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Groups > comp.theory > #35634 > unrolled thread
| Started by | olcott <NoOne@NoWhere.com> |
|---|---|
| First post | 2021-07-03 10:19 -0500 |
| Last post | 2021-07-04 14:14 +0100 |
| Articles | 20 on this page of 130 — 13 participants |
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Could H correctly decide that P never halts? olcott <NoOne@NoWhere.com> - 2021-07-03 10:19 -0500
Re: Could H correctly decide that P never halts? wij <wyniijj@gmail.com> - 2021-07-03 08:25 -0700
Re: Could H correctly decide that P never halts? olcott <NoOne@NoWhere.com> - 2021-07-03 10:32 -0500
Re: Could H correctly decide that P never halts? Richard Damon <Richard@Damon-Family.org> - 2021-07-03 11:56 -0400
Re: Could H correctly decide that P never halts? olcott <NoOne@NoWhere.com> - 2021-07-03 11:19 -0500
Re: Could H correctly decide that P never halts? Bonita Montero <Bonita.Montero@gmail.com> - 2021-07-03 18:28 +0200
Re: Could H correctly decide that P never halts? olcott <NoOne@NoWhere.com> - 2021-07-03 11:51 -0500
Re: Could H correctly decide that P never halts? Bonita Montero <Bonita.Montero@gmail.com> - 2021-07-03 19:18 +0200
Re: Could H correctly decide that P never halts? olcott <NoOne@NoWhere.com> - 2021-07-03 12:19 -0500
Re: Could H correctly decide that P never halts? Bonita Montero <Bonita.Montero@gmail.com> - 2021-07-03 19:33 +0200
Re: Could H correctly decide that P never halts? Ben Bacarisse <ben.usenet@bsb.me.uk> - 2021-07-03 19:10 +0100
Re: Could H correctly decide that P never halts? olcott <NoOne@NoWhere.com> - 2021-07-03 13:49 -0500
Re: Could H correctly decide that P never halts? Richard Damon <Richard@Damon-Family.org> - 2021-07-03 14:29 -0400
Re: Could H correctly decide that P never halts? Ben Bacarisse <ben.usenet@bsb.me.uk> - 2021-07-03 17:25 +0100
Re: Could H correctly decide that P never halts? olcott <NoOne@NoWhere.com> - 2021-07-03 11:49 -0500
Re: Could H correctly decide that P never halts? Richard Damon <Richard@Damon-Family.org> - 2021-07-03 14:17 -0400
Re: Could H correctly decide that P never halts? [ Why can H ignore its own behavior? ] olcott <NoOne@NoWhere.com> - 2021-07-03 14:08 -0500
Re: Could H correctly decide that P never halts? [ Why can H ignore its own behavior? ] Richard Damon <Richard@Damon-Family.org> - 2021-07-03 15:43 -0400
Re: Could H correctly decide that P never halts? [ Why can H ignore its own behavior? ](my thanks to Richard) olcott <NoOne@NoWhere.com> - 2021-07-03 15:37 -0500
Re: Could H correctly decide that P never halts? [ Why can H ignore its own behavior? ](my thanks to Richard) Richard Damon <Richard@Damon-Family.org> - 2021-07-03 17:58 -0400
Re: Could H correctly decide that P never halts? [ Why can H ignore its own behavior? ](my thanks to Richard) olcott <NoOne@NoWhere.com> - 2021-07-03 17:20 -0500
Re: Could H correctly decide that P never halts? [ Why can H ignore its own behavior? ](my thanks to Richard) Richard Damon <Richard@Damon-Family.org> - 2021-07-03 18:55 -0400
Re: Could H correctly decide that P never halts? [ Why can H ignore its own behavior? ](my thanks to Richard) Mr Flibble <flibble@reddwarf.jmc> - 2021-07-03 23:57 +0100
Re: Could H correctly decide that P never halts? [ Why can H ignore its own behavior? ](my thanks to Richard) olcott <NoOne@NoWhere.com> - 2021-07-03 18:13 -0500
Re: Could H correctly decide that P never halts? [ Why can H ignore its own behavior? ](my thanks to Richard) Richard Damon <Richard@Damon-Family.org> - 2021-07-03 19:04 -0400
Re: Could H correctly decide that P never halts? [ prerequisites to understanding me] olcott <NoOne@NoWhere.com> - 2021-07-03 18:34 -0500
Re: Could H correctly decide that P never halts? [ prerequisites to understanding me] Richard Damon <Richard@Damon-Family.org> - 2021-07-03 20:06 -0400
Re: Could H correctly decide that P never halts? [ prerequisites to understanding me] olcott <NoOne@NoWhere.com> - 2021-07-03 19:26 -0500
Re: Could H correctly decide that P never halts? [ prerequisites to understanding me] Richard Damon <Richard@Damon-Family.org> - 2021-07-03 21:21 -0400
Re: Could H correctly decide that P never halts? [ prerequisites to understanding me] olcott <NoOne@NoWhere.com> - 2021-07-03 20:41 -0500
Re: Could H correctly decide that P never halts? [ prerequisites to understanding me] Richard Damon <Richard@Damon-Family.org> - 2021-07-03 22:13 -0400
Re: Could H correctly decide that P never halts? [ prerequisites to understanding me] olcott <NoOne@NoWhere.com> - 2021-07-03 21:22 -0500
Re: Could H correctly decide that P never halts? [ prerequisites to understanding me] Richard Damon <Richard@Damon-Family.org> - 2021-07-03 23:24 -0400
Re: Could H correctly decide that P never halts? [ prerequisites to understanding me] olcott <NoOne@NoWhere.com> - 2021-07-03 22:32 -0500
Re: Could H correctly decide that P never halts? [ prerequisites to understanding me] Richard Damon <Richard@Damon-Family.org> - 2021-07-04 06:40 -0400
Re: Could H correctly decide that P never halts? [ prerequisites to understanding me] olcott <NoOne@NoWhere.com> - 2021-07-04 09:24 -0500
Re: Could H correctly decide that P never halts? [ prerequisites to understanding me] Richard Damon <Richard@Damon-Family.org> - 2021-07-04 14:50 -0400
Re: Could H correctly decide that P never halts? [ Why can H ignore its own behavior? ](my thanks to Richard) André G. Isaak <agisaak@gm.invalid> - 2021-07-03 19:14 -0600
Re: Could H correctly decide that P never halts? [ Why can H ignore its own behavior? ](my thanks to Richard) olcott <NoOne@NoWhere.com> - 2021-07-03 20:38 -0500
Re: Could H correctly decide that P never halts? [ Why can H ignore its own behavior? ](my thanks to Richard) André G. Isaak <agisaak@gm.invalid> - 2021-07-03 22:14 -0600
Re: Could H correctly decide that P never halts? [ Why can H ignore its own behavior? ](my thanks to Richard) André G. Isaak <agisaak@gm.invalid> - 2021-07-03 22:18 -0600
Re: Could H correctly decide that P never halts? [ Why can H ignore its own behavior? ](my thanks to Richard) olcott <NoOne@NoWhere.com> - 2021-07-03 23:18 -0500
Re: Could H correctly decide that P never halts? [ Why can H ignore its own behavior? ](my thanks to Richard) André G. Isaak <agisaak@gm.invalid> - 2021-07-04 00:50 -0600
Re: Could H correctly decide that P never halts? [ Why can H ignore its own behavior? ](my thanks to Richard) olcott <NoOne@NoWhere.com> - 2021-07-04 09:15 -0500
Re: Could H correctly decide that P never halts? [ Why can H ignore its own behavior? ](my thanks to Richard) André G. Isaak <agisaak@gm.invalid> - 2021-07-04 10:31 -0600
Re: Could H correctly decide that P never halts? [ Why can H ignore its own behavior? ](my thanks to Richard) olcott <NoOne@NoWhere.com> - 2021-07-05 08:07 -0500
Re: Could H correctly decide that P never halts? [ Why can H ignore its own behavior? ](my thanks to Richard) Malcolm McLean <malcolm.arthur.mclean@gmail.com> - 2021-07-04 11:24 -0700
Re: Could H correctly decide that P never halts? [ Why can H ignore its own behavior? ](my thanks to Richard) André G. Isaak <agisaak@gm.invalid> - 2021-07-04 13:00 -0600
Re: Could H correctly decide that P never halts? [ Why can H ignore its own behavior? ](my thanks to Richard) olcott <NoOne@NoWhere.com> - 2021-07-04 15:09 -0500
Re: Could H correctly decide that P never halts? [ Why can H ignore its own behavior? ](my thanks to Richard) Malcolm McLean <malcolm.arthur.mclean@gmail.com> - 2021-07-04 13:30 -0700
