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From: immibis <news@immibis.com>
Newsgroups: comp.theory,sci.logic
Subject: Re: Refutation of the Peter Linz Halting Problem proof 2024-03-05
 --partial agreement--
Date: Fri, 8 Mar 2024 03:11:02 +0100
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On 8/03/24 03:02, olcott wrote:
> On 3/7/2024 7:35 PM, immibis wrote:
>> On 7/03/24 18:05, olcott wrote:
>>> On 3/7/2024 8:47 AM, immibis wrote:
>>>> On 7/03/24 03:40, olcott wrote:
>>>>> On 3/6/2024 8:22 PM, immibis wrote:
>>>>>> On 7/03/24 01:12, olcott wrote:
>>>>>>> On 3/6/2024 5:59 PM, immibis wrote:
>>>>>>>> On 7/03/24 00:55, olcott wrote:
>>>>>>>>> Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn
>>>>>>>>> Correctly reports that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ must abort its simulation.
>>>>>>>>>
>>>>>>>>> H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qy
>>>>>>>>> Correctly reports that H ⟨Ĥ⟩ ⟨Ĥ⟩ need not abort its simulation.
>>>>>>>>
>>>>>>>> What are the exact steps which the exact same program with the 
>>>>>>>> exact same input uses to get two different results?
>>>>>>>> I saw x86utm. In x86utm there is a mistake because Ĥ.H is not 
>>>>>>>> defined to do exactly the same steps as H, which means you 
>>>>>>>> failed to do the Linz procedure.
>>>>>>>
>>>>>>> Both H(D,D) and H1(D,D) answer the exact same question:
>>>>>>> Can I continue to simulate my input without ever aborting it?
>>>>>>>
>>>>>>
>>>>>> Both H(D,D) and H1(D,D) are computer programs (or Turing 
>>>>>> machines). They execute instructions (or transitions) in sequence, 
>>>>>> determined by their programming and their input.
>>>>>
>>>>> Yet because they both know their own machine address
>>>>> they can both correctly determine whether or not they
>>>>> themselves are called in recursive simulation.
>>>>
>>>> They cannot do anything except for exactly what they are programmed 
>>>> to do.
>>>
>>> H1(D,D) and H(D,D) are programmed to do this.
>>> Because H1(D,D) simulates D(D) that calls H(D,D) that
>>> aborts its simulation of D(D). H1 can see that its
>>> own simulated D(D) returns from its call to H(D,D).
>>>
>>>>>
>>>>> An Olcott machine can perform an equivalent operation.
>>>>>
>>>>> Because Olcott machines are essentially nothing more than
>>>>> conventional UTM's combined with Conventional Turing machine
>>>>> descriptions their essence is already fully understood.
>>>>>
>>>>> The input to Olcott machines can simply be the conventional
>>>>> space delimited Turing Machine input followed by four spaces.
>>>>>
>>>>> This is followed by the machine description of the machine
>>>>> that the UTM is simulating followed by four more spaces.
>>>>
>>>> To make the Linz proof work properly with Olcott machines, Ĥ should 
>>>> search for 4 spaces, delete its own machine description, and then 
>>>> insert the description of the original H. Then the Linz proof works 
>>>> for Olcott machines.
>>>
>>> That someone can intentionally break an otherwise correct
>>> halt decider
>>
>> It always gives exactly the same answer as the working one, so how is 
>> it possibly broken?
>
> [non-answer removed]

Perhaps you didn't understand the question. How is a machine broken if 
it always gives the right answer?