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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory
Subject: Re: 197 page execution trace of DDD correctly simulated by HHH
Date: Fri, 28 Jun 2024 12:05:18 -0500
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On 6/28/2024 11:26 AM, joes wrote:
> Am Fri, 28 Jun 2024 10:25:36 -0500 schrieb olcott:
>> On 6/28/2024 8:14 AM, joes wrote:
>>> Am Thu, 27 Jun 2024 12:30:38 -0500 schrieb olcott:
> 
>>>> When this is construed as non-halting criteria then simulating
>>>> termination analyzer H0 is correct to reject these inputs as
>>>> non-halting by returning 0 to its caller.
>>> To the caller DDD, which then returns to its own caller H0, which
>>> returns „halting” to main… hold on.
> Where do you disagree?
> 
>>>> Simulating termination analyzers must report on the behavior that
>>>> their finite string input specifies thus H0 must report that DDD
>>>> correctly emulated by H0 remains stuck in recursive simulation.
>>> H0 must not report on itself, only on DDD. Which you’ve proven halts.
>>> We don’t care how H0 deviates (i.e. is incorrect) in its simulation.
>>> That would be main {H0(H0(DDD))}.
> 
>> The behavior of the directly executed DDD() is irrelevant because that
>> is not the behavior of the input.
> What is the difference here?
> 
>> Deciders compute the mapping from
>> their actual finite string input to an output by a sequence of finite
>> string transformations.
> And should get the right answer.
> 
>> In this case the sequence is the line-by-line execution trace of the
>> behavior of DDD correctly emulated by HHH.
> No, the sequence is the behaviour of DDD, period.
> 
>> The behavior of this input must include and cannot ignore the recursive
>> emulation specified by the fact that DDD is calling its own emulator.
> Yes, and the behaviour of H0 is that it produces the exact same behaviour
> as DDD.
> 

You don't seem to understand basic facts.

_DDD()
[00002172] 55               push ebp      ; housekeeping
[00002173] 8bec             mov ebp,esp   ; housekeeping
[00002175] 6872210000       push 00002172 ; push DDD
[0000217a] e853f4ffff       call 000015d2 ; call HHH(DDD)
[0000217f] 83c404           add esp,+04
[00002182] 5d               pop ebp
[00002183] c3               ret
Size in bytes:(0018) [00002183]

The call from DDD to HHH(DDD) when N steps of DDD are
correctly emulated by any pure function x86 emulator
HHH cannot possibly return.

That you assume that it does against the facts is ridiculous.

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