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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory,sci.logic
Subject: Re: Simulating termination analyzers by dummies --- What does halting
 mean?
Date: Wed, 19 Jun 2024 08:20:28 -0500
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On 6/19/2024 3:57 AM, joes wrote:
> Am Tue, 18 Jun 2024 21:30:43 -0500 schrieb olcott:
>> 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:
> 
>>>> 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.

> You are confusing the machines with their simulators. No longer simulating
> has nothing to do with the simulatee. It does not "know" it is being
> simulated. That is entirely in the power of the simulator. Only it can
> freely choose to simulate more steps. The simulated machine then proceeds.
> 
>>> 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.
> 

I am establishing the notion of abnormal termination for Turing
machines within the standard terms of the art.

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

> Oh, they can. It's just that repeating states don't halt in a finite
> number of steps.
> 

void DDD()
{
   H0(DDD);
}

Richard cannot understand that H0 can correctly determine in a
finite number of steps that DDD correctly emulated by H0 cannot
halt in a finite number of steps. I used to think that he was
lying about this. It seems that the actual case is that I
overestimated his skill level.

Now that he says that when Bonita said "Everything correct"
he is taking this to mean something is wrong I am back to
thinking that he might be a liar.

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