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From: olcott <polcott2@gmail.com>
Newsgroups: comp.theory,sci.logic
Subject: Re: We finally know exactly how H1(D,D) derives a different result
 than H(D,D)
Date: Fri, 8 Mar 2024 00:00:17 -0600
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On 3/7/2024 9:28 PM, Richard Damon wrote:
> On 3/7/24 4:00 PM, olcott wrote:
>> On 3/7/2024 5:18 PM, Richard Damon wrote:
>>> On 3/7/24 3:02 PM, olcott wrote:
>>>> On 3/7/2024 4:32 PM, Richard Damon wrote:
>>>>> On 3/7/24 1:05 PM, olcott wrote:
>>>>>> H1(D,D) maps its input + its own machine address 00001422 to its 
>>>>>> output.
>>>>>>   H(D,D) maps its input + its own machine address 00001522 to its 
>>>>>> output.
>>>>>> Thus both H1 and H are computable functions of their input.
>>>>>
>>>>> And thus you are admitting that Neither H or H1 are actually 
>>>>> correct Halt Deciders, as Halt Deciders must be only a function of 
>>>>> the description of the Compuation to be decided.
>>>>>
>>>>
>>>> It turns out that both H(D,D) and H1(D,D) do correctly determine
>>>> whether or not they must abort the simulation of their input.
>>>
>>> Which isn't the halting question, so you are LYING.
>>
>> As I completely explained yet you persistently ignore the
>> halting question can only be correctly answered indirectly
>> otherwise inputs that contradict the decider that is embedded
>> within these inputs have no answer at all.
> 
> In other words you argue by lying.
> 
> The QUESTION is, and always will be, does the computation described by 
> the input Halt when run.
> 
> The Computation so described is FIXED and UNCHANGING reguards of what 
> the decider that is deciding does, as is the behavior of the H that it 
> is built on.
> 
> That was FIXED and made UNCHANGING when it was defined.
> 
> Thus, the question does THIS H^(H^) halt? HAS a definite and fixed 
> answer. SO you LIE when you said it doesn't.
> 
> Your problem seems to be that you think "Get the Right Answer?" is a 
> valid program instruction, or that H can somehow "change" itself after 
> H^ gets defined. IT CAN'T.
> 
> YOU persistently ignore this fact, likely because you are too stupid and 
> ignorant to understand that fundamental nature of programs, that they 
> will do what their programming says they will do, and that programming 
> doesn't change, EVER, with out the creation of some NEW program that is 
> different from its predicesor.
> 
> YOU *NEVER* have the right to change the question for a problem.
> 
> You can try to point out the the problem is inconsistant, and propose a 
> NEW PROBLEM, but that doesn't change the old.
> 
> You can talk about your new problem that you think is more useful than 
> the actual Halting Problem, after all, someone might be more interested 
> in the incorrect opinion of an admittedly faulty "Olcott-Halt Decider" 
> than the actual behavior of the Computation they are interested in.
> 
> NOT.
> 
> What you can't to is say you are working on one problem, while trying to 
> change it to mean something different. That is just call LYING, and you 
> seem to know that you doing it (you might feel you have justified 
> reasons to talk about a different problem) so the lie is DELIBERATE.
> 
>>
>>>>
>>>> That you or others consider this somehow improper does not change
>>>> the verified fact that they both correctly determine whether or
>>>> not they must abort their simulation.
>>>
>>> Which isn't the Halting Question, which you claim you are working on, 
>>> so you are just LYING.
>>>
>>
>> Already fully explained many many times (including above)
>> yet your ignorance is very persistent.
> 
> So, you think you can change the question and still be talking about the 
> same question.
> 
> You ARE the LIAR PARADOX.
> 
>>
>>>>
>>>> It is also the case that both H1(D,D) and H(D,D) are a pure function
>>>> of their inputs when we construe their own machine address to be an
>>>> element of these inputs.
>>>
>>> Which means they are not computing the Halting Function, which isn't 
>>> a function of the decider, so again, you are LYING.
>>>
>>
>> Already fully explained many many times (including above)
>> yet your ignorance is very persistent.
> 
> Yes, you have ADMITTED that you are LYING about working on the Halting 
> Problem.
> 
>>
>>>>
>>>>>>
>>>>>> Turing machines don't even have the idea of their own machine
>>>>>> address so this exact same thing cannot be Turing computable.
>>>>>
>>>>> And it isn't a Halt Decider even in Olcott machines as the 
>>>>> algorithm is shown to vary by a parameter that it isn't allowed to 
>>>>> vary to be a Halt Decider.
>>>>>
>>>>>>
>>>>>> Olcott machines entirely anchored in Turing machine notions
>>>>>> can compute the equivalent of H1(D,D) and H(D,D).
>>>>>>
>>>>>> Because Olcott machines are essentially nothing more than
>>>>>> conventional UTM's combined with Conventional Turing machine
>>>>>> descriptions their essence is already fully understood.
>>>>>>
>>>>>> The input to Olcott machines can simply be the conventional
>>>>>> space delimited Turing Machine input followed by four spaces.
>>>>>>
>>>>>> This is followed by the machine description of the machine
>>>>>> that the UTM is simulating followed by four more spaces.
>>>>>>
>>>>>> When this input is ignored Olcott machines compute the
>>>>>> exact same set as Turing machines.
>>>>>>
>>>>>> Unlike Turing machines, Olcott machines have the basis to
>>>>>> determine that they have been called with copies of their
>>>>>> own TMD.
>>>>>>
>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn   // Ĥ applied to ⟨Ĥ⟩ does not 
>>>>>> halt
>>>>>>
>>>>>> With Olcott machines Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ <Ĥ> and H ⟨Ĥ⟩ ⟨Ĥ⟩ <H> do
>>>>>> not have the same inputs thus can compute different outputs
>>>>>> when they do not ignore their own TMD.
>>>>>>
>>>>>>
>>>>>
>>>>> THen you build H^ wrong. Of course with your change in mechanics, 
>>>>> the H^ that needs to be generated will be a bit different.
>>>>>
>>>>
>>>> That Olcott machines always know their own TMD is unconventional.
>>>
>>> And breaks much of the background of Turing Machines, 
>>
>> Not at all. Not in the least little bit.
>> Olcott machines are 100% fully specified
>> in terms of Turing machines.
> 
> Yes, BUT if you talk about an Olcott machine, you MUST include the added 
> data as part of the description of that machine, or you are just LYING.
> 
>>
>>> so if you what to use ANY establish property of Turing Machine, you 
>>> must include that now extra data EXPLICITLY.
>>>
>>
>> It is already explicitly included in the definition of an Olcott machine.
> 
> Which means that you can't actually write an Olcott-Machine that matches 
> the requirements for a Halt Decider.
> 
> A Halt Decider MUST be able to defined as taking JUST the description of 
> the Computation to be decided (the Allgorithm and the Data). In general 
> this also holds, to be a Foo decider, the decider must be give JUST the 
> information about the thing that we are deciding Foo on.
> 

If Olcott machines that are essentially Turing machines configured in
a certain way can detect when they themselves are called in recursive
========== REMAINDER OF ARTICLE TRUNCATED ==========