Re: Could H correctly decide that P never halts? [ Why can H ignore its own behavior? ](my thanks to Richard) olcott <NoOne@NoWhere.com> - 2021-07-05 08:23 -0500
Re: Could H correctly decide that P never halts? [ Why can H ignore its own behavior? ](my thanks to Richard) Malcolm McLean <malcolm.arthur.mclean@gmail.com> - 2021-07-05 07:18 -0700
Re: Could H correctly decide that P never halts? [ Why can H ignore its own behavior? ](my thanks to Richard) olcott <NoOne@NoWhere.com> - 2021-07-05 09:31 -0500
Re: Could H correctly decide that P never halts? [ Why can H ignore its own behavior? ](my thanks to Richard) Richard Damon <Richard@Damon-Family.org> - 2021-07-05 11:10 -0400
Re: Could H correctly decide that P never halts? [ Why can H ignore its own behavior? ](my thanks to Richard) olcott <NoOne@NoWhere.com> - 2021-07-05 10:18 -0500
Re: Could H correctly decide that P never halts? [ Why can H ignore its own behavior? ](my thanks to Richard) Richard Damon <Richard@Damon-Family.org> - 2021-07-05 11:29 -0400
Re: Could H correctly decide that P never halts? [ Why can H ignore its own behavior? ](my thanks to Richard) Richard Damon <Richard@Damon-Family.org> - 2021-07-05 12:53 -0400
Re: Could H correctly decide that P never halts? [ Why can H ignore its own behavior? ](my thanks to Richard) olcott <NoOne@NoWhere.com> - 2021-07-05 12:13 -0500
Re: Could H correctly decide that P never halts? [ Why can H ignore its own behavior? ](my thanks to Richard) André G. Isaak <agisaak@gm.invalid> - 2021-07-04 14:55 -0600
Re: Could H correctly decide that P never halts? [ Why can H ignore its own behavior? ](my thanks to Richard) "dklei...@gmail.com" <dkleinecke@gmail.com> - 2021-07-04 18:03 -0700
Re: Could H correctly decide that P never halts? [ Why can H ignore its own behavior? ](my thanks to Richard) olcott <NoOne@NoWhere.com> - 2021-07-04 20:42 -0500
Re: Could H correctly decide that P never halts? [ Why can H ignore its own behavior? ](my thanks to Richard) Richard Damon <Richard@Damon-Family.org> - 2021-07-04 22:31 -0400
Re: Could H correctly decide that P never halts? [ Why can H ignore its own behavior? ](my thanks to Richard) Richard Damon <Richard@Damon-Family.org> - 2021-07-05 08:05 -0400
Re: Could H correctly decide that P never halts? [ Why can H ignore its own behavior? ](my thanks to Richard) olcott <NoOne@NoWhere.com> - 2021-07-05 08:40 -0500
Re: Could H correctly decide that P never halts? [ Why can H ignore its own behavior? ](my thanks to Richard) Richard Damon <Richard@Damon-Family.org> - 2021-07-05 12:13 -0400
Re: Could H correctly decide that P never halts? [ Why can H ignore its own behavior? ](my thanks to Richard) Jeff Barnett <jbb@notatt.com> - 2021-07-04 23:20 -0600
Re: Could H correctly decide that P never halts? [ Why can H ignore its own behavior? ](my thanks to Richard) Mike Terry <news.dead.person.stones@darjeeling.plus.com> - 2021-07-05 18:49 +0100
Re: Could H correctly decide that P never halts? [ Why can H ignore its own behavior? ](my thanks to Richard) Jeff Barnett <jbb@notatt.com> - 2021-07-05 15:17 -0600
Re: Could H correctly decide that P never halts? olcott <NoOne@NoWhere.com> - 2021-07-05 17:38 -0500
Re: Could H correctly decide that P never halts? [ Why can H ignore its own behavior? ](my thanks to Richard) olcott <NoOne@NoWhere.com> - 2021-07-04 23:52 -0500
Re: Could H correctly decide that P never halts? [ Why can H ignore its own behavior? ](my thanks to Richard) Richard Damon <Richard@Damon-Family.org> - 2021-07-05 10:36 -0400
Re: Could H correctly decide that P never halts? [ Why can H ignore its own behavior? ](my thanks to Richard) olcott <NoOne@NoWhere.com> - 2021-07-05 09:51 -0500
Re: Could H correctly decide that P never halts? [ Why can H ignore its own behavior? ](my thanks to Richard) Richard Damon <Richard@Damon-Family.org> - 2021-07-05 11:36 -0400
Re: Could H correctly decide that P never halts? [ Why can H ignore its own behavior? ](my thanks to Richard) olcott <NoOne@NoWhere.com> - 2021-07-05 08:18 -0500
Re: Could H correctly decide that P never halts? [ Why can H ignore its own behavior? ](my thanks to Richard) Richard Damon <Richard@Damon-Family.org> - 2021-07-05 12:11 -0400
Re: Could H correctly decide that P never halts? [ Why can H ignore its own behavior? ](my thanks to Richard) Richard Damon <Richard@Damon-Family.org> - 2021-07-04 15:02 -0400
Re: Could H correctly decide that P never halts? Ben Bacarisse <ben.usenet@bsb.me.uk> - 2021-07-03 21:20 +0100
Re: Could H correctly decide that P never halts? olcott <NoOne@NoWhere.com> - 2021-07-03 16:21 -0500
Re: Could H correctly decide that P never halts? Ben Bacarisse <ben.usenet@bsb.me.uk> - 2021-07-04 01:23 +0100
Re: Could H correctly decide that P never halts? [ already agreed ] olcott <NoOne@NoWhere.com> - 2021-07-03 19:30 -0500
Re: Could H correctly decide that P never halts? [ already agreed ] Ben Bacarisse <ben.usenet@bsb.me.uk> - 2021-07-04 17:46 +0100
Re: Could H correctly decide that P never halts? [ already agreed ] olcott <NoOne@NoWhere.com> - 2021-07-04 12:00 -0500
Re: Could H correctly decide that P never halts? [ already agreed ] Ben Bacarisse <ben.usenet@bsb.me.uk> - 2021-07-05 02:04 +0100
Re: Could H correctly decide that P never halts? [ already agreed ] olcott <NoOne@NoWhere.com> - 2021-07-04 20:57 -0500
Re: Could H correctly decide that P never halts? [ already agreed ] Ben Bacarisse <ben.usenet@bsb.me.uk> - 2021-07-05 03:14 +0100
Re: Could H correctly decide that P never halts? [ already agreed ] olcott <NoOne@NoWhere.com> - 2021-07-04 21:28 -0500
Re: Could H correctly decide that P never halts? [ already agreed ] Richard Damon <Richard@Damon-Family.org> - 2021-07-04 22:33 -0400
Re: Could H correctly decide that P never halts? [ already agreed ] olcott <NoOne@NoWhere.com> - 2021-07-04 22:09 -0500
Re: Could H correctly decide that P never halts? [ already agreed ] Richard Damon <Richard@Damon-Family.org> - 2021-07-05 07:33 -0400
Re: Could H correctly decide that P never halts? [ already agreed ] olcott <NoOne@NoWhere.com> - 2021-07-05 08:38 -0500
Re: Could H correctly decide that P never halts? [ already agreed ] Richard Damon <Richard@Damon-Family.org> - 2021-07-05 10:57 -0400
Re: Could H correctly decide that P never halts? [ already agreed ] olcott <NoOne@NoWhere.com> - 2021-07-05 09:59 -0500
Re: Could H correctly decide that P never halts? [ already agreed ] Richard Damon <Richard@Damon-Family.org> - 2021-07-05 11:34 -0400
Re: Could H correctly decide that P never halts? [ already agreed ] Richard Damon <Richard@Damon-Family.org> - 2021-07-05 13:16 -0400
Re: Could H correctly decide that P never halts? [ already agreed ] olcott <NoOne@NoWhere.com> - 2021-07-05 12:48 -0500
Re: Could H correctly decide that P never halts? [ already agreed ] Richard Damon <Richard@Damon-Family.org> - 2021-07-05 14:36 -0400
Re: Could H correctly decide that P never halts? [ already agreed ] Richard Damon <Richard@Damon-Family.org> - 2021-07-05 12:18 -0400
Re: Could H correctly decide that P never halts? [ already agreed ] Ben Bacarisse <ben.usenet@bsb.me.uk> - 2021-07-05 23:58 +0100
Re: Could H correctly decide that P never halts? [ already agreed ] olcott <NoOne@NoWhere.com> - 2021-07-05 19:54 -0500
Re: Could H correctly decide that P never halts? [ already agreed ] Richard Damon <Richard@Damon-Family.org> - 2021-07-05 21:29 -0400
Re: Could H correctly decide that P never halts? [ already agreed ] Ben Bacarisse <ben.usenet@bsb.me.uk> - 2021-07-06 03:35 +0100
Re: Could H correctly decide that P never halts? [ already agreed ] olcott <NoOne@NoWhere.com> - 2021-07-05 22:30 -0500
