Sujet : Re: Simulation vs. Execution in the Halting Problem
De : richard (at) *nospam* damon-family.org (Richard Damon)
Groupes : comp.theoryDate : 12. Jun 2025, 23:58:46
Autres entêtes
Organisation : i2pn2 (i2pn.org)
Message-ID : <d63ae91d8844447cfe0c3bec2421573228061e34@i2pn2.org>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22
User-Agent : Mozilla Thunderbird
On 6/12/25 11:46 AM, olcott wrote:
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
}
But is the repesentation of it,
I guess you mind just don't understand representation, and since almost everything we do is based on that concept, you are showing you don't understand much about anything.
Re: Simulation vs. Execution in the Halting Problem
Re: Simulation vs. Execution in the Halting Problem
