On 6/8/2022 7:58 PM, Richard Damon wrote:
> On 6/8/22 8:51 PM, olcott wrote:
>> On 6/8/2022 7:46 PM, Richard Damon wrote:
>>>
>>> On 6/8/22 8:17 PM, olcott wrote:
>>>> On 6/8/2022 7:13 PM, Mr Flibble wrote:
>>>>> On Wed, 8 Jun 2022 19:10:39 -0500
>>>>> olcott <NoOne@NoWhere.com> wrote:
>>>>>
>>>>>> On 6/8/2022 7:04 PM, Mr Flibble wrote:
>>>>>>> On Wed, 8 Jun 2022 18:56:28 -0500
>>>>>>> olcott <NoOne@NoWhere.com> wrote:
>>>>>>>> On 6/8/2022 6:37 PM, Richard Damon wrote:
>>>>>>>>> On 6/8/22 11:53 AM, olcott wrote:
>>>>>>>>>> On 6/8/2022 10:37 AM, wij wrote:
>>>>>>>>>>> On Wednesday, 8 June 2022 at 08:43:00 UTC+8, Mr Flibble wrote:
>>>>>>>>>>>> Hi,
>>>>>>>>>>>>
>>>>>>>>>>>>    From discussion with Olcott in comp.lang.c++ I have
>>>>>>>>>>>> determined that his so called refutation of the HP proofs is
>>>>>>>>>>>> based around the behaviour of his simulation-based decider, H:
>>>>>>>>>>>>
>>>>>>>>>>>> void Q(u32 x)
>>>>>>>>>>>> {
>>>>>>>>>>>> if (H(x, x))
>>>>>>>>>>>> FUBAR();
>>>>>>>>>>>> return;
>>>>>>>>>>>> }
>>>>>>>>>>>>
>>>>>>>>>>>> int main()
>>>>>>>>>>>> {
>>>>>>>>>>>> Output("Input_Halts = ", H((u32)Q, (u32)Q));
>>>>>>>>>>>> }
>>>>>>>>>>>>
>>>>>>>>>>>> He asserts H(Q,Q)=0 based on a nested simulation being detected
>>>>>>>>>>>> (i.e. Q invoking H) irregardless of whether FUBAR halts or not.
>>>>>>>>>>>>
>>>>>>>>>>>> If FUBAR halts H gives the wrong answer.
>>>>>>>>>>>>
>>>>>>>>>>>> He claims H(Q,Q)=0 as it gets stuck in a recursive (nested)
>>>>>>>>>>>> simulation however that wouldn't be the case for non-simulating
>>>>>>>>>>>> decider for which there would be no such recursion.
>>>>>>>>>>>>
>>>>>>>>>>>> Can we finally put this to bed and change the fucking topic?
>>>>>>>>>>>>
>>>>>>>>>>>> /Flibble
>>>>>>>>>>>
>>>>>>>>>>> +-----------------------------------------------------------------+
>>>>>>>>>>>
>>>>>>>>>>> | The HP proof has nothing to do with how the 'H' is
>>>>>>>>>>> constructed. |
>>>>>>>>>>> +-----------------------------------------------------------------+
>>>>>>>>>>>
>>>>>>>>>>> Many such liar's-paradox-like examples are for easy
>>>>>>>>>>> comprehension (for educational purpose). The real 'H' inside P
>>>>>>>>>>> is an algorithm computationally equivalent to 'H' (so, no
>>>>>>>>>>> any 'call' is there, and the pattern matching tech. is very
>>>>>>>>>>> difficult, by now to say.
>>>>>>>>>>> And, this 'H' is also allowed given by some super intelligent
>>>>>>>>>>> god.... whatever).
>>>>>>>>>>
>>>>>>>>>> It is the pathological self reference(Olcott 2004) relationship
>>>>>>>>>> between H and P that has previously been considered to make P
>>>>>>>>>> undecidable for H.
>>>>>>>>>>
>>>>>>>>>>         For any program H that might determine if programs
>>>>>>>>>> halt, a
>>>>>>>>>> "pathological"
>>>>>>>>>>         program P, called with some input, can pass its own
>>>>>>>>>> source
>>>>>>>>>> and its input to
>>>>>>>>>>         H and then specifically do the opposite of what H
>>>>>>>>>> predicts P will do. No H
>>>>>>>>>>         can exist that handles this case.
>>>>>>>>>> https://en.wikipedia.org/wiki/Halting_problem
>>>>>>>>>>> +------------------------------------------------+
>>>>>>>>>>> | olcott's brain is incapable of logic function. |
