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From: dbush <dbush.mobile@gmail.com>
Newsgroups: comp.theory
Subject: Re: Incorrect requirements --- Computing the mapping from the input
 to HHH(DD)
Date: Sun, 11 May 2025 13:44:12 -0400
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In-Reply-To: <vvqnev$i5d0$3@dont-email.me>

On 5/11/2025 1:40 PM, olcott wrote:
> On 5/11/2025 12:21 PM, wij wrote:
>> On Sun, 2025-05-11 at 12:00 -0500, olcott wrote:
>>> On 5/11/2025 11:28 AM, wij wrote:
>>>> On Sun, 2025-05-11 at 10:38 -0500, olcott wrote:
>>>>> On 5/11/2025 9:34 AM, wij wrote:
>>>>>> On Sat, 2025-05-10 at 21:19 -0500, olcott wrote:
>>>>>>> On 5/10/2025 9:09 PM, wij wrote:
>>>>>>>> On Sat, 2025-05-10 at 20:56 -0500, olcott wrote:
>>>>>>>>> On 5/10/2025 8:44 PM, wij wrote:
>>>>>>>>>> On Sat, 2025-05-10 at 20:26 -0500, olcott wrote:
>>>>>>>>>>> On 5/10/2025 8:17 PM, wij wrote:
>>>>>>>>>>>> On Sat, 2025-05-10 at 17:03 -0500, olcott wrote:
>>>>>>>>>>>>> On 5/10/2025 4:44 PM, wij wrote:
>>>>>>>>>>>>>> On Sat, 2025-05-10 at 14:29 -0500, olcott wrote:
>>>>>>>>>>>>>>> On 5/10/2025 2:02 PM, wij wrote:
>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>> You don't know the counter example in the HP proof, your D 
>>>>>>>>>>>> is not the case what HP
>>>>>>>>>>>> says.
>>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>> Sure I do this is it! (as correctly encoded in C)
>>>>>>>>>>>
>>>>>>>>>>> typedef void (*ptr)();
>>>>>>>>>>> int HHH(ptr P);
>>>>>>>>>>>
>>>>>>>>>>> int DD()
>>>>>>>>>>> {
>>>>>>>>>>>         int Halt_Status = HHH(DD);
>>>>>>>>>>>         if (Halt_Status)
>>>>>>>>>>>           HERE: goto HERE;
>>>>>>>>>>>         return Halt_Status;
>>>>>>>>>>> }
>>>>>>>>>>>
>>>>>>>>>>> int main()
>>>>>>>>>>> {
>>>>>>>>>>>         HHH(DD);
>>>>>>>>>>> }
>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> Try to convert it to TM language to know you know nothing.
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>> I spent 22 years on this. I started with the Linz text
>>>>>>>>>
>>>>>>>>> When Ĥ is applied to ⟨Ĥ⟩
>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
>>>>>>>>>        or
>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>>>>
>>>>>>>>> (a) Ĥ copies its input ⟨Ĥ⟩
>>>>>>>>> (b) Ĥ invokes embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩
>>>>>>>>> (c) embedded_H simulates ⟨Ĥ⟩ ⟨Ĥ⟩ ...
>>>>>>>>>
>>>>>>>>> Thus ⟨Ĥ⟩ ⟨Ĥ⟩ correctly simulated by embedded_H
>>>>>>>>> cannot possibly reach its simulated final halt state
>>>>>>>>> ⟨Ĥ.qn⟩
>>>>>>>>>
>>>>>>>>>> To refute the HP, you need to understand what it exactly means 
>>>>>>>>>> in TM.
>>>>>>>>>
>>>>>>>>> I have known this for 22 years.
>>>>>>>>
>>>>>>>> A working TM. Build it explicitly from transition function, then 
>>>>>>>> explain
>>>>>>>> your derivation. You know nothing.
>>>>>>>>
>>>>>>>
>>>>>>> That would be like examining how an operating system
>>>>>>> works entirely from its machine code.
