On 11/08/2025 20:00, olcott wrote:

<snip>

Likewise I want to know the radius of a square circle
that has a length of 2.0 of one of its equal length
four sides.

Then you'll need to provide the area and the other radius.

$A_{\text{octorad}} = 4R^{2} + 2\pi R^{2} + \pi r^{2} - 4r^{2}$

--
Richard Heathfield
Email: rjh at cpax dot org dot uk
"Usenet is a strange place" - dmr 29 July 1999
Sig line 4 vacant - apply within

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On 8/11/25 2:50 PM, olcott wrote:

On 8/11/2025 1:45 PM, dbush wrote:

On 8/11/2025 2:40 PM, olcott wrote:

On 8/11/2025 1:33 PM, Richard Heathfield wrote:

On 11/08/2025 18:48, olcott wrote:

On 8/11/2025 11:42 AM, Richard Heathfield wrote:

On 11/08/2025 17:29, olcott wrote:

<snip>

I have proven that DD correctly simulated by HHH

No, you haven't. You have asserted that the simulation is correct, >>>>>> but we all know that it derives a different result from that
produced by direct execution, and therefore we all know that the
simulation is not correct.

You can only fully know that your assumption is incorrect
when you notice that no specific error exists.

The specific error is that you get the wrong answer.

It is incorrect to say it *is* a wrong answer until after
a specific error in the basis of this answer is found.

It's the wrong answer because it doesn't meet the required specification:

That merely presumes that the required specification
is correct. Reasoning from first principles would never
make this mistake. https://jamesclear.com/first-principles#

But to reason from first principles, you must KNOW those first principles.

That would be like what is a "Function" (in the general mathematical
sense used in computability theory)

and like what a program is.

What Halting and Non-Halting Actually mean.

How correct simulation is actually defined.

How representations work, and mean.


For example, in that paper you quote, Musk didn't ignore or not bother
learning about fundamental like gravity or thrust.

He just ignored classical ideas of how you would build a rocket.

So, to do that for computability theory, you start from the basic
definition of what a computation is, and the functions that they try to compute.

This is part of your problem, because computations take their "meaning"
from the functions they are trying to compute, and thus you claim that a
Turing Machine can't be asked to compute the halting-behavior of another
Turing Machine, is saying that the halting-behavior of a Turing machine
isn't a "function" of that Turing Machine, which is just a denial of the
basic properties of Turing Machines and computations.

That or you are claiming that your idea of programs can't actually
represent a Turing Machine to another Turing Machine, but that would
imply that Universal Turing Machines don't exist, which means your
concept of programs is significantly less powerful than the system that
Has actual Turing Machines and UTMs.

Your problem is you didn't work from "first-principles" but from "sero-princples" and thus is just a system of ignorance, like flat-earthers.

Sorry, but that *IS* the basic facts of what you have been trying to
say, which has just made it clear to the world how stupid you are and
that you just don't care about the truth, so you cries about trying to
protect the world from climate change liars is just a pathetic lie.

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Am Mon, 11 Aug 2025 14:00:14 -0500 schrieb olcott:

On 8/11/2025 1:51 PM, dbush wrote:

It's correct because I want to know if any arbitrary algorithm X with
input Y will halt when executed directly.

Likewise I want to know the radius of a square circle that has a length
of 2.0 of one of its equal length four sides.

Come on. Every algorithm either halts or doesn't.

It would be *very* useful to me if I had an algorithm H that could tell
me that in *all* possible cases.

It would also be very useful if wishes really were horses.

Yeah, it would.
--
Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
It is not guaranteed that n+1 exists for every n.

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On 12/08/2025 09:14, joes wrote:

Am Mon, 11 Aug 2025 14:00:14 -0500 schrieb olcott:

On 8/11/2025 1:51 PM, dbush wrote:

It's correct because I want to know if any arbitrary algorithm X with
input Y will halt when executed directly.

Likewise I want to know the radius of a square circle that has a length
of 2.0 of one of its equal length four sides.

Come on. Every algorithm either halts or doesn't.

We can do better than that. Every algorithm halts, because if it
doesn't halt it isn't an algorithm: "An algorithm must always
terminate after a finite number of steps ... a very finite
number, a reasonable number" - DEK

--
Richard Heathfield
Email: rjh at cpax dot org dot uk
"Usenet is a strange place" - dmr 29 July 1999
Sig line 4 vacant - apply within

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On 12/08/2025 09:56, Chris M. Thomasson wrote:

On 8/12/2025 1:54 AM, Richard Heathfield wrote:

On 12/08/2025 09:14, joes wrote:

Am Mon, 11 Aug 2025 14:00:14 -0500 schrieb olcott:

On 8/11/2025 1:51 PM, dbush wrote:

It's correct because I want to know if any arbitrary
algorithm X with
input Y will halt when executed directly.

