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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: Mon, 15 Jul 2024 10:19:23 +0300
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On 2024-07-14 14:26:21 +0000, olcott said:

> On 7/14/2024 2:56 AM, Mikko wrote:
>> On 2024-07-13 11:57:59 +0000, olcott said:
>> 
>>> On 7/13/2024 2:46 AM, Mikko wrote:
>>>> On 2024-07-12 13:07:59 +0000, olcott said:
>>>> 
>>>>> On 7/12/2024 2:55 AM, Mikko wrote:
>>>>>> On 2024-07-11 14:16:34 +0000, olcott said:
>>>>>> 
>>>>>>> On 7/11/2024 1:50 AM, Mikko wrote:
>>>>>>>> On 2024-07-10 13:37:30 +0000, olcott said:
>>>>>>>> 
>>>>>>>>> On 7/10/2024 2:18 AM, Mikko wrote:
>>>>>>>>>> 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.
>>>>>>>>> 
>>>>>>>>> A correct emulator can correctly any correct x86 instructions.
>>>>>>>>> When it emulates non-halting code then itself does not halt.
>>>>>>>> 
>>>>>>>> Not quite right but should be easy to fix. There should be a verb before "any",
>>>>>>>> for example "execute". Of course there still is a probelm with the meaning
>>>>>>>> "any correct x86 instructions". Intel may publish a new x86 processor that has
>>>>>>>> instructios that the emulator cannot know but are nevertheless correct x86
>>>>>>>> instructions because Intel says so. In the second sentence "it" should be used
>>>>>>>> istead of "itself".
>>>>>>>> 
>>>>>>> 
>>>>>>> Intel has already done this and they call this x64.
>>>>>>> A 1907 Model-T Ford cannot have upgrades and still
>>>>>>> be a 1907 model-T Ford. Likewise for the x86 language.
>>>>>> 
>>>>>> A new version of a 1907 Model-T Ford is possible and can have the same name
>>>>>> except that the "1907" must be replaced as it refers to the year. That the
>>>>>> "Model-T" is also replaced is a free chioce of Ford.
>>>>>> 
>>>>>> Likewise Intel is free to call a new processor whatever they want to call it.
>>>>>> 
>>>>> 
>>>>> The x86 language is a fixed constant.
>>>> 
>>>> Where has Intel promised so?
>>>> 
>>> 
>>> Backward compatibility requires it.
>>> https://en.wikipedia.org/wiki/X86_assembly_language#:~:text=x86%20assembly%20language%20is%20the,was%20launched%20in%20April%201972. 
>>> 
>> 
>> That is about the assembly language, not about the machine instruction set.
>> It does not say much about backwards compatibitlity and does not say that
>> Intel has promised anything about that.
>> 
>> Policy of backwards compatibilty does not mean that every x86 proscessor
>> has exactly the same machine language as 8086. Later version have introduced
>> new instructions and new processor modes and Intel is not prohibited from
>> adding still more in future models. It only means that the meanings of the
>> existing instructions are not changed when executed in existing modes.
>> 
> 
> When I refer to the x86 language I am referring to the minimal
> subset that runs the same way on compatible Intel 32-bit processors.

Then you should specify a specific instruction set, preferably the one
that your compiler uses.

-- 
Mikko