Re: Simulating Termination Analyzer HHH correctly rejects input DDD

On 10/11/24 8:41 AM, olcott wrote:

On 10/11/2024 4:47 AM, Mikko wrote:

On 2024-10-11 01:55:37 +0000, olcott said:
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On 10/9/2024 6:48 PM, Richard Damon wrote:

On 10/9/24 2:46 PM, olcott wrote:

On 10/9/2024 6:46 AM, Richard Damon wrote:

On 10/9/24 7:01 AM, olcott wrote:

On 10/9/2024 1:08 AM, Jeff Barnett wrote:

On 10/8/2024 6:49 AM, Andy Walker wrote:

... after a short break.
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     Richard -- no-one sane carries on an extended discussion with
someone they [claim to] consider a "stupid liar".  So which are you?
Not sane?  Or stupid enough to try to score points off someone who is
incapable of conceding them?  Or lying when you describe Peter? You
must surely have better things to do.  Meanwhile, you surely noticed
that Peter is running rings around you.
>
     Peter -- you surely have better things to do.  No-one sensible
is reading the repetitive stuff.  Decades, and myriads of articles, ago
people here tried to help you knock your points into shape, but anything
sensible is swamped by the insults.  Free advice, worth roughly what you
are paying for it:  step back, and summarise [from scratch, not using HHH
and DDD (etc) without explanation] (a) what it is you think you are trying
to prove and (b) what progress you claim to have made.  No more than one
side of paper.  Assume that people who don't actively insult you are, in
fact, trying to help.

>
And this approach has been tried many times. It makes no more progress than the ones you are criticizing. Just assume the regulars are lonesome, very lonesome and USENET keeps everybody off the deserted streets at night.

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HHH is an emulating termination analyzer that takes the machine
address of DDD as input then emulates the x86 machine language
of DDD until a non-terminating behavior pattern is recognized.

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But fails, because you provided it with a proven incorrect pattern
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HHH recognizes this pattern when HHH emulates itself emulating DDD
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void DDD()
{
   HHH(DDD);
   return;
}
>

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Which isn't a correct analysis (but of course, that is just what you do)
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Since we know that HHH(DDD) returns 0, it can not be a non- terminating behaivor, but that claim is just a lie.
>

One cannot simply ignore the actual behavior specified by the
finite string x86 machine language of DDD such that
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Right, one can not ignore the fact that HHH(DDD) is determined to return 0.
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DDD emulated by each corresponding HHH that can possibly
exist never returns

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More lies. It has been determined that EVERY DDD that calls an HHH(DDD) that returns 0 will halt.
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The DDDs that don't return are the ones that call an HHH that never returns an answer.
>

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*Your weasel words are in incorrect paraphrase of this*

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WHAT PARAPHARSE.
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DDD emulated by each corresponding HHH that can possibly
exist never returns

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No, that means the behavior of the code of DDD when directly executed. or youy are lying about working on the Halting Problem.
>

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It seems to me that you just said that:
the behavior of DDD emulated by HHH
<is not>
the behavior of DDD emulated by HHH.

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At least one could say so because the exptession "the behaviour of DDD
emulated by HHH" can be interpreted in two ways.

 It can be interpreted an infinite number of ways when the requirement
that the interpretation be correct is dropped.

And, the only CORRECT interpretation goes by the DEFINITIONS of the words, which means that "non-termination" is a property of a complete program (which your "finite-string" for DDD does not express) and that said program never reaches a terminal state even after an unbounded number of steps, which this HHH's emulation doesn't do.
So, you are just proving yourself to be a blatant liar.

 The x86 machine code of DDD and HHH provides the single correct
way to interpret DDD emulated by HHH.
 

Right, and that machine code needs to INCLUDE the machine code of HHH, since your doesn't, it is IMPOSSIBLE for your HHH to do what you claim, as it CAN NOT correctly emulate the call to HHH.

_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]
 Fully operational code is here.
https://github.com/plolcott/x86utm/blob/master/Halt7.c

Which proves you wrong, as you can show that DDD returns since HHH(DDD) return 0

 

By the usual rules
"emulated by HHH" is an attribute of DDD with the intent to prevent
cofusion with some other function that happens to have the same name.
But it could also be an attribute of "the behaviour" with the intent
to prevent confision with the true behaviour of DDD if the behaviour
emulated by HHH is different from the behaviour specified by the input
to HHH.
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