Re: HHH(DDD) is correct to reject its input as non-halting --- EVIDENCE THAT I AM CORRECT

On 2025-06-29 14:37:40 +0000, olcott said:

On 6/29/2025 4:14 AM, Mikko wrote:

On 2025-06-28 13:02:02 +0000, olcott said:
 

On 6/28/2025 3:50 AM, Fred. Zwarts wrote:

Op 28.jun.2025 om 01:30 schreef olcott:

On 6/26/2025 4:16 AM, Fred. Zwarts wrote:

Op 25.jun.2025 om 16:09 schreef olcott:

On 6/25/2025 2:59 AM, Fred. Zwarts wrote:

Op 24.jun.2025 om 16:06 schreef olcott:

 None of the code in HHH can possibly cause DDD correctly
simulated by HHH to reach its own simulated "return" statement.

 Yes, exactly, that is the bug.
 

 Recursive emulation is only a tiny bit more complicated
than recursion yet no one here seems to have a clue.
Do you know what recursion is?
(If you don't that would explain a lot)

As usual irrelevant claims without evidence. No rebuttal.

 Ah so you don't know what recursion is.
 

HHH has a bug that makes that it does not recognise the halting behaviour of the program specified in the input.

 If you don't even know what recursion is then
you are totally unqualified to review these things.
 

Even a beginner can see that the input is a pointer to code, including the code to abort and halt. But HHH is programmed to ignore the conditional branch instructions, when simulating itself, so it thinks that there is an infinite loop when there are only a finite number of recursions.
But Olcott does not understand that not all recursions are infinite.

 When the measure is whether or not DDD correctly
simulated by HHH can possibly reach its own "return"
instruction final halt state nothing inside HHH can
possibly have any effect on this.

 Programs that report about their own behaviour are not useful and
are interesting only if you can derive someting like a paradox.

 Because Turing machines cannot possibly ever take
other actual directly executed Turing Machines as
inputs they use the proxy of finite string Turing
Machine descriptions.

Seems to be sufficient at least for an universal Turing machine.

The behavior of the correctly simulated input to
a partial halt decider is always the same as the
behavior of the directly executed Turing Machine
except when this input calls its own decider.

For a partial decider that should not be a problem as it is permitted
to answer nothing if it cannot determine the ocorrect answer about a
direct execution. But you have a big problem if you want to construct
a complete halting oracle.
--
Mikko