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From: "Fred. Zwarts" <F.Zwarts@HetNet.nl>
Newsgroups: comp.theory
Subject: Re: DDD simulated by HHH cannot possibly halt (Halting Problem) ---
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Date: Mon, 14 Apr 2025 19:15:59 +0200
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Op 14.apr.2025 om 13:56 schreef olcott:
> On 4/14/2025 4:25 AM, joes wrote:
>> Am Sun, 13 Apr 2025 21:11:56 -0500 schrieb olcott:
>>> On 4/13/2025 6:11 PM, Richard Damon wrote:
>>>> On 4/13/25 5:00 PM, olcott wrote:
>>>>> On 4/13/2025 3:00 PM, dbush wrote:
>>>>>> On 4/13/2025 3:59 PM, olcott wrote:
>>>>>>> On 4/13/2025 3:54 AM, joes wrote:
>>>>>>>> Am Fri, 11 Apr 2025 10:56:32 -0500 schrieb olcott:
>>>>>>>>> On 4/11/2025 3:24 AM, Richard Heathfield wrote:
>>>>>>>>>> On 11/04/2025 08:57, Mikko wrote:
>>
>>>>>>>>>> Mr Olcott can have his principle if he likes, but only by EITHER
>>>>>>>>>> proving it (which, as you say, he has not yet done) OR by taking
>>>>>>>>>> it as axiomatic, leaving the world of mainstream computer science
>>>>>>>>>> behind him,
>>>>>>>>>> constructing his own computational 'geometry' so to speak, and
>>>>>>>>>> abandoning any claim to having overturned the Halting Problem.
>>>>>>>>>> Navel contemplation beckons.
>>>>>>>>>> Axioms are all very well, and he's free to invent as many as he
>>>>>>>>>> wishes, but nobody else is obliged to accept them.
>>>>>>>>>>
>>>>>>>>> *Simulating termination analyzer Principle*
>>>>>>>>> It is always correct for any simulating termination analyzer to
>>>>>>>>> stop simulating and reject any input that would otherwise prevent
>>>>>>>>> its own termination.
>>>>>>>> Sure. Why doesn’t the STA simulate itself rejecting its input?
>>>>>>>>
>>>>>>> Because that is a STUPID idea and categorically impossible because
>>>>>>> the outermost HHH sees its needs to stop simulating before any inner
>>>>>>> HHH can possibly see this.
>>>>>>>
>>>>>> In other words, you agree that Linz and others are correct that no H
>>>>>> exists that satisfies these requirements:
>>>>>> Given any algorithm (i.e. a fixed immutable sequence of instructions)
>>>>>> X described as <X> with input Y:
>>>>>> A solution to the halting problem is an algorithm H that computes the
>>>>>> following mapping:
>>>>>> (<X>,Y) maps to 1 if and only if X(Y) halts when executed directly
>>>>>> (<X>,Y) maps to 0 if and only if X(Y) does not halt when executed
>>>>>> directly
>>>>>>
>>>>> No stupid! Those freaking requirements are wrong and*
>>>>> anchored in the ignorance  of rejecting the notion of a simulating
>>>>> termination analyzer OUT-OF-HAND WITHOUT REVIEW.
>>>> No, those "freeking requirement" *ARE* the requirements
>>> AND AS STUPID AS {REQUIRING} A GEOMETRIC SQUARE CIRCLE IN THE SAME
>>> TWO-DIMENSIONAL PLANE.
>> Nothing is stupid about wanting a halt decider. It’s just not obvious
>> that it’s impossible.
>>
> 
> When people insist that a termination analyzer reports
> on behavior other than the behavior that its finite string
> input specifies this is isomorphic to requiring a perfectly
> geometric square circle in the same two dimensional plane,
> simply logically impossible, thus an incorrect requirement.

It is very clear what its finite string input specifies: when exactly 
this same finite string input is used in direct execution or in 
world-class simulators, we see that it specifies a halting program 
according to the unique semantics of the x86 language.
It is not clear what a geometric square circle is. So, your comparison 
fails.
But I think we agree that there is no algorithm that can determine for 
all possible inputs whether the input specifies a program that 
(according to the semantics of the machine language) halts when directly 
executed. Correct?