Re: Could H correctly decide that P never halts? [ already agreed ] Richard Damon <Richard@Damon-Family.org> - 2021-07-06 06:47 -0400
Re: Could H correctly decide that P never halts? [ already agreed ] Daniel Pehoushek <pehoushek1@gmail.com> - 2021-07-06 04:15 -0700
Re: Could H correctly decide that P never halts? [ already agreed ] Ben Bacarisse <ben.usenet@bsb.me.uk> - 2021-07-06 13:23 +0100
Re: Could H correctly decide that P never halts? [ already agreed ] Richard Damon <Richard@Damon-Family.org> - 2021-07-05 07:21 -0400
Re: Could H correctly decide that P never halts? [ halting criteria ] olcott <NoOne@NoWhere.com> - 2021-07-05 08:26 -0500
Re: Could H correctly decide that P never halts? [ halting criteria ] Richard Damon <Richard@Damon-Family.org> - 2021-07-05 10:50 -0400
Re: Could H correctly decide that P never halts? [ halting criteria ] olcott <NoOne@NoWhere.com> - 2021-07-05 09:56 -0500
Re: Could H correctly decide that P never halts? [ already agreed ] Richard Damon <Richard@Damon-Family.org> - 2021-07-05 10:30 -0400
Re: Could H correctly decide that P never halts? [ already agreed ] olcott <NoOne@NoWhere.com> - 2021-07-05 09:33 -0500
Re: Could H correctly decide that P never halts? [ already agreed ] Richard Damon <Richard@Damon-Family.org> - 2021-07-05 11:02 -0400
Re: Could H correctly decide that P never halts? [ already agreed ] (correction) olcott <NoOne@NoWhere.com> - 2021-07-05 09:35 -0500
Re: Could H correctly decide that P never halts? [ already agreed ] (correction) Richard Damon <Richard@Damon-Family.org> - 2021-07-05 12:38 -0400
Re: Could H correctly decide that P never halts? [ already agreed ] (correction) olcott <NoOne@NoWhere.com> - 2021-07-05 12:10 -0500
Re: Could H correctly decide that P never halts? [ already agreed ] (correction) Richard Damon <Richard@Damon-Family.org> - 2021-07-05 13:43 -0400
Re: Could H correctly decide that P never halts? [ Richard's excellent summation ] olcott <NoOne@NoWhere.com> - 2021-07-05 19:12 -0500
Re: Could H correctly decide that P never halts? Siri Cruise <chine.bleu@yahoo.com> - 2021-07-03 11:01 -0700
Re: Could H correctly decide that P never halts? olcott <NoOne@NoWhere.com> - 2021-07-03 13:15 -0500
Re: Could H correctly decide that P never halts? Ben Bacarisse <ben.usenet@bsb.me.uk> - 2021-07-03 21:09 +0100
Re: Could H correctly decide that P never halts? [ incorrect question ] olcott <NoOne@NoWhere.com> - 2021-07-03 16:06 -0500
Re: Could H correctly decide that P never halts? [ incorrect question ] Ben Bacarisse <ben.usenet@bsb.me.uk> - 2021-07-04 01:44 +0100
Re: Could H correctly decide that P never halts? [ incorrect question ] olcott <NoOne@NoWhere.com> - 2021-07-03 19:59 -0500
Re: Could H correctly decide that P never halts? [ incorrect question ] Mr Flibble <flibble@reddwarf.jmc> - 2021-07-04 02:34 +0100
Re: Could H correctly decide that P never halts? [ incorrect question ] olcott <NoOne@NoWhere.com> - 2021-07-03 20:46 -0500
Re: Could H correctly decide that P never halts? [ incorrect question ] Ben Bacarisse <ben.usenet@bsb.me.uk> - 2021-07-04 17:45 +0100
Re: Could H correctly decide that P never halts? Ben Bacarisse <ben.usenet@bsb.me.uk> - 2021-07-03 19:15 +0100
Re: Could H correctly decide that P never halts? olcott <NoOne@NoWhere.com> - 2021-07-03 13:58 -0500
Re: Could H correctly decide that P never halts? Siri Cruise <chine.bleu@yahoo.com> - 2021-07-03 22:37 -0700
Re: Could H correctly decide that P never halts? Ben Bacarisse <ben.usenet@bsb.me.uk> - 2021-07-04 14:14 +0100
Page 5 of 7 — ← Prev page 1 2 3 4 [5] 6 7 Next page →
| From | Ben Bacarisse <ben.usenet@bsb.me.uk> |
|---|---|
| Date | 2021-07-04 17:46 +0100 |
| Subject | Re: Could H correctly decide that P never halts? [ already agreed ] |
| Message-ID | <877di6c9ii.fsf@bsb.me.uk> |
| In reply to | #35673 |
olcott <NoOne@NoWhere.com> writes:
> On 7/3/2021 7:23 PM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>> On 7/3/2021 3:20 PM, Ben Bacarisse wrote:
>>>> olcott <NoOne@NoWhere.com> writes:
>>>>
>>>>> On 7/3/2021 11:25 AM, Ben Bacarisse wrote:
>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>
>>>>>>> void P(u32 x)
>>>>>>> {
>>>>>>> u32 Input_Halts = H(x, x);
>>>>>>> if (Input_Halts)
>>>>>>> HERE: goto HERE;
>>>>>>> }
>>>>>> For H(P,P) to be correct one of these must apply:
>>>>>>
>>>>>> H(P,P) == 0 and P(P) does not halt, or
>>>>>> H(P,P) != 0 and P(P) halts.
>>>>>>
>>>>>> Neither is the case.
>>>>>
>>>>> Superficially it may seem that way until you realize (as you have
>>>>> already realized) that the halt criteria must be adapted for a
>>>>> simulating halt decider.
>>>>>
>>>> The two criteria I gave are for the halting problem. No one is
>>>> interested in your "other kind of halting". It seems you agree that H
>>>> is wrong about the input given above as far as the criterion of actual
>>>> halting goes.
>>>
>>> Because a simulating halt decider must always abort the simulation of
>>> every input that never halts its halt deciding criteria must be
>>> adapted.
>> If you "adapted" the criterion for being a prime number you could
>> "refute" many more theorems.
>
> You already agreed that this adaptation of the halt status criteria is
> correct:
A halting computation is one that halts. There is no other criterion.
I can't believe this needs to be said.
For H(P,P) to be correct one of these must apply:
H(P,P) == 0 and P(P) does not halt, or
H(P,P) != 0 and P(P) halts.
Neither is the case for your H and P.
--
Ben.
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| From | olcott <NoOne@NoWhere.com> |
|---|---|
| Date | 2021-07-04 12:00 -0500 |
| Subject | Re: Could H correctly decide that P never halts? [ already agreed ] |
| Message-ID | <O-mdndOK8KzRdHz9nZ2dnUU7-X_NnZ2d@giganews.com> |
| In reply to | #35699 |
On 7/4/2021 11:46 AM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> On 7/3/2021 7:23 PM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>> On 7/3/2021 3:20 PM, Ben Bacarisse wrote:
>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>
>>>>>> On 7/3/2021 11:25 AM, Ben Bacarisse wrote:
>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>
>>>>>>>> void P(u32 x)
>>>>>>>> {
>>>>>>>> u32 Input_Halts = H(x, x);
>>>>>>>> if (Input_Halts)
>>>>>>>> HERE: goto HERE;
>>>>>>>> }
>>>>>>> For H(P,P) to be correct one of these must apply:
>>>>>>>
>>>>>>> H(P,P) == 0 and P(P) does not halt, or
>>>>>>> H(P,P) != 0 and P(P) halts.
>>>>>>>
>>>>>>> Neither is the case.
>>>>>>
>>>>>> Superficially it may seem that way until you realize (as you have
>>>>>> already realized) that the halt criteria must be adapted for a
>>>>>> simulating halt decider.
>>>>>>
>>>>> The two criteria I gave are for the halting problem. No one is
>>>>> interested in your "other kind of halting". It seems you agree that H
>>>>> is wrong about the input given above as far as the criterion of actual
>>>>> halting goes.
>>>>
>>>> Because a simulating halt decider must always abort the simulation of
>>>> every input that never halts its halt deciding criteria must be
>>>> adapted.
>>> If you "adapted" the criterion for being a prime number you could
>>> "refute" many more theorems.
>>
>> You already agreed that this adaptation of the halt status criteria is
>> correct:
>
> A halting computation is one that halts. There is no other criterion.
> I can't believe this needs to be said.
>
> For H(P,P) to be correct one of these must apply:
>
> H(P,P) == 0 and P(P) does not halt, or
> H(P,P) != 0 and P(P) halts.
>
> Neither is the case for your H and P.
