Path: news.eternal-september.org!eternal-september.org!.POSTED!not-for-mail
From: "Fred. Zwarts" <F.Zwarts@HetNet.nl>
Newsgroups: comp.theory
Subject: Re: HHH(DD) --- COMPUTE ACTUAL MAPPING FROM INPUT TO OUTPUT --- Using
 Finite String Transformations
Date: Thu, 24 Apr 2025 22:05:16 +0200
Organization: A noiseless patient Spider
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Op 24.apr.2025 om 21:47 schreef olcott:
> On 4/24/2025 2:15 PM, Fred. Zwarts wrote:
>> Op 24.apr.2025 om 19:46 schreef olcott:
>>> On 4/24/2025 3:11 AM, Fred. Zwarts wrote:
>>>> Op 24.apr.2025 om 05:34 schreef olcott:
>>>>> On 4/23/2025 7:31 PM, Mike Terry wrote:
>>>>>>
>>>>>> ...because HHH stops simulating before reaching that step in the 
>>>>>> computation.  Note that I said
>>>>>>
>>>>>> MT:  Both traces were of course /identical/,
>>>>>>       *up to the point where HHH stops simulating*
>>>>>>
>>>>>> So I was factually correct.
>>>>>>
>>>>>>
>>>>>> Mike.
>>>>>>
>>>>>
>>>>> It *is not* up to the point where HHH stops simulating.
>>>>>
>>>>> It is up to the point where the simulated versus directly
>>>>> executed calls HHH(DD).
>>>>>
>>>> That is exactly the same point. If not, show the difference in the 
>>>> traces before that point.
>>>
>>> As soon as the directly executed DD calls HHH(DD) this
>>> call immediately returns.
>>>
>>> When DD emulated by HHH calls HHH(DD) then HHH emulates
>>> DD and also emulates itself emulating DD. This is one
>>> whole recursive emulation than the directly executed
>>> DD can possibly get to.
> 
>> Again a lot of words, which hide that you cannot show where the traces 
>> differ up to that point.
> 
> THEY DIFFER BY THE EMULATED DD REACHES RECURSIVE EMULATION
> AND THE DIRECTLY EXECUTED DD NEVER DOES.
> 

No, they both reach a finite recursion, but the HHH aborts and the 
direct execution does not abort. That difference, however, is not 
visible in the trace up to that point.