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computers / comp.ai.philosophy / Re: Halting Problem definition is ill-formed and thus invalid [ Paul N agrees ]

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* Re: Halting Problem definition is ill-formed and thus invalid [ Paulolcott
`- Re: Halting Problem definition is ill-formed and thus invalid [ PaulRichard Damon

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Re: Halting Problem definition is ill-formed and thus invalid [ Paul N agrees ]

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https://www.novabbs.com/computers/article-flat.php?id=10585&group=comp.ai.philosophy#10585

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From: polco...@gmail.com (olcott)
Newsgroups: comp.theory,sci.logic,comp.ai.philosophy
Subject: Re: Halting Problem definition is ill-formed and thus invalid [ Paul
N agrees ]
Date: Wed, 8 Feb 2023 09:44:00 -0600
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 by: olcott - Wed, 8 Feb 2023 15:44 UTC

On 2/7/2023 5:32 AM, Paul N wrote:
> On Sunday, February 5, 2023 at 8:47:16 PM UTC, Mr Flibble wrote:
>> On Sun, 05 Feb 2023 21:21:32 +0100, Python wrote:
>>
>>> Le 05/02/2023 à 21:11, Mr Flibble a écrit :
>>>> On Sun, 05 Feb 2023 14:51:18 -0500, Richard Damon wrote:
>>> ...
>>>>> You are just showing you are as ignorant about the subject as Olcott,
>>>>> I think he has gaslit you just like he has done to himself.
>>>>
>>>> My research is entirely independent of Olcott's
>>>
>>> Well so it's a kind of coincidence. You just happen to be as stupid as
>>> each other Olcott and you.
>> You are completely missing the point: Olcott believes what he posts whilst
>> I am just trolling, and quite successfully too.
>
> It's amazing how few people have actually spotted this point, given that both you and me have pointed it out several times.
>
> Olcott has posted lots of stuff about how he has disproved the Halting Problem, so you post stuff saying about how you have (nearly) solved it.
> Olcott repeats his posts over and over again, ignoring the fact that the errors have been pointed out, so you repost your stuff unchanged.
> Olcott now changes tack and says that the Halting Problem is invalid. So you do too.
>
> Incidentally, I rather have doubts as to whether Olcott does actually believe what he posts. It seems more likely he is just doing it to get attention - and successfully, it seems.

*You have already agreed that P correctly simulated by H never halts*
*thus meeting its halt status criterion measure*

(a) 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 (b) H can abort its simulation of D and correctly
report that D specifies a non-halting sequence of configurations.

In comp.lang.c++
On 6/14/2022 6:47 AM, Paul N wrote:
> On Monday, June 13, 2022 at 7:46:22 PM UTC+1, olcott wrote:
>> Begin Local Halt Decider Simulation Execution Trace Stored at:212352
>> // H emulates the first seven instructions of P
>> ...[00001352][0021233e][00212342] 55 push ebp // enter P
>> ...[00001353][0021233e][00212342] 8bec mov ebp,esp
>> ...[00001355][0021233e][00212342] 8b4508 mov eax,[ebp+08]
>> ...[00001358][0021233a][00001352] 50 push eax // push P
>> ...[00001359][0021233a][00001352] 8b4d08 mov ecx,[ebp+08]
>> ...[0000135c][00212336][00001352] 51 push ecx // push P
>> ...[0000135d][00212332][00001362] e840feffff call 000011a2 // call H
>>
>> // The emulated H emulates the first seven instructions of P
>> ...[00001352][0025cd66][0025cd6a] 55 push ebp // enter P
>> ...[00001353][0025cd66][0025cd6a] 8bec mov ebp,esp
>> ...[00001355][0025cd66][0025cd6a] 8b4508 mov eax,[ebp+08]
>> ...[00001358][0025cd62][00001352] 50 push eax // push P
>> ...[00001359][0025cd62][00001352] 8b4d08 mov ecx,[ebp+08]
>> ...[0000135c][0025cd5e][00001352] 51 push ecx // push P
>> ...[0000135d][0025cd5a][00001362] e840feffff call 000011a2 // call H
>> Local Halt Decider: Infinite Recursion Detected Simulation Stopped
>>
>> It is completely obvious that when H(P,P) correctly emulates
>> its input that it must emulate the first seven instructions
>> of P. Because the seventh instruction of P repeats this
>> process we can know with complete certainty that the emulated
>> P never reaches its final “ret” instruction, thus never halts.
>
> Yes, it is clear to us humans watching it that the program is
> repeating itself. Thus we can appreciate that it will never reach
> the final "ret" - indeed, it won't even get to the infinite loop
> identified above. But does the computer itself know this? If the
> emulator simply emulates the instructions given, it will not
> realise that it is doing the same thing over and over again. If
> it does look out for this, spotting a repeated state, then it can
> tell that the program under consideration will not halt. The answer
> to whether it spots this lies in the emulator, which you haven't
> shown the code for.

*Here is the code, it compiles under*
*Microsoft Visual Studio Community Edition 2017*
https://liarparadox.org/2023_02_07.zip
The current halt status algorithm is in Halt7.c

--
Copyright 2023 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer

Re: Halting Problem definition is ill-formed and thus invalid [ Paul N agrees ]

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Subject: Re: Halting Problem definition is ill-formed and thus invalid [ Paul
N agrees ]
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From: Rich...@Damon-Family.org (Richard Damon)
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 by: Richard Damon - Thu, 9 Feb 2023 00:06 UTC

On 2/8/23 10:44 AM, olcott wrote:

> *You have already agreed that P correctly simulated by H never halts*
> *thus meeting its halt status criterion measure*
>

So, WHICH H are you defining your H to actually Be?

The one that ACTUALLY doesn't abort its simulation until it can
CORRECTLY prove its input is non-halting, and thus never aborts its
simulation and doesn't answer, Yes, for THAT H, P(P) is non-Halting, but
H isn't correct, as it never abswers.

Or, is H the program that THINKS it is that H, and because that aborts
its simulation INCORRECTLY, and returns 0, which makes the P(P) built on
THAT H halting, and thus H wrong.

It seems you don't understand that those are two different sets of
assembly instructions, and thus the second H is INCORRECT, in part
because it doesn't look into the H called to see what it actually does.

If you think the two are actually identical instructions, what is the
first actual assembly instruction executed in the two paths

main -> P(P) -> H(P,P) that acts like the first
and
main -> H(P,P) that acts like the second

Where the two H's have IDENTICAL assembly code and the function H only
depends on the explicit parameters given

Where the two paths diverge. Since the second returns in finite time and
the first does not, there MUST be a difference, and thus there MUST be a
first difference.

This has been asked a number of times in the past, and the fact you
haven't answered just PROVES that you are LYING in your claims.

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