Re: Simulation vs. Execution in the Halting Problem --- Linz

On 5/31/2025 12:05 PM, Mr Flibble wrote:

On Sat, 31 May 2025 11:45:21 -0500, olcott wrote:
 

On 5/31/2025 7:35 AM, Richard Damon wrote:

On 5/31/25 2:41 AM, olcott wrote:

On 5/30/2025 8:16 PM, Richard Damon wrote:

On 5/30/25 11:41 AM, olcott wrote:

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.

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Nope. Where does it say that?
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>

_DDD()
[00002192] 55             push ebp [00002193] 8bec           mov
ebp,esp [00002195] 6892210000     push 00002192 [0000219a]
e833f4ffff     call 000015d2  // call HHH [0000219f] 83c404
add esp,+04 [000021a2] 5d             pop ebp [000021a3]
c3             ret Size in bytes:(0018) [000021a3]
>
DDD emulated by HHH must be aborted.   // otherwise infinite recursion
DDD emulated by HHH1 need not be aborted.
>
>

No, nether HHH or HHH1 can correctly emulate this input, as it is
incomplete, andt eh call HHH is IMPOSSIBLE to emulate in this input, as
its target is not in the input, so any action that looks elsewhere for
it is just not emulating THIS INPUT.
>
Sorry, your acknoldegement that you DDD just isn't a program means that
its representation is NOT correctly emulatable.
>
Fix that issue by including in the input the code of the HHH that
calling it, and we find that HHH just doesn't correctly emulate it, as
it has been defined, in halt7.c, to chose to abort at a spot where the
program does not stop.
>
Of course, in your mind HHH isn't a program either, as you refuse to
accept that it must be a single program. Note, an infinite set of
programs is not A program.
>
>
Thus, you have admitted that your whole system of logic is based on
using incorrect definitions and lies.
>

Your problem is you don't understand that the only things you can
correctly simulate are PROGRAMS, which mean they include all of their
code, which is also expressed in the input given to the simulator.
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>

Whether it actually is depends on the quality of the simulator.
There is no exception for the case when the simulator is called. If
the behaviour in the simulation is different from a real execution
then the simulation is wrong.
>
>

A function that calls its own simulator specifies different behavior
than a function that does not call its own simulator.

>
No it doesn't, as we can only be talking about programs, and programs
don't know who "their simulator" is, only what simulator they were
built to use.
>
>

One of the advantages of Turing machines is that there is no
possibility to call anything so the effect of calling the simulator
need not be considered.
>
>

The same issue occurs in the Linz proof, it is merely more difficult
to see. The correctly simulated ⟨Ĥ⟩ ⟨Ĥ⟩ cannot possibly reach

its

own simulated final halt state ⟨Ĥ.qn⟩

>
What "Issue"? Your problme is you don't understand what you are
taling about.
>
>

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

embedded_H ⟨Ĥ⟩

⟨Ĥ⟩ ⊢* Ĥ.qn

>
Why did the copy of H change to being called "embedded_H"?
>
>

I am using cleaner notational conventions.

>
How is using two names for the same thing cleaner?
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>

>

(a) Ĥ copies its input ⟨Ĥ⟩
(b) Ĥ invokes embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩
(c) embedded_H simulates ⟨Ĥ⟩ ⟨Ĥ⟩
(d) simulated ⟨Ĥ⟩ copies its input ⟨Ĥ⟩
(e) simulated ⟨Ĥ⟩ invokes simulated embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩
(f) simulated embedded_H simulates ⟨Ĥ⟩ ⟨Ĥ⟩
(g) goto (d) with one more level of simulation
>
>

Which then gets stop when the embedded_H invoked at the first
invocation of (b) doing step (c) decides to abort its emulation.
>
>

I have it emulate one more level before stopping.
It would stop at (e).

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Rigth, so you go (a) (b) (c) (e) (h) return to H^.qn (i) H^ Halts
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Thus showing the input is non-halting.
>
>

*EVERYONE ALWAYS GETS THIS WRONG*
>
int main()
{
    DDD(); // The HHH that DDD calls is not supposed to
}        // report on the behavior of its caller, nitwit
>
Likewise embedded_H is not supposed to report on the behavior of the
computation that itself is embedded within.

 I think I see where you are going wrong: HHH has to call DDD in main();
HHH has to be the outermost context for a halting result of NON-HALTING to
be valid.
 /Flibble
 /Flibble

It is always correct for every HHH(DDD) to reject its
input as specifying a non-halting sequence of
configurations.
The fact that the directly executed DDD() that calls
HHH(DDD) does halt is none-of-the-business of HHH.
HHH is only accountable for the behavior that its
actual input actually specifies.
*This seems to be brand new computer science*
--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer