Re: Turing Machine computable functions apply finite string transformations to inputs VERIFIED FACT

On 5/1/2025 8:07 PM, olcott wrote:

On 5/1/2025 10:32 AM, Ben Bacarisse wrote:

Richard Heathfield <rjh@cpax.org.uk> writes:
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On 30/04/2025 19:30, Mike Terry wrote:

On 30/04/2025 16:46, Richard Heathfield wrote:

On 30/04/2025 16:15, olcott wrote:

On 4/29/2025 5:03 PM, Richard Heathfield wrote:

On 29/04/2025 22:38, olcott wrote:
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<snip>
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int DD()
{
    int Halt_Status = HHH(DD);
    if (Halt_Status)
      HERE: goto HERE;
    return Halt_Status;
}
>
HHH is correct DD as non-halting BECAUSE THAT IS
WHAT THE INPUT TO HHH(DD) SPECIFIES.

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You're going round the same loop again.
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Either your HHH() is a universal termination analyser or it isn't.

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The domain of HHH is DD.

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Then it is attacking not the Halting Problem but the Olcott Problem,
which is of interest to nobody but you.

It would be (if correct) attacking the common proof for HP theorem as it
occurs for instance in the Linz book which PO links to from time to time.

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Yes. That's what I call the Olcott Problem.
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De gustibus non est disputandum, but I venture to suggest that (correctly)
overturning Turing's proof would be of cosmos-rocking interest to the world
of computer science, compared to which pointing out a minor flaw in a
minor[1] proof would, even if correct, have no more effect on our field
than lobbing a pebble into the swash at high tide.
>
I suspect that the only reason we bother to argue with Mr Olcott so much is
because (even if he does so unwittingly) he manages to convey the
appearance of attacking the Halting Problem, and arguing about the Halting
Problem is a lot more fun than arguing about the Olcott Problem.
>
To be of any interest, solving the Olcott Problem would have to have
important consequences. But does it? Let's see.
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Dr Linz Theorem 12.1 (Halting Problem is Undecidable): There does not exist
any Turing machine H that behaves as required by Linz Definition 12.1. Thus
the halting problem is undecidable.
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Dr Linz has a proof for this claim, which can be found here:
<https://john.cs.olemiss.edu/~hcc/csci311/notes/chap12/ch12.pdf>
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If the proof is flawless, the conclusion stands and Mr Olcott is simply
wrong.

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There is peculiar irony here.  The proof is /not/ flawless.  It has, in
fact, a flaw that PO pointed out (although in passing).  PO does not
care about the flaw because it is easily fixed, but it's there none the
less[1].
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Anyway, Linz only gives this argument because it is of historical
interest.  His "real" proof immediately follows this argument (in the
book) and is a trivial corollary of the fact, proved in chapter 11, that
not all recursively enumerable languages are recursive.  But no crank
ever addresses that proof.  I wonder why...
>

 You wasted fifteen years of my life by your change-the-subject
form of rebuttal so I no longer tolerate that from an anyone.
 The flaw in all of the conventional proofs

Which start with the assumption that an H exists that satisfies these requirements:
Given any algorithm (i.e. a fixed immutable sequence of instructions) X described as <X> with input Y:
A solution to the halting problem is an algorithm H that computes the following mapping:
(<X>,Y) maps to 1 if and only if X(Y) halts when executed directly
(<X>,Y) maps to 0 if and only if X(Y) does not halt when executed directly

is that the
mapping they they DO specify IS NOT the direct execution
of their input.

In other words, a contradiction is reached, thereby proving the above assumption false, as Linz and other have shown and as you have *explicitly* agreed is correct.

 Everyone moronically ignores how the pathological relationship
between the INPUT and its TERMINATION ANALYZER CHANGES the
behavior OF THE INPUT.
 

No, everyone just understands proof by contradiction.

*When HHH emulates DD correctly the emulated DD cannot possibly halt*
THIS IS THE BEHAVIOR OF THE FREAKING INPUT.

In other words, when the input is changed such that function HHH is a pure simulator, HHH(DD) does not halt.
Changing the input is not allowed.

 That you saw Bill's identical twin brother rob a liquor
store DOES NOT PROVE THAT BILL IS GUILTY.
 If anyone would pay any freaking attention THEY WOULD
ALL KNOW THAT I HAVE BEEN CORRECT ABOUT THIS FOR SEVERAL YEARS.
 

If the proof is flawed through some error of reasoning, *either* it merely
fails to correctly support its conclusion *or* a duly corrected proof
/overturns/ the conclusion.

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... or a duly corrected proof /supports/ the conclusion.  Maybe you
intended this possibility to be include in your first "merely fails to
support" alternative.
>

The latter would be /extremely/ interesting, but it would also mean that we
have two proofs proving opposite things, and so it would effectively be a
cataclysmic sideways attack on Turing's reasoning.

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It would be a cataclysmic attack on all of conventional mathematics and
logic as two proofs, one of T and another of ~T, makes the system of
proof oneself inconsistent (and everything becomes a theorem).
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[1] It centres on the naming of states in H and the derived machines H'
and H^.  Details available if you really care.
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