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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory
Subject: Re: DDD specifies recursive emulation to HHH and halting to HHH1
Date: Sat, 29 Mar 2025 21:35:48 -0500
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On 3/29/2025 8:12 PM, Richard Damon wrote:
> On 3/29/25 6:44 PM, olcott wrote:
>> On 3/29/2025 5:08 PM, dbush wrote:
>>> On 3/29/2025 5:46 PM, olcott wrote:
>>>> On 3/29/2025 3:14 PM, dbush wrote:
>>>>> On 3/29/2025 4:01 PM, olcott wrote:
>>>>>> On 3/29/2025 2:26 PM, dbush wrote:
>>>>>>> On 3/29/2025 3:22 PM, olcott wrote:
>>>>>>>> On 3/29/2025 2:06 PM, dbush wrote:
>>>>>>>>> On 3/29/2025 3:03 PM, olcott wrote:
>>>>>>>>>> On 3/29/2025 10:23 AM, dbush wrote:
>>>>>>>>>>> On 3/29/2025 11:12 AM, olcott wrote:
>>>>>>>>>>>> On 3/28/2025 11:00 PM, dbush wrote:
>>>>>>>>>>>>> On 3/28/2025 11:45 PM, olcott wrote:
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> It defines that it must compute the mapping from
>>>>>>>>>>>>>> the direct execution of a Turing Machine
>>>>>>>>>>>>>
>>>>>>>>>>>>> Which does not require tracing an actual running TM, only 
>>>>>>>>>>>>> mapping properties of the TM described. 
>>>>>>>>>>>>
>>>>>>>>>>>> The key fact that you continue to dishonestly ignore
>>>>>>>>>>>> is the concrete counter-example that I provided that
>>>>>>>>>>>> conclusively proves that the finite string of machine
>>>>>>>>>>>> code input is not always a valid proxy for the behavior
>>>>>>>>>>>> of the underlying virtual machine.
>>>>>>>>>>>
>>>>>>>>>>> In other words, you deny the concept of a UTM, which can take 
>>>>>>>>>>> a description of any Turing machine and exactly reproduce the 
>>>>>>>>>>> behavior of the direct execution.
>>>>>>>>>>
>>>>>>>>>> I deny that a pathological relationship between a UTM and
>>>>>>>>>> its input can be correctly ignored.
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>> In such a case, the UTM will not halt, and neither will the 
>>>>>>>>> input when executed directly.
>>>>>>>>
>>>>>>>> It is not impossible to adapt a UTM such that it
>>>>>>>> correctly simulates a finite number of steps of an
>>>>>>>> input.
>>>>>>>>
>>>>>>>
>>>>>>> 1) then you no longer have a UTM, so statements about a UTM don't 
>>>>>>> apply
>>>>>>
>>>>>> We can know that when this adapted UTM simulates a
>>>>>> finite number of steps of its input that this finite
>>>>>> number of steps were simulated correctly.
>>>>>
>>>>> And therefore does not do a correct UTM simulation that matches the 
>>>>> behavior of the direct execution as it is incomplete.
>>>>>
>>>>
>>>> It is dishonest to expect non-terminating inputs to complete.
>>>
>>> An input that halts when executed directly is not non-terminating
>>>
>>>>
>>>>>>
>>>>>>> 2) changing the input is not allowed
>>>>>>
>>>>>> The input is unchanged. There never was any
>>>>>> indication that the input was in any way changed.
>>>>>>
>>>>>
>>>>> False, if the starting function calls UTM and UTM changes, you're 
>>>>> changing the input.
>>>>>
>>>>
>>>> When UTM1 is a UTM that has been adapted to only simulate
>>>> a finite number of steps 
>>>
>>> And is therefore no longer a UTM that does a correct and complete 
>>> simulation
>>>
>>>> and input D calls UTM1 then the
>>>> behavior of D simulated by UTM1 
>>>
>>>
>>> Is not what I asked about.  I asked about the behavior of D when 
>>> executed directly.
>>>
>>
>> Off topic for this thread.
>> UTM1 D DOES NOT HALT
>> UTM2 D HALTS
>> D is the same finite string in both cases.
>>
> 
> No it isn't, not if it is the definition of a PROGRAM.
> 

_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]

The behavior that these machine code bytes specify:
558bec6872210000e853f4ffff83c4045dc3
as an input to HHH is different than these
same bytes as input to HHH1 as a verified fact.

> Or, are you admitting you don't understand the meaning of a program?
> 

It seems that you "just don't believe in" verified facts.

> If D doesn't include the machine it calls, then NOTHING can emulate it 
> past the call instruction without violating the definition of a 
> computation/pure program, which you have admitted is a core requirement 
> of your decider (which it turns out it never met).
> 

The Peter Linz proof explicitly includes the halt
decider embedded within it. The principle is the same.

When Ĥ is applied to ⟨Ĥ⟩ it reaches Ĥ.qn
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

When embedded_H is applied to ⟨Ĥ⟩ ⟨Ĥ⟩ then ⟨Ĥ⟩
does not reach either ⟨Ĥ.qy⟩ or ⟨Ĥ.qn⟩

> Sorry, you are just proving that you don't understand what you are 
> talking about.


-- 
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer