Re: Simulation vs. Execution in the Halting Problem

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:

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🧠 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.

 Nope. Where does it say that?
 

_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.

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.
 

>

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.

>
(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).

Of courese, if it never does, then H can nover do it either (since they are DEFINED to be the exact same algorithm) and thus H failea to be a decider.
 You are just showing that your "logic" is based on LYING about what you are doing, and that you are too stupid to understand the requirements, and you don't care to learn them (or are just mentally incapable).
 

Rejected as non-halting at (e) can't reach simulated ⟨Ĥ.qn⟩ state.
--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer