Re: Who here understands that the last paragraph is Necessarily true?

Op 19.jul.2024 om 16:23 schreef olcott:

On 7/19/2024 2:57 AM, Mikko wrote:

On 2024-07-17 13:27:08 +0000, olcott said:
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On 7/17/2024 2:43 AM, Mikko wrote:

On 2024-07-16 18:24:49 +0000, olcott said:
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On 7/16/2024 3:12 AM, Mikko wrote:

On 2024-07-15 02:33:28 +0000, olcott said:
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On 7/14/2024 9:04 PM, Richard Damon wrote:

On 7/14/24 9:27 PM, olcott wrote:

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Any input that must be aborted to prevent the non termination
of simulating termination analyzer HHH necessarily specifies
non-halting behavior or it would never need to be aborted.

>
Excpet, as I have shown, it doesn't.
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Your problem is you keep on ILEGALLY changing the input in your argument because you have misdefined what the input is.
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>
_DDD()
[00002163] 55         push ebp      ; housekeeping
[00002164] 8bec       mov ebp,esp   ; housekeeping
[00002166] 6863210000 push 00002163 ; push DDD
[0000216b] e853f4ffff call 000015c3 ; call HHH(DDD)
[00002170] 83c404     add esp,+04
[00002173] 5d         pop ebp
[00002174] c3         ret
Size in bytes:(0018) [00002174]
>
The input *is* the machine address of this finite
string of bytes: 558bec6863210000e853f4ffff83c4045dc3

>
You have already said that a decider is not allowed to answer anything
other than its input. Now you say that the the program at 15c3 is not
a part of the input. Therefore a decider is not allowed consider it
even to the extent to decide whether it ever returns. But without that
knowledge it is not possible to determine whether DDD halts.
>

>
It maps the finite string 558bec6863210000e853f4ffff83c4045dc3
to non-halting behavior because this finite string calls HHH(DDD)
in recursive simulation.

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That mapping is not a part of the finite string and not a part of the
problem specification.

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decider/input pairs <are> a key element of the specification.

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 Computable functions are the formalized analogue of the intuitive notion of algorithms, in the sense that a function is computable if there exists an algorithm that can do the job of the function, i.e. given an input of the function domain it can return the corresponding output. https://en.wikipedia.org/wiki/Computable_function
 

Not of any specification of any interesting problem.
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 Everyone here seems to think that they can stupidly ignore
the fact that an input calls its own decider and make pretend
that this pathological relationship does not exist.
 

The finite string does not reveal what is the
effect of calling whatever that address happens to contain.

>
A simulating termination analyzer proves this.

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Irrelevant, as you just said it is not a part of the input.
>

 It is not part of the input in that we already know that
HHH halts and we only need to find out whether or not DDD halts.
 

How do you know that HHH halts? The direct execution? But then we already know that DDD halts as well, because its direct execution halts as well.
But since you seem to think that a correct simulation can change the behaviour of a program, you should also show that the behaviour of HHH does not change when simulated.
DDD is a misleading and unneeded complication. It is easy to eliminate DDD:
        int main() {
          return HHH(main);
        }
This has the same problem. This proves that the problem is not in DDD, but in HHH, which halts when it aborts the simulation, but it decides that the simulation of itself does not halt.
HHH is simply unable to decide about finite recursions.
void Finite_Recursion (int N) {
   if (N > 0) Finite_Recursion (N - 1);
}
It decides after N recursions that there is an infinite recursion, which is incorrect.
Olcott's HHH is programmed to abort the simulation after N cycles of recursive simulations. Therefore, it is incorrect to abort the simulation of HHH when the simulated HHH has performed only N-1 cycles, because that changes the behaviour of HHH.
Since the simulated HHH always runs one cycle behind the simulating HHH, it is clear that HHH can never simulate enough cycles for a correct simulation, as is required by the x86 language.
Therefore, the simulation is incorrect according to the criteria olcott stipulated.
The conclusion is simple:
HHH cannot possibly simulate itself correctly.
No matter how much olcott wants it to be correct, or how many times olcott repeats that it is correct, it does not change the fact that such a simulation is incorrect, because it is unable to reach the end.
Olcott's own claim that the simulated HHH does not reach its end confirms it. The trace he has shown also proves that HHH cannot reach the end of its own simulation. So, his own claims prove that it is true that HHH cannot possibly simulate itself up to the end, which makes the simulation incorrect.
Sipser would agree that this incorrect simulation cannot be used to detect a non-halting behaviour.
Olcott could not point to an error, but prefers to ignore it. So, I will repeat it, until either an error is found, or olcott admits that HHH cannot possibly simulate itself correctly.