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From: olcott <polcott2@gmail.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 11:25:42 -0600
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On 3/8/2024 1:47 AM, Richard Damon wrote:
> On 3/7/24 11:29 PM, olcott wrote:
>> On 3/8/2024 12:48 AM, Richard Damon wrote:
>>> On 3/7/24 10:30 PM, olcott wrote:
>>>> On 3/8/2024 12:09 AM, Richard Damon wrote:
>>>>> On 3/7/24 9:38 PM, olcott wrote:
>>>>>> On 3/7/2024 9:53 PM, Richard Damon wrote:
>>>>>>> On 3/7/24 7:29 PM, olcott wrote:
>>>>>>>> On 3/7/2024 8:34 PM, Richard Damon wrote:
>>>>>>>>> On 3/7/24 6:02 PM, 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?
>>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
>>>>>>>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn   // Ĥ applied to ⟨Ĥ⟩ does 
>>>>>>>>>> not halt
>>>>>>>>>>
>>>>>>>>>> When this is executed in an Olcott machine then
>>>>>>>>>> Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ <Ĥ> is a different computation than H ⟨Ĥ⟩ ⟨Ĥ⟩ <H>
>>>>>>>>>
>>>>>>>>> WHY?
>>>>>>>>>
>>>>>>>>> The Master UTM will not change the usage at H^.H, because that 
>>>>>>>>> is just an internal state of the Machine H^
>>>>>>>>>
>>>>>>>>
>>>>>>>> The ONLY thing that the master UTM does differently is append
>>>>>>>> the TMD to the TMD's own tape.
>>>>>>>
>>>>>>> Right, so it gives H and H^ that input.
>>>>>>>
>>>>>>> H^ can erase that input and replace it.
>>>>>>>
>>>>>>>>
>>>>>>>>> At that point we have the IDENTICAL set of transitions (with 
>>>>>>>>> just an equivalence mapping of state numbers) as H will have, 
>>>>>>>>> and the EXACT same input as H
>>>>>>>>
>>>>>>>> it is stipulated by the definition of Olcott machines
>>>>>>>> Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ <Ĥ> // last element is <Ĥ> (not H)
>>>>>>>
>>>>>>> Nope.
>>>>>>>
>>>>>>> H^.H isn't a "Olcott Machine" it is a sub-machine of H^
>>>>>>
>>>>>> You already know that there is no such thing as sub-machines of
>>>>>> Turing machines. Turing machines have a set of states and a tape.
>>>>>> They have no sub-machines.
>>>>>
>>>>> A "Sub-Machine" of a Turing Machine is where you put a copy of 
>>>>> another Turing Machine into the state space of the overall/parent 
>>>>> machine,
>>>>>
>>>>> When the machine H^ reaches the state H^.H, then it encounters the 
>>>>> EXACT sequence of states (after the eqivalence mapping generated 
>>>>> when inserting them H's q0 -> H^.Hqo and so on)
>>>>>
>>>>> It is called a sub-machine because it acts largely as if that 
>>>>> Turing Machihe "called" the equivalent Turing Machine as a 
>>>>> subroutine, only the machine code was expanded "inline".
>>>>
>>>> There is no way that any conventional Turing machine can tell
>>>> that the states of another Turing machine are embedded withing it.
>>>
>>> It doesn't need to "KNOW", but it HAS THEM there.
>>>
>>> It was DESIGNED that way, so the programmer knows what he did.
>>>
>>
>> immibis thought that Ĥ could copy external <H> on top of internal <Ĥ>
>> Your words seemed to agree with this.
> 
> It can't get the <H> from an external source, but can have a copy of 
> that inside of it and use that to replace it.
> 

Not unless it is yet another parameter, thus diverges
too far from the original Linz.

> Think of it as copying the description from its const data segment.
> 

-- 
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer