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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory
Subject: Re: DDD correctly emulated by HHH is Correctly rejected as
 non-halting V2
Date: Mon, 15 Jul 2024 07:22:19 -0500
Organization: A noiseless patient Spider
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On 7/15/2024 3:49 AM, joes wrote:
> Am Sun, 14 Jul 2024 19:30:27 -0500 schrieb olcott:
>> On 7/14/2024 7:20 PM, joes wrote:
>>> Am Sun, 14 Jul 2024 09:00:55 -0500 schrieb olcott:
>>>> On 7/14/2024 3:29 AM, joes wrote:
>>>>> Am Sat, 13 Jul 2024 18:33:53 -0500 schrieb olcott:
>>>>>> On 7/13/2024 6:26 PM, joes wrote:
>>>>>>> Can you elaborate? All runtime instances share the same static
>>>>>>> code.
>>>>>>> I am talking about the inner HHH which is called by the simulated
>>>>>>> DDD. That one is, according to you, aborted. Which is wrong,
>>>>>>> because by virtue of running the same code, the inner HHH aborts
>>>>>>> ITS simulation of DDD calling another HHH.
>>>>> What are the twins and what is their difference? Do you disagree with
>>>>> my tracing?
> 
>>>> The directly executed DDD is like the first call of infinite
>>>> recursion. The emulated DDD is just like the second call of infinite
>>>> recursion. When the second call of infinite recursion is aborted then
>>>> the first call halts.
>>> Not really. Execution does not continue.
> 
>>>> void Infinite_Recursion()
>>>> {
>>>>      Infinite_Recursion();
>>>> }
>>>> The above *is* infinite recursion.
>>>> A program could emulate the above code and simply skip line 3 causing
>>>> Infinite_Recursion() to halt.
>>> That would be incorrect.
> 
>>>> When DDD calls HHH(DDD) HHH returns.
>>> Therefore it does not need to be aborted.
> 
>>>> When DDD correctly emulated by HHH the call never returns as is proven
>>>> below. The executed DDD() has HHH(DDD) skip this call.
>>> I do not see this below.
> 
>>>> HHH(DDD) must skip this call itself by terminating the whole DDD
>>>> process.
>>>
>>>> Because this HHH does not know its own machine address HHH only sees
>>>> that DDD calls a function that causes its first four steps to be
>>>> repeated. HHH does not know that this is recursive simulation. To HHH
>>>> it looks just like infinite recursion.
>>>
>>>> New slave_stack at:1038c4 -- create new process context for 1st DDD
>>>> Begin Local Halt Decider Simulation   Execution Trace Stored at:1138cc
>>>
>>>> [0000217a][001138b4][0000217f] e853f4ffff call 000015d2 ; call
>>>> HHH(DDD) New slave_stack at:14e2ec -- create new process context for
>>>> 2nd DDD
>>>
>>>> [0000217a][0015e2dc][0000217f] e853f4ffff call 000015d2 ; call
>>>> HHH(DDD) Local Halt Decider: Infinite Recursion Detected Simulation
>>>> Stopped
>>> How is this detected? Is it also triggered when calling a function in a
>>> loop?
> 
>> You never bothered to answer whether or not you have 100% understanding
>> of infinite recursion. If you don't then you can never understand what I
>> am saying. If you do that I already proved my point. Here is the proof
>> again:
> As if you would believe me.
> You never bothered to answer my questions (see above).
> That only proves that HHH and DDD halt.
> 

This *is* an answer too difficult for you to understand.

New slave_stack at:1038c4
Begin Local Halt Decider Simulation   Execution Trace Stored at:1138cc
[00002172][001138bc][001138c0] 55         push ebp      ; housekeeping
[00002173][001138bc][001138c0] 8bec       mov ebp,esp   ; housekeeping
[00002175][001138b8][00002172] 6872210000 push 00002172 ; push DDD
[0000217a][001138b4][0000217f] e853f4ffff call 000015d2 ; call HHH(DDD)
New slave_stack at:14e2ec
[00002172][0015e2e4][0015e2e8] 55         push ebp      ; housekeeping
[00002173][0015e2e4][0015e2e8] 8bec       mov ebp,esp   ; housekeeping
[00002175][0015e2e0][00002172] 6872210000 push 00002172 ; push DDD
[0000217a][0015e2dc][0000217f] e853f4ffff call 000015d2 ; call HHH(DDD)
Local Halt Decider: Infinite Recursion Detected Simulation Stopped

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