>>>>>>
>>>>>> You are refuting a CS foundamental theorem (i.e. HP) officially.
>>>>>> So, yes, and actually MORE need to be done (beyond your imagination).
>>>>>>
>>>>>> Knowing a car or smart phone,... is far different from making one.
>>>>>> Knowing E=mc^2 is far from knowing relativity, making A-bomb 
>>>>>> (actually, making
>>>>>> A-bomb don't need to know E=mc^2, people are often fooled by 
>>>>>> popular saying)
>>>>>> Every chapter of Linz's book, C text textbook has exercises, you 
>>>>>> need to those
>>>>>> exercises AT LEAST to comment CS (and computation theory is more 
>>>>>> advanced topic
>>>>>> than TM). Saying so is because we know you can't do the exercise 
>>>>>> and boast lots
>>>>>> about TM stuff (and pretty much anything else from just reading 
>>>>>> words), even
>>>>>> about theorem.
>>>>>>
>>>>>
>>>>> When Ĥ is applied to ⟨Ĥ⟩
>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
>>>>>      or
>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>
>>>>> (a) Ĥ copies its input ⟨Ĥ⟩
>>>>> (b) Ĥ invokes embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩
>>>>> (c) embedded_H simulates ⟨Ĥ⟩ ⟨Ĥ⟩
>>>>>
>>>>> All that I need to know is that I proved that
>>>>> embedded_H correctly recognizes the repeating
>>>>> pattern where its correctly simulated ⟨Ĥ⟩ ⟨Ĥ⟩
>>>>> cannot possibly reach its own simulated final
>>>>> halt state of ⟨Ĥ.qn⟩
>>>>>
>>>>> https://www.liarparadox.org/Linz_Proof.pdf
>>>>>
>>>>>>> We only have to actually know one detail:
>>>>>>> Every counter-example input encoded in any model
>>>>>>> of computation always specifies recursive simulation
>>>>>>> that never halts to its corresponding simulating
>>>>>>> termination analyzer.
>>>>>>
>>>>>> More example here that you don't understand nearly all CS terms.
>>>>>>
>>>>>
>>>>> Mere empty rhetoric entirely bereft of any supporting
>>>>> reasoning. The x86 language is comparable to a RASP
>>>>> machine that is equivalent to a Turing machine.
>>>>
>>>> Question:
>>>> 1. Do you understand that you can't do the exercises in Linz's book?
>>>
>>> Everything is 100% irrelevant besides the fact that
>>> I have shown that ⟨Ĥ⟩ ⟨Ĥ⟩ correctly simulated by
>>> embedded_H cannot possibly reach its own simulated
>>> final halt state ⟨Ĥ.qn⟩. Thus when embedded_H reports
>>> on the behavior that its input specifies it can
>>> correctly transition to Ĥ.qn.
>>>
>>>> 2. Do you understand your ability of C/assembly/TM is less than 1 
>>>> year CS level?
>>>>
>>>
>>> I construe C as high level assembly language thus
>>> disregard any inessentials. No change since K & R
>>> is of any use to me. I write C++ the same way. I
>>> use it as C with classes. I also use std::vector a lot.
>>
>> Q3. If people know the capability of the author of POOH is less than 1 
>> year CS
>>      level. How persuasive and reliable of POOH do you think it would be?
>>
>> Q4: Why no one can reproduce the result of POOH for these 22? years?
>>
>>
> 
> _DDD()
> [00002172] 55         push ebp      ; housekeeping
> [00002173] 8bec       mov  ebp,esp  ; housekeeping
> [00002175] 6872210000 push 00002172 ; push DDD
> [0000217a] e853f4ffff call 000015d2 ; call HHH(DDD)
> [0000217f] 83c404     add  esp,+04
> [00002182] 5d         pop  ebp
> [00002183] c3         ret
> Size in bytes:(0018) [00002183]
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