On 2/8/2022 5:56 AM, Richard Damon wrote:
> On 2/8/22 12:28 AM, olcott wrote:
>> On 2/7/2022 8:03 PM, Richard Damon wrote:
>>>
>>> On 2/7/22 8:52 PM, olcott wrote:
>>>> On 2/7/2022 7:26 PM, Richard Damon wrote:
>>>>> On 2/7/22 8:08 PM, olcott wrote:
>>>>>> On 2/7/2022 5:46 PM, Richard Damon wrote:
>>>>>>> On 2/7/22 9:59 AM, olcott wrote:
>>>>>>>> On 2/7/2022 5:47 AM, Richard Damon wrote:
>>>>>>>>> On 2/6/22 11:30 PM, olcott wrote:
>>>>>>>>>> On 2/6/2022 10:05 PM, Richard Damon wrote:
>>>>>>>>>>>
>>>>>>>>>>> On 2/6/22 10:04 PM, olcott wrote:
>>>>>>>>>>>> On 2/6/2022 3:39 PM, Richard Damon wrote:
>>>>>>>>>>>>>
>>>>>>>>>>>>> On 2/6/22 3:53 PM, olcott wrote:
>>>>>>>>>>>>>> On 2/6/2022 2:33 PM, Richard Damon wrote:
>>>>>>>>>>>>>>> On 2/6/22 3:15 PM, olcott wrote:
>>>>>>>>>>>>>>>> On 2/6/2022 1:43 PM, dklei...@gmail.com wrote:
>>>>>>>>>>>>>>>>> On Sunday, February 6, 2022 at 8:31:41 AM UTC-8, olcott
>>>>>>>>>>>>>>>>> wrote:
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> H determines [halting] on the basis of matching
>>>>>>>>>>>>>>>>>> infinite behavior patterns.
>>>>>>>>>>>>>>>>>> When an infinite behavior pattern is matched H aborts
>>>>>>>>>>>>>>>>>> its simulation and
>>>>>>>>>>>>>>>>>> transitions to its final reject state. Otherwise H
>>>>>>>>>>>>>>>>>> transitions to its
>>>>>>>>>>>>>>>>>> accept state when its simulation ends.
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> This is incomplete because it does not cover the case
>>>>>>>>>>>>>>>>> where the
>>>>>>>>>>>>>>>>> machine neither halts nor matches an "infinite behavior
>>>>>>>>>>>>>>>>> pattern".
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> It covers the case that had previously been considered
>>>>>>>>>>>>>>>> to be proof that the halting problem is undecidable.
>>>>>>>>>>>>>>>> That is all that I need to refute these proofs.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> You need to prove a theorem: There is a finite set of
>>>>>>>>>>>>>>>>> patterns such
>>>>>>>>>>>>>>>>> that every Turing machine either halts or matches one
>>>>>>>>>>>>>>>>> of these
>>>>>>>>>>>>>>>>> patterns.
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> But I feel sure that theorem is not true.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> To solve the halting problem my program must be all
>>>>>>>>>>>>>>>> knowing. To refute the proofs I merely need to show that
>>>>>>>>>>>>>>>> their counter-example can be proved to never halt.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> And you just ignore the fact that if H applied to <H^>
>>>>>>>>>>>>>>> <H^> goes to H.Qn, then by construction H^ <H^> goes to
>>>>>>>>>>>>>>> H^.Qn, and halts, and since H, to be an accurate Halt
>>>>>>>>>>>>>>> Decider, must only go to H,Qn if the machine its input
>>>>>>>>>>>>>>> represents will never halt. They you also don't seem to
>>>>>>>>>>>>>>> understand that the computaton that <H^> <H^> represents
>>>>>>>>>>>>>>> IS H^ applied to <H^>. So, H was just wrong.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> So, you haven't actually proved the thing you claim
>>>>>>>>>>>>>>> youhave, but only that you have amassed an amazing pile
>>>>>>>>>>>>>>> of unsound logic based on wrong definitions that have
>>>>>>>>>>>>>>> hoodwinked yourself into thinking you have shown
>>>>>>>>>>>>>>> something useful.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> You are so good at doing this that you have gaslighted
>>>>>>>>>>>>>>> yourself so you can't actually understand what actual
