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aa:pcp [2019/10/17 08:07] pdmatei |
aa:pcp [2021/01/17 22:48] (current) cata.chiru |
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| * for this proof, we consider that Turing Machines have only two final states $math[F=\{s_{yes},s_{no}\}], which model the output of $math[0/1]. It is straightforward how one can take an arbitrary Turing Machine an turn it into a //yes/no//-Turing Machine. | * for this proof, we consider that Turing Machines have only two final states $math[F=\{s_{yes},s_{no}\}], which model the output of $math[0/1]. It is straightforward how one can take an arbitrary Turing Machine an turn it into a //yes/no//-Turing Machine. | ||
| * We require that the execution of $math[M] on $math[w] never moves the head to the left of the first symbol of the word. Making sure this happens is a little more technically involved, however algorithmically possible. We shall skip these details. | * We require that the execution of $math[M] on $math[w] never moves the head to the left of the first symbol of the word. Making sure this happens is a little more technically involved, however algorithmically possible. We shall skip these details. | ||
| - | * We require that the word $w\neq \epsilon$, hence we do not allow simulations of the empty word. To accommodate for this, we can easily add a new symbol to the alphabet, which is used specifically for encoding the empty word. | + | * We require that the word $ w \neq \epsilon $, hence we do not allow simulations of the empty word. To accommodate for this, we can easily add a new symbol to the alphabet, which is used specifically for encoding the empty word. |