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From: Richard Damon <richard@damon-family.org>
Newsgroups: comp.theory
Subject: Re: DDD specifies recursive emulation to HHH and halting to HHH1
Date: Sat, 29 Mar 2025 18:26:44 -0400
Organization: i2pn2 (i2pn.org)
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On 3/29/25 5:46 PM, olcott wrote:
> 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.

Why? Since a UTM exactly reproduces the behavior of the machine 
described to it, if that won't terminate, then neither can the UTM.

Running forever *IS* exactly reproducing that behavior.

> 
>>>
>>>> 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.

ANd thus UTM1 is not a UTM. PERIOD, and to call it one is just a LIE, 
and it not reaching a final state is no longer a proof of non-halting.

You just want to call your non-street-legal racing car as if it was 
street-legal because that is what you started with.

Telling that to the judge, might get you thrown into jail for contempt 
or perjery.

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

Becuase UTM2 is a real UTM that gives the right answer.

> 
>> 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.
>