Re: DDD specifies recursive emulation to HHH and halting to HHH1

Am Thu, 27 Mar 2025 12:50:12 -0500 schrieb olcott:

On 3/27/2025 2:18 AM, Fred. Zwarts wrote:

Op 27.mrt.2025 om 04:09 schreef olcott:

On 3/26/2025 8:22 PM, Richard Damon wrote:

Non-Halting is that the machine won't reach its final staste even if
an unbounded number of steps are emulated. Since HHH doesn't do that,
it isn't showing non-halting.

DDD emulated by any HHH will never reach its final state in an
unbounded number of steps.
DDD emulated by HHH1 reaches its final state in a finite number of
steps.

It is not very interesting to know whether a simulator reports that it
is unable to reach the end of the simulation of a program that halts in
direct execution.

That IS NOT what HHH is reporting.

That is exactly what it does, and you have said so before(tm).

HHH correctly rejects DDD because DDD correctly emulated by HHH cannot
possibly reach its own final halt state.

DDD doesn't *do* anything, it is being simulated. HHH can't reach
DDD's existing halt state.

It is interesting to know:
'Is there an algorithm that can determine for all possible inputs
whether the input specifies a program that [...]
halts when directly executed?'
This question seems undecidable for Olcott.

It is the halts while directly executed that is impossible for all
inputs. No TM can ever report on the behavior of the direct execution of
any other TM.

The direct execution of a TM is obviously computable from its description.

A TM can only report on the behavior that the machine code of another TM
specifies. When it specifies a pathological relationship then the
behavior caused by the pathological relationship MUST BE REPORTED.

No, the machine code doesn't "specify a pathological relationship", that
is purely a feature of trying to simulate it with the included simulator.

--
Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
It is not guaranteed that n+1 exists for every n.