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Jan 29, 2022, 3:29:48 PMJan 29

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On 1/29/2022 11:34 AM, Ben Bacarisse wrote:

> Richard Damon <Ric...@Damon-Family.org> writes:

>

>> And the 'actual behavior of its actual inputs' is DEFINED to be what

>> the computation the input actually does when run as an independent

>> machine, or what a UTM will do when simulating that input.

>>

>> If that isn't the meaning you are using, then you are just lying that

>> you are working on the halting problem, which is what seems to be the

>> case. (That you are lying that is).

>

> It is certainly true that PO is not addressing the halting problem. He

> has been 100% clear that false is, in his "opinion", the correct result

> for at least one halting computation. This is not in dispute (unless

> he's retracted that and I missed it).

>

THIS POINT ADDRESSES THE KEY QUESTION

Which state does Ĥ applied to ⟨Ĥ⟩ transition to correctly ?

The following simplifies the syntax for the definition of the Linz

Turing machine Ĥ, it is now a single machine with a single start state.

A copy of Linz H is embedded at Ĥ.qx.

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

There are no finite number of steps of the pure simulation of ⟨Ĥ⟩

applied to ⟨Ĥ⟩ by embedded_H such that this simulated input meets the

Linz definition of halting:

computation that halts … the Turing machine will halt whenever it enters

a final state. (Linz:1990:234)

Therefore it is correct to say that the input to embedded_H specifies a

sequence of configurations that never halts.

https://www.researchgate.net/publication/358009319_Halting_problem_undecidability_and_infinitely_nested_simulation_V3

> To you and I, this means that he's not working on the halting problem,

> but I am not sure you can say he is lying about that. For one thing,

> how can he be intending to deceive (a core part of lying) when he's been

> clear the he accepts the wrong answer as being the right one? If

> someone claims to be working on "the addition problem", and also claims

> that 2+2=5 is correct, it's hard to consider either claim to be a lie.

> The person is just deeply confused.

>

> But what sort of confused can explain this nonsense? I think the answer

> lies in PO's background. The "binary square root" function is not

> computable as far as a mathematician is concerned because no TM can halt

> with, say, sqrt(0b10) on the tape. But to an engineer, the function

> poses no problem because we can get as close as we like. If

> 0b1.01101010000 is not good enough, just add more digits.

>

> The point is I think PO does not know what a formal, mathematical

> problem really is. To him, anything about code, machines or programs is

> about solving an engineering problem "well enough" -- with "well enough"

> open to be defined by PO himself.

>

> More disturbing to me is that he is not even talking about Turing

> machines, again as evidenced by his own plain words. It is not in

> dispute that he claims that two (deterministic) TMs, one an identical

> copy of the other, can transition to different states despite both being

> presented with identical input. These are not Turing machines but Magic

> machines, and I can't see how any discussion can be had while the action

> of the things being considered is not a simple function of the input and

> the state transition graph.

>

THIS POINT ADDRESSES A SIDE ISSUE NOT RELEVANT TO THE KEY QUESTION:

What are the details of how ⟨Ĥ⟩ applied to ⟨Ĥ⟩ behaves?

H and embedded_H are not identical, one has an infinite loop appended to

its accept state.

I will not tolerate digression into this side issue until after mutual

agreement is achieved on the first point. Until then these side issues

are no more than a dishonest dodge distraction away from the main point.

> This is why I stopped replying. While there are things to say about

> PO's Other Halting problem (principally that even the POOH problem can't

> be solved), I had nothing more to say while the "machines" being

> discussed are magic.

>

--

Copyright 2021 Pete Olcott

Talent hits a target no one else can hit;

Genius hits a target no one else can see.

Arthur Schopenhauer

> Richard Damon <Ric...@Damon-Family.org> writes:

>

>> And the 'actual behavior of its actual inputs' is DEFINED to be what

>> the computation the input actually does when run as an independent

>> machine, or what a UTM will do when simulating that input.

>>

>> If that isn't the meaning you are using, then you are just lying that

>> you are working on the halting problem, which is what seems to be the

>> case. (That you are lying that is).

>

> It is certainly true that PO is not addressing the halting problem. He

> has been 100% clear that false is, in his "opinion", the correct result

> for at least one halting computation. This is not in dispute (unless

> he's retracted that and I missed it).

>

THIS POINT ADDRESSES THE KEY QUESTION

Which state does Ĥ applied to ⟨Ĥ⟩ transition to correctly ?

The following simplifies the syntax for the definition of the Linz

Turing machine Ĥ, it is now a single machine with a single start state.

A copy of Linz H is embedded at Ĥ.qx.

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

There are no finite number of steps of the pure simulation of ⟨Ĥ⟩

applied to ⟨Ĥ⟩ by embedded_H such that this simulated input meets the

Linz definition of halting:

computation that halts … the Turing machine will halt whenever it enters

a final state. (Linz:1990:234)

Therefore it is correct to say that the input to embedded_H specifies a

sequence of configurations that never halts.

https://www.researchgate.net/publication/358009319_Halting_problem_undecidability_and_infinitely_nested_simulation_V3

> To you and I, this means that he's not working on the halting problem,

> but I am not sure you can say he is lying about that. For one thing,

> how can he be intending to deceive (a core part of lying) when he's been

> clear the he accepts the wrong answer as being the right one? If

> someone claims to be working on "the addition problem", and also claims

> that 2+2=5 is correct, it's hard to consider either claim to be a lie.

> The person is just deeply confused.

>

> But what sort of confused can explain this nonsense? I think the answer

> lies in PO's background. The "binary square root" function is not

> computable as far as a mathematician is concerned because no TM can halt

> with, say, sqrt(0b10) on the tape. But to an engineer, the function

> poses no problem because we can get as close as we like. If

> 0b1.01101010000 is not good enough, just add more digits.

>

> The point is I think PO does not know what a formal, mathematical

> problem really is. To him, anything about code, machines or programs is

> about solving an engineering problem "well enough" -- with "well enough"

> open to be defined by PO himself.

>

> More disturbing to me is that he is not even talking about Turing

> machines, again as evidenced by his own plain words. It is not in

> dispute that he claims that two (deterministic) TMs, one an identical

> copy of the other, can transition to different states despite both being

> presented with identical input. These are not Turing machines but Magic

> machines, and I can't see how any discussion can be had while the action

> of the things being considered is not a simple function of the input and

> the state transition graph.

>

THIS POINT ADDRESSES A SIDE ISSUE NOT RELEVANT TO THE KEY QUESTION:

What are the details of how ⟨Ĥ⟩ applied to ⟨Ĥ⟩ behaves?

H and embedded_H are not identical, one has an infinite loop appended to

its accept state.

I will not tolerate digression into this side issue until after mutual

agreement is achieved on the first point. Until then these side issues

are no more than a dishonest dodge distraction away from the main point.

> This is why I stopped replying. While there are things to say about

> PO's Other Halting problem (principally that even the POOH problem can't

> be solved), I had nothing more to say while the "machines" being

> discussed are magic.

>

--

Copyright 2021 Pete Olcott

Talent hits a target no one else can hit;

Genius hits a target no one else can see.

Arthur Schopenhauer

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