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From: "Fred. Zwarts" <F.Zwarts@HetNet.nl>
Newsgroups: comp.theory
Subject: Re: Computable Functions --- OUTPUTS MUST CORRESPOND TO INPUTS
Date: Wed, 23 Apr 2025 19:44:05 +0200
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Op 23.apr.2025 om 17:32 schreef olcott:
> On 4/23/2025 6:25 AM, joes wrote:
>> Am Tue, 22 Apr 2025 13:51:48 -0500 schrieb olcott:
>>> On 4/22/2025 1:07 PM, Fred. Zwarts wrote:
>>>> Op 22.apr.2025 om 18:28 schreef olcott:
>>>>> On 4/22/2025 7:57 AM, joes wrote:
>>>>>> Am Tue, 15 Apr 2025 15:44:06 -0500 schrieb olcott:
>>
>>>>>>> You continue to stupidly insist that int sum(int x, int y) {return x
>>>>>>> + y; }
>>>>>>> returns 7 for sum(3,2) because you incorrectly understand how these
>>>>>>> things fundamentally work.
>>>>>>>
>>>>>>> It is stupidly wrong to expect HHH(DD) report on the direct
>>>>>>> execution of DD when you are not telling it one damn thing about
>>>>>>> this direct execution.
>>>>>> What else is it missing that the processor uses to execute it?
>>>>>>
>>>>> libx86emu <is> a correct x86 processor and does emulate its inputs
>>>>> correctly.
>>>>
>>>> The key thing here is that Olcott consistently does not understand that
>>>> HHH is given a finite string input that according to the semantics of
>>>> the x86 language specifies a halting program,
>>>
>>> That is stupidly incorrect.
>> No, DD halts (when executed directly). HHH is not a halt decider, not 
>> even
>> for DD only.
>>
>>> People here stupidly assume that the outputs are not required to
>>> correspond to the inputs.
>> But the direct execution of DD is computable from its description.
>>
> 
> Not as an input to HHH.
> When HHH computes halting for DD is is only allowed
> to apply the finite string transformations specified
> by the x86 language to the machine code of DD.
> 
> Because DD DOES CALL ITS OWN EMULATOR this does require
> the behavior measured is DD emulating by HHH including
> HHH emulating itself emulating DDD.

Indeed, HHH fails to see the finite recursion completely, because HHH 
aborts and therefore does not reach the end of the halting program.
This makes HHH blind for the behaviour of the program specified by the 
finite string.
Indeed, a blind program cannot see the correct behaviour of the input.