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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory
Subject: Re: Simulation vs. Execution in the Halting Problem
Date: Mon, 2 Jun 2025 10:23:15 -0500
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On 6/2/2025 1:56 AM, Mikko wrote:
> On 2025-06-01 21:41:36 +0000, olcott said:
> 
>> On 6/1/2025 6:30 AM, Mikko wrote:
>>> On 2025-05-30 15:41:59 +0000, olcott said:
>>>
>>>> On 5/30/2025 3:45 AM, Mikko wrote:
>>>>> On 2025-05-29 18:10:39 +0000, olcott said:
>>>>>
>>>>>> On 5/29/2025 12:34 PM, Mr Flibble wrote:
>>>>>>>
>>>>>>> 🧠 Simulation vs. Execution in the Halting Problem
>>>>>>>
>>>>>>> In the classical framework of computation theory (Turing machines),
>>>>>>> simulation is not equivalent to execution, though they can 
>>>>>>> approximate one
>>>>>>> another.
>>>>>>
>>>>>> To the best of my knowledge a simulated input
>>>>>> always has the exact same behavior as the directly
>>>>>> executed input unless this simulated input calls
>>>>>> its own simulator.
>>>>>
>>>>> The simulation of the behaviour should be equivalent to the real
>>>>> behaviour.
>>>>
>>>> That is the same as saying a function with infinite
>>>> recursion must have the same behavior as a function
>>>> without infinite recursion.
>>>
>>> A function does not have a behaviour. A function has a value for
>>> every argument in its domain.
>>>
>>> A function is not recursive. A definition of a function can be
>>> recursive. There may be another way to define the same function
>>> without recursion.
>>>
>>> A definition of a function may use infinite recursion if it is also
>>> defined how that infinite recursion defines a value.
>>>
>>> Anyway, from the meaning of "simulation" follows that a simulation
>>> of a behaviour is (at least in some sense) similar to the real
>>> behaviour. Otherwise no simulation has happened.
>>>
>>
>> void DDD()
>> {
>>    HHH(DDD);
>>    return;
>> }
>>
>> The *input* to simulating termination analyzer HHH(DDD)
>> specifies recursive simulation that can never reach its
>> *simulated "return" instruction final halt state*
> 
> It does not matter whether a particular simulation does or does not
> reach its "return" instruction.

It completely matters. DDD correctly simulated by HHH
proves the exact behavior that the input to HHH(DDD)
actually specifies.

> It only matters whether whether the
> beahaviour specified by the input (which in this case is DDD) will
> reach its own "return", and it does.
> 

The behavior specified by the input never reaches
its own "return" instruction.

You are confusing the behavior specified by the input
with the behavior of its caller.

int main()
{
   DDD(); // calls HHH(DDD) that cannot report on the behavior
}        // of its caller because its caller is *not* its input.


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