Re: H ⟨Ĥ⟩ ⟨Ĥ⟩ is correct when reports on the actual behavior that it sees

On 3/14/24 1:52 PM, olcott wrote:

On 3/14/2024 6:37 AM, Mikko wrote:

On 2024-03-12 19:47:09 +0000, olcott said:
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On 3/12/2024 2:29 PM, Richard Damon wrote:

On 3/12/24 12:16 PM, olcott wrote:

On 3/12/2024 1:58 PM, Richard Damon wrote:

On 3/12/24 9:45 AM, olcott wrote:

On 3/12/2024 11:31 AM, Richard Damon wrote:

On 3/12/24 7:02 AM, olcott wrote:

On 3/12/2024 3:49 AM, Mikko wrote:

On 2024-03-11 15:34:04 +0000, olcott said:
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On 3/11/2024 10:17 AM, Mikko wrote:

On 2024-03-11 14:31:37 +0000, olcott said:
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On 3/11/2024 4:51 AM, Mikko wrote:

On 2024-03-10 14:29:20 +0000, olcott said:
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On 3/10/2024 7:25 AM, Mikko wrote:

On 2024-03-09 15:49:39 +0000, olcott said:
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On 3/9/2024 3:07 AM, Mikko wrote:

On 2024-03-08 16:09:58 +0000, olcott said:
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On 3/8/2024 9:29 AM, Mikko wrote:

On 2024-03-08 05:23:34 +0000, Yaxley Peaks said:
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With all of these extra frills, aren't you working outside the premise
of the halting problem? Like how Andre pointed out.

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Yes, he is.
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The halting problem concerns itself with turing machines and what you
propose is not a turing machine.

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That is true. However, we can formulate similar problems and proofs
for other classes of machines.
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I am working on the computability of the halting problem
(the exact same TMD / input pairs) by a slightly augmented
notion of Turing machines as elaborated below:
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Olcott machines are entirely comprised of a UTM + TMD and one
extra step that any UTM could perform, append the TMD to the
end of its own tape.

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An important question to answer is whether a Turing machine can
simulate your machines.

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Olcott machines are entirely comprised of a UTM + TMD and one
extra step that any UTM could perform, append the TMD to the end
of its own tape.

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Then a Turing machine can simulate your machine.
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Yes, except the TM doing the simulating cannot be an Olcott machine.

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That is not "ecept", that is containted in what the word "Truring machine"
means.
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Anway, a Truing machine can, with a simulation of your machine, compute
everything your machine can, so your machine cannot compute anyting a
Turing machine cannot.
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Turing Machines, Olcott Machines, RASP machines and my C functions
can always correctly report on the behavior of their actual input
When they report on this question:
Will you halt if you never abort your simulation?

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If they only talk about themselves they are not useful.
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When every simulating halt decider reports on the actual behavior
that it actually sees, then the pathological input does not
thwart it.

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If it is not useful then nobody cares whether some input can thwart it.
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Best selling author of Theory of Computation textbooks:
*Introduction To The Theory Of Computation 3RD, by sipser*
https://www.amazon.com/Introduction-Theory-Computation-Sipser/dp/8131525295/
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Date 10/13/2022 11:29:23 AM
*MIT Professor Michael Sipser agreed this verbatim paragraph is correct*
(He has neither reviewed nor agreed to anything else in this paper)
(a) 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
(b) H can abort its simulation of D and correctly report that D specifies a non-halting sequence of configurations.
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*When we apply this criteria* (elaborated above)
Will you halt if you never abort your simulation?
*Then the halting problem is solved*
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No, that isn't what you asked,
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You asked if the CORRECT SIMULATION of the input won't stop, can H abort its simlation.
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Of course, if H aborts it simulation, then BY DEFINITION, its simulation doesn't go on forever and isn't a correct simulation, so that doesn't apply.
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It is your persistently repeated mistake that has been corrected
hundreds of times that a correct simulation requires a complete
simulation.
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The words that Professor Sipser agreed to clearly mean that
when a correct partial simulation of D proves that a correct
complete simulation of D would never stop running then H
has both its abort criteria and its halt status criteria.

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Nope, because there is no such thing in Computation theory.
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Only in your incorrect reconstruction from lack of principles.
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You didn't say "PARTIAL", so he wasn't even thinking of it.
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"simulates its input D until"
clearly does not mean simulate forever
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But you said:
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... until H correctly determines that its simulated D would never stop running unless aborted
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The "until" clearly does not mean to simulate forever.

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But "would never stop" clearly does.
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 Only when you ignore the rest of the sentence.
"simulated D would never stop running {unless aborted}
 

But it DOES, The simulated D uses the H that is your final decison on what H does, which will abort and return 0, so any correct simulation of D will reach an end and stop running.
You aren't allowed to talk about changing D to someother machine, as that ism't the input, so even if you imagine H becoming a non-aborting version, D will still refer to the original aborting version and the simulation will halt.
You just don't seem to understand this about programs.
Remember, H^ / D are defined to use a SPECIFIC decider, which has a PRECISE definiton, any talk about changing H doesn't affect what is in H^ / D. Of course, if you do decider to change the H to something else, it can't claim to have defeated the proof, as the H^ / D it looked at wasn't the one designed for it, so you now need to look at that DIFFERENT QUESTION.