XPost: sci.math, comp.theory
From: mikko.levanto@iki.fi
olcott kirjoitti 28.11.2025 klo 16.54:
> On 11/28/2025 2:29 AM, Mikko wrote:
>> olcott kirjoitti 27.11.2025 klo 17.17:
>>> On 11/27/2025 1:40 AM, Mikko wrote:
>>>> olcott kirjoitti 26.11.2025 klo 17.15:
>>>>> On 11/26/2025 3:22 AM, Mikko wrote:
>>>>>> olcott kirjoitti 26.11.2025 klo 5.17:
>>>>>>> On 11/25/2025 9:09 PM, Kaz Kylheku wrote:
>>>>>>>> On 2025-11-26, olcott wrote:
>>>>>>>>> On 11/25/2025 8:36 PM, Kaz Kylheku wrote:
>>>>>>>>>> On 2025-11-26, olcott wrote:
>>>>>>>>>>> None of them ever had the slightest clue about Montague
>>>>>>>>>>> Grammar. Except for one they all had very severe math
>>>>>>>>>>> phobia.
>>>>>>>>>>
>>>>>>>>>> So do you; you are terribly afraid of the mathematical idea that
>>>>>>>>>> simulations that are paused still exist and have future
>>>>>>>>>> states.
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>> I am not going to discuss your psychotic nonsense.
>>>>>>>>
>>>>>>>> In all honesty, you and your therapist /should/ be laser focused
>>>>>>>> on your
>>>>>>>> own psychotic nonsense.
>>>>>>>>
>>>>>>>>> You already agreed that I am correct so this subject
>>>>>>>>> is closed.
>>>>>>>>
>>>>>>>> Whaaat ...
>>>>>>>>
>>>>>>>>> news://news.eternal-september.org/20251104183329.967@kylheku.com
>>>>>>>>> On 11/4/2025 8:43 PM, Kaz Kylheku wrote:
>>>>>>>>>> On 2025-11-05, olcott wrote:
>>>>>>>>>>>
>>>>>>>>>>> The whole point is that D simulated by H
>>>>>>>>>>> cannot possbly reach its own simulated
>>>>>>>>>>> "return" statement no matter what H does.
>>>>>>>>>>
>>>>>>>>>> Yes; this doesn't happen while H is running.
>>>>>>>>>>
>>>>>>>>>> So while H does /something/, no matter what H does,
>>>>>>>>>> that D simulation won't reach the return statement.
>>>>>>>>
>>>>>>>> But we know that. If H is nonreturning, of course D is.
>>>>>>>> Since D calls H(D), D is suspended until H(D) returns,
>>>>>>>> which means forever if H(D) is nonterminating.
>>>>>>>>
>>>>>>>> I have no idea what you are trying to milk out of this;
>>>>>>>> it is completely uncontroversial.
>>>>>>>
>>>>>>> I really did figure out how to determine the
>>>>>>> correct halt status that the halting problem's
>>>>>>> counter-example input specifies to it decider.
>>>>>>
>>>>>> The basic halting problem is about Turing machines. A Turing machine
>>>>>> specifies only one bhavour. It does not specify anything else to the
>>>>>> decider. An ambiguous program is outside of the domain of the halting
>>>>>> problem.
>>>>>
>>>>> That is inaccurate.
>>>>
>>>> No, it is not. Of course there are many ways to formulate the problem
>>>> but what I said is true about the basic formulation. All formulations
>>>> restrict the domain to unambiguous specifications.
>>>>
>>>
>>> It is a perfectly unambiguous ultimately
>>> self-contradictory specification.
>>
>> If the specification of D is perfectly unambiguous there is no point
>> to say "specifies to its decider".
>
> That DD simulated by HHH never stops running
> unless aborted by HHH proves that the input
> to HHH(DD) specifies non halting behavior.
That DD halts when executed by a C conforming execution environment
proves that the input to HHH(DD) specifies a halting behaviour. If
the DD simulated by HHH specifies a different behaviour then the DD
simulated by HHH is not the DD given as the input. If the answer
returned by HHH(DD) does not mean "halts" then HHH is not a halt
decider nor a partial halt decider.
> The caller of a function is never an argument to
> this same function.
A C function cannot have any function as an argument, only a pointer
to a function of a particular type. A pointer to any function of the
right type can be given as the argument, including the a pointer to
the calling function.
> The DD executed in main that
> calls HHH(DD) is not the same DD as the one that
> HHH simulates.
However, it is the one that is geven as the input in HHH(DD).
--
Mikko
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