Likewise I want to know the radius of a square circle that
has a length
of 2.0 of one of its equal length four sides.

Come on. Every algorithm either halts or doesn't.

We can do better than that. Every algorithm halts, because if
it doesn't halt it isn't an algorithm: "An algorithm must
always terminate after a finite number of steps ... a very
finite number, a reasonable number" - DEK

Think if the only reason an algorithm halted was because the
Earth got destroyed? Otherwise, it would still be running? ;^o

Algorithms don't run.

It's Astronomy that tells us all programs halt, but it's Computer
Science that tells us all algorithms terminate.

--
Richard Heathfield
Email: rjh at cpax dot org dot uk
"Usenet is a strange place" - dmr 29 July 1999
Sig line 4 vacant - apply within

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On 12/08/2025 09:54, Richard Heathfield wrote:

[...] Every algorithm halts, because if it doesn't halt it isn't an algorithm: "An algorithm must always terminate after a finite number
of steps ... a very finite number, a reasonable number" - DEK

Just possibly worth pointing out that "finite" does not mean
"bounded", nor even "bounded after a bounded number of steps". I enter
into evidence the game "Sylver Coinage" [qv], which includes games that
go on for arbitrarily many moves, and where the "arbitrarily many" can
be re-set arbitrarily many times, but nevertheless the game always
terminates [usually very quickly]. [FTAOD, I have not previously used
the word "algorithm" in this article ....]

--
Andy Walker, Nottingham.
Andy's music pages: www.cuboid.me.uk/andy/Music
Composer of the day: www.cuboid.me.uk/andy/Music/Composers/Duphly

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On 12/08/2025 15:06, Andy Walker wrote:

On 12/08/2025 09:54, Richard Heathfield wrote:

[...] Every algorithm halts, because if it doesn't halt it
isn't an algorithm: "An algorithm must always terminate
after a finite number of steps ... a very finite number, a
reasonable number" - DEK

Just possibly worth pointing out that "finite" does not mean
"bounded",

Uh, yes it does. That's /precisely/ what it means. (but see below)

https://en.wiktionary.org/wiki/finite (but see below)

Adjective (but see below)

finite (comparative more finite, superlative most finite)

Having an end or limit; (of a quantity) constrained by bounds

(but see below)

nor even "bounded after a bounded number of steps". I enter
into evidence the game "Sylver Coinage" [qv], which includes
games that go on for arbitrarily many moves, and where the
"arbitrarily many" can be re-set arbitrarily many times, but
nevertheless the game always terminates [usually very
quickly].

How is that evidence for "infinite" not being bounded? (but see
below)

I don't know the game, but I suspect it's more likely to be
evidence for the game's rules being designed to ensure that it's
likely to end at some point. A solitaire game that ends on a
double 1 dice roll could in principle go on for ever, but in
practice 99% of games will end within ~165 rolls, and most a lot
sooner. Theoretically unbounded but in practice very bound indeed...

BBBBBB U U TTTTTTT
B B U U T
B B U U T
BBBBBB U U T
B B U U T
B B U U T ... ... ...
BBBBBB UUUUU T ... ... ...



I learned the word "infinite" when I was about six years old, and
it was /defined/ in the (probably Concise Oxford) dictionary as
"boundless".

And that's fine as far as it goes. But to me a double is not a
couple of sixes; it's a data type. Likewise union is not a trade
association, static doesn't mean screen snow, float is not a
buoy, and char has little to do with burning.

"Finite" and "infinite" (and indeed "bounded") have acquired
relevant technical meanings to which my attention has now been
drawn and which it would be disingenuous for me to ignore.

--- to wit ---

Finite means having a definite number of elements or steps.

Bounded means there is some limit you cannot exceed (upper
bound), but that limit could be infinite in terms of steps or
possibilities.


And by those definitions they do indeed mean very different things.


