Sujet : Re: Simulation vs. Execution in the Halting Problem
De : polcott333 (at) *nospam* gmail.com (olcott)
Groupes : comp.theoryDate : 12. Jun 2025, 16:46:48
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <102esp9$2ohps$7@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
User-Agent : Mozilla Thunderbird
On 6/12/2025 3:50 AM, joes wrote:
Am Wed, 11 Jun 2025 19:20:30 -0500 schrieb olcott:
On 6/11/2025 7:03 PM, wij wrote:
On Wed, 2025-06-11 at 18:45 -0500, olcott wrote:
On 6/11/2025 6:25 PM, wij wrote:
On Wed, 2025-06-11 at 17:33 -0500, olcott wrote:
On 6/11/2025 4:57 PM, wij wrote:
On Wed, 2025-06-11 at 16:44 -0500, olcott wrote:
On 6/11/2025 4:23 PM, wij wrote:
On Wed, 2025-06-11 at 16:10 -0500, olcott wrote:
On 6/11/2025 3:59 PM, wij wrote:
On Wed, 2025-06-11 at 15:30 -0500, olcott wrote:
On 6/11/2025 2:45 PM, wij wrote:
On Wed, 2025-06-11 at 14:39 -0500, olcott wrote:
On 6/11/2025 2:31 PM, wij wrote:
On Wed, 2025-06-11 at 14:14 -0500, olcott wrote:
On 6/11/2025 1:25 PM, wij wrote:
On Wed, 2025-06-11 at 12:59 -0500, olcott wrote:
It turns out that no one ever noticed that simulating
halt deciders nullify the HP counter-example input in
that this input cannot possibly reach its contradictory
part.
DDD does reach that part; HHH doesn't. When HHH simulates DDD, DDD is
not running (on the processor), it is passive data executed by HHH.
Which requires H(D) to report on the behavior of its
caller instead of reporting on the behavior that its
input actually specifies.
What do you mean by "actually specifies"?
There is no *input* to any termination analyzer that can do the
opposite of whatever value that this termination analyzer
returns
>
Your reinterpretation of of HP case is wrong.
Your D or H is not the case mention in the HP proof.
>
There cannot possibly exist any D mine or anyone else's that is
encoded to do the opposite of whatever value that H returns.
What does your DDD do? Do as HHH says?
Yes and we have the exact same issue with TM's it is merely more
difficult to see.
I am not going to get into that until after you totally understand
this at the C level. I am unwilling to talk about this endlessly in
circles.
lol.
D has to be able to perform exactly H's function (if D is a TM and if
H exists).
Otherwise, that D is not the counter-example mentioned in the HP
proof.
>
I have to covered too. Unless you understand that D cannot be both an
input to H and its caller there is no sense going there.
>
If it (D) cannot be both an input to H and its caller, that D is no
resemble of the counter-example mentioned in the HP proof. You made a
crippled D.
>
No D that anyone in the universe can define can simultaneously be the
caller of a function and the input to the same function.
If you think that it can then provide such a D.
Oh, how you are wrong. It is an elementary part of CS that data can be
interpreted as code and code has a data representation.
int main()
{
DDD(); // calls HHH(DDD) its parameter is not its caller
}
-- Copyright 2025 Olcott "Talent hits a target no one else can hit; Geniushits a target no one else can see." Arthur Schopenhauer
Re: Simulation vs. Execution in the Halting Problem
Re: Simulation vs. Execution in the Halting Problem
