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From: Mikko <mikko.levanto@iki.fi>
Newsgroups: comp.theory
Subject: Re: DDD correctly emulated by HHH is Correctly rejected as non-halting V2
Date: Tue, 16 Jul 2024 10:20:31 +0300
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On 2024-07-15 12:22:19 +0000, olcott said:

> 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

The trace does not show that HHH returns so there is no basis to
think that HHH is a decider.

-- 
Mikko