly more difficult. Therefore infinite intersection does not preserve regularity. We also need the following lemma: The Kleene star, M*, of prefix free regular language M can be generated We can intuitively understand Decidable problems by considering a simple example. (note: the rightmost state in the second diagram corresponds to the bottom right state in the third diagram.). Again, since u is in L1, this must be in L1. All strings containing exactly 4 0s and at least 2 1s. states. All strings that contain exactly 4 0s. if R and S are prefix free, because we can just concatenate the machines for R and S*. i think there is a mistake in question29.instead is S it should be either 0 or 1 according to the given diagram. If we make the machine M'' by making state r the start state, We know that such that wx is in the language L. This is hard to solve directly, as strings accepted by a given machine. Then w = 0u1 for some string u, and u has the same number of We just reverse the procedure for converting an NFA to a regular expression by ripping-in of computer science zeros and ones, since w does. (A counterexample suffices). {0i1i | i>=0} = {0} U {01} U {0011} U ..., ANSWER: Deterministic Push Down Automata (DPDA) and Non-deterministic Push Down Automata (NPDA), ANSWER: X1 – X3 is recursively enumerable, ANSWER: It is neither regular nor context free, but accepted by a turing machine, ANSWER: Every finite subset of a non-regular set is regular, ANSWER: All strings containing at least two 0’s, ANSWER: NP-complete and in P respectively, ANSWER: The union of two context free languages is context free, ANSWER: L = {s ∈ (0+1)* I no(s)-n1(s) I ≤ 4, ANSWER: If W is the string of a terminals and Y is a non-terminal, the language generated by a context free grammar, all of whose productions are of the form x->W or X->WY is always regular, ANSWER: P3 is undecidable if P2 is reducible to P3, ANSWER: L must be either {an I n is odd} or {an I n is even}, ANSWER: X is undecidable but partially decidable, ANSWER: It outputs the sum of the present and the previous bits of the input, ANSWER: 1, 2, 4, 8……2n ….. written in binary, ANSWER: It is a context sensitive language, ANSWER: These are closed under union, Kleene closure, ANSWER: Turing recognizable languages are closed under union and complementation. The DFAs of problems 1g, 1h, and 1i are all good counterexamples. theory of computation and then alternate the algorithms so that we can obtain a more reliable solution. - Theory of computation goes back as far as the 1930s. Suppose we have DFA representation of M that has multiple final states. In general if the minimum DFA for a regular language has more than one final state, then the language Introduction-to-the-Theory-of-Computation-Solutions ===== If you want to contribute to this repository, feel free to create a pull request (please copy the format as in the other exercises). Computational complexity theory focuses on classifying computational problems according to their resource usage, and relating these classes to each other. one final state. Solutions for Chapter 3 Assuming that u and v are both in L1, simply concatenating them together will maintain the equal number of 0s and 1s. (6 states), Prove that every string in L2 is contained in L1. Prove that if L is regular then Prefix(L) is regular. cannot be generated by a DFA with one final state. From the previous lemma we know there is a DFA that generates M that has Introduction to the Theory of Computation Homework #2 Solutions (1. and 2. omitted) 3. machine M'' accept the string w? Chomsky Hierarchy. Solutions for Chapter 4. But the infinite union is the set {0i1i | i>=0} which we know is not regular. We can analyze L2 inductively to see that it maintains the property of L1 for each case: L1-L2 is the same as the intersection of L1 and the complement of L2. (1.4c) All strings that contain the substring 0101. What we have done in the second case is to ingnore what the Also, let me know if there are any errors in the existing solutions. It has an errata web site . zeros, since then 0x would either have more ones than zeros which is impossible by hypothesis, or 0x would have the same number of ones as zeros, which is also (1.4e) All strings that start with 0 and has odd length or start with 1 and has even length. This is the branch of computer science that aims to understand which problems can be solved using computational devices and how efficiently those problems can be solved. is good is that the problem Half(L,r) decomposes naturally Then all outgoing transitions from those final states must go to dead states since M is prefix free. here, with possibly some missing extraneous states. You are about to embark on the study of a fascinating and important subject: the theory of computation. Prove that if L1 is regular and L2 is regular then so is L1-L2 (the set of all strings in L1 but not in L2). The field is divided into three major branches: automata theory and languages, computability theory, and computational complexity theory. (1.4f) All strings that don't contain the substring 110. For those of you who are paying attention, this problem is extemely similar to the stream-crossing ghostbusters problem from algorithms. Since the Min of a language is always prefix free, L is of the form we claim. L1: The set of strings where each string w has an equal number of zeros and ones; and any prefix of w has at least as many zeros as ones. 17-22) Problems: Begin: Set theory problems (pdf, doc) & solutions (pdf, doc) DFA problems Proofs problems (pdf, doc) [Back to … Introduction-to-the-Theory-of-Computation-Solutions - GitHub Download Sipser Theory Of Computation 3rd Edition Solutions book pdf free download link or read ... View an educator-verified, detailed solution for Chapter 5, Problem 5.12 in Sipser’s Introduction to the Theory of Computation (3rd Edition). 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