>
Yes and by this same reasoning we know that when the simulation of an
infinite loop is aborted this proves that the infinite loop that only
jumps to its own machine address is indeed a halting computation.
void Infinite_Loop()
{
HERE: goto HERE;
}
int main()
{
u32 Input_Would_Halt2 = H((u32)Infinite_Loop, (u32)Infinite_Loop);
Output("Input_Would_Halt2 = ", Input_Would_Halt2);
}
--
Copyright 2021 Pete Olcott
"Great spirits have always encountered violent opposition from mediocre
minds." Einstein
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| From | Ben Bacarisse <ben.usenet@bsb.me.uk> |
|---|---|
| Date | 2021-07-05 02:04 +0100 |
| Subject | Re: Could H correctly decide that P never halts? [ already agreed ] |
| Message-ID | <87pmvxbmg3.fsf@bsb.me.uk> |
| In reply to | #35700 |
olcott <NoOne@NoWhere.com> writes:
> On 7/4/2021 11:46 AM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>> On 7/3/2021 7:23 PM, Ben Bacarisse wrote:
>>>> olcott <NoOne@NoWhere.com> writes:
>>>>
>>>>> On 7/3/2021 3:20 PM, Ben Bacarisse wrote:
>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>
>>>>>>> On 7/3/2021 11:25 AM, Ben Bacarisse wrote:
>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>
>>>>>>>>> void P(u32 x)
>>>>>>>>> {
>>>>>>>>> u32 Input_Halts = H(x, x);
>>>>>>>>> if (Input_Halts)
>>>>>>>>> HERE: goto HERE;
>>>>>>>>> }
>>>>>>>> For H(P,P) to be correct one of these must apply:
>>>>>>>>
>>>>>>>> H(P,P) == 0 and P(P) does not halt, or
>>>>>>>> H(P,P) != 0 and P(P) halts.
>>>>>>>>
>>>>>>>> Neither is the case.
>>>>>>>
>>>>>>> Superficially it may seem that way until you realize (as you have
>>>>>>> already realized) that the halt criteria must be adapted for a
>>>>>>> simulating halt decider.
>>>>>>>
>>>>>> The two criteria I gave are for the halting problem. No one is
>>>>>> interested in your "other kind of halting". It seems you agree that H
>>>>>> is wrong about the input given above as far as the criterion of actual
>>>>>> halting goes.
>>>>>
>>>>> Because a simulating halt decider must always abort the simulation of
>>>>> every input that never halts its halt deciding criteria must be
>>>>> adapted.
>>>> If you "adapted" the criterion for being a prime number you could
>>>> "refute" many more theorems.
>>>
>>> You already agreed that this adaptation of the halt status criteria is
>>> correct:
>> A halting computation is one that halts. There is no other criterion.
>> I can't believe this needs to be said.
>> For H(P,P) to be correct one of these must apply:
>> H(P,P) == 0 and P(P) does not halt, or
>> H(P,P) != 0 and P(P) halts.
>> Neither is the case for your H and P.
>
> Yes and by this same reasoning we know that when the simulation of an
> infinite loop is aborted this proves that the infinite loop that only
> jumps to its own machine address is indeed a halting computation.
>
> void Infinite_Loop()
> {
> HERE: goto HERE;
> }
I presume this is supposed to be sarcasm, but your position (that a
halting computation is halting one) is so absurd that I can't be 100%
sure. Whatever you intend by it, you make sure, as always, to avoid the
facts:
- Every instance of the halting problem has a correct yes/no answer.
- Yes is the correct answer for those strings that represent halting
computations. No is the correct answer for all other instances.
- You don't accept this and are claiming that and algorithm that reports
no for a halting computation is correct.
For some reason you think this is interesting. What's interesting is
that you appear to honestly believe it. It's a shame you won't honestly
stand up for it.
--
Ben.
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| From | olcott <NoOne@NoWhere.com> |
|---|---|
| Date | 2021-07-04 20:57 -0500 |
| Subject | Re: Could H correctly decide that P never halts? [ already agreed ] |
| Message-ID | <NrGdnXPuYcSG-n_9nZ2dnUU7-XednZ2d@giganews.com> |
| In reply to | #35699 |
On 7/4/2021 11:46 AM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> On 7/3/2021 7:23 PM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>> On 7/3/2021 3:20 PM, Ben Bacarisse wrote:
>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>
>>>>>> On 7/3/2021 11:25 AM, Ben Bacarisse wrote:
>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>
>>>>>>>> void P(u32 x)
>>>>>>>> {
>>>>>>>> u32 Input_Halts = H(x, x);
>>>>>>>> if (Input_Halts)
>>>>>>>> HERE: goto HERE;
>>>>>>>> }
>>>>>>> For H(P,P) to be correct one of these must apply:
>>>>>>>
>>>>>>> H(P,P) == 0 and P(P) does not halt, or
>>>>>>> H(P,P) != 0 and P(P) halts.
>>>>>>>
>>>>>>> Neither is the case.
>>>>>>
>>>>>> Superficially it may seem that way until you realize (as you have
>>>>>> already realized) that the halt criteria must be adapted for a
>>>>>> simulating halt decider.
>>>>>>
>>>>> The two criteria I gave are for the halting problem. No one is
>>>>> interested in your "other kind of halting". It seems you agree that H
>>>>> is wrong about the input given above as far as the criterion of actual
>>>>> halting goes.
>>>>
>>>> Because a simulating halt decider must always abort the simulation of
>>>> every input that never halts its halt deciding criteria must be
>>>> adapted.
>>> If you "adapted" the criterion for being a prime number you could
>>> "refute" many more theorems.
>>
>> You already agreed that this adaptation of the halt status criteria is
>> correct:
>
> A halting computation is one that halts. There is no other criterion.
> I can't believe this needs to be said.
>
void Infinite_Loop()
{
HERE: goto HERE;
}
int main()
{
u32 Input_Would_Halt2 = H((u32)Infinite_Loop, (u32)Infinite_Loop);
Output("Input_Would_Halt2 = ", Input_Would_Halt2);
}
Then Infinite_Loop() is a halting computation on the basis that H aborts
its simulation of Infinite_Loop().
--
Copyright 2021 Pete Olcott
"Great spirits have always encountered violent opposition from mediocre
minds." Einstein
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| From | Ben Bacarisse <ben.usenet@bsb.me.uk> |
|---|---|
| Date | 2021-07-05 03:14 +0100 |
| Subject | Re: Could H correctly decide that P never halts? [ already agreed ] |
| Message-ID | <87fswtbj84.fsf@bsb.me.uk> |
| In reply to | #35712 |
olcott <NoOne@NoWhere.com> writes:
> On 7/4/2021 11:46 AM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>> On 7/3/2021 7:23 PM, Ben Bacarisse wrote:
>>>> olcott <NoOne@NoWhere.com> writes:
>>>>
>>>>> On 7/3/2021 3:20 PM, Ben Bacarisse wrote:
>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>
>>>>>>> On 7/3/2021 11:25 AM, Ben Bacarisse wrote:
>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>
>>>>>>>>> void P(u32 x)
>>>>>>>>> {
>>>>>>>>> u32 Input_Halts = H(x, x);
>>>>>>>>> if (Input_Halts)
>>>>>>>>> HERE: goto HERE;
>>>>>>>>> }
>>>>>>>> For H(P,P) to be correct one of these must apply:
>>>>>>>>
>>>>>>>> H(P,P) == 0 and P(P) does not halt, or
>>>>>>>> H(P,P) != 0 and P(P) halts.
>>>>>>>>
>>>>>>>> Neither is the case.
>>>>>>>
>>>>>>> Superficially it may seem that way until you realize (as you have
>>>>>>> already realized) that the halt criteria must be adapted for a
>>>>>>> simulating halt decider.
>>>>>>>
>>>>>> The two criteria I gave are for the halting problem. No one is
>>>>>> interested in your "other kind of halting". It seems you agree that H
>>>>>> is wrong about the input given above as far as the criterion of actual
>>>>>> halting goes.
>>>>>
>>>>> Because a simulating halt decider must always abort the simulation of
>>>>> every input that never halts its halt deciding criteria must be
>>>>> adapted.
>>>> If you "adapted" the criterion for being a prime number you could
>>>> "refute" many more theorems.
>>>
>>> You already agreed that this adaptation of the halt status criteria is
>>> correct:
>> A halting computation is one that halts. There is no other criterion.
>> I can't believe this needs to be said.
>>
>
> void Infinite_Loop()
> {
> HERE: goto HERE;
> }
>
> int main()
> {
> u32 Input_Would_Halt2 = H((u32)Infinite_Loop, (u32)Infinite_Loop);
> Output("Input_Would_Halt2 = ", Input_Would_Halt2);
> }
>
> Then Infinite_Loop() is a halting computation on the basis that H
> aborts its simulation of Infinite_Loop().
I think you've lost the plot. Infinite_Loop() is a not a halting
computation. H(Infinite_Loop, Infinite_Loop) must halt if there is to
be a result, but Infinite_Loop(Infinite_Loop) (technically the
computation represented by the arguments to H) is not a halting
computation.
Your H(P,P) == 0 is wrong because P(P) halts. There is only one
definition of halting and it is not your "adapted" one.
