Re: How the requirements that Professor Sipser agreed to are exactly met --- Mike's key mistake

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Sujet : Re: How the requirements that Professor Sipser agreed to are exactly met --- Mike's key mistake
De : richard (at) *nospam* damon-family.org (Richard Damon)
Groupes : comp.theory
Date : 17. May 2025, 03:50:48
Autres entêtes
Organisation : i2pn2 (i2pn.org)
Message-ID : <74025a25d23cae7e6590f32656d7423dd3c4f956@i2pn2.org>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26
User-Agent : Mozilla Thunderbird
On 5/16/25 8:49 PM, olcott wrote:
On 5/16/2025 7:38 PM, Richard Damon wrote:
On 5/16/25 6:39 PM, olcott wrote:
On 5/16/2025 5:03 PM, Richard Damon wrote:
On 5/16/25 4:29 PM, olcott wrote:
On 5/16/2025 3:06 PM, Fred. Zwarts wrote:
Op 16.mei.2025 om 07:29 schreef olcott:
*Not at all. I am following these exact words*
<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>
>
On 5/14/2025 7:36 PM, Mike Terry wrote:
Shows exactly how to implement those words to implement
a correct Simulating Termination Analyzer. Mike provides
a complete example of how this works.
>
>
>
>
Sipser agreed to a vacuous statement, because the condition 'correctly simulates' was not met.
>
And by this you mean that when the spec requires
a partial simulation
>
*until H correctly determines that its simulated D*
*would never stop running unless aborted*
>
You "interpret" this to mean that it must
infinitely simulate non-terminating inputs.
>
Which means, as I explained else, if H, after doing a partial simulation, can determine that a COMPLETE simulation of this exact input would be non-halting, it can abort.
>
>
Not quite. One key detail is missing.
*H correctly determines that its simulated D*
*would never stop running unless aborted*
>
Is referring to what the behavior of D would be
(in the hypothetical case) where this very same
H never aborted.
>
Nope, since D must stay D, and D must be a fully encoded program and thus doesn't change when you make the hypothetical H.
>
 *Click here to get the whole article*
https://al.howardknight.net/? STYPE=msgid&MSGI=%3C1003cu5%242p3g1%241%40dont-email.me%3E
 Mike perfectly explains all of this with a concrete
example. In this case H determines that its infinite
loop input would never stop running unless aborted
so it aborts it and correctly rejects it.
 H is not being asked what is the behavior of this
infinite loop after H aborts it. It is being asked
what its behavior would be if H never aborted it.
 HHH is not being asked what is the behavior of
DDD after HHH aborts it. It is being asked
what its behavior would be if HHH never aborted it.
 
What makes you think I haven't.
And, since DDD needs to be a PROGRAM to do any of this, as non-leaf functions can't be correctly emulated, that DDD DOES include the code of the HHH it was built for, which is the HHH that aborts and returns 0.
Give that input to your hypothetical HHH that doesn't abort, and it will reach the end.
Of course, you can't actualy do that, as it requires that you computer can store two different programs in the same memory locations, so you "Hypothetical HHH" needs to be put somewhere else in memory, like HHH1 was.
That shows that THIS HHH (the one that aborts) if it is imagined to not abort, while DDD is considered to be the program it always was, will reach the end of its simulation, and thus show that DDD is halting.
Of course, your HHH gives up before that so doesn't know that is the proper behavior, but just assumes incorrectly that it is ok to abort, so does.
To quote the part you seem to ignore:

 PO's problem is his misinterpretation of "its simulated input would never stop running unless aborted".  In the case of his HHH/DD, the simulated input (DD) /does/ stop running if simulated far enough, but HHH simply /doesn't/ go far enough because PO has mistakenly decided he's seen some pattern that implies non-halting in the trace.  [A pattern akin to the "tight loop" pattern, except that the tight loop pattern is sound, while his pattern is unsound, matching on a halting input. Simples!]
He says this because it is true. The simulation of the input to HHH WILL reach the final state if the simulating HHH is imagined and hypothetically changed (with out changing the input) to not abort.
Note, that is the key, the input must be complete, and the input not changed.
If you can't meet those requirements, you can't even hypothosize your HHH, and thus can't use that criteria.

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