Sujet : Re: Bad faith and dishonesty
De : polcott333 (at) *nospam* gmail.com (olcott)
Groupes : comp.theoryDate : 02. Jun 2025, 04:23:19
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <101j5f7$2urhr$1@dont-email.me>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28
User-Agent : Mozilla Thunderbird
On 6/1/2025 8:24 PM, Richard Damon wrote:
On 6/1/25 12:36 PM, olcott wrote:
On 6/1/2025 11:22 AM, Mike Terry wrote:
On 01/06/2025 10:48, Mikko wrote:
On 2025-05-31 15:12:57 +0000, olcott said:
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On 5/31/2025 8:37 AM, Mike Terry wrote:
On 31/05/2025 11:03, Richard Heathfield wrote:
On 31/05/2025 10:42, Mikko wrote:
On 2025-05-30 17:27:05 +0000, olcott said:
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On 5/30/2025 12:06 PM, Richard Heathfield wrote:
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<snip>
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As far as I can recall, Olcott's ramblings never go within discus- throwing distance of a potentially erroneous step.
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There is no *INPUT* D to termination analyzer H
that can possibly do the opposite of whatever
value that H returns.
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We can apply Truing'c construction to every Truing machine and
prove that no Turing machine is a halting decider. With a
similar construction many other problems about Turing machine
behaviour can be proven uncomputable, too.
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Indeed. In fact, if I'm reading Mr Olcott's retort correctly...
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Turing: imagine a program P...
Turing: ...argument goes here...
Turing: ...and it turns out that P can't exist.
Olcott: Turing is wrong, because P can't exist.
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I /seeeeee.../
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More like:
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Turing: imagine a program P...
Turing: ...modify P as follows to create D...
Turing: ...argument goes here...
Turing: ...so P decides D's halting incorrectly
Turing: ...and so P is not a halt decider.
Olcott: Turing is wrong, because *D* can't exist.
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I /seeeeee.../
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Mike.
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Turing wasn't even dealing with halting.
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No, but he proved that there are problems that are not Turing decidable.
That halting is one of them can be proven with an adaptation of Turing's
proof.
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The question Turing wanted to solve is whether a computation writes
infinitely many unerasable symbols. A computation that halts doesn't.
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*Here is the orignal Linz proof*
https://www.liarparadox.org/Linz_Proof.pdf
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Linz' presentation of the proof is not as good as it could be.
Linz is not careful with details. For example, the formulas in
the definition 12.1 specify to configurations (and a relation
between them) but does not specify which tape location is under
the head.
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Linz's notation does specify the tape head location. In his notation
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aQb,
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where a,b are strings of the tape alphabet, and Q is a TM state, the meaning is that the TM is in state Q, and the tape contains string ab (concatenation of a and b), and the tape head is about to read the first character of b.
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You're right that Linz's presentation is not completely careful. He is aiming at CS students specifically without any special mathematical background. Where I see lack of care, I also see that it would be clear how to address that issue if required, so it is not a big issue for me.
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Mike.
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He made the big mistake of have two start states
in the same sequence of configurations.
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No, you are forgetting that which one of the two lines you use is based
This is the error that Ben said that I caught.
His definition of Ĥ has two q0 start states one after the other.
https://www.liarparadox.org/Linz_Proof.pdf-- Copyright 2025 Olcott "Talent hits a target no one else can hit; Geniushits a target no one else can see." Arthur Schopenhauer
Re: Bad faith and dishonesty
Re: Bad faith and dishonesty