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From: "Fred. Zwarts" <F.Zwarts@HetNet.nl>
Newsgroups: comp.theory
Subject: Re: What it would take...
Date: Sat, 10 May 2025 17:25:40 +0200
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Op 10.mei.2025 om 17:07 schreef olcott:
> On 5/9/2025 9:18 PM, Richard Damon wrote:
>> On 5/9/25 9:11 PM, Richard Heathfield wrote:
>>> The HHH code doesn't exactly invite confidence in its author, and his 
>>> theory is all over the place, but a thought experiment suggests itself.
>>>
>>> If we were not all wasting our time bickering with a career 
>>> bickerer... if we were to really /really/ try, could we patch up his 
>>> case and send him on to his Turing Award? And if so, how?
>>>
>>> ISTR that there is suspected to be a theoretical window for him, so I 
>>> suppose what I'm asking is what sort of boathook we would need to 
>>> poke that window a little wider.
>>>
>>> Can he even get there from here? Evidence would suggest that 
>>> simulation is a dead end unless he can find a way to get the 
>>> simulated program to include its own simulation in its behaviour, 
>>> which he has not yet managed to do - but /is/ there a way?
>>>
>>> Or could he abandon simulation completely and instead write a TM 
>>> parser that builds an AST and walks it looking for evidence of 
>>> terminating or looping? If he could, would that turn the trick?
>>>
>>> Or do we have a latter day Cantor waiting in the wings to close the 
>>> window once and for all?
>>>
>>> Is there, in short, any way of putting out this un-halting flame war 
>>> and turning this group to better use?
>>>
>>
>> If he was willing to include the code for HHH in the input 
>> representing DDD, then HHH would be able to atempt to correctly 
>> emulate this input.
>>
> 
> DDD and HHH have always been in the same memory space.
> DDD is the program under test and HHH is the test program.

And because DDD calls HHH, HHH (with the conditional abort) is part of 
the input.

> 
>> There have been methods put forward, that given an acceptace of the 
>> detectability of DDD calling HHH, which can only be done it seems if 
>> we make the system non-turing complete by saying that the input 
>> program and the decider are put into the same memory space, and we are 
>> not allowed to "copy" an algorithm to make a new copy, but only call 
>> the origianal version so HHH can detect the recursion by reference to 
>> that address that some versions of programs that do this "recursive 
>> simulation" can be correctly decider (but not all, like the 
>> pathological version).
>>
> 
> HHH is essentially a UTM thus EVERYTHING is in the memory space
> of its UTM tape.

Except that HHH has a conditional abort, which makes that it is blind 
for the most important part of the input: the conditional abort. This 
bugs makes that HHH is unable to see the specified behaviour.

> 
>> In this method, the Decider detecting the recursion, tries emulating 
>> the code in two parrallel branches based on both possible answers, and 
>> if one branch matches the behavior of the answer, it can return that 
>> answ3er.
>>
> 
> We can emulate a branch as if a conditional expression
> was true and as if it was false. HHH determines the
> behavior of DDD on the basis of what would happen if
> HHH never aborted.

Which is not the input, because the input specifies a HHH that does 
abort. This change of input makes that HHH computes an incorrect mapping.

> 
> <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
> 

Which is a vacuous statement, because it does not correctly simulate its 
input, but it simulates another input. The input specified a finite 
recursion, so the simulator is wrong when it decides that its simulated 
D would never stop.

>> Thus, inputs like DDD() that always halt on the return of HHH(DDD) 
>> will be correctly determined to be halting, and a varient that just 
>> goes into an infinite loop can be such detected by well know 
>> procedures as non- halting.
>>
> 
> When HHH continues to emulate DDD until HHH sees the
> repeating pattern that would prevent DDD from ever
> terminating

No, because it is only a finite repeating pattern, which would 
terminate, because the simulated HHH aborts.

> 
>      H can abort its simulation of D and correctly report that D
>      specifies a non-halting sequence of configurations.
> </MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>

A vacuous statement, because the required conditions are not met.

> 
>> The pathological DD() is detected as pathological, and we can perhaps 
>> extend the definition to allow it to respiond with an "I give up, I 
>> don't know the answer" response, but such an extention explicitly does 
>> NOT meet the requirements, but is better than not answering or giving 
>> a wrong answer.
>>
>> The problem is this result doesn't meet Peter's Goal, as it isn't 
>> really the Halting Problem that he has his major problem with, but the 
>> fact that Turings Proof became the basis for a number of broader 
>> proofs of incompleteness and undeciability that fills formal logic.
>>
>> This is what he can't handle, and this is because he just doesn't 
>> really understand how all of the works because his mental models of 
>> logic is just way to small and simple.
> 
> Dodge and weave is all you know.
> 
> int DD()
> {
>    int Halt_Status = HHH(DD);
>    if (Halt_Status)
>      HERE: goto HERE;
>    return Halt_Status;
> }
> 
> It is a verified fact that DD correctly emulated
> by any simulating termination analyzer HHH specifies
> a non-terminating sequence of configurations.

Only if HHH does not abort. If HHH aborts, it returns to DD and DD 
halts. A correct simulation should see that.