--
Ben.
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| From | olcott <NoOne@NoWhere.com> |
|---|---|
| Date | 2021-07-04 21:28 -0500 |
| Subject | Re: Could H correctly decide that P never halts? [ already agreed ] |
| Message-ID | <bJCdnQAh7eLG83_9nZ2dnUU7-cfNnZ2d@giganews.com> |
| In reply to | #35714 |
On 7/4/2021 9:14 PM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> On 7/4/2021 11:46 AM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>> On 7/3/2021 7:23 PM, Ben Bacarisse wrote:
>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>
>>>>>> On 7/3/2021 3:20 PM, Ben Bacarisse wrote:
>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>
>>>>>>>> On 7/3/2021 11:25 AM, Ben Bacarisse wrote:
>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>>
>>>>>>>>>> void P(u32 x)
>>>>>>>>>> {
>>>>>>>>>> u32 Input_Halts = H(x, x);
>>>>>>>>>> if (Input_Halts)
>>>>>>>>>> HERE: goto HERE;
>>>>>>>>>> }
>>>>>>>>> For H(P,P) to be correct one of these must apply:
>>>>>>>>>
>>>>>>>>> H(P,P) == 0 and P(P) does not halt, or
>>>>>>>>> H(P,P) != 0 and P(P) halts.
>>>>>>>>>
>>>>>>>>> Neither is the case.
>>>>>>>>
>>>>>>>> Superficially it may seem that way until you realize (as you have
>>>>>>>> already realized) that the halt criteria must be adapted for a
>>>>>>>> simulating halt decider.
>>>>>>>>
>>>>>>> The two criteria I gave are for the halting problem. No one is
>>>>>>> interested in your "other kind of halting". It seems you agree that H
>>>>>>> is wrong about the input given above as far as the criterion of actual
>>>>>>> halting goes.
>>>>>>
>>>>>> Because a simulating halt decider must always abort the simulation of
>>>>>> every input that never halts its halt deciding criteria must be
>>>>>> adapted.
>>>>> If you "adapted" the criterion for being a prime number you could
>>>>> "refute" many more theorems.
>>>>
>>>> You already agreed that this adaptation of the halt status criteria is
>>>> correct:
>>> A halting computation is one that halts. There is no other criterion.
>>> I can't believe this needs to be said.
>>>
>>
>> void Infinite_Loop()
>> {
>> HERE: goto HERE;
>> }
>>
>> int main()
>> {
>> u32 Input_Would_Halt2 = H((u32)Infinite_Loop, (u32)Infinite_Loop);
>> Output("Input_Would_Halt2 = ", Input_Would_Halt2);
>> }
>>
>> Then Infinite_Loop() is a halting computation on the basis that H
>> aborts its simulation of Infinite_Loop().
>
> I think you've lost the plot. Infinite_Loop() is a not a halting
> computation. H(Infinite_Loop, Infinite_Loop) must halt if there is to
> be a result, but Infinite_Loop(Infinite_Loop) (technically the
> computation represented by the arguments to H) is not a halting
> computation.
>
Every computation that would never halt unless its simulation is aborted
is equally not a halting computation.
There is no exception to this rule in the same way and for the same
reason that the numeric integer 5 is always numerically greater than the
numeric integer 3.
On 5/11/2021 11:10 AM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> Truism:
>> Every simulation that would never stop unless Halts() stops
>> it at some point specifies infinite execution.
>
> Any algorithm that implements this truism is, of course, a halting
> decider.
--
Copyright 2021 Pete Olcott
"Great spirits have always encountered violent opposition from mediocre
minds." Einstein
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| From | Richard Damon <Richard@Damon-Family.org> |
|---|---|
| Date | 2021-07-04 22:33 -0400 |
| Subject | Re: Could H correctly decide that P never halts? [ already agreed ] |
| Message-ID | <VbuEI.28080$Vj7.5662@fx46.iad> |
| In reply to | #35715 |
On 7/4/21 10:28 PM, olcott wrote:
> On 7/4/2021 9:14 PM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>> On 7/4/2021 11:46 AM, Ben Bacarisse wrote:
>>>> olcott <NoOne@NoWhere.com> writes:
>>>>
>>>>> On 7/3/2021 7:23 PM, Ben Bacarisse wrote:
>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>
>>>>>>> On 7/3/2021 3:20 PM, Ben Bacarisse wrote:
>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>
>>>>>>>>> On 7/3/2021 11:25 AM, Ben Bacarisse wrote:
>>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>>>
>>>>>>>>>>> void P(u32 x)
>>>>>>>>>>> {
>>>>>>>>>>> u32 Input_Halts = H(x, x);
>>>>>>>>>>> if (Input_Halts)
>>>>>>>>>>> HERE: goto HERE;
>>>>>>>>>>> }
>>>>>>>>>> For H(P,P) to be correct one of these must apply:
>>>>>>>>>>
>>>>>>>>>> H(P,P) == 0 and P(P) does not halt, or
>>>>>>>>>> H(P,P) != 0 and P(P) halts.
>>>>>>>>>>
>>>>>>>>>> Neither is the case.
>>>>>>>>>
>>>>>>>>> Superficially it may seem that way until you realize (as you have
>>>>>>>>> already realized) that the halt criteria must be adapted for a
>>>>>>>>> simulating halt decider.
>>>>>>>>>
>>>>>>>> The two criteria I gave are for the halting problem. No one is
>>>>>>>> interested in your "other kind of halting". It seems you agree
>>>>>>>> that H
>>>>>>>> is wrong about the input given above as far as the criterion of
>>>>>>>> actual
>>>>>>>> halting goes.
>>>>>>>
>>>>>>> Because a simulating halt decider must always abort the
>>>>>>> simulation of
>>>>>>> every input that never halts its halt deciding criteria must be
>>>>>>> adapted.
>>>>>> If you "adapted" the criterion for being a prime number you could
>>>>>> "refute" many more theorems.
>>>>>
>>>>> You already agreed that this adaptation of the halt status criteria is
>>>>> correct:
>>>> A halting computation is one that halts. There is no other criterion.
>>>> I can't believe this needs to be said.
>>>>
>>>
>>> void Infinite_Loop()
>>> {
>>> HERE: goto HERE;
>>> }
>>>
>>> int main()
>>> {
>>> u32 Input_Would_Halt2 = H((u32)Infinite_Loop, (u32)Infinite_Loop);
>>> Output("Input_Would_Halt2 = ", Input_Would_Halt2);
>>> }
>>>
>>> Then Infinite_Loop() is a halting computation on the basis that H
>>> aborts its simulation of Infinite_Loop().
>>
>> I think you've lost the plot. Infinite_Loop() is a not a halting
>> computation. H(Infinite_Loop, Infinite_Loop) must halt if there is to
>> be a result, but Infinite_Loop(Infinite_Loop) (technically the
>> computation represented by the arguments to H) is not a halting
>> computation.
>>
>
> Every computation that would never halt unless its simulation is aborted
> is equally not a halting computation.
>
> There is no exception to this rule in the same way and for the same
> reason that the numeric integer 5 is always numerically greater than the
> numeric integer 3.
A P will halt even if the instance of the simulator that is simulating
it is change to a UTM that will never stop. This has even been proven by
your.
Thus, YOU have proven that by this rule, P is a Halting Simulation.
>
> On 5/11/2021 11:10 AM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>> Truism:
>>> Every simulation that would never stop unless Halts() stops
>>> it at some point specifies infinite execution.
>>
>> Any algorithm that implements this truism is, of course, a halting
>> decider.
>
>
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| From | olcott <NoOne@NoWhere.com> |
|---|---|
| Date | 2021-07-04 22:09 -0500 |
| Subject | Re: Could H correctly decide that P never halts? [ already agreed ] |
| Message-ID | <ZNednaw0HfVu6n_9nZ2dnUU7-X_NnZ2d@giganews.com> |
| In reply to | #35715 |
On 7/4/2021 9:28 PM, olcott wrote:
> On 7/4/2021 9:14 PM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>> On 7/4/2021 11:46 AM, Ben Bacarisse wrote:
>>>> olcott <NoOne@NoWhere.com> writes:
>>>>
>>>>> On 7/3/2021 7:23 PM, Ben Bacarisse wrote:
>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>
>>>>>>> On 7/3/2021 3:20 PM, Ben Bacarisse wrote:
>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>
>>>>>>>>> On 7/3/2021 11:25 AM, Ben Bacarisse wrote:
>>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>>>
>>>>>>>>>>> void P(u32 x)
>>>>>>>>>>> {
>>>>>>>>>>> u32 Input_Halts = H(x, x);
>>>>>>>>>>> if (Input_Halts)
>>>>>>>>>>> HERE: goto HERE;
>>>>>>>>>>> }
>>>>>>>>>> For H(P,P) to be correct one of these must apply:
>>>>>>>>>>
>>>>>>>>>> H(P,P) == 0 and P(P) does not halt, or
>>>>>>>>>> H(P,P) != 0 and P(P) halts.
>>>>>>>>>>
>>>>>>>>>> Neither is the case.
>>>>>>>>>
>>>>>>>>> Superficially it may seem that way until you realize (as you have
>>>>>>>>> already realized) that the halt criteria must be adapted for a
>>>>>>>>> simulating halt decider.
>>>>>>>>>
>>>>>>>> The two criteria I gave are for the halting problem. No one is
>>>>>>>> interested in your "other kind of halting". It seems you agree
>>>>>>>> that H
>>>>>>>> is wrong about the input given above as far as the criterion of
>>>>>>>> actual
>>>>>>>> halting goes.
>>>>>>>
>>>>>>> Because a simulating halt decider must always abort the
>>>>>>> simulation of
>>>>>>> every input that never halts its halt deciding criteria must be
>>>>>>> adapted.
>>>>>> If you "adapted" the criterion for being a prime number you could
>>>>>> "refute" many more theorems.
>>>>>
>>>>> You already agreed that this adaptation of the halt status criteria is
>>>>> correct:
>>>> A halting computation is one that halts. There is no other criterion.
>>>> I can't believe this needs to be said.
>>>>
>>>
>>> void Infinite_Loop()
>>> {
>>> HERE: goto HERE;
>>> }
>>>
>>> int main()
>>> {
>>> u32 Input_Would_Halt2 = H((u32)Infinite_Loop, (u32)Infinite_Loop);
>>> Output("Input_Would_Halt2 = ", Input_Would_Halt2);
>>> }
>>>
>>> Then Infinite_Loop() is a halting computation on the basis that H
>>> aborts its simulation of Infinite_Loop().
>>
>> I think you've lost the plot. Infinite_Loop() is a not a halting
>> computation. H(Infinite_Loop, Infinite_Loop) must halt if there is to
>> be a result, but Infinite_Loop(Infinite_Loop) (technically the
>> computation represented by the arguments to H) is not a halting
>> computation.
>>
>
> Every computation that would never halt unless its simulation is aborted
> is equally not a halting computation.
>
> There is no exception to this rule in the same way and for the same
> reason that the numeric integer 5 is always numerically greater than the
> numeric integer 3.
>
> On 5/11/2021 11:10 AM, Ben Bacarisse wrote:
> > olcott <NoOne@NoWhere.com> writes:
> >
> >> Truism:
> >> Every simulation that would never stop unless Halts() stops
> >> it at some point specifies infinite execution.
> >
> > Any algorithm that implements this truism is, of course, a halting
> > decider.
>
>
It can be easily verified that if no H ever aborts any P that P(P) will
never halt. If any H must abort any P then this H does correctly decide
that this P does not halt.
--
Copyright 2021 Pete Olcott
"Great spirits have always encountered violent opposition from mediocre
minds." Einstein
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| From | Richard Damon <Richard@Damon-Family.org> |
|---|---|
| Date | 2021-07-05 07:33 -0400 |
| Subject | Re: Could H correctly decide that P never halts? [ already agreed ] |
| Message-ID | <i6CEI.114095$ytM9.39518@fx35.iad> |
| In reply to | #35718 |
On 7/4/21 11:09 PM, olcott wrote: > It can be easily verified that if no H ever aborts any P that P(P) will > never halt. If any H must abort any P then this H does correctly decide > that this P does not halt. > Wrong. If NO H ever aborts any P then this H will not answer the question H(P,P) and thus isn't a valid decider. Once H is written so that it will be able to answer the question, then the question that matches the halting problem is does THIS instance of H need to abort the copy of the representation of P and I in order for the simulation to reach an end. (Not Any). This can be shown by changing that one instance to be a UTM, which will simulate P to its natural end. It can be shown that P(P) will complete on its own by simply running it, P will call H(P,P) which will decide incorrectly that P(P) is non-halting, return that answer to P and P will halt. Please show what is wrong with that analysis, an actual PROOF (not just claim by meaning, you don't use valid meanings), or someone with repute that uses that definition the way you do. Note fundamental rule, Words mean what the community has them mean, an individual can not validly give them a different meaning.
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| From | olcott <NoOne@NoWhere.com> |
|---|---|
| Date | 2021-07-05 08:38 -0500 |
| Subject | Re: Could H correctly decide that P never halts? [ already agreed ] |
| Message-ID | <mL-dnd_DlOr7ln79nZ2dnUU7-IHNnZ2d@giganews.com> |
| In reply to | #35725 |
On 7/5/2021 6:33 AM, Richard Damon wrote: > On 7/4/21 11:09 PM, olcott wrote: > >> It can be easily verified that if no H ever aborts any P that P(P) will >> never halt. If any H must abort any P then this H does correctly decide >> that this P does not halt. >> > > Wrong. > > If NO H ever aborts any P then this H will not answer the question > H(P,P) and thus isn't a valid decider. An answer that would not be a damned lie would be: An answer that would not be a damned lie would be: An answer that would not be a damned lie would be: An answer that would not be a damned lie would be: An answer that would not be a damned lie would be: Right you are correct if no H ever aborted any P then P(P) would never halt and this would also prevent H from being a decider. -- Copyright 2021 Pete Olcott "Great spirits have always encountered violent opposition from mediocre minds." Einstein
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| From | Richard Damon <Richard@Damon-Family.org> |
|---|---|
| Date | 2021-07-05 10:57 -0400 |
| Subject | Re: Could H correctly decide that P never halts? [ already agreed ] |
| Message-ID | <R5FEI.30871$_X3.20480@fx40.iad> |
| In reply to | #35734 |
On 7/5/21 9:38 AM, olcott wrote: > On 7/5/2021 6:33 AM, Richard Damon wrote: >> On 7/4/21 11:09 PM, olcott wrote: >> >>> It can be easily verified that if no H ever aborts any P that P(P) will >>> never halt. If any H must abort any P then this H does correctly decide >>> that this P does not halt. >>> >> >> Wrong. >> >> If NO H ever aborts any P then this H will not answer the question >> H(P,P) and thus isn't a valid decider. > > An answer that would not be a damned lie would be: > An answer that would not be a damned lie would be: > An answer that would not be a damned lie would be: > An answer that would not be a damned lie would be: > An answer that would not be a damned lie would be: > > Right you are correct if no H ever aborted any P then P(P) would never > halt and this would also prevent H from being a decider. > But the way you interpret that sentence, it would be a damned lie. Maybe Literally. The Devil loves to use statements that are mostly true but have a ting of falsehood in it to corrupt the hearer. Are you working for him? But Yes, if H never aborted any P then H would not be a decider. That does NOT mean that H is correct in aborting P and calling it non-Halting, since when we change H we change P. Hn non aborting -> Hn(Pn,Pn) will be none halting. Ha aborting -> Ha(Pa,Pa) will answer, by your stipulation non-halting. This means that Pa(Pa) will get the non-Halting answer from its call to Ha(Pa,Pa) and will halt and thus Pa(Pa) is a Halting computaiton, and Ha was wrong. All you have shown is that Ha(Pn,Pn) can say non-halting and be right about that answer, which doesn't mean a thing, as Pn never claimed to be able to confound Ha, only Hn, and only if Hn could actually answer. The machine that confounds Ha is Pa, and it DOES succeed, as Ha says Pa(Pa) is non-Halting when even as you have confiremed Pa(Pa) does Halt. You just get stuck in your 'logic' and want non-Halting to be the right answer because you confuse Pa and Pn because you don't seem to understand the concept of complete programs.
