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

On 5/31/25 12:45 PM, 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:
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On 5/29/2025 12:34 PM, Mr Flibble wrote:

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🧠 Simulation vs. Execution in the Halting Problem
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In the classical framework of computation theory (Turing machines),
simulation is not equivalent to execution, though they can approximate one
another.

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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.

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The simulation of the behaviour should be equivalent to the real
behaviour.

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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]
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DDD emulated by HHH must be aborted.   // otherwise infinite recursion
DDD emulated by HHH1 need not be aborted.
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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.
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Sorry, your acknoldegement that you DDD just isn't a program means that its representation is NOT correctly emulatable.
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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.
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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.
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Thus, you have admitted that your whole system of logic is based on using incorrect definitions and lies.
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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.
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A function that calls its own simulator specifies different
behavior than a function that does not call its own simulator.

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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.
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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.
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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⟩

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What "Issue"? Your problme is you don't understand what you are taling about.
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When Ĥ is applied to ⟨Ĥ⟩
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

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Why did the copy of H change to being called "embedded_H"?
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I am using cleaner notational conventions.

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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
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Which then gets stop when the embedded_H invoked at the first invocation of (b) doing step (c) decides to abort its emulation.
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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.
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 *EVERYONE ALWAYS GETS THIS WRONG*

How, since it is. Or are you just showing you think lying is ok

 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.
 

Right, it reports on the behavior of the computation that is represented to it, even if that just happens to be its caller.
You are just showing that you don't understand the basic meaning of the terms you are using, and that you recklessly refuse to learn them.
It seems your idea of "logic" isn't based on actual logic, but on your ability to lie.
Sorry, but that *IS* what you are telling the world by your actions.