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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory
Subject: Re: Incorrect requirements --- Computing the mapping from the input
 to HHH(DD)
Date: Sun, 11 May 2025 15:19:41 -0500
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On 5/11/2025 1:38 PM, wij wrote:
> On Sun, 2025-05-11 at 12:40 -0500, 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]
>>
>> All anyone need do to show that I am wrong
>> is provide the steps where DDD emulated by
>> HHH according to the rules of the x86 language
>> reaches its own emulated "ret" instruction.
>>
>> Because no one can actually correctly show any
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