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| From | olcott <NoOne@NoWhere.com> |
|---|---|
| Date | 2021-07-05 09:59 -0500 |
| Subject | Re: Could H correctly decide that P never halts? [ already agreed ] |
| Message-ID | <682dnVEYTI7Dg379nZ2dnUU7-V-dnZ2d@giganews.com> |
| In reply to | #35745 |
On 7/5/2021 9:57 AM, Richard Damon wrote: > On 7/5/21 9:38 AM, olcott wrote: >> On 7/5/2021 6:33 AM, Richard Damon wrote: >>> On 7/4/21 11:09 PM, olcott wrote: >>> >>>> It can be easily verified that if no H ever aborts any P that P(P) will >>>> never halt. If any H must abort any P then this H does correctly decide >>>> that this P does not halt. >>>> >>> >>> Wrong. >>> >>> If NO H ever aborts any P then this H will not answer the question >>> H(P,P) and thus isn't a valid decider. >> >> An answer that would not be a damned lie would be: >> An answer that would not be a damned lie would be: >> An answer that would not be a damned lie would be: >> An answer that would not be a damned lie would be: >> An answer that would not be a damned lie would be: >> >> Right you are correct if no H ever aborted any P then P(P) would never >> halt and this would also prevent H from being a decider. >> > > But the way you interpret that sentence, it would be a damned lie. Maybe > Literally. The Devil loves to use statements that are mostly true but > have a ting of falsehood in it to corrupt the hearer. > > Are you working for him? > > But Yes, if H never aborted any P then H would not be a decider. That > does NOT mean that H is correct in aborting P and calling it > non-Halting, since when we change H we change P. > > Hn non aborting -> Hn(Pn,Pn) will be none halting. > Ha aborting -> Ha(Pa,Pa) will answer, by your stipulation non-halting. > > This means that Pa(Pa) will get the non-Halting answer from its call to > Ha(Pa,Pa) and will halt and thus Pa(Pa) is a Halting computaiton, and Ha > was wrong. > > All you have shown is that Ha(Pn,Pn) can say non-halting and be right > about that answer, which doesn't mean a thing, as Pn never claimed to be > able to confound Ha, only Hn, and only if Hn could actually answer. > > The machine that confounds Ha is Pa, and it DOES succeed, as Ha says > Pa(Pa) is non-Halting when even as you have confiremed Pa(Pa) does Halt. > You just get stuck in your 'logic' and want non-Halting to be the right > answer because you confuse Pa and Pn because you don't seem to > understand the concept of complete programs. > It can be easily verified that if no H(n) ever aborts any P(m) that P(P) will never halt. If any H(n) must abort any P(m) then this H(n) does correctly decide that this P(m) does not halt. -- Copyright 2021 Pete Olcott "Great spirits have always encountered violent opposition from mediocre minds." Einstein
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| From | Richard Damon <Richard@Damon-Family.org> |
|---|---|
| Date | 2021-07-05 11:34 -0400 |
| Subject | Re: Could H correctly decide that P never halts? [ already agreed ] |
| Message-ID | <YDFEI.419$9p5.339@fx19.iad> |
| In reply to | #35746 |
On 7/5/21 10:59 AM, olcott wrote: > It can be easily verified that if no H(n) ever aborts any P(m) that P(P) > will never halt. If any H(n) must abort any P(m) then this H(n) does > correctly decide that this P(m) does not halt. > > FALSE STATEMENT AS PREVIOUS PROVEN. Yes, if we have an Hn that never aborts, then Pn(Pn) will never halt, but then Hn never gives an answer. If we have an Ha that will abort, then Pa(Pa) will be a halting computation, and it can be shown that the Ha simulating a Pa(Pa) incorrectly decides that Pa is non-halting because it fails to take into account that this Pa includes a Ha that will incorrectly decide that Pa(Pa) is non-halting, and that simulating Ha could have not aborted its simulation and seen that Pa(Pa) does halt. You logic confounds Pa with Pn and comes to the wrong conclusion.
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| From | Richard Damon <Richard@Damon-Family.org> |
|---|---|
| Date | 2021-07-05 13:16 -0400 |
| Subject | Re: Could H correctly decide that P never halts? [ already agreed ] |
| Message-ID | <28HEI.7299$mR.3487@fx33.iad> |
| In reply to | #35746 |
On 7/5/21 10:59 AM, olcott wrote: > On 7/5/2021 9:57 AM, Richard Damon wrote: >> On 7/5/21 9:38 AM, olcott wrote: >>> On 7/5/2021 6:33 AM, Richard Damon wrote: >>>> On 7/4/21 11:09 PM, olcott wrote: >>>> >>>>> It can be easily verified that if no H ever aborts any P that P(P) >>>>> will >>>>> never halt. If any H must abort any P then this H does correctly >>>>> decide >>>>> that this P does not halt. >>>>> >>>> >>>> Wrong. >>>> >>>> If NO H ever aborts any P then this H will not answer the question >>>> H(P,P) and thus isn't a valid decider. >>> >>> An answer that would not be a damned lie would be: >>> An answer that would not be a damned lie would be: >>> An answer that would not be a damned lie would be: >>> An answer that would not be a damned lie would be: >>> An answer that would not be a damned lie would be: >>> >>> Right you are correct if no H ever aborted any P then P(P) would never >>> halt and this would also prevent H from being a decider. >>> >> >> But the way you interpret that sentence, it would be a damned lie. Maybe >> Literally. The Devil loves to use statements that are mostly true but >> have a ting of falsehood in it to corrupt the hearer. >> >> Are you working for him? >> >> But Yes, if H never aborted any P then H would not be a decider. That >> does NOT mean that H is correct in aborting P and calling it >> non-Halting, since when we change H we change P. >> >> Hn non aborting -> Hn(Pn,Pn) will be none halting. >> Ha aborting -> Ha(Pa,Pa) will answer, by your stipulation non-halting. >> >> This means that Pa(Pa) will get the non-Halting answer from its call to >> Ha(Pa,Pa) and will halt and thus Pa(Pa) is a Halting computaiton, and Ha >> was wrong. >> >> All you have shown is that Ha(Pn,Pn) can say non-halting and be right >> about that answer, which doesn't mean a thing, as Pn never claimed to be >> able to confound Ha, only Hn, and only if Hn could actually answer. >> >> The machine that confounds Ha is Pa, and it DOES succeed, as Ha says >> Pa(Pa) is non-Halting when even as you have confiremed Pa(Pa) does Halt. >> You just get stuck in your 'logic' and want non-Halting to be the right >> answer because you confuse Pa and Pn because you don't seem to >> understand the concept of complete programs. >> > > It can be easily verified that if no H(n) ever aborts any P(m) that P(P) > will never halt. If any H(n) must abort any P(m) then this H(n) does > correctly decide that this P(m) does not halt. > > Except as I explained, this logic shows that Pn doesn't halt, not Pa, so Ha still gets Pa wrong, as Pa does Halt. Pn not halting isn't a problem, as Hn doesn't answer either so doesn't need to be confounded.
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| From | olcott <NoOne@NoWhere.com> |
|---|---|
| Date | 2021-07-05 12:48 -0500 |
| Subject | Re: Could H correctly decide that P never halts? [ already agreed ] |
| Message-ID | <2e-dnbgWnNJj2H79nZ2dnUU7-X3NnZ2d@giganews.com> |
| In reply to | #35762 |
On 7/5/2021 12:16 PM, Richard Damon wrote:
> On 7/5/21 10:59 AM, olcott wrote:
>> On 7/5/2021 9:57 AM, Richard Damon wrote:
>>> On 7/5/21 9:38 AM, olcott wrote:
>>>> On 7/5/2021 6:33 AM, Richard Damon wrote:
>>>>> On 7/4/21 11:09 PM, olcott wrote:
>>>>>
>>>>>> It can be easily verified that if no H ever aborts any P that P(P)
>>>>>> will
>>>>>> never halt. If any H must abort any P then this H does correctly
>>>>>> decide
>>>>>> that this P does not halt.
>>>>>>
>>>>>
>>>>> Wrong.
>>>>>
>>>>> If NO H ever aborts any P then this H will not answer the question
>>>>> H(P,P) and thus isn't a valid decider.
>>>>
>>>> An answer that would not be a damned lie would be:
>>>> An answer that would not be a damned lie would be:
>>>> An answer that would not be a damned lie would be:
>>>> An answer that would not be a damned lie would be:
>>>> An answer that would not be a damned lie would be:
>>>>
>>>> Right you are correct if no H ever aborted any P then P(P) would never
>>>> halt and this would also prevent H from being a decider.
>>>>
>>>
>>> But the way you interpret that sentence, it would be a damned lie. Maybe
>>> Literally. The Devil loves to use statements that are mostly true but
>>> have a ting of falsehood in it to corrupt the hearer.
>>>
>>> Are you working for him?
>>>
>>> But Yes, if H never aborted any P then H would not be a decider. That
>>> does NOT mean that H is correct in aborting P and calling it
>>> non-Halting, since when we change H we change P.
>>>
>>> Hn non aborting -> Hn(Pn,Pn) will be none halting.
>>> Ha aborting -> Ha(Pa,Pa) will answer, by your stipulation non-halting.
>>>
>>> This means that Pa(Pa) will get the non-Halting answer from its call to
>>> Ha(Pa,Pa) and will halt and thus Pa(Pa) is a Halting computaiton, and Ha
>>> was wrong.
>>>
>>> All you have shown is that Ha(Pn,Pn) can say non-halting and be right
>>> about that answer, which doesn't mean a thing, as Pn never claimed to be
>>> able to confound Ha, only Hn, and only if Hn could actually answer.
>>>
>>> The machine that confounds Ha is Pa, and it DOES succeed, as Ha says
>>> Pa(Pa) is non-Halting when even as you have confiremed Pa(Pa) does Halt.
>>> You just get stuck in your 'logic' and want non-Halting to be the right
>>> answer because you confuse Pa and Pn because you don't seem to
>>> understand the concept of complete programs.
>>>
>>
>> It can be easily verified that if no H(n) ever aborts any P(m) that P(P)
>> will never halt. If any H(n) must abort any P(m) then this H(n) does
>> correctly decide that this P(m) does not halt.
