Re: Linz's proofs. --- Good catch !

On 3/7/2024 1:57 PM, Richard Damon wrote:

On 3/7/24 11:37 AM, olcott wrote:

On 3/7/2024 1:15 PM, Richard Damon wrote:

On 3/7/24 11:11 AM, olcott wrote:

On 3/7/2024 12:46 PM, Richard Damon wrote:

On 3/7/24 10:30 AM, olcott wrote:

On 3/7/2024 10:44 AM, immibis wrote:

On 7/03/24 17:16, Ben Bacarisse wrote:

immibis <news@immibis.com> writes:
>

On 7/03/24 12:32, Ben Bacarisse wrote:

The students I taught seemed to have no problem with this sort of case
analysis.  But the "assume H does X" argument lead to lots of "but H1
could be better" arguments.

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They aren't satisfied with "we can do the exact same thing with H1 to prove
that H1 doesn't work either"?

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In the vast majority of cases, yes, but even then there is a logical
problem with going down that route -- there is no H so there can't be an
H1 that does better.  Once this objection is properly examined, it turns
out to be the argument I ended up preferring anyway.  H isn't a halt
decider, it's just any old TM and we show it can't be halt decider for
one reason or another.
>

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Unless your students are extremely pedantic... maybe they are... I don't see what's illogical with:
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"I think H is a halt decider."
"But it doesn't: see this proof."
"Oh. Well, even though H isn't a halt decider, how do we know there isn't a program H1 which is a halt decider?"
"The proof would still work for H1, or H2, or any other program you think is a halt decider."
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>

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It is an easily verified fact that:
H(D,D) Sees that D(D) is calling H(D,D) at machine address 00001522
H1(D,D) Sees that D(D) is NOT calling H1(D,D) at machine address 00001422
*different machine addresses is the reason for different return values*
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>

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Which proves that H1 and H are different computation and thus different Turing Machines, so H1 getting the right answer doesn't "fix" H's getting the wrong answer.
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Good catch !!!
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For Olcott machines Linz H ⟨Ĥ⟩ ⟨Ĥ⟩ would map its input with
its own TMD concatenated to this input to its Boolean result.
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Linz Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ would map its input with its own TMD
concatenated to this input to its Boolean result.
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thus finally explaining how Linz H ⟨Ĥ⟩ ⟨Ĥ⟩ can correctly
determine the halt status of its input while Linz Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩
cannot.

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Which only indicates that either you built your H^ incorrectly, as H^.H is supposed to do exactly the same thing as H itself,

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Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ is supposed to do exactly the same things as H ⟨Ĥ⟩ ⟨Ĥ⟩
only if they have the same input.
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When Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ are Olcott machines they always
have an additional input that makes the input to Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩
and H ⟨Ĥ⟩ ⟨Ĥ⟩ different.

 And a machone that depends on anythohng other than the description of the input is provvably NOT a Halt Decider,

When it correctly determines the actual halt status of an
actual input TMD+Finite_String then it correctly decided
this TMD+Finite_String.
No one can say that it gets the wrong answer when it
gets the right answer.
The most than anyone can say is that this answer may
not be Turing computable.
When an Olcott machine maps TMD+Finite_String+Own_TMD
to a Boolean value then this Olcott machine derives
its output as a pure function of its inputs.

so you are just admitting that H isn't a Halt Decider, since for some values of description, it gives the wrong answer.
 

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or if that is impossible to do, that your modification to Turing Machine rules have made your system not-Turing Complete, as you are admitting that the is a Computation that could be made with Turing Machines (H^ from a given H) that you can't make in Olcott-Machines.
>

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You are not evaluating Turing Complete correctly. Olcott machines
exactly compute all of the same decision problems that Turing
machines compute when the Olcott machines ignore their own TMD.

 Right, but if they don't then they can't be said to compute the mapping that ignores that input.
 So, since H doesn't ignore its extra input, it doesn't compute that mapping called Halting, that isn't dependent on the machine deciding.

The set of Olcott machines that compute the mapping from
their inputs (ignoring their own TMD) to an output is
exactly the same set as the set of Turing machines that
compute the mapping from their inputs to their own output.
This analysis proves that a subset of Olcott machines
compute the same set as the set of Turing machines.
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer