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From: "Fred. Zwarts" <F.Zwarts@HetNet.nl>
Newsgroups: comp.theory
Subject: Re: Defining a correct halt decider
Date: Sat, 7 Sep 2024 12:31:26 +0200
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Op 06.sep.2024 om 13:41 schreef olcott:
> On 9/6/2024 6:12 AM, Mikko wrote:
>> On 2024-09-05 13:39:14 +0000, olcott said:
>>
>>> On 9/5/2024 2:39 AM, Mikko wrote:
>>>> On 2024-09-03 13:17:56 +0000, olcott said:
>>>>
>>>>> On 9/3/2024 3:44 AM, Mikko wrote:
>>>>>> On 2024-09-02 16:06:11 +0000, olcott said:
>>>>>>
>>>>>>> A correct halt decider is a Turing machine T with one accept 
>>>>>>> state and one reject state such that:
>>>>>>>
>>>>>>> If T is executed with initial tape contents equal to an encoding 
>>>>>>> of Turing machine X and its initial tape contents Y, and 
>>>>>>> execution of a real machine X with initial tape contents Y 
>>>>>>> eventually halts, the execution of T eventually ends up in the 
>>>>>>> accept state and then stops.
>>>>>>>
>>>>>>> If T is executed with initial tape contents equal to an encoding 
>>>>>>> of Turing machine X and its initial tape contents Y, and 
>>>>>>> execution of a real machine X with initial tape contents Y does 
>>>>>>> not eventually halt, the execution of T eventually ends up in the 
>>>>>>> reject state and then stops.
>>>>>>
>>>>>> Your "definition" fails to specify "encoding". There is no standard
>>>>>> encoding of Turing machines and tape contents.
>>>>>
>>>>> That is why I made the isomorphic x86utm system.
>>>>> By failing to have such a concrete system all kinds
>>>>> of false assumptions cannot be refuted.
>>>>
>>>> If it were isnomorphic the same false assumtipns would apply to both.
>>>
>>> They do yet I cannot provide every single details of
>>> the source-code of the Turing machine because these
>>> details would be too overwhelming.
>>>
>>> So instead every author makes a false assumption that
>>> is simply believed to be true with no sufficient basis
>>> to show that it isn't true.
>>>
>>> Once I prove my point as the x86 level I show how the
>>> same thing applies to the Peter Linz proof.
>>
>> Your recent presentations are so far from Linz' proof that they
>> look totally unrelated.
>>
> 
> I must begin where people are so far no one even understands
> the concept of recursive emulation.
> 
> _DDD()
> [00002172] 55         push ebp      ; housekeeping
> [00002173] 8bec       mov ebp,esp   ; housekeeping
> [00002175] 6872210000 push 00002172 ; push DDD
> [0000217a] e853f4ffff call 000015d2 ; call HHH(DDD)
> [0000217f] 83c404     add esp,+04
> [00002182] 5d         pop ebp
> [00002183] c3         ret
> Size in bytes:(0018) [00002183]
> 
> Show the details of how DDD emulated by HHH
> reaches its own machine address 0000217f.
> 
> 00002172, 00002173, 00002175, 0000217a calls HHH(DDD)
> then
> 00002172, 00002173, 00002175, 0000217a calls HHH(DDD)...
> 
> WHAT SHOULD THE NEXT STEPS BE?
> 

That has been told now several times. A correct simulation (as by HHH1, 
or the unmodified world class simulator) show that the next instruction 
is at 000015d2 (in HHH) and when HHH returns the next instruction in DDD 
is 0000217f, 00002182, 00002183 and DDD halts.
But, indeed, HHH fails to reach that part of the simulation, because it 
decided to stop the simulation too soon.
The problem is the code that tries to detect an infinite recursion. It 
fails to see that there are only two recursion for each HHH.
Olcott is still dreaming of a HHH that will not see this 'special 
condition'. He does not realise that by adding the code to recognize 
this 'special condition' and stop the simulation, the other HHH, that 
does not see this 'special condition', disappeared and remains only in 
his dreams. HHH, when simulating *itself* should now decide about the 
DDD that uses this new HHH that sees this 'special condition' and aborts.
*That code* is where the problem is, not the incomplete simulation itself.
In fact, if Olcott would have found a solution for the halting problem, 
then it would not be in the simulation, but in the detection of the 
'special condition'. The correctness of the simulation is only relevant 
because it is used as input for the code to detect the 'special 
condition'. The proof that the halting problem cannot be solved with a 
computation implies that Olcott's detection of the 'special condition' 
cannot be correct.
It is probable that Olcott understands that it cannot be correct and 
therefore he hides the code for the recognition of this 'special condition'.
He probably knows that if he would publish this part of the decider, 
people would spot many errors in it immediately.
He has only shown some trivial examples as evidence that this detection 
of the 'special condition' is correct, such as Infinite_Loop and 
Infinite_Recursion, but, of course, a few trivial examples do not prove 
the correctness of this algorithm.
Therefore, he keeps pulling the attention away from the correctness of 
the detection of the 'special condition', to the correctness of the 
simulation.