Re: Proof that DDD specifies non-halting behavior --- point by point

On 8/14/24 12:24 AM, olcott wrote:

On 8/13/2024 11:04 PM, Richard Damon wrote:

On 8/13/24 11:48 PM, olcott wrote:

On 8/13/2024 10:21 PM, Richard Damon wrote:

On 8/13/24 10:38 PM, olcott wrote:

On 8/13/2024 9:29 PM, Richard Damon wrote:

On 8/13/24 8:52 PM, olcott wrote:

void DDD()
{
   HHH(DDD);
   return;
}
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_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]
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A simulation of N instructions of DDD by HHH according to
the semantics of the x86 language is necessarily correct.
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Nope, it is just the correct PARTIAL emulation of the first N instructions of DDD, and not of all of DDD,

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That is what I said dufuss.

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Nope. You didn't. I added clairifying words, pointing out why you claim is incorrect.
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For an emulation to be "correct" it must be complete, as partial emulations are only partially correct, so without the partial modifier, they are not correct.
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A complete emulation of one instruction is
a complete emulation of one instruction

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A correct simulation of N instructions of DDD by HHH is
sufficient to correctly predict the behavior of an unlimited
simulation.

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Nope, if a HHH returns to its caller,

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*Try to show exactly how DDD emulated by HHH returns to its caller*
(the first one doesn't even have a caller)
Use the above machine language instructions to show this.
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Remember how English works:
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When you ask "How DDD emulated by HHH returns to its callers".

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Show the exact machine code trace of how DDD emulated
by HHH (according to the semantics of the x86 language)
reaches its own machine address 00002183

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No. The trace is to long,

 Show the Trace of DDD emulated by HHH
and show the trace of DDD emulated by HHH
emulated by the executed HHH
Just show the DDD code traces.
 

First you need to make a DDD that meets the requirements, and that means that it calls an HHH that meets the requirements.
Since you HHH doesn't, I can't show what isn't there yet.
The biggest problem with HHH is that it isn't pure function of just its declared inputs, it detects via a side channel if it is the "root" emulator, and changes its behavior because of that.
So, since we don't have an actual DDD that meets its requirements, we can't make a correct trace of it to show what it does.
Second, you have an incoherent requiremnet in your challenge, the "trace of DDD emulated by HHH" will not be created by HHH, since your HHH doesn't do a complete emulation, and your requirement is for a complete trace, so you are just creating a strawman deception.
It shouldn't be too hard for a decent programmer to actually make the required HHH, but from what I see of your code, you are not one of those. Of course, getting it to behave like you want would be tougher, as you need to program in a BUG make it ignore the conditionals inside of HHH when it emulates the input.