On 10/17/2022 10:47 AM, Dennis Bush wrote:
> On Monday, October 17, 2022 at 11:44:10 AM UTC-4, olcott wrote:
>> On 10/17/2022 10:40 AM, Dennis Bush wrote:
>>> On Monday, October 17, 2022 at 11:31:41 AM UTC-4, olcott wrote:
>>>> On 10/17/2022 10:16 AM, Dennis Bush wrote:
>>>>> On Monday, October 17, 2022 at 11:06:45 AM UTC-4, olcott wrote:
>>>>>> On 10/17/2022 9:49 AM, Dennis Bush wrote:
>>>>>>> On Monday, October 17, 2022 at 10:38:09 AM UTC-4, olcott wrote:
>>>>>>>> On 10/17/2022 7:47 AM, Dennis Bush wrote:
>>>>>>>>> On Monday, October 17, 2022 at 5:30:59 AM UTC-4, Fred. Zwarts wrote:
>>>>>>>>>> I have been following the discussions about Halt deciders with interest.
>>>>>>>>>> As a retired software designer and developer, I have a lot of practical
>>>>>>>>>> experience, but not much theoretical education, although the theoretical
>>>>>>>>>> background is very interesting. I learned a lot. I would like to verify
>>>>>>>>>> that I understand it correctly. Could you point out any errors in the
>>>>>>>>>> summary below?
>>>>>>>>>>
>>>>>>>>>> 1) (Definition of halt) A program X with input Y is said to halt if it
>>>>>>>>>> reaches its end condition after a finite number of steps. It does not
>>>>>>>>>> halt if it continues to execute infinitely.
>>>>>>>>>> (So, X(Y) either halts, or it does not halt.)
>>>>>>>>>> (It is irrelevant whether the end condition is reached in the 'normal'
>>>>>>>>>> way, or by other means, e.g. an unhandled 'exception'.)
>>>>>>>>>>
>>>>>>>>>> 2) (Definition of halt decider) A halt decider H is a program that,
>>>>>>>>>> given a program X with input Y decides, after a finite number of steps,
>>>>>>>>>> whether X(Y) halts or not.
>>>>>>>>>> (H(X,Y) itself must halt after a finite number of steps. It must return
>>>>>>>>>> either 1 if X(Y) halts, or 0 if X(Y) does not halt, where 1 and 0 are a
>>>>>>>>>> convention, which could also be two other arbitrary values.)
>>>>>>>>>>
>>>>>>>>>> From 1 and 2 it follows:
>>>>>>>>>>
>>>>>>>>>> 3) If X(Y) halts, then H must return 1. If H does not return 1 in a
>>>>>>>>>> finite number of steps, it might return another interesting result, but
>>>>>>>>>> it is not a halt decider. (Not returning 1 includes returning other
>>>>>>>>>> values, not halting, or throwing 'exceptions'.)
>>>>>>>>>>
>>>>>>>>>> 4) If X(Y) does not halt, then H must return 0. If it does not return 0
>>>>>>>>>> in a finite number of steps, it might return another interesting result,
>>>>>>>>>> but it is not a halt decider. (Not returning 0 includes returning other
>>>>>>>>>> values, not halting, or throwing 'exceptions'.)
>>>>>>>>>>
>>>>>>>>>> Paradoxical program:
>>>>>>>>>>
>>>>>>>>>> 5) It is always possible to construct a program P, that uses code with
>>>>>>>>>> the same logic as H, in order to do the opposite of what H(P,P) returns.
>>>>>>>>>> (P does not necessarily need to use the exact same code as H does,
>>>>>>>>>> amongst others it could use a modified copy of H, or a simulation of H.)
>>>>>>>>>>
>>>>>>>>>> From 5 it follows that a general halt decider that works for any X and
>>>>>>>>>> Y does not exist:
>>>>>>>>>>
>>>>>>>>>> From 3, 4 and 5 it follows:
>>>>>>>>>>
>>>>>>>>>> 6) If P(P) halts, then H should return 1, but if H would do so, P(P)
>>>>>>>>>> would not halt.
>>>>>>>>>>
>>>>>>>>>> 7) If P(P) does not halt, H should return 0, but if H would do so, P(P)
>>>>>>>>>> would halt.
>>>>>>>>>>
