Re: This first time anyone In the entire history of the halting problem derived a correct return value for HHH(DD)

On 12/8/2024 4:55 AM, Mikko wrote:

On 2024-12-05 04:20:50 +0000, olcott said:
 

There is an 80% chance that I will be alive in one month.
There may be an extended pause in my comments.
I will try to bring a computer to the out of town hospital.
>
On 12/4/2024 8:16 PM, Richard Damon wrote:

On 12/4/24 8:50 PM, olcott wrote:

On 12/4/2024 7:45 PM, Richard Damon wrote:

On 12/4/24 8:41 PM, olcott wrote:

On 12/4/2024 7:31 PM, Richard Damon wrote:

On 12/4/24 8:06 PM, olcott wrote:

On 12/4/2024 6:11 PM, Richard Damon wrote:

On 12/4/24 9:27 AM, olcott wrote:

On 12/3/2024 5:39 PM, Richard Damon wrote:

On 12/3/24 6:08 PM, olcott wrote:

On 12/3/2024 3:03 AM, joes wrote:

Am Mon, 02 Dec 2024 20:48:49 -0600 schrieb olcott:

On 11/28/2024 1:48 PM, joes wrote:

>

You said:
  >>> HHH can't simulate itself.
That is WRONG !!!
HHH DOES EMULATE ITSELF PROVING THAT IT CAN EMULATE ITSELF.

We know that HHH halts. It doesn't simulate itself halting.
>

>
Please try and explain how you are not dishonest what you
try to change the subject from my rebuttal of your statement:
>
 >>> HHH can't simulate itself.
>
That HHH does emulate itself emulating DDD proves
THAT IT CAN DO THIS.
>

>
But only if your think that wrong answer can be right.

>
I did not mention anything about answers my entire
scope is that HHH does emulate itself emulating DDD
thus conclusively proving that HHH can emulated itself
emulating DDD.
>
Whenever you go out-of-scope like this it surely
seems dishonest to me.
>

>
But the behaivor that HHH shows that it has *IS* an "answer",

>
DDD emulated by any HHH according to the semantics of
the x86 language cannot possibly reach its "ret" instruction
whether HHH aborts this emulation after N steps or never aborts.
>
>

>
Just a nonsense sentence, since HHH can't emulate HHH as it isn't given it,

>
Why do you have to keep fucking lying about this?
I could die on the operating table in two weeks!
>

>
What's the lie?
>
Can you point to what I say that is wrong, and a reliable reference that show it?
>
All you have is your own lies to call it a lie.
>
And yes, you might die in two weeks, and the only thing you will have left behind is all your lies.

>
Yes you fucking jackass this conclusively proves that
HHH does emulate itself emulating DDD.
>
https://github.com/plolcott/x86utm/blob/master/Halt7.c
>
>

>
Nope.
>
It proves that your HHH fails to meet its requirement to be pure function

>
It proves that HHH does emulate itself emulating DDD.
>
Once we get through this point then we know that DDD
does not halt:
>
DDD emulated by any HHH according to the semantics of
the x86 language cannot possibly reach its "ret" instruction
whether HHH aborts this emulation after N steps or never aborts.
*This tells us that DDD emulated by HHH DOES NOT HALT*
>
We do not begin to examine whether or not HHH found this
answer as a pure function until after we agree with the
prior point.
>
*In all of the history of the halting problem there*
*have never been a correct return value for this*
>
typedef void (*ptr)();
int HHH(ptr P);
>
int DD()
{
   int Halt_Status = HHH(DD);
   if (Halt_Status)
     HERE: goto HERE;
   return Halt_Status;
}
>
int main()
{
   HHH(DD);
}

 This is not a useful main. It is sufficient to determine whether HHH
returns but not to determine whther it returns the correct value.
 

When we understand that the first point is correct
then we know that HHH returning 0 is correct.
*This has much has never ever been done before*

 This is one of the well known methods to prove that the value HHH returns
is incorrect. If HHH returns 0 then DD returns 0 but the meaning of 0 in
this context is that DD does not halt. THerefore the value returned by
HHH is incorrect.
 

Every expert in the C programming language has agreed that DD
simulated by HHH cannot possibly return. Everyone disagreeing
with this has dishonestly used to strawman deception to refer to different behavior than DD simulated by HHH.
void DDD()
{
   HHH(DDD);
   return;
}
The behavior of DDD simulated by HHH is isomorphic.
I provide this example for people that are only novices
with C programming.

Then after we know that HHH(DD) is correct to return
0 for its input we get to applying Mike's idea to
make HHH a pure function.

 We know otherwise.
 

--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer