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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory
Subject: Re: not identical deciders
Date: Thu, 25 Jul 2024 09:39:21 -0500
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On 7/25/2024 9:24 AM, joes wrote:
> Am Thu, 25 Jul 2024 08:56:37 -0500 schrieb olcott:
>> On 7/24/2024 10:29 PM, Mike Terry wrote:
>>> On 23/07/2024 14:31, olcott wrote:
>>>> On 7/23/2024 1:32 AM, 0 wrote:
>>>>> On 2024-07-22 13:46:21 +0000, olcott said:
>>>>>> On 7/22/2024 2:57 AM, Mikko wrote:
>>>>>>> On 2024-07-21 13:34:40 +0000, olcott said:
>>>>>>>> On 7/21/2024 4:34 AM, Mikko wrote:
>>>>>>>>> On 2024-07-20 13:11:03 +0000, olcott said:
>>>>>>>>>> On 7/20/2024 3:21 AM, Mikko wrote:
>>>>>>>>>>> On 2024-07-19 14:08:24 +0000, olcott said:
>>>>>>>>>>>
>>>>>>>>>>>> When we use your incorrect reasoning we would conclude that
>>>>>>>>>>>> Infinite_Loop() is not an infinite loop because it only
>>>>>>>>>>>> repeats until aborted and is aborted.
> 
>>>>>>>> (b) We know that a decider is not allowed to report on the
>>>>>>>> behavior computation that itself is contained within.
>>>>>>> No, we don't. There is no such prohibition.
> 
>>>>> If a Turing machine can take a description of a TM as its input or as
>>>>> a part of its input it can also take its own description.
>>>>> Every Turing machine can be given its own description as input but a
>>>>> Turing machine may interprete it as something else.
>>>>>
>>>> In this case we have two x86utm machines that are identical except
>>>> that DDD calls HHH and DDD does not call HHH1.
> So they are not identical.
> 
>>>> It is empirically proven that this changes their behavior and the
>>>> behavior of DDD.
>> It is empirically proven according to the semantics of the x86 machine
>> code of DDD that DDD correctly emulated by HHH has different behavior
>> than DDD correctly emulated by HHH1.
> So they can't be identical.
> 
>> If you care study the code that I just provided
>> you can see that when DDD is correctly emulated by HHH that DDD does
>> correctly have different behavior than DDD correctly emulated by HHH1.
> 
>> HHH1 and HHH are essentially identical and the only reason why DDD
>> correctly emulated by HHH has different behavior than DDD correctly
>> emulated by HHH1 is that DDD does call HHH in recursive emulation and
>> DDD does not call HHH1 in recursive emulation.
> 
>> When DDD calls HHH in recursive emulation (as I have proven that it
>> does**) and DDD does not call HHH1 in recursive simulation (as I have
>> proven that it does not**)
>> then DDD will have different behavior.
> 
>> I don't believe that someone of your intelligence and knowledge could
>> possibly actually fail to understand that the behavior of function DDD
>> emulated by a function that it calls in recursive emulation would not be
>> different than the behavior of this same function DDD emulated by a
>> function that it does not call in recursive emulation.
> 
>> HHH does see recursive emulation that will never stop unless aborted.
>> HHH1 does not see this.
> 
>> HHH does see recursive emulation that will never stop unless aborted.
>> HHH1 does not see this, because HHH has already aborted its DDD. When
>> any one call of infinite recursion has been aborted then it is infinite
>> recursion that abnormally terminates.
> 
> One function calls itself, one doesn't.
> 

typedef void (*ptr)();
int HHH(ptr P);
int HHH1(ptr P);

void DDD()
{
   HHH(DDD);
}

int main()
{
   HHH(DDD);
   HHH1(DDD);
}

HHH and HHH1 implement the exact same algorithm.
The difference in behavior is the simple fact that DDD
calls HHH(DDD) in recursive emulation and does not call
HHH1(DDD) in recursive emulation.

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