XPost: comp.theory, comp.lang.c++
From: 643-408-1753@kylheku.com

On 2025-11-19, olcott  wrote:
> On 11/18/2025 10:52 PM, Kaz Kylheku wrote:
>> On 2025-11-19, olcott  wrote:
>>> On 11/18/2025 10:31 PM, Kaz Kylheku wrote:
>>>> On 2025-11-19, olcott  wrote:
>>>>> On 11/18/2025 8:53 PM, Kaz Kylheku wrote:
>>>>>> On 2025-11-19, olcott  wrote:
>>>>>>> On 11/18/2025 7:01 PM, Kaz Kylheku wrote:
>>>>>>>> On 2025-11-18, olcott  wrote:
>>>>>>>>> On 11/18/2025 3:21 PM, Kaz Kylheku wrote:
>>>>>>>>>> On 2025-11-18, olcott  wrote:
>>>>>>>>>>> If you ask a decider to determine if my
>>>>>>>>>>> sister's name is "Sally" and I don't tell
>>>>>>>>>>> it who I am then the information contained
>>>>>>>>>>> in the input is insufficient. This does not
>>>>>>>>>>> in any way limit computation itself.
>>>>>>>>>>
>>>>>>>>>> The problem is that UTM(D) can work out the fact that
>>>>>>>>>> D halts. Why is it that UTM knows that D's sister's
>>>>>>>>>> name is Sally, but H does not?
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>> UTM(D) is answering a different question.
>>>>>>>>> (a) It is not providing any answer at all.
>>>>>>>>
>>>>>>>> Well, of course, by "UTM" we mean a /decider/ that purely simulates:
>>>>>>>>
>>>>>>>>      bool UTM(ptr P) {
>>>>>>>>        sim S = sim_create(P);
>>>>>>>>        sim_step_exhaustively(S);
>>>>>>>>        return true;
>>>>>>>>      }
>>>>>>>>
>>>>>>>> All deciders applied to D are tasked with answering exactly the same
>>>>>>>> question.
>>>>>>>>
>>>>>>>> Pretending that a different question was asked is nonproductive;
>>>>>>>> the answer will be interpreted to the original question.
>>>>>>>>
>>>>>>>> All the information needed to answer is positively contained in D.
>>>>>>>>
>>>>>>>> It is just too complex relative to H.
>>>>>>>>
>>>>>>>
>>>>>>> What The F does UTM decide when DD calls UTM(DD)?
>>>>>>
>>>>>> That doesn't happen; DD calls HHH(DD).
>>>>>>
>>>>>> A diagonal functon set against UTM, call it DDUTM,
>>>>>> cannot be decided by UTM(DDUTM).
>>>>>>
>>>>>> That call simply does not return.
>>>>>>
>>>>>
>>>>> Yes, and the other one does return proving the
>>>>> whole point that I have been making for three
>>>>> years that everyone (besides Ben) was too damned
>>>>> dishonest to acknowledge has been true all along.
>>>>
>>>> What "other one"? Is that referring to HHH(DD)?
>>>>
>>>> HHH(DD) returns; UTM(DDUTM) does not return.
>>>>
>>>> That's four functions; HHH isn't UTM; DD isn't DDUTM.
>>>>
>>>> HHH and DDUTM are unrelated; UTM and DD are unrelated.
>>>>
>>>
>>> void DDD()
>>> {
>>>     HHH(DDD);
>>>     return;
>>> }
>>>
>>> A simulating termination analyzer that must
>>> abort the interpretation of the above ASCII
>>> string to prevent its own non-termination
>>> has different behavior than a simulating
>>> termination analyzer that need not abort
>>> its interpretation of the above exact same
>>> ASCII string.
>>
>> 1. Up to that abort point, UTM(DDD) and HHH(DDD) conduct an
>> absolutely identical simulation. The only difference is
>> that the simulation continues under UTM, and is abandoned
>> under HHH.
>>
>
> No that is counter-factual please try again.

Iron-clad, unvarnished truth.

I mean that the simulaton states are identicdal, not that absolutely
all calculations in the scenario are the same.

You may be thinking that the internals of the simulation are different
because at every step HHH tests some conditions whch UTM does not.
However, this has no effect on the simulation state.

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