Re: Simulation vs. Execution in the Halting Problem

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Sujet : Re: Simulation vs. Execution in the Halting Problem
De : polcott333 (at) *nospam* gmail.com (olcott)
Groupes : comp.theory
Date : 02. Jun 2025, 18:49:49
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <101ko7t$3e8mq$1@dont-email.me>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
User-Agent : Mozilla Thunderbird
On 6/2/2025 12:42 PM, Mike Terry wrote:
On 02/06/2025 18:02, olcott wrote:
On 6/2/2025 11:34 AM, Mike Terry wrote:
On 02/06/2025 04:16, olcott wrote:
On 6/1/2025 9:43 PM, Mike Terry wrote:
On 01/06/2025 17:23, olcott wrote:
On 6/1/2025 10:42 AM, Mike Terry wrote:
On 01/06/2025 05:01, olcott wrote:
On 5/31/2025 10:32 PM, Mike Terry wrote:
On 01/06/2025 02:31, olcott wrote:
On 5/31/2025 7:44 PM, Mike Terry wrote:
On 01/06/2025 01:18, olcott wrote:
[..snip..]
>
Your response below basically says that you don't understand what "simulation" is.
>
That's not at all surprising, but given this, you are in no position to make assertions about whether two simulations are the same or not, up to the point where one of them is aborted.
>
I'll probably go back up the thread to where I asked you questions, and just give the answers myself for the record.
>
>
We cannot do a separate side-by-side execution trace of
HHH(DDD) and HHH1(DDD) because the DDD simulated by HHH1
calls HHH(DDD) as a part of this same simulation.
>
Duh!  The DDD simulated by HHH ALSO calls HHH(DDD) as a part of the same simulation.
>
They BOTH call HHH(DDD) as part of the simulation.  Duuuuuh....
>
I've presented the two traces to you side by side on more than one occasion.  Do you really have no recollection of that? Your explanation of why we supposedly can't put them side by side is literally gibberish!
>
>
 From the trace shown below we can see that HHH simulates
DDD one whole execution trace more than HHH1 does.
>
Really?  That's not at all what I see - but perhaps you can explain what you're saying.
>
Mark on the trace below where you think HHH1's simulation [i.e. the simulation /performed/ by HHH1] starts and ends.  Also mark where you think HHH's simulation starts and ends.
>
Then to save me the trouble, try to put them side by side to see if they match up...
>
>
Mike.
>
>
I really appreciate your sincere honesty and the great
diligence that you have shown evaluating my work. No
one else on the planet has put nearly the same effort
as you in carefully evaluating the key details of my work.
>
>
There is a terminology issue here to resolve.
>
If A simulates B and B simulates C, what should a "trace" of A's simulation look like?
a)   it includes both B's and C's instructions, interlaced.
b)   it is just B's instructions.
>
>
HHH1(DDD) simulates one instance of DDD.
HHH(DDD) simulates DDD and simulates itself simulating DDD
and then aborts after it has already simulated DDD one more
time than HHH1 ever does.
>
HHH1(DDD) simulates one instance of DDD, then its nested simulations.
>
HHH(DDD) simulates one instance of DDD, then its nested simulations.
>
IT'S THE SAME, UP TO THE POINT WHERE DDD IS ABORTED, EXACTLY AS I CLAIMED.
>
>
(1) HHH simulates DDD and then simulates itself simulating DDD.
(2) HHH1 never ever simulates itself.
These two are not the same.
>
They are sooo the same!
(1) HHH simulates DDD and then simulates HHH simulating DDD.
(2) HHH1 simulates DDD and then simulates HHH simulating DDD.
>
Dude, we are talking about the /simulations/, not the simulators.
>
>
Because the simulators do simulate other simulators we
cannot simply ignore this.
 The identity of the outer simulators is not relevent as it does not affect the course of the simulation, only how far the simulation is allowed to proceed.  So we CAN ignore it - all that matters is how far it continues the simulation.
 Obviously if a simulation includes an inner simulator, that simulator is a part of the simulation and we cannot ignore /that/.
 
We much keep track of some
of the details.
>
HHH1(DDD) simulates DDD that calls a simulated HHH(DDD)
that simulates DDD and then simulates itself simulating DDD.
>
This is getting boring...
 HHH1(DDD) simulates DDD that calls a simulated HHH(DDD)
that simulates DDD that calls a simulated simulated HHH
that simulates DDD simulating DDD...
 HHH(DDD) simulates DDD that calls a simulated HHH(DDD)
that simulates DDD that calls a simulated simulated HHH
AND THEN OUTER HHH ABORTS ITS SIMULATION.
  *THE SIMULATIONS ARE EXACTLY THE SAME UP TO THE POINT WHERE HHH ABORTS ITS SIMULATION*
  Mike.
 
When you fail to pay attention to all the key details
it may seem this way.
How many times does HHH1(DDD) simulate itself simulating DDD?
How many times does HHH(DDD) simulate itself simulating DDD?
*Saying that you don't think it makes a difference dodges the question*
--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer

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