Re: Simulation vs. Execution in the Halting Problem

On Wed, 2025-06-11 at 19:20 -0500, olcott wrote:

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:

 
Yes all other people (especially Dennis Bush) are saying
that H(D) is required to report on the behavior of the
direct execution of D() never noticing that this stupidly
requires H(D) to report on the behavior of its caller.

 
If the H above means the H that the HP refers to. The H is
required to
report its argument's behavior (ie. by H(D)). But NOT required by
simulation.

 
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.
 

The HP does not care what D does (simply to say).
 

 
Everyone says that H(D) must re[port on the behavior of
the direct execution of D().

 
That is what the HP asks.
 

The HP only requires: H(D)==1 iff D() halts
 
 

 
int main()
{
           D(); // calls H(D)
}
 
Which requires H(D) to report on the behavior of its
caller instead of reporting on the behavior that its
input actually specifies.

 
That is no problem. H does not care what D does inside (simply to
say).
The HP simply asks for a H that "H(D)==1 iff D() halts".
 

 
Which requires H to report on something that it cannot possibly see.

 
On the contrary, what the HP proves is very useful.
 

 
I am not talking about the halting problem, I have always
been talking about the conventional halting problem proof.
THIS PROOF IS WRONG

 
When talking about proof, we say it is valid or not.. By doing so, we have
to unambiguously pose the problem and the derivation to the conclusion.
The HP proof just did that.
 

 
It may seem that way if you pay less than 100%
complete attention.
 
The HP proof depends on an *INPUT* that does
the opposite of whatever value that H returns
and no such *INPUT* can possibly exist.

 
That is absolutely correct. No such *INPUT* (i.e. D) can possible exit is because
the H inside D does not exist at all.
So, if the H is assumed to exist, then D will exist to make H undecidable.
 

 
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.

 
Why not? D and H are supposed to be TM (or C function).
If the D cannot do the opposite of whatever value that H returns, then
that D is not powerful enough to be a TM, not an interesting case.
 

 
Can you be your biological mother's biological father?

 
What is the same reason? What's the relationship of 1+1=2 relates to HP?
 

It is for this same reason that the function's caller
cannot simultaneously be its input.

 
D and H belong to the same set of TM equivalent stuff.

 
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.

 
The problem is that you don't know TM and C as 1-year CS student does.
All the people here have problem to get the answer fits your level of understanding.
 

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.

Then, what does your example mean?

void D() {
  H(D);
}

D cannot simultaneously be the caller of a function
and the input to the same function?
Are you refuting what you said?

If you think that it can then provide such a D.