Re: A state transition diagram proves ... GOOD PROGRESS

Am Fri, 18 Oct 2024 12:11:17 -0500 schrieb olcott:

On 10/18/2024 12:00 PM, joes wrote:

Am Fri, 18 Oct 2024 11:39:52 -0500 schrieb olcott:

On 10/18/2024 9:41 AM, joes wrote:

Am Fri, 18 Oct 2024 09:10:04 -0500 schrieb olcott:

On 10/18/2024 6:17 AM, Richard Damon wrote:

On 10/17/24 11:47 PM, olcott wrote:

On 10/17/2024 10:27 PM, Richard Damon wrote:

On 10/17/24 9:47 PM, olcott wrote:

On 10/17/2024 8:13 PM, Richard Damon wrote:

On 10/17/24 7:31 PM, olcott wrote:

>

When DDD is correctly emulated by HHH according to the
semantics of the x86 language DDD cannot possibly reach its
own machine address [00002183] no matter what HHH does.
+-->[00002172]-->[00002173]-->[00002175]-->[0000217a]--+

>

Except that 0000217a doesn't go to 00002172, but to 000015d2

>

The Emulating HHH sees those addresses at its begining and then
never again.
Then the HHH that it is emulating will see those addresses, but not
the outer one that is doing that emulation of HHH.
And so on.
Which HHH do you think EVER gets back to 00002172?
What instruction do you think that it emulates that would tell it
to do so?
At best the trace is:
00002172 00002173 00002175 0000217a conditional emulation of
00002172 conditional emulation of 00002173 conditional emulation of
00002175 conditional emulation of 0000217a CE of CE of 00002172 ...

OK great this is finally good progress.

The more interesting part is HHH simulating itself, specifically the
if(Root) check on line 502.

That has nothing to do with any aspect of the emulation until HHH has
correctly emulated itself emulating DDD.

What? That is part of HHH, not DDD.

Until if (root) is true it has no effect on DDD emulated by HHH.

The existence of the check has an effect right from the start;
besides, it is true the first time it is executed.

Message-ID: <rLmcnQQ3-N_tvH_4nZ2dnZfqnPGdnZ2d@brightview.co.uk>
On 3/1/2024 12:41 PM, Mike Terry wrote:
   > Obviously a simulator has access to the internal state (tape
   > contents etc.) of the simulated machine. No problem there.
This seems to indicate that the Turing machine UTM version of HHH
can somehow see each of the state transitions of the DDD resulting
from emulating its own Turing machine description emulating DDD.

Of course. It needs to, in order to simulate it. Strictly speaking it
has no idea of its simulation of a simulation two levels down, only
of the immediate simulation; the rest is just part of whatever
program the simulated simulator is simulating, which happens to be
itself.

  From the concrete execution trace of DDD emulated by HHH
according to the semantics of the x86 language people with sufficient
technical competence can see that the halt status criteria that
professor Sipser agreed to has been met.
      If emulating termination analyzer HHH emulates its input DDD
      until HHH determines that its emulated DDD would never stop
      running unless aborted ...

But it would.

*Joes can't seem to understand this*
Only the outer-most HHH meets its abort criteria first, thus unless
it aborts as soon as it meets this criteria none of them will ever
abort.

This is very simple to understand. Almost as simple as: even if only
the outermost HHH didn't abort, it would still halt,

Yet that is based on the factually incorrect assumption that every
instance of HHH does not use the exact same machine code.

Same as the outer HHH returning that the inner ones wouldn't.

--
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.