>>>>>>>>>>> | (other kind of functions seem quite excellent) |
>>>>>>>>>>> +------------------------------------------------+
>>>>>>>>>>> It should be safe to say any concept involves logical operation,
>>>>>>>>>>> olcott cannot
>>>>>>>>>>> make it formally correct (he relies on his "language's logic").
>>>>>>>>>>> For example, I doubt he can handle several lines of
>>>>>>>>>>> inter-connected codes.
>>>>>>>>>>> ...
>>>>>>>>>>> All should be fine... except olcott now insists "H(P,P)==0" is
>>>>>>>>>>> correct while
>>>>>>>>>>> there is no definition of H shown.
>>>>>>>>>>
>>>>>>>>>> I am not claiming that H(P,P) correctly determines the halt
>>>>>>>>>> status of its input. I am claiming that non-halting is the
>>>>>>>>>> correct halt status of its input.
>>>>>>>>>
>>>>>>>>> Ok, so if you are JUST claiming that Non-Halting is the right
>>>>>>>>> answer, but H doesn't actually return that answer, you can be
>>>>>>>>> correct on that.
>>>>>>>>
>>>>>>>> THIS BY ITSELF IS AN ENORMOUS BREAKTHROUGH
>>>>>>>> I am claiming that H(P,P)==0 is the correct answer to the
>>>>>>>> "impossible" input that previously had no correct answer at all.
>>>>>>>>
>>>>>>>> Because people on this forum are trying to be as disagreeable as
>>>>>>>> they possibly can be I must move one point at a time.
>>>>>>>>
>>>>>>>> It has taken at least six months to get agreement on the totally
>>>>>>>> obvious fact that H(P,P)==0 is correct. With an actual honest
>>>>>>>> dialogue there is no way that this should have taken more than
>>>>>>>> three days to get everyone to agree.
>>>>>>>>
>>>>>>>> As soon as we achieve a consensus on this point we can move on to
>>>>>>>> the next point.
>>>>>>> Your H is not a halt decider as it gets the answer wrong if P would
>>>>>>> have halted; instead what you have is an S, a simulation detector.
>>>>>>>
>>>>>>> So if you wish to continue you should be asserting S(P,P)==0 and not
>>>>>>> H(P,P)==0.
>>>>>>>
>>>>>>> /Flibble
>>>>>>
>>>>>> So the concept of a simulating halt decider is beyond your
>>>>>> intellectual capacity?
>>>>> It isn't a halt decider if it gets the answer to the question of
>>>>> whether or not P halts wrong.  Your H gets the answer wrong if P halts
>>>>> so it isn't a halt decider.
>>>>>
>>>>> /Flibble
>>>>>
>>>>
>>>>
>>>> (1) Deciders(computer science) compute the mapping from their inputs
>>>> to an accept or reject state.
>>>>
>>>> (2) The actual behavior of the actual input to H(P,P) is proven to
>>>> never halt.
>>>
>>> ONLY if H(P,P) never returns 0.
>>>
>>
>> So when H aborts the emulation of its input this dead process that is
>> not even still in memory manages to leap to its "ret" instruction even
>> though it no longer exists?
>
> Excpet the question isn't does this (partial) simulation halt, but does
> the amchine it represents halt, and the machine NEVER stops until it
> finishes.
>

In other words you are saying although the a decider must compute the
mapping from its input to an accept reject state on the basis of the
actual behavior specified by this input

EXCEPT WHEN YOU WANT TO BE DISAGREEABLE THEN YOUR OPINION OVER-RIDES AND
SUPERSEDES COMPUTER SCIENCE.

--
Copyright 2022 Pete Olcott

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