--
Richard Heathfield
Email: rjh at cpax dot org dot uk
"Usenet is a strange place" - dmr 29 July 1999
Sig line 4 vacant - apply within

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On 12/08/2025 20:27, Richard Heathfield wrote:

[...] Every algorithm halts, because if it doesn't halt it isn't
an algorithm: "An algorithm must always terminate
after a finite number of steps ... a very finite number, a>>> reasonable number" - DEK

Just possibly worth pointing out that "finite" does not mean "bounded",

Uh, yes it does. That's /precisely/ what it means. (but see below) https://en.wiktionary.org/wiki/finite (but see below)
Adjective (but see below)
finite (comparative more finite, superlative most finite)
    Having an end or limit; (of a quantity) constrained by bounds
 (but see below)

Perhaps you shouldn't believe everything you read in dictionaries?
What words would you use to describe a game which /must/ terminate [after, therefore, a finite number of moves] -- not /probably/ terminate --, but
in which that finite number cannot [in general] be bounded part-way into
the game?

nor even "bounded after a bounded number of steps". I enter into
evidence the game "Sylver Coinage" [qv], which includes games that
go on for arbitrarily many moves, and where the>> "arbitrarily many" can be re-set arbitrarily many times, but
nevertheless the game always terminates [usually very
quickly].

How is that evidence for "infinite" not being bounded?  (but see below)

"Infinite" is a word you have introduced. It's "arbitrarily many"
that isn't bounded. Eg, I can [easily] construct a game of SC in which
[say] 100 moves have been played at which point I invite you to give me
a number -- say, 1000000 -- and I can then construct a game continuation
with that number of moves. [I don't claim the game thus constructed to
be /well/ played.]

I don't know the game,

GIYF.

but I suspect it's more likely to be evidence for the game's rules being designed to ensure that it's likely to end at some point.

It will definitely end, but the number of moves after which that happens is not bounded. But it is easy to see that progress is being made.
[SC is not the only game of this type, but it has a lot of known theory.]

--
Andy Walker, Nottingham.
Andy's music pages: www.cuboid.me.uk/andy/Music
Composer of the day: www.cuboid.me.uk/andy/Music/Composers/Handel

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On 8/12/25 4:54 AM, Richard Heathfield wrote:

On 12/08/2025 09:14, joes wrote:

Am Mon, 11 Aug 2025 14:00:14 -0500 schrieb olcott:

On 8/11/2025 1:51 PM, dbush wrote:

It's correct because I want to know if any arbitrary algorithm X with
input Y will halt when executed directly.

Likewise I want to know the radius of a square circle that has a length
of 2.0 of one of its equal length four sides.

Come on. Every algorithm either halts or doesn't.

We can do better than that. Every algorithm halts, because if it doesn't
halt it isn't an algorithm: "An algorithm must always terminate after a finite number of steps ... a very finite number, a reasonable number" - DEK

Which is one of the great philosophical debates in Computation theory,
are things that some people want to call algorithms, but which don't
halt on some inputs non-hatling algorithms, or something else that needs
a new name.

If they need a new name, the the Halting Problem also needs a new name,
as you have define that all algorithms must halt, it is a test if
something IS an algorithm. It also means that there are somethings we
don't know if they are algorithms or not, as we can't determine if they
will halt or not. To me, this definition leads to more problems that it
is worth.

Especially since some fully defined sequence of deterministic steps will
halt for some inputs, but not for others, so does that mean that
sequence of steps is sometimes an algorithm, and sometimes not?

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On 8/12/25 10:11 AM, olcott wrote:

On 8/12/2025 3:14 AM, joes wrote:

Am Mon, 11 Aug 2025 14:00:14 -0500 schrieb olcott:

On 8/11/2025 1:51 PM, dbush wrote:

It's correct because I want to know if any arbitrary algorithm X with
input Y will halt when executed directly.

Likewise I want to know the radius of a square circle that has a length
of 2.0 of one of its equal length four sides.

Come on. Every algorithm either halts or doesn't.

Machine M contains simulating halt decider H
M.q0 ⟨M⟩ ⊢* M.H ⟨M⟩ ⟨M⟩ ⊢* M.qy
M.q0 ⟨M⟩ ⊢* M.H ⟨M⟩ ⟨M⟩ ⊢* M.qn

*Repeats until aborted proving non-halting*

Nope, at least not if H does abort.> (a) M copies its input ⟨M⟩

(b) M invokes M.H ⟨M⟩ ⟨M⟩
(c) M.H simulates ⟨M⟩ ⟨M⟩

then M.H ⟨M⟩ ⟨M⟩ transitions to M.qn
causing M applied to ⟨M⟩ halt

and the correct simutation of (M) (M) continues past the point that H as stopped observing the correct simulation (since the correct simulation
isn't dependent on H to do it) for another cycle where the H whose
simulation was started in the first (c) will abort its simulation, goto
M.qn and M will halt.

