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

On 10/18/24 12:39 PM, olcott wrote:

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.
 

and if HHH decides to abort its emulation, it also should know that
every level of condition emulation it say will also do the same thing,

If I understand his words correctly Mike has already disagreed with
this.

He hasn't.
>

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.

Nope.
Proven previously and you accepted by default by not pointing out an error.
Your HHH neither "correctly simulated" per his definitions or correctly predicts the behavior of such a simulation, and thus never acheived the required criteria.
All you have done is proved you lie.

 <MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
     If simulating halt decider H correctly simulates its input D
     until H correctly determines that its simulated D would never
     stop running unless aborted then
      H can abort its simulation of D and correctly report that D
     specifies a non-halting sequence of configurations.
</MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
 I will paraphrase this to use clearer language that directly applies
to HHH and DDD.
      If emulating termination analyzer HHH emulates its input DDD
     according to the semantics of the x86 language (including HHH
     emulating itself emulating DDD) until HHH correctly determines
     that its emulated DDD would never stop running unless aborted
     then ...
      HHH can abort its emulation of DDD and correctly report that DDD
     specifies a non-terminating sequence of x86 instructions.
 

*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.
 Since you should know that this assumption is factually
incorrect I could it as flat out dishonestly on your part.
 

since it is
simulating a halting program: the nested version will abort.
>

and thus the call HHH at 0000217a will be returned from, > and HHH has
no idea what will happen after that, so it KNOWS it is ignorant of the
answer.