I have seen this statement in my studies and I cannot figure out why it is true.

We know that $P_{HALT} \leq_T \overline{P_{HALT}}$, but $P_{HALT} \leq_m \overline{P_{HALT}}$ does not hold.

I know, that If the many-one-reduction was possible, then both problems were semi-decidable, which makes the Halting problem decidable. But I would prefer a more intuitive answer.

  • 1
    $\begingroup$ Do you see why $P_{HALT}\le_T\overline{P_{HALT}}$? $\endgroup$ Aug 15 at 19:46
  • $\begingroup$ not really, tbh $\endgroup$
    – thenlaw
    Aug 15 at 20:17

If $P_{\mathrm{HALT}} \leq_m \overline{P_{\mathrm{HALT}}}$ then given a Turing machine $M$, you can come up with another Turing machine $M'$ such that $M$ halts iff $M'$ doesn't halt (in both cases, we run the machine on a blank tape). Let's break this into two conditions:

  1. If $M$ halts then $M'$ doesn't halt.
  2. If $M$ doesn't halt then $M'$ halts.

It is easy to satisfy the first condition: $M'$ can simulate $M$, and if $M$ halts, enter an infinite loop. However, it is not clear how to satisfy the second condition — how can you tell that $M$ doesn't halt? This is, intuitively, the reason that no such reduction exists.

In contrast, $P_{\mathrm{HALT}} \leq_T \overline{P_{\mathrm{HALT}}}$ trivially: you can solve the halting problem given an oracle for the not-halting problem. If you don't see it immediately (as you indicate in the comments), then you haven't internalized the definition of a Turing reduction.


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