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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory
Subject: Re: DDD specifies recursive emulation to HHH and halting to HHH1
Date: Sat, 29 Mar 2025 16:46:26 -0500
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On 3/29/2025 3:14 PM, dbush wrote:
> On 3/29/2025 4:01 PM, olcott wrote:
>> On 3/29/2025 2:26 PM, dbush wrote:
>>> On 3/29/2025 3:22 PM, olcott wrote:
>>>> On 3/29/2025 2:06 PM, dbush wrote:
>>>>> On 3/29/2025 3:03 PM, olcott wrote:
>>>>>> On 3/29/2025 10:23 AM, dbush wrote:
>>>>>>> On 3/29/2025 11:12 AM, olcott wrote:
>>>>>>>> On 3/28/2025 11:00 PM, dbush wrote:
>>>>>>>>> On 3/28/2025 11:45 PM, olcott wrote:
>>>>>>>>>>
>>>>>>>>>> It defines that it must compute the mapping from
>>>>>>>>>> the direct execution of a Turing Machine
>>>>>>>>>
>>>>>>>>> Which does not require tracing an actual running TM, only 
>>>>>>>>> mapping properties of the TM described. 
>>>>>>>>
>>>>>>>> The key fact that you continue to dishonestly ignore
>>>>>>>> is the concrete counter-example that I provided that
>>>>>>>> conclusively proves that the finite string of machine
>>>>>>>> code input is not always a valid proxy for the behavior
>>>>>>>> of the underlying virtual machine.
>>>>>>>
>>>>>>> In other words, you deny the concept of a UTM, which can take a 
>>>>>>> description of any Turing machine and exactly reproduce the 
>>>>>>> behavior of the direct execution.
>>>>>>
>>>>>> I deny that a pathological relationship between a UTM and
>>>>>> its input can be correctly ignored.
>>>>>>
>>>>>
>>>>> In such a case, the UTM will not halt, and neither will the input 
>>>>> when executed directly.
>>>>
>>>> It is not impossible to adapt a UTM such that it
>>>> correctly simulates a finite number of steps of an
>>>> input.
>>>>
>>>
>>> 1) then you no longer have a UTM, so statements about a UTM don't apply
>>
>> We can know that when this adapted UTM simulates a
>> finite number of steps of its input that this finite
>> number of steps were simulated correctly.
> 
> And therefore does not do a correct UTM simulation that matches the 
> behavior of the direct execution as it is incomplete.
> 

It is dishonest to expect non-terminating inputs to complete.

>>
>>> 2) changing the input is not allowed
>>
>> The input is unchanged. There never was any
>> indication that the input was in any way changed.
>>
> 
> False, if the starting function calls UTM and UTM changes, you're 
> changing the input.
> 

When UTM1 is a UTM that has been adapted to only simulate
a finite number of steps and input D calls UTM1 then the
behavior of D simulated by UTM1 never reaches its final
halt state.

When D is simulated by ordinary UTM2 that D does not call
Then D reaches its final halt state.

> Changing the input is not allowed.
I never changed the input. D always calls UTM1.
thus is the same input to UTM1 as it is to UTM2.

-- 
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer