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Select all true statements1 pointThere is a TM that loops on all inputsLet A be any decidable language. There is a TM that loops on exactly the strings not in A.Let A be any decidable language. There is a TM that loops on exactly the strings in A.There is an undecidable language A such that there is a TM that halts on exactly the strings in A.There is an undecidable language A such that the complement of A is decidable.There is a decidable language A such that the complement of A is undecidable.

Question

Select all true statements1 pointThere is a TM that loops on all inputsLet A be any decidable language. There is a TM that loops on exactly the strings not in A.Let A be any decidable language. There is a TM that loops on exactly the strings in A.There is an undecidable language A such that there is a TM that halts on exactly the strings in A.There is an undecidable language A such that the complement of A is decidable.There is a decidable language A such that the complement of A is undecidable.

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Solution 1

  1. True. There is a Turing Machine (TM) that loops on all inputs. This is simply a TM that does not have any accepting states, so it will never halt and thus loop on all inputs.

  2. False. If A is a decidable language, then there is a TM that decides it, meaning it halts on all inputs. Therefore, there cannot be a TM that loops on exactly the strings not in A, because the TM that decides A must halt on these strings.

  3. False. Similar to the previous statement, if A is a decidable language, then there is a TM that decides it, meaning it halts on all inputs. Therefore, there cannot be a TM that loops on exactly the strings in A, because the TM that decides A must halt on these strings.

  4. False. If A is an undecidable language, then there is no TM that decides it, meaning there is no TM that halts on exactly the strings in A.

  5. False. If A is an undecidable language, then its complement is also undecidable. This is because if the complement of A were decidable, then we could decide A by running the TM for the complement of A and then flipping the answer, which contradicts the assumption that A is undecidable.

  6. False. If A is a decidable language, then its complement is also decidable. This is because if A is decidable, then there is a TM that decides it, and we can construct a TM that decides the complement of A by running the TM for A and then flipping the answer.

This problem has been solved

Solution 2

  1. True. There is a Turing Machine (TM) that loops on all inputs. This is simply a TM that does not have any accepting states, so it will never halt and thus loop on all inputs.

  2. False. If A is a decidable language, then there is a TM that decides it, meaning it halts on all inputs. Therefore, there cannot be a TM that loops on exactly the strings not in A, because the TM that decides A must halt on these strings.

  3. False. Similar to the previous point, if A is a decidable language, then there is a TM that decides it, meaning it halts on all inputs. Therefore, there cannot be a TM that loops on exactly the strings in A, because the TM that decides A must halt on these strings.

  4. False. If there is a TM that halts on exactly the strings in A, then A is decidable, which contradicts the assumption that A is undecidable.

  5. False. The complement of an undecidable language is also undecidable. This is because if the complement of A was decidable, then we could decide A by running the TM for the complement of A and then flipping the answer, which contradicts the assumption that A is undecidable.

  6. False. The complement of a decidable language is also decidable. This is because if A is decidable, then we can decide the complement of A by running the TM for A and then flipping the answer.

This problem has been solved

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