WebTheorem. ETM = {hMi M is a TM and L(M) = ∅} is undecidable. Proof. We show ATM is reducible to ETM. First, given hM,wi, a Turing machine can modify the encoding of M, to transform M into M′, which recognises L(M)∩{w}. This is what the new machine M′ does on input x: 1. If x 6= w, reject. 2. If x = w, run M on w and accept if M accepts. WebTM is not recognizable, then E TM is not recognizable. 3 E TM is co-recognizable 3.1 By construction We will build a co-recognizer C for E TM. C = \ On input hMi: (0) If hMiis not an encoding of a Turing machine, then reject. (1) For i = 1;2;3;:::: (a) Run the enumerator for …
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WebProve that the language EQTM = {< A,B > A and B are two Turing Machines with L (A) = L (B)} is neither Turing-recognizable nor Co-Turing-Recognizable. This problem has been solved! You'll get a detailed solution from a subject matter expert … WebAt least one of them must be not recognizable. E TM is recognizable. E TM is not recognizable. 11. EQ TM = fM 1;M 2 jM 1 and M 2 are TMs and L(M 1) = L(M 2)g EQ TM is undecidable. Is EQ TM recognizable? See the text book. 12. Language is Context-free CONTEXT FREE TM = fM jM is a TM and L(M) is context freeg third grade math state test practice
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WebOct 29, 2024 · 1 Answer Sorted by: 0 L (M) = < M> means that the only string the M accepts is its own description < M> , hopefully you can see now that S is the language of TMs that accept only their own descriptions Proof : We provide a reduction from ATM to S , ATM ≤m S to show that s is not Turing-recognizable. WebShow that ETM is co-Turing-recognizable. (A language Lis co-Turing-recognizable if its complement Lis Turing-recognizable.) Note that the complement of ETM is ETM = {hMi … WebThe ETM file extension indicates to your device which app can open the file. However, different programs may use the ETM file type for different types of data. While we do not … third grade math printables