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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory
Subject: Re: Simulating termination analyzers by dummies --- criteria is met
Date: Sun, 23 Jun 2024 08:13:42 -0500
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On 6/23/2024 2:57 AM, Mikko wrote:
> On 2024-06-22 14:11:28 +0000, olcott said:
> 
>> On 6/22/2024 8:27 AM, Richard Damon wrote:
>>> On 6/22/24 9:04 AM, olcott wrote:

>>>>
>>>> I am the sole inventor of the simulating halt decider.
>>>>
>>>> Ben Bacarisse contacted professor Sipser to verify that he
>>>> really did says this. The details are in this forum about
>>>> the same date.
>>>>
>>>> https://www.amazon.com/Introduction-Theory-Computation-Michael-Sipser/dp/113318779X/
>>>>
>>>> <MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
>>>>    If simulating halt decider H correctly simulates its input D
>>>>    until H correctly determines that its simulated D would never
>>>>    stop running unless aborted then
>>>>
>>>>    H can abort its simulation of D and correctly report that D
>>>>    specifies a non-halting sequence of configurations.
>>>> </MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
>>>
>>> And, as I remember, he also verified that he disagrees with your 
>>> definition of correct simulation.
>>>
>>>>
>>>> *Ben also verified that the criteria have been met*
>>>> On 10/14/2022 7:44 PM, Ben Bacarisse wrote:
>>>>  > I don't think that is the shell game. PO really /has/ an H
>>>>  > (it's trivial to do for this one case) that correctly determines
>>>>  > that P(P) *would* never stop running *unless* aborted.
>>>
>>> Right, Ben was willing to do what I am not that you can prove that, 
>>> by your definition, H can show that it "must" abort its simulation or 
>>> the input will run forever.
>>>
>>> But, just like me, he also agrees that this is NOT the defintion of 
>>> Halting, so H is just shown to be a correct (partial) POOP decider 
>>> but ot a Halt Decider, not even for that one input.
>>>
>>
>> On 10/14/2022 7:44 PM, Ben Bacarisse wrote:
>>  > I don't think that is the shell game. PO really /has/ an H
>>  > (it's trivial to do for this one case) that correctly determines
>>  > that P(P) *would* never stop running *unless* aborted.
>>  >
>>  > He knows and accepts that P(P) actually does stop. The
>>  > wrong answer is justified by what would happen if H
>>  > (and hence a different P) where not what they actually are.
>>  >
>> *Ben agrees that the criteria is met for the input*
>>
>> Computable functions are the formalized analogue of the
>> intuitive notion of algorithms, in the sense that a function
>> is computable if there exists an algorithm that can do the
>> job of the function, i.e. *given an input of the function*
>> *domain it can return the corresponding output*
>> https://en.wikipedia.org/wiki/Computable_function
>>
>> *Ben disagrees that the criteria is met for the non-input*
>> Yet no one here can stay focused on the fact that non-inputs
>> *DO NOT COUNT*
> 
> In particular, you can't. You have insisted that your "decider"
> or "anlyzer" (or whatever word you happen to use) H or HH (or
> hwatever name you happen to use) must return false because a
> non-input (where instead of the actually called function another
> function that does not halt is called) does not halt.
> 

You said it backwards. When I say that I am not guilty and did
not rob the liquor store you cannot paraphrase this as he admitted
that he robbed the liquor store.

H performs a sequence of finite string transformations on
its finite input of x86 machine code. These transformations
include that D calls H(D,D) while being simulated by H. In
such a case the call from D to H(D,D) cannot possibly return.

>> void DDD()
>> {
>>    HHH0(DDD);
>> }
>>
>> int main()
>> {
>>    Output("Input_Halts = ", HHH0(DDD));
>>    Output("Input_Halts = ", HHH1(DDD));
>> }
>>
>> It is a verified fact that the behavior that finite string DDD
>> presents to HH0 is that when DDD correctly simulated by HH0
>> calls HH0(DDD) that this call DOES NOT RETURN.
>>
>> It is a verified fact that the behavior that finite string DDD
>> presents to HH1 is that when DDD correctly simulated by HH1
>> calls HH0(DDD) that this call DOES RETURN.
> 

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