Is the language of TMs that halt on some string recognizable?

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Problem Detail: I would like to show that the following language is recognizable: $$L:= { langle M rangle mid M text{ is a TM that halts on some string}}.$$ How do I go about showing that this language is recognizable? I know that all recognizable languages are reducible to $HALT_epsilon$, so I figure if I can show that this language reduces to $HALT_epsilon$, then I am all set. I am defining $HALT_epsilon$ as follows: $$HALT_epsilon:= { langle M rangle mid M text{ is a TM that halts on } epsilon },$$ where $epsilon$ is the empty string. We can reduce $HALT$ on $x$ to $HALT_epsilon$ by a reduction $F(langle M, xrangle) = langle M’ rangle $, where $M'(y) = M(x)$. For this reduction, we just ignore the input string $y$, which we know will be $epsilon$ and just run $M$ on $x$ instead. Here, $HALT$ is defined as $$HALT:= { langle M,x rangle mid M text{ is a TM that halts on } x }.$$ I tried leveraging a similar technique to show that $L$ is recognizable, but I could not come up with anything better than this (somewhat crazy) TM that has a $HALT_epsilon$ oracle: $ D^{HALT_epsilon} =$ On input $langle M rangle:$

  1. Construct $N = $ “On input $x:$
    1. Run $M$ in parallel on all inputs $yin Sigma^*$.
    2. If $M$ halts on any $y$ then accept, otherwise loop.“
  2. Query the oracle to determine whether $langle N rangle in HALT_epsilon$.
  3. If the oracle answers YES, accept; if NO, reject.

Note: My notation for TM algorithms is based on “Theory of Computation” by Sipser. Step 2 for the definition of $N$ is a bit redundant, but in this type of context, is it okay to say something like “If $M$ halts on any $y$, then halt?” I think all I have shown here is that $L$ is decidable relative to $HALT_epsilon$. I don’t know if this implies that $L$ is recognizable. Can a Turing reduction be used in this manner to show that a language is recognizable? I’m confused as to what it means for a language to be recognizable. The task seems obvious if we go back to the definition: If some TM $R$ accepts strings in $L$, then $R$ recognizes $L$. So what if $R=D^{HALT_epsilon}$, and in the body of $D^{HALT_epsilon}$ we use some crazy reduction like $N$? In general, to show recognizability, can we just come up with a reduction like $N$ that may or may not halt? Is it a problem that $N$ will never halt if $langle M rangle notin L$?

Asked By : baffld

Answered By : Raphael

You can construct a recognizer following the same principle used for the recognizer for HALT. The only extra bit is how you check “all” inputs without getting stuck in a non-terminating computation. An important technique you can use here is called dovetailing (expressed with inputs from $mathbb{N}$):

  1. Simulate one step of $M$ on $1$.
  2. Simulate two steps of $M$ on $1$ and $2$ each.
  3. Simulate three steps on $1$, $2$ and $3$ each.

$quadvdots$

  • Terminate and accept once any of the simulated computations terminates and accepts.

If there is a halting input of $M$, this dovetailed simulation certainly finds it after finite time. If there is none, it loops and is correct in doing so. This is, in essence, your $N$ (with an explanation why it’s actually a computable function). You don’t need the rest of the reduction in order to show that $L$ is semi-decidable.

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Question Source : http://cs.stackexchange.com/questions/22929