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From: joes <noreply@example.org>
Newsgroups: comp.theory
Subject: Re: The execution trace of HHH1(DDD) shows the divergence
Date: Sat, 7 Jun 2025 16:43:29 -0000 (UTC)
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Am Sat, 07 Jun 2025 11:02:55 -0500 schrieb olcott:
> On 6/7/2025 10:54 AM, wij wrote:
>> On Sat, 2025-06-07 at 10:35 -0500, olcott wrote:
>>> On 6/7/2025 10:31 AM, wij wrote:
>>>> On Sat, 2025-06-07 at 09:57 -0500, olcott wrote:
>>>>> On 6/7/2025 9:54 AM, wij wrote:
>>>>>> On Sat, 2025-06-07 at 09:32 -0500, olcott wrote:

>>>>>>> The execution trace of HHH1(DDD) shows the divergence of DDD
>>>>>>> emulated by HHH from DDD emulated by HHH1.
>>>>>>> Shows that DDD emulated by HHH and DDD emulated by HHH1 diverges
>>>>>>> as soon as HHH begins emulating itself emulating DDD.
>>>>>>>
>>>>>>> *From the execution trace of HHH1(DDD) shown below*
>>>>>>> DDD emulated by HHH1              DDD emulated by HHH [00002183]
>>>>>>> push ebp               [00002183] push ebp [00002184] mov
>>>>>>> ebp,esp            [00002184] mov ebp,esp [00002186] push 00002183
>>>>>>> ; DDD    [00002186] push 00002183 ; DDD [0000218b] call 000015c3 ;
>>>>>>> HHH    [0000218b] call 000015c3 ; HHH *HHH1 emulates DDD once then
>>>>>>> HHH emulates DDD once, these match*

What do you mean by "then" and "once"? That implies completion and
succession, however we are only ever simulating a single call, and
only HHH1 simulates any returns of DDD/HHH. 
HHH1 simulates DDD completely, HHH recurses and aborts *inside*. HHH
does not simulate DDD once, it only enters the call, but never exits.

>>>>>>> The next instruction of DDD that HHH emulates is at the machine
>>>>>>> address of 00002183.
>>>>>>> The next instruction of DDD that HHH1 emulates is at the machine
>>>>>>> address of 00002190.

At those addresses we have the first instruction of DDD and the one after
the call, respectively. 


[main -> HHH1(DDD) -> HHH(DDD) -> HHH(DDD)]
>>>>>>> New slave_stack at:198d21  DDD emulated by HHH *This is the
>>>>>>> beginning of the divergence of the behavior*
>>>>>>> *HHH is emulating itself emulating DDD, HHH1 never does that*

HHH1 does simulate HHH simulating DDD.

>>>>>> The HP is asking for such a H that H(D)==1 iff D() halts.
>>>>>> You are always solving POO Problem.
>>>>>>
>>>>> int main()
>>>>> {
>>>>>      DDD(); // The HHH(DDD) that DDD calls cannot report
>>>>> }          // on the behavior of its caller.
>>>>
>>>> That is what the HP theorem says, the halting decider is not
>>>> possible.
>>>>
>>> The HP theorem never bothered to notice that it has contradictory
>>> axioms. HHH(DDD) IS NOT ALLOWED TO REPORT ON THE BEHAVIOR OF ITS
>>> CALLER.

Which axiom contradicts which?

>> Nope. It you who don't understand English.
>> 
> The theory of computation does not allow a halt decider to report on the
> behavior of its caller. Cite the chapter and verse where it does allow
> this.
And lo, no program shall be forbidden to call the mighty halt decider.
Recursions 22:22

-- 
Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
It is not guaranteed that n+1 exists for every n.