On 10/17/2022 11:25 AM, Dennis Bush wrote:
> On Monday, October 17, 2022 at 12:15:27 PM UTC-4, olcott wrote:
>> On 10/17/2022 11:00 AM, Dennis Bush wrote:
>>> On Monday, October 17, 2022 at 11:57:18 AM UTC-4, olcott wrote:
>>>> 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.
>>>
>>> No one is denying that you're able to compute a subset of the PO-halting function. The halting problem proofs are about the halting function.
>>>
>>>>
>>>> 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.
>>>
>>> The conventional proofs are making claims about the halting function, not the PO-halting function, therefore claims about the PO-halting function are irrelevant.
>>
>> [ repeat of previously refuted statement ]
>>
>> int Sipser_D(int (*M)())
>> {
>> if ( Sipser_H(M, M) )
>> return 0;
>> return 1;
>> }
>> This notion of a simulating halt decider is proven to correctly
>> determine the halt status of Sipser_D by Sipser_H.
>> *Rebutting the Sipser Halting Problem Proof*
>> https://www.researchgate.net/publication/364302709_Rebutting_the_Sipser_Halting_Problem_Proof
>
> In other words, you can compute a subset of the PO-halting function. And since the halting problem proofs make claims about the halting function, claims about the PO-halting function are irrelevant.

The halting problem proofs make claims about the halting function on the
basis that the halt status of Sipser_D cannot be correctly determined by
Sipser_H. The notion of a simulating halt decider removes that basis
thus causing all of these conventional proofs to fail.

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