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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: Wed, 19 Jun 2024 20:24:05 -0400
Organization: i2pn2 (i2pn.org)
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On 6/19/24 9:20 AM, olcott wrote:
> 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.
> 

L     IIIII EEEEE
L       I   E
L       I   EEEE
L       I   E
LLLLL IIIII EEEEE

I fully understand how a decider can correctly determine in a finite 
number of steps how inputs like infinite_loop or infinite_recursion can 
be detected.

What is an illogical statement is about DDD correctly simulated by H0, 
since it doesn't DO an actual correct simulation, only a partial one, 
and you can't even show it does that (sincd you never show it simulating 
into H0).

Then you try to LIE with invalid logic of looking at DIFFERENT inputs of 
DDD's built on DIFFERENT version of H0, whoich just shows you don't 
understand the elementary basics of the field you are trying to talk about.


> 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.
> 

Nope, they are just showing they are as ignorant of the topic as you are.

You seemed to have missed the class on Logical Fallacies, and the error 
of Arguemnt by (False) Authority.