Question

Show that the language A = {<M₁> | the language accepted by the Turing Machine M₁ is 1^*} is not decidable. Present your proof in the style of the proof of Th. 5.3, which shows below.

PROOF We let R be a TM that decides REGULARTm and construct TM S to decide ATM. Then S works in the following manner. S - "On input (M, w), where M is a TM and w is a string: 1. Construct the following TM M2 M2"On input: 1. If r has the form 0"1", accept. 2. If ar does not have this form, run M on input w and accept if M accepts w." 25 2. 3. Run R on input ?M-? If R accepts, accept; if R rejects, reject."

Answer 1

The proof for given problem is shown below: Let R be a TM that decides H_TM and construct TM K such that it decides A_TM. The algorithm is shown below: On input < M1, wl > , Ml denotes a TM & wl is a string. Construct TM M2. o M2 = On input x If x has form l^n, accept. If x is not in this form, run Ml with input wl & accept if Ml would accept wl. Run R with input M2. If R would accept, accept it, if R rejects then reject. Explanation: If an algorithm exists, then another algorithm A could be written so that when B is given as input decides if B accepts all strings. If B cannot accept all strings, A outputs some program A (B) that doesn't accept all strings. If B doesn't accept all, A omits some program that would accept all strings. There is a TM which doesn't allow any string. There is a TM that accepts every string.