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From: joes <noreply@example.org>
Newsgroups: comp.theory
Subject: Re: 197 page execution trace of DDD correctly simulated by HHH
Date: Sun, 30 Jun 2024 20:16:58 -0000 (UTC)
Organization: i2pn2 (i2pn.org)
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Am Sun, 30 Jun 2024 12:25:55 -0500 schrieb olcott:
> On 6/30/2024 3:42 AM, joes wrote:
>> Am Sat, 29 Jun 2024 15:03:02 -0500 schrieb olcott:
>>> On 6/29/2024 2:44 PM, joes wrote:
>>>> Am Fri, 28 Jun 2024 14:28:20 -0500 schrieb olcott:
>>>>> On 6/28/2024 2:18 PM, joes wrote:
>>>>>> Am Fri, 28 Jun 2024 12:53:46 -0500 schrieb olcott:
>>>>>>> On 6/28/2024 12:41 PM, joes wrote:
>>>>>>>> Thanks for leaving the unanswered questions in place, though I’d
>>>>>>>> rather have you answer them.

>>>>>> Why doesn’t the first recursive H return?
>> It should abort, just like the outer one.
> The outer HHH meets its abort criteria one execution trace sooner than
> the next inner one because HHH needs to see two complete execution
> traces before its abort criteria has been met.
But it sees that the inner one would abort, so it can let it end normally.

> As soon as the outer HHH sees the inner one complete one full execution
> trace then the outer one has its abort criteria.
Same as the inner does, so the inner one does not in fact repeat.

>> No, I mean: why does the inner simulator repeat instead of aborting,
>> the same as the outer one does?
Repeat.

>> My point is: all recursive calls both enter and detect a
>> repeating state.
> The inner ones always see one less execution trace than the next outer
> one, thus could only meet their abort criteria after they have already
> been aborted.
Why are they aborted?

-- 
Am Fri, 28 Jun 2024 16:52:17 -0500 schrieb olcott:
Objectively I am a genius.