>>>>>>>>>> 8) If P(P) halts and H does not return 1 after a finite number of steps,
>>>>>>>>>> then H is not a halt decider.
>>>>>>>>>> (The result could nevertheless be interesting for other purposes.)
>>>>>>>>>> (It is irrelevant what causes P(P) to halt.)
>>>>>>>>>>
>>>>>>>>>> 9) If P(P) does not halt and H does not return 0 after a finite number
>>>>>>>>>> of steps, then H is not a halt decider.
>>>>>>>>>> (The result could nevertheless be interesting for other purposes.)
>>>>>>>>>
>>>>>>>>> Your understanding is correct. To sum things up, the halting function (using the mathematical notion of a function), performs the following mapping:
>>>>>>>>>
>>>>>>>>> For *any* algorithm (i.e. a fixed immutable sequence of instructions) X and input Y:
>>>>>>>>> H(X,Y)==1 if and only if X(Y) halts, and
>>>>>>>>> H(X,Y)==0 if and only if X(Y) does not halt
>>>>>>>>>
>>>>>>>>> And the halting problem proofs show that this mapping is not computable, i.e. it is impossible for an algorithm to compute this mapping.
>>>>>>>>>
>>>>>>>> *Professor Sipser has agreed to these verbatim words* (and no more)
>>>>>>>> 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.
>>>>>>>
>>>>>>> And he agreed to those words based on their commonly known meanings, not your alternate weasel-word meanings.
>>>>>>>
>>>>>>> The conventional definition of "correctly simulating" means that the simulated behavior EXACTLY matches the behavior of direct execution.
>>>>>> I have proven an exception to this rule:
>>>>>
>>>>> That's not a rule. It's a definition.
>>>>>
>>>>>
>>>>>>
>>>>>> int Sipser_D(int (*M)())
>>>>>> {
>>>>>> if ( Sipser_H(M, M) )
>>>>>> return 0;
>>>>>> return 1;
>>>>>> }
>>>>>>
>>>>>> For the infinite set of H/D pairs:
>>>>>> Every correct simulation of D by H will never reach the final state of D
>>>>>> because D specifies recursive simulation to H.
>>>>>
>>>>> So in other words your Sipser_H is computing the PO-halting function:
>>>>>
>>>> *The PO-halting function is now Sipser approved*
>>>
>>> No it's not, because he used the actual meaning of the words and not your weasel-worded definitions. Using the real definitions,
>>
>> *Professor Sipser has agreed to these verbatim words* (and no more)
>> 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.
>> *A paraphrase of a portion of the above paragraph*
>> Would D correctly simulated by H ever stop running if not aborted?
>>
>> The answer of "no" is proved on page 3 of this paper.
>>
>> *Rebutting the Sipser Halting Problem Proof*
>> https://www.researchgate.net/publication/364302709_Rebutting_the_Sipser_Halting_Problem_Proof
>> *Still no rebuttal of page 3 because you know that page 3 is correct*
>
> You still seem to think that because you have an H that partially computes the PO-halting function that it has anything to do with the halting function. It does not.
>
> So anything that does not address whether the halting function is computable is irrelevant.

Anyone that is sufficiently technically competent can verify that H does
correctly determine the halt status of D correctly simulated by H.

This proves that the conventional proofs that rely on D doing the
opposite of whatever H decides have been refuted by the notion of a
simulating halt decider.

This does not prove that the halting problem has been solved.

--
Copyright 2022 Pete Olcott "Talent hits a target no one else can hit;
Genius hits a target no one else can see." Arthur Schopenhauer