Liste des Groupes | Revenir à c theory |
On 5/6/25 10:40 PM, olcott wrote:I still have the original email.On 5/6/2025 6:00 PM, Richard Damon wrote:Right, how is HHH correct to abort its emulation?On 5/6/25 1:54 PM, olcott wrote:>On 5/6/2025 6:06 AM, Richard Damon wrote:>On 5/5/25 10:29 PM, olcott wrote:>On 5/5/2025 8:06 PM, Richard Damon wrote:>On 5/5/25 11:51 AM, olcott wrote:>On 5/5/2025 10:17 AM, Mr Flibble wrote:>What constitutes halting problem pathological input:>
>
Input that would cause infinite recursion when using a decider of the
simulating kind.
>
Such input forms a category error which results in the halting problem
being ill-formed as currently defined.
>
/Flibble
I prefer to look at it as a counter-example that refutes
all of the halting problem proofs.
>
int DD()
{
int Halt_Status = HHH(DD);
if (Halt_Status)
HERE: goto HERE;
return Halt_Status;
}
Which isn't a program until you include the SPECIFIC HHH that it refutes, and thus your talk about correctly emulated by HHH is just a lie.
>>>
https://github.com/plolcott/x86utm
>
The x86utm operating system includes fully
operational HHH and DD.
https://github.com/plolcott/x86utm/blob/master/Halt7.c
>
When HHH computes the mapping from *its input* to
the behavior of DD emulated by HHH this includes
HHH emulating itself emulating DD. This matches
the infinite recursion behavior pattern.
>
And *ITS INPUT*, for the HHH that answers 0, is the representation of a program
Not at all. This has always been stupidly wrong.
The input is actually a 100% perfectly precise
sequence of steps. With pathological self-reference
some of these steps are inside the termination analyzer.
>
Can't be, as the input needs to be about a program, which must, by the definition of a program, include all its algorithm.
>
Yes, there are steps that also occur in the termination analyzer, but they have been effectively copied into the program the input describes.
>
Note, nothing says that the representation of the program has to be an assembly level description of it. It has to be a complete description, that 100% defines the results the code will generate (and if it will generate) but it doesn't need to be the exact assembly code,
>
YOU even understand that, as you present the code as "C" code, which isn't assembly.
>
What you forget is that the input program INCLUDES as its definiton, all of the code it uses, and thus the call to the decider it is built on includes that code into the decider, and that is a FIXED and DETERMINDED version of the decider, the one that THIS version of the input is designed to make wrong.
>
This doesn't change when you hypothosize a different decider looking at THIS input.
>
<MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
If simulating halt decider H correctly simulates its
input D until H correctly determines that its simulated D
*would never stop running unless aborted* then
>
*would never stop running unless aborted*
Refers to a hypothetical HHH/DD pair of the same HHH that
DD calls except that this hypothetical HHH never aborts.
>
Right, but a correct simulation of D does halt,
How the Hell is breaking the rules specified
by the x86 language possibly correct?
>Yes, and you do, because most of what you say IS "random gibberish"
I could say that the sum of 5 + 7 is a dirty sock
according to the rules of random gibberish.
>But the problem is you don't do that, but think you are because you don't know the rules.
When I go by the rules of arithmetic I am proved
wrong.
>No it isn't.
DD <is> emulated by HHH according to the rules
of the x86 language that specify the HHH also
emulates itself emulating DD
>Which isn't part of the rules.
until HHH determines that for the hypothetical
HHH/DD pair where the hypothetical HHH never
aborts DD would never stop running.
>That second paragraph is a lie and a misquote.
<MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
If simulating halt decider H correctly simulates its
input D until H correctly determines that its simulated D
*would never stop running unless aborted* then
>
*would never stop running unless aborted*
refers to the hypothetical HHH/DD pair where
HHH never aborts its simulation.
>
Sorry, you are just makeing it obvious to all how stupid you are, and that you consider lying a correct form of logic.Not quite.
Professor Sipser said H could abort it simulation of D when it can prove that the correct simulation of D (by whoever) would not halt.
That is the original (as it is the only) D.If you only glance at the words before artificially
Your problem is you think constants can be changed, and that things that are the same can be different, and that things that are defined different can be treated the same.The correct way for simulating halt deciders to work
This is because your logic is built on fantasy and lies.--
Les messages affichés proviennent d'usenet.