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From: Mikko <mikko.levanto@iki.fi>
Newsgroups: comp.theory
Subject: Re: Liar detector: Fred, Richard, Joes and Alan --- Ben's agreement
Date: Wed, 10 Jul 2024 10:18:49 +0300
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On 2024-07-09 14:14:16 +0000, olcott said:

> On 7/9/2024 1:14 AM, Mikko wrote:
>> On 2024-07-08 17:36:58 +0000, olcott said:
>> 
>>> On 7/8/2024 11:16 AM, Fred. Zwarts wrote:
>>>> Op 08.jul.2024 om 18:07 schreef olcott:
>>>>> 
>>>>> Try to show how infinity is one cycle too soon.
>>>>> 
>>>> You believe that two equals infinity.
>>> 
>>> void Infinite_Loop()
>>> {
>>>    HERE: goto HERE;
>>> }
>>> 
>>> void Infinite_Recursion()
>>> {
>>>    Infinite_Recursion();
>>> }
>>> 
>>> void DDD()
>>> {
>>>    HHH(DDD);
>>> }
>>> 
>>> Two cycles is enough to correctly determine that none
>>> of the above functions correctly emulated by HHH can
>>> possibly halt.
>>> 
>>> That you don't see this is ignorance or deception.
>> 
>> There is an important detail that determines whether an infinite
>> execution can be inferred. That is best illustrated by the following
>> examples:
>> 
>> void Finite_Loop()
>> {
>>   int x = 10000;
>> HERE:
>>   if (x > 0) {
>>     x--;
>>     goto HERE;
>>   }
>> }
>> 
>> void Finite_Recursion(int n)
>> {
>>   if (n > 0) {
>>     Finite_Recursion(n + 1);
>>   }
>> }
>> 
>> void DDD()
>> {
>>   HHH(DDD); // HHH detects recursive simulation and then simulates no more
>> }
>> 
>> The important difference is that in my examples there is a conditional
>> instruction that can (and does) prevent infinite exectuion.
>> 
> 
> When we ask:
> Does the call from DDD emulated by HHH to HHH(DDD) return?

Why would anyone ask that? A question should make clear its topic.
Instead one could ask whether HHH can fully emulate DDD if that is
what one wants to know. Or one may think that HHH and DDD are so
unimteresting that there is no point to ask anyting about them.

-- 
Mikko