**Problem Detail:**

The proof in my textbook, that $E_{TM}$ can be decided by oracle machine $O^{A_{TM}}$, uses a Turing machine $P$ such that for an input $w$:

- $P$ runs the Turing machine $M$ on all strings of $\Sigma^*$
- If $M$ accepts the string, $P$ accepts

$O^{A_{TM}}$ then asks the oracle if $<P,w> \in A_{TM}$

I can't seem to understand why $P$ can run $M$ for all strings of $\Sigma^*$ because this is an infinite amount of strings. I do, however, understand that $P$ actually never has to run and is purely constructed to ask the oracle if it accepts.

###### Asked By : Barney Kelly

#### Answered By : Yuval Filmus

The technique is knows as *dovetailing*. The machine $P$ keeps a two-dimensional tape and uses the following algorithm:

- Run $M$ on the first string for one step.
- Run $M$ on the first two strings for one step.
- Run $M$ on the first three strings for one step.
- ...

(We imagine that all possible strings are arranged in some arbitrary computable order, say $\Lambda,0,1,00,01,10,11,000,\ldots$.)

Each simulation of $M$ runs on its own "row" of the two-dimensional tape of $P$, which also stores the location of the head and the current state. (A two-dimensional tape can be simulated by a one-dimensional tape.)

If $M$ ever halts on one of the strings, then $P$ halts. Therefore $P$ halts iff $M$ halts on *some* input.

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Question Source : http://cs.stackexchange.com/questions/18887

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