>>>>>>>>>>>>>>> Truth is.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> You simply do know know enough computer science to
>>>>>>>>>>>>>> understand that you are wrong and never will because you
>>>>>>>>>>>>>> believe that you are right.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>
>>>>>>>>>>>>> And you clearly don't know enough Computation Theory to
>>>>>>>>>>>>> talk about it.
>>>>>>>>>>>>>
>>>>>>>>>>>>> Since the is a Theorm in Computation Theory, using
>>>>>>>>>>>>> Computation Theory Deffinitions, that is your problem.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> Because all simulating halt deciders are deciders they are
>>>>>>>>>>>>>> only accountable for computing the mapping from their
>>>>>>>>>>>>>> input finite strings to an accept or reject state on the
>>>>>>>>>>>>>> basis of whether or not their correctly simulated input
>>>>>>>>>>>>>> could ever reach its final state: ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* ⟨Ĥ⟩.qn.
>>>>>>>>>>>>>
>>>>>>>>>>>>> And if you are working on the Halting Problem of
>>>>>>>>>>>>> Computation Theory, BY DEFINITION, the meaning of 'correcty
>>>>>>>>>>>>> simulted' is simulation by a REAL UTM which BY DEFINITION
>>>>>>>>>>>>> exactly matches the behavior of Computation that it is
>>>>>>>>>>>>> representation of, which for <H^> <H^> is H^ applied to <H^>
>>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>> If an infinite number is steps is not enough steps for the
>>>>>>>>>>>> correct simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H to transition to
>>>>>>>>>>>> ⟨Ĥ⟩.qn then the input to embedded_H meets the Linz
>>>>>>>>>>>> definition of a sequence of configurations that never halts.
>>>>>>>>>>>
>>>>>>>>>>> WRONG.
>>>>>>>>>>>
>>>>>>>>>>> If embedded_H DOES an infinite number of steps and doesn't
>>>>>>>>>>> reach a final state, then it shows its input never halts.
>>>>>>>>>> When embedded_H matches this infinite pattern in the same
>>>>>>>>>> three iterations:
>>>>>>>>>>
>>>>>>>>>> Then these steps would keep repeating:
>>>>>>>>>>    Ĥ1 copies its input ⟨Ĥ2⟩ to ⟨Ĥ3⟩ then embedded_H simulates
>>>>>>>>>> ⟨Ĥ2⟩ ⟨Ĥ3⟩
>>>>>>>>>>    Ĥ2 copies its input ⟨Ĥ3⟩ to ⟨Ĥ4⟩ then embedded_H simulates
>>>>>>>>>> ⟨Ĥ3⟩ ⟨Ĥ4⟩
>>>>>>>>>>    Ĥ3 copies its input ⟨Ĥ4⟩ to ⟨Ĥ5⟩ then embedded_H simulates
>>>>>>>>>> ⟨Ĥ4⟩ ⟨Ĥ5⟩...
>>>>>>>>>>
>>>>>>>>>> that you agreed show the simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H
>>>>>>>>>> will never reach ⟨Ĥ⟩.qn in any number of steps, which proves
>>>>>>>>>> that this input cannot possibly meet the Linz definition of
>>>>>>>>>> halting:
>>>>>>>>>>
>>>>>>>>>> computation that halts … the Turing machine will halt whenever
>>>>>>>>>> it enters a final state. (Linz:1990:234)
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>> OK, so the only computatiopn that you show that does not halt
>>>>>>>>> is H, so H can not be a decider.
>>>>>>>>
>>>>>>>> In the above example embedded_H simulates three iterations of
>>>>>>>> nested simulation to match the infinitely nested simulation
>>>>>>>> pattern.
>>>>>>>> In reality it needs less than this to match this pattern.
>>>>>>>>
>>>>>>>>
>>>>>>>
>>>>>>> And if it doesn't do an infinite number, the H^ that is using it
>>>>>>> will Halt,
>>>>>>
>>>>>> embedded_H only examines the actual behavior of its inputs as if
>>>>>> its was a guard assigned to watch the front. If someone comes in
>>>>>> the back door (non-inputs) embedded_H is not even allowed to pay
>>>>>> attention.
