Re: Simulation vs. Execution in the Halting Problem

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