>>
>>
>
> Except as I explained, this logic shows that Pn doesn't halt, not Pa, so
> Ha still gets Pa wrong, as Pa does Halt. Pn not halting isn't a problem,
> as Hn doesn't answer either so doesn't need to be confounded.
>
// Simplified Linz Ĥ (Linz:1990:319)
void P(u32 x)
{
u32 Input_Halts = H(x, x);
if (Input_Halts)
HERE: goto HERE;
}
int main()
{
P((u32)P);
}
If any H(n) must abort any P(m) then this H(n) does correctly decide
that this P(m) does not halt.
In the above computation (zero based addressing) H(1) aborts P(2).
Halting problem undecidability and infinitely nested simulation
https://www.researchgate.net/publication/351947980_Halting_problem_undecidability_and_infinitely_nested_simulation
--
Copyright 2021 Pete Olcott
"Great spirits have always encountered violent opposition from mediocre
minds." Einstein
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| From | Richard Damon <Richard@Damon-Family.org> |
|---|---|
| Date | 2021-07-05 14:36 -0400 |
| Subject | Re: Could H correctly decide that P never halts? [ already agreed ] |
| Message-ID | <1jIEI.537$Q75.114@fx24.iad> |
| In reply to | #35765 |
On 7/5/21 1:48 PM, olcott wrote:
> On 7/5/2021 12:16 PM, Richard Damon wrote:
>> On 7/5/21 10:59 AM, olcott wrote:
>>>
>>> It can be easily verified that if no H(n) ever aborts any P(m) that P(P)
>>> will never halt. If any H(n) must abort any P(m) then this H(n) does
>>> correctly decide that this P(m) does not halt.
>>>
>>>
>>
>> Except as I explained, this logic shows that Pn doesn't halt, not Pa, so
>> Ha still gets Pa wrong, as Pa does Halt. Pn not halting isn't a problem,
>> as Hn doesn't answer either so doesn't need to be confounded.
>>
>
> // Simplified Linz Ĥ (Linz:1990:319)
> void P(u32 x)
> {
> u32 Input_Halts = H(x, x);
> if (Input_Halts)
> HERE: goto HERE;
> }
>
> int main()
> {
> P((u32)P);
> }
>
> If any H(n) must abort any P(m) then this H(n) does correctly decide
> that this P(m) does not halt.
>
> In the above computation (zero based addressing) H(1) aborts P(2).
Right, and then it returns the non-halting answer to P(0) which then
Halts, showing that All these P will halt, and thus H(1) was incorrect
in deciding that P(2) was non-Halting.
If you want to claim that H(1) doesn't return its answer to P(0) then
when it is H(0) it also won't return its answer to main (or it has
proved itself not to be a computation) and thus it proves itself to not
be a decider.
This PROVES that H(1) made an error in deciding that P(2) was
non-halting, and the reason it made the error is that it negected to
consider that H(3) would ALSO abort its simulation of P(4), and thus
made an error in its logic.
>
> Halting problem undecidability and infinitely nested simulation
>
> https://www.researchgate.net/publication/351947980_Halting_problem_undecidability_and_infinitely_nested_simulation
>
>
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| From | Richard Damon <Richard@Damon-Family.org> |
|---|---|
| Date | 2021-07-05 12:18 -0400 |
| Subject | Re: Could H correctly decide that P never halts? [ already agreed ] |
| Message-ID | <7hGEI.13233$SMMb.13036@fx17.iad> |
| In reply to | #35734 |
On 7/5/21 9:38 AM, olcott wrote: > On 7/5/2021 6:33 AM, Richard Damon wrote: >> On 7/4/21 11:09 PM, olcott wrote: >> >>> It can be easily verified that if no H ever aborts any P that P(P) will >>> never halt. If any H must abort any P then this H does correctly decide >>> that this P does not halt. >>> >> >> Wrong. >> >> If NO H ever aborts any P then this H will not answer the question >> H(P,P) and thus isn't a valid decider. > > An answer that would not be a damned lie would be: > An answer that would not be a damned lie would be: > An answer that would not be a damned lie would be: > An answer that would not be a damned lie would be: > An answer that would not be a damned lie would be: > > Right you are correct if no H ever aborted any P then P(P) would never > halt and this would also prevent H from being a decider. > > Right, the Pn built on the Hn that doesn't abort Pn will never halt, but that Hn doesn't meet the requirements of being a decider so doesn't need to be refuted. Just because some Ha needs (and thus all Ha will) to abort some Pa doesn't mean that it is correct in assuming that this Pa is also non-halting. Remember Ha != Hn and thus Pa != Pn, so the proof about Pn doesn't apply to Pa. FAIL
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| From | Ben Bacarisse <ben.usenet@bsb.me.uk> |
|---|---|
| Date | 2021-07-05 23:58 +0100 |
| Subject | Re: Could H correctly decide that P never halts? [ already agreed ] |
| Message-ID | <87tul89xms.fsf@bsb.me.uk> |
| In reply to | #35715 |
olcott <NoOne@NoWhere.com> writes:
| void P(u32 x)
| {
| u32 Input_Halts = H(x, x);
| if (Input_Halts)
| HERE: goto HERE;
| }
> Every computation that would never halt unless its simulation is
> aborted is equally not a halting computation.
For H(P,P) to be correct one of these must apply:
H(P,P) == 0 and P(P) does not halt, or
H(P,P) != 0 and P(P) halts.
Neither is the case. Simple facts that you deny.
> There is no exception to this rule in the same way and for the same
> reason that the numeric integer 5 is always numerically greater than
> the numeric integer 3.
There are, indeed, no exceptions. For H(P,P) to be correct one of these
must apply:
H(P,P) == 0 and P(P) does not halt, or
H(P,P) != 0 and P(P) halts.
Neither is the case for your P. In fact, as you have clearly stated,
P(P) halts and H(P,P) == 0.
This is what halting is about. No one cares about PO "Other-Halting" or
whatever you want to call it.
--
Ben.
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| From | olcott <NoOne@NoWhere.com> |
|---|---|
| Date | 2021-07-05 19:54 -0500 |
| Subject | Re: Could H correctly decide that P never halts? [ already agreed ] |
| Message-ID | <v7KdnXhXO_0xNH79nZ2dnUU7-KvNnZ2d@giganews.com> |
| In reply to | #35777 |
On 7/5/2021 5:58 PM, Ben Bacarisse wrote:
>
> olcott <NoOne@NoWhere.com> writes:
>
> | void P(u32 x)
> | {
> | u32 Input_Halts = H(x, x);
> | if (Input_Halts)
> | HERE: goto HERE;
> | }
>
>> Every computation that would never halt unless its simulation is
>> aborted is equally not a halting computation.
>
> For H(P,P) to be correct one of these must apply:
>
> H(P,P) == 0 and P(P) does not halt, or
> H(P,P) != 0 and P(P) halts.
>
> Neither is the case. Simple facts that you deny.
>
That the prefix to infinitely nested simulation halts because its suffix
is aborted cannot reasonably be construed as a computation that halts in
the same way that that following computation can be construed as a
computation that halts:
int Add(int N, int M)
{
return N + M;
}
That every aspect of the following computation can be classified as a
computation that never halts unless its simulation is aborted
unequivocally places it in the non-halting category.
int main()
{
u32 Input_Halts = H((u32)P, (u32)P);
Output("Input_Halts = ", Input_Halts);
}
Every computation that never halts unless is simulation is aborted
defines the exact same set of non-halting computations as the halting
problem, except for the halting problem counter-example templates.
Because it defines the exact same halting and non-halting set of
computations as the halting problem it is proved to be an equivalent
criterion measure.
--
Copyright 2021 Pete Olcott
"Great spirits have always encountered violent opposition from mediocre
minds." Einstein
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| From | Richard Damon <Richard@Damon-Family.org> |
|---|---|
| Date | 2021-07-05 21:29 -0400 |
| Subject | Re: Could H correctly decide that P never halts? [ already agreed ] |
| Message-ID | <IlOEI.3$uj5.0@fx03.iad> |
| In reply to | #35783 |
On 7/5/21 8:54 PM, olcott wrote: > hat the prefix to infinitely nested simulation halts because its suffix > is aborted cannot reasonably be construed as a computation that halts in > the same way that that following computation can be construed as a > computation that halts: What you miss is that the suffix of the chain would be aborted as part of the algorithm of the computation that is being analyized. If that doesn't lead to Halting-Behavior then you need to consider this as non-halting: for(int i=0; i<5; i++) continue; as the loop stops due to a decision inside the algorithm. Rememver the sequence is: P0 is stated. P0 calls H0 (P,P) H0 starts simulating P(P), this is now computation P1 at some point H0, which is part of P0 decided to abort its simulation of P1 H0 returns its answer to P0 P0 then Halts Thus that top level P is shown to be a Halting Computation. It stopped TOTALLY under its own decision. This is because the machine P fully contains ALL of the algorithm of H.
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