On Monday, October 17, 2022 at 12:29:34 PM UTC-4, olcott wrote:
> 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.

Correct: the halting function maps D to halting but Sipser_H maps D to non-halting, so it is unable to compute the halting function.

> The notion of a simulating halt decider removes that basis
> thus causing all of these conventional proofs to fail.

Answering a different question doesn't make the original question answerable.