>>>>>>
>>>>>
>>>>> If the 'actual behavior' of the input <H^> <H^> is not the behavior
>>>>> of H^ applied to <H^> you are lying about doing the Halting Problem.
>>>>>
>>>>
>>>> If it is true that the simulated input to embedded_H cannot possibly
>>>> ever reach its final state of ⟨Ĥ⟩.qn, then nothing in the universe
>>>> can possibly contradict the fact that the input specifies a
>>>> non-halting sequences of configurations. If God himself said
>>>> otherwise then God himself would be a liar.
>>>>
>>>
>>> Except that if H/embedded_H aborts its simulation and goes to H.Qn,
>>> then the CORRECT simulation of its input (that done by a REAL UTM)
>>> will show that it will go to H^.Qn.
>>>
>>> All you have proven is that if H doesn't abort, and thus doesn't go
>>> to H.Qn, and thus fails to be a correct decider, then H^ applied to
>>> <H^> is non-halting.
>>>
>>> You keep on thinking that a simulation that aborts its simulation is
>>> a 'correct' simulation. By the definition in Computation Theory, this
>>> is not true. If you think it is, it just proves that you don't
>>> understand the field.
>>>
>>> FAIL.
>>>
>>>> If we know that we have a black cat then we know that we have a cat.
>>>
>>> Except that if you DON'T have a black cat but think you do then you
>>> are wrong. If H aborts its simulation, it isn't a UTM and doesn't
>>> 'correctly' simulate.
>>>
>>>>
>>>> If we know that we have a sequence of configurations that cannot
>>>> possibly ever reach its final state then we know that we have a
>>>> non-halting sequence of configurations.
>>>>
>>>
>>> Except that is has been PROVEN that if H -> H.Qn then the pattern
>>> WILL reach the final state.
>>>
>>> The fact that H can't ever reach that state proves just proves that
>>> if H is a UTM, which don't abort, then H^ will be non-halting, but H
>>> is still wrong for not answering. If H does abort, then it hasn't
>>> proven anything, and it has been proven that it is wrong.
>>>
>>> FAIL
>>
>> You are either not bright enough to get this or dishonest.
>> I don't care which, I need to up my game to computer scientists.
>>
>
> So, can't refute what I say so you go to arguing by insults, classic
> Olcott logical fallicy.
>

Fundamentally you seem to lack the intellectual capacity to understand
what I am saying. This is proven on the basis that what I am saying can
be verified as true entirely on the basis of the meaning of its words.

> Face it, you are just WRONG about your assertions, maybe because you
> just don't know the field, so don't have any idea what is legal or not.
>
> Also note, you keep talking about needing 'Computer Scientists' to
> understand, that is really incorrect, you need to be able to explain it
> to someone who understands Computation Theory, which is a fairly
> specialized branch of Mathematics. Yes, it is part of the foundation of
> Computer Science, but isn't the sort of thing that a normal Computer
> Scientist will deal with day to day.

I need someone to analyze what I am saying on the deep meaning of what I
am saying instead of mere rote memorized meanings from textbooks.

The key mistake that my reviewers are making is that they believe that
the halt decider is supposed to evaluate its input on the basis of some
proxy for the actual behavior of this actual input rather than the
actual behavior specified by this actual input.

--
Copyright 2021 Pete Olcott "Talent hits a target no one else can hit;
Genius hits a target no one else can see." Arthur Schopenhauer