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From: Richard Damon <richard@damon-family.org>
Newsgroups: comp.theory
Subject: Re: DDD specifies recursive emulation to HHH and halting to HHH1
Date: Sun, 30 Mar 2025 07:18:28 -0400
Organization: i2pn2 (i2pn.org)
Message-ID: <a62cf45cf74098558ed56143f702b4cad179f59b@i2pn2.org>
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On 3/29/25 10:35 PM, olcott wrote:
> 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.

But that input isn't a "Program" and can't be "correctly emulated" 
without information from outside the input.

As I said, you don't understand the definition of a program.

R

> 
>> Or, are you admitting you don't understand the meaning of a program?
>>
> 
> It seems that you "just don't believe in" verified facts.

it seems  you don't understand what a "fact" is.

> 
>> 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

Which isn't his Ĥ as Ĥ went into its copy of H, not something else that 
is your embedded_H.

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

But it does, since you say you H does.

if H (Ĥ) (Ĥ) goes to H.qn then embedded_H, being that exact same code 
when used as embedded_H (Ĥ) (Ĥ) will go to embedded_H.qn which is the 
same state as Ĥ.qn and thus Ĥ (Ĥ) will halt, and H and embedded_H are wrong.

You want to claim that H and embedded_H do something different, but they 
can't per the definition of a program. To make you point, you need to 
show the exact first step where H and embedded_H, being the IDENTICAL 
program, with IDENTICAL inputs differ in there behavior.

You can't, and you have demonstrated that by failing to actually answer 
this question for years, but just trying to sdetrack with irrelevant 
details for year.

Sorry, but you are showing that you "proof" is just based on false 
claims and that you are just a pathological liar.
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