Theory of computation homework help
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The document discusses proofs related to regular languages, Turing machines, and decidability. 1) It proves that a single-tape TM that cannot write over the input recognizes only regular languages by showing there are a finite number of states the TM can be in after reading the input. 2) It uses diagonalization to show that for any Turing-recognizable language A of TM descriptions, there exists a decidable language not decided by any TM in A. 3) It proves the language of DFAs accepting a string with more 1s than 0s is decidable by reducing the problem to testing if the intersection of a CFL and regular language is empty. 4) It proves a
Theory of computation homework help
- 1. 1. Show that single-tape TMs that cannot write on the portion of the tape containing the input string recognize only regular languages. Answer: Let M = (Q, Σ, Γ, q0, qaccept, qreject) be a single-tape TM that cannot write on the input portion of the tape. A typical case when M works on an input string x is as follows: the tape head will stay in the input portion for some time, and then enter the non-input portion (i.e., the portion of the tape on the right of the |x|th cells) and stay there for some time, then go back to the input portion, and stay there for some time, and then enter the non-input portion, and so on. We call the event that the tape head switches from input portion to non-input portion an out event, and the event that the tape head switches from non-input portion to input-portion an in event. Let firstx denote the state that M is in just after its first “out” event (i.e., the state of M when it first enters the non-input portion). In case M never enters the non-input portion, we assign firstx = qaccept if M accepts x, and assign firstx = qreject if M does not accept x. Next, we define a characteristic function fx such that for any q ∈ Q, fx(q) = q implies that if M is at state q and about to perform an “in” event, the next “out” event will change M in state q ; in case M never enters the non-input portion again, we assign fx(q) = qaccept if M enters the accept state inside the input portion, and qreject otherwise. It is easy to check that if for two strings x and y, if firstx = firsty and for all q, fx(q) = fy(q), we have x and y are indistinguishable by M. (That is, M accepts xz if and only if M accepts yz.) As there are finite choices of firstx and fx (precisely, |Q||Q|+1 such choices), the number of distinguishable strings are finite. By Myhill-Nerode theorem, the language recognized by M is regular. 2. Let A be a Turing-recognizable language consisting of descriptions of Turing machines, { M1 , M2 , . . .}, where every Mi is a decider. Prove that some decidable language D is not decided by any decider Mi whose description appears in A.† (Hint: You may find it helpful to consider an enumerator for A, and re-visit the diagonalization technique.) Answer: Since A is Turing-recognizable, there exists an enumerator E that enumerates it. In particular, we let Mi be the ith output of E (note: Mi may not be distinct). Let s1, s2, s3 . . . be the list of all possible strings in {0, 1}∗ . Now, we define a TM D as follows: D = “On input w: 1. If w /∈ {0, 1}∗ , reject. 2. Else, w is equal to si for a specific i. 3. Use E to enumerate M1 , M2 , . . . until Mi . 4. Run Mi on input w. 5. If Mi accepts, reject. Otherwise, accept.” 1 THEORY OF COMPUTATION Our online Tutors are available 24*7 to provide Help with Theory Of Computation Homework/Assignment or a long term Graduate/Undergraduate Theory Of Computation Project. Our Tutors being experienced and proficient in Theory Of Computation sensure to provide high quality Theory Of Computation Homework Help. Upload your Theory Of Computation Assignment at ‘Submit Your Assignment’ button or email it to [email protected]. You can use our ‘Live Chat’ option to schedule an Online Tutoring session with our Theory Of Computation Tutors.
- 2. Clearly, D is a decider (why??). However, D is different from any Mi (why??), so that D is not in A. 3. Let E = { M | M is a DFA that accepts some string with more 1s than 0s}. Show that E is decidable. (Hint: Theorems about CFLs are helpful here.) Answer: Let A = {x | x has more 1s than 0s}. The language A is context-free, as we can easily construct a PDA to recognize A. Now, we construct the TM M below to decide E as follows: M = “On input M where M is a DFA: 1. Construct B = A ∩ L(M). Note that B is CFL, since L(M) is regular and A is CFL. 2. Test whether B is empty. 3. If yes, reject. Otherwise, accept. 4. Let C be a language. Prove that C is Turing-recognizable if and only if a decidable language D exists such that C = {x | ∃y( x, y ∈ D)}. Answer: If D exists, we can construct a TM M such that we search each possible string y, and testing whether x, y ∈ D. If such y exists, accept. Such a machine M will accept any string in C in finite steps, so C is Turing-recognizable. If C is recognized by some TM M, we define D = { x, y | M accepts x within |y| steps }. Clearly, D is decidable. Also, x ∈ C if and only if there exists y such that x, y ∈ D. Thus, C = {x | ∃y( x, y ∈ D)}. 5. (Bonus Question) Show that the problem of determining whether a CFG generates all string in 1∗ is decidable. In other words, show that { G | G is a CFG over {0, 1} and 1∗ ⊆ L(G)} is a decidable language. Answer: Please discussed the solution with Yu-Han directly. 2visit us at www.assignmentpedia.com or email us at [email protected] or call us at +1 520 8371215