Scan the tape to see if there is any unmarked 0 found. This machine is identical to 3. If no unmarked 0 is found, accept. If no such unmarked 0 is found, accept.
Obviously, be an ordinary Turing machine. If recognizes the class of Turing- can simulate of can without problems. How simulates of is described as follows. We assume the following during the shifting for Step 3: We use states to remember the symbol to be shifted right. The first symbol of the tape after the shift is a symbol not used by the original M.
If the current symbol is c and the next symbol is the marked b, after the shifting, c becomes marked but b is not marked. The shifting stops when we see a blank symbol.
Let and be two decidable languages and and be the corresponding TMs. The aim is to construct a TM based on and such that the concatenation is also decidable. Since a given input concatenation of strings and has finite possible partitions, a nondeterministic TM is chosen to simplifies the description. Nondeterministically split into and 2. Besides, iff there exists a split of eventually halts because and is decidable since there exists an NTM Therefore, such that accepts are both deciders.
Let be a decidable language and be the corresponding TM. The aim is to construct a TM based on such that is also decidable. Since a given input has finite possible combinations of strings where and , a nondeterministic TM is chosen to simplify the description. Besides, eventually halts because is a decider. Therefore, is decidable since there exists an NTM which decides. Let be a decidable language and based on such that is described as follows.
TM be the corresponding TM. The aim is to construct a TM , the complement of , is also decidable. Besides, eventually halts because is decidable since there exists a TM which decides. The Let aim is to construct a TM based on and such that the intersection is is described as follows. Besides, is decidable 3. The aim is to construct a TM based on and such that the concatenation is also Turing-recognizable. Since a given input concatenation of strings and has finite possible partitions, a nondeterministic TM is chosen to simplify the description.
However, are not deciders. Therefore, which recognizes iff there exists a partition of such that may loop forever on some input because is Turing-recognizable since there exists an NTM. The aim is to is also Turing-recognizable. Since a given input has finite possible combinations of strings where nondeterministic TM is chosen to simplify the description. However, , where is not empty. The aim is to construct a TM based on and such that the intersection is also Turing-recognizable.
The resulting TM is described as follows. Convert G to an equivalent grammar in Chomsky normal form. Let which is a CFL from Problem 2. If TM accepts, produces no strings of lengths at least string of length at least p. If not, produces a 4. The goal is that is decidable iff the emptiness of is also decidable. Since is a CFL from in theorem 4. Let be the CFG of Problem 2. Suppose that every decider is in A. Since A is Turingrecognizable, A is also enumerable.
Let be the ith decider in A. We may construct the following decider as follows: Then can be derived by applying the diagonalization method as illustrated by the following table. The table below demonstrates an example. Obviously, reject accept accept Theory of Computation Homework 5 Solution.
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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. Experts of Theory of Computation Assignment at Assignments Help Tutors are available 24x7 to help students. To get the solution of your Theory of Computation Assignment Help you just need to submit your Theory of Computation Assignment at our '' Submit your Assignments/Homework here '' form.