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From: Richard Damon <richard@damon-family.org>
Newsgroups: comp.theory,sci.logic
Subject: Re: Simulating termination analyzers by dummies --- What does halting
 mean?
Date: Tue, 18 Jun 2024 22:41:11 -0400
Organization: i2pn2 (i2pn.org)
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On 6/18/24 10:30 PM, olcott wrote:
> On 6/18/2024 9:16 PM, Richard Damon wrote:
>> On 6/18/24 1:25 PM, olcott wrote:
>>> On 6/18/2024 12:06 PM, joes wrote:
>>>
>>> void DDD()
>>> {
>>>    H0(DDD);
>>> }
>>>
>>> DDD correctly simulated by any H0 cannot possibly halt.
>>>
>>>> DDD halts iff H0 halts.
>>>
>>> Halting is a technical term-of-the-art that corresponds
>>> to terminates normally. Because Turing machines are
>>> abstract mathematical objects there has been no notion
>>> of abnormal termination for a Turing machine.
>>
>> No "normally" as Turing Machine have no "abnormal terminatiom"
>>
>> You just don't understand what they are.
>>
>>>
>>> We can derive a notion of abnormal termination for Turing
>>> machines from the standard terms-of-the-art.
>>
>> How?
>>
>>>
>>> Some TM's loop and thus never stop running, this is classical
>>> non-halting behavior. UTM's simulate Turing machine descriptions.
>>> This is the same thing as an interpreter interpreting the
>>> source-code of a program.
>>>
>>> A UTM can be adapted so that it only simulates a fixed number
>>> of iterations of an input that loops. When this UTM stops
>>> simulating this Turing machine description we cannot correctly
>>> say that this looping input halted.
>>>
>>
>> And then are no longer UTMs, and YES, if a machine based on such am 
>> modifed UTM (so it is no long a UTM) when the UTM stops simulating, we 
>> can not say the input halted, nor can we say it didn't halt.
>>
> 
> When such a UTM has been adapted to only simulate
> the first ten states of its input TMD, then every
> simulated TMD with more than ten states did not
> terminate normally.

Then it is no longer a UTM.

And its simulation say NOTHING about the machine not terminating, 
normally nor not.

Terminating is a property of the actual machine, and not a sumulation of it.

You could say the SIMULATION didn't terminate normally, but you can't 
say the machine didn't or even the Turing Machine Description, as you 
could give that exact same TMD to a real UTM and find out the actual 
behaviof or the input.

You just have lost track of the defintions of what is REALITY (the 
actual behavior of the machine) and what is just imagination.

> 
>> The not-a-UTM just came to a no-answer state.
>>
> 
> I have to go one-step-at-a-time with everyone or
> they get overwhelmed and leap to the conclusion
> that I am wrong.

Nope, you just forget about what is defined to be real.

> 
>> The answer will be provided by useing an ACTUAL UTM that keeps on 
>> going, or the direct execution of the machine,
>>
>> You are just stuck in your idea that Lies are sometimes ok.
> 
> You are stuck on the idea that repeating states cannot
> be recognized in a finite number of steps.

No, ACTUAL REPEATING states can be recognised, but I guess you are too 
stupid to understand my description of it.

> 
> void Infinite_Loop()
> {
>    HERE: goto HERE;
> }
> 
> void Infinite_Recursion()
> {
>    Infinite_Recursion();
> }
> 
> void DDD()
> {
>    H0(DDD);
> }
> 
> Every C programmer that knows what an x86 emulator is knows
> that when H0 emulates the machine language of Infinite_Loop,
> Infinite_Recursion, and DDD that it must abort these emulations
> so that itself can terminate normally.

Which doesn't mean the program DDD needs to be abort to have it halt.

That is just a figment of your imagination that doesn't understand the 
difference between truth and lies.

> 
> This was recently confirmed in the C group.
> 

Yes, by Bonita, whose confirmation is, if anything a mark against the 
statement.