Your problem is you don't have the right definition for correct
simulation, becuase you work off of a zero-knowledge based set of
principles, and thus show your ignorance.

It would be *very* useful to me if I had an algorithm H that could tell >>>> me that in *all* possible cases.

It would also be very useful if wishes really were horses.

Yeah, it would.

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On 13/08/2025 01:48, Andy Walker wrote:

On 12/08/2025 20:27, Richard Heathfield wrote:

[...] Every algorithm halts, because if it doesn't halt it isn't
an algorithm: "An algorithm must always terminate
after a finite number of steps ... a very finite number, a>>>
reasonable number" - DEK

Just possibly worth pointing out that "finite" does not mean
"bounded",

Uh, yes it does. That's /precisely/ what it means. (but see below)
https://en.wiktionary.org/wiki/finite (but see below)
Adjective (but see below)
finite (comparative more finite, superlative most finite)
     Having an end or limit; (of a quantity) constrained by bounds
  (but see below)

    Perhaps you shouldn't believe everything you read in
dictionaries?

What the dictionary told a six-year-old in a single word was
true... but lacked nuance. Perhaps we shouldn't blame
dictionaries for failing to define everything /and/ being too
heavy to carry conveniently in one hand.

What words would you use to describe a game which /must/
terminate [after,
therefore, a finite number of moves] -- not /probably/ terminate
--, but
in which that finite number cannot [in general] be bounded
part-way into
the game?

Hmm. Maybe you didn't but see below after all. Might have been a
plan.

--
Richard Heathfield
Email: rjh at cpax dot org dot uk
"Usenet is a strange place" - dmr 29 July 1999
Sig line 4 vacant - apply within

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On 13/08/2025 03:19, Richard Damon wrote:

On 8/12/25 4:54 AM, Richard Heathfield wrote:

On 12/08/2025 09:14, joes wrote:

Am Mon, 11 Aug 2025 14:00:14 -0500 schrieb olcott:

On 8/11/2025 1:51 PM, dbush wrote:

It's correct because I want to know if any arbitrary
algorithm X with
input Y will halt when executed directly.

Likewise I want to know the radius of a square circle that
has a length
of 2.0 of one of its equal length four sides.

Come on. Every algorithm either halts or doesn't.

We can do better than that. Every algorithm halts, because if
it doesn't halt it isn't an algorithm: "An algorithm must
always terminate after a finite number of steps ... a very
finite number, a reasonable number" - DEK

Which is one of the great philosophical debates in Computation
theory, are things that some people want to call algorithms, but
which don't halt on some inputs non-hatling algorithms, or
something else that needs a new name.

If they need a new name, the the Halting Problem also needs a new
name, as you have define that all algorithms must halt, it is a
test if something IS an algorithm.

Could such a test even exist? I suspect not.

It also means that there are
somethings we don't know if they are algorithms or not, as we
can't determine if they will halt or not. To me, this definition
leads to more problems that it is worth.

Fair.

Especially since some fully defined sequence of deterministic
steps will halt for some inputs, but not for others, so does that
mean that sequence of steps is sometimes an algorithm, and
sometimes not?

Maybe it means that the world does not always so easily fall into
two neat categories (tis/tisn't) as we might prefer. NB "maybe".
I am far too uneasy about such a claim to actually pin it to my
jacket and make it; I merely place it on a long stick, hold it
high above the parapet, and await the shot from a safe distance.

--
Richard Heathfield
Email: rjh at cpax dot org dot uk
"Usenet is a strange place" - dmr 29 July 1999
Sig line 4 vacant - apply within

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On 13/08/2025 04:10, Richard Heathfield wrote:
[...]

Hmm. Maybe you didn't but see below after all. Might have been a plan.

I did, and it wasn't; but I had forgotten what you had said by
the time I was half-way through composing my reply. Had I remembered
in time, I would have edited my response to be more appropriate [and conciliatory].

I also forgot to mention, as I had intended, the Angels and
Devils game

https://en.wikipedia.org/wiki/Angel_problem

which also has aspects relating to infinity and boundedness, and is
[arguably] more fun than Sylver Coinage.

--
Andy Walker, Nottingham.
Andy's music pages: www.cuboid.me.uk/andy/Music
Composer of the day: www.cuboid.me.uk/andy/Music/Composers/Handel

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On 13/08/2025 10:38, Andy Walker wrote:

Had I remembered
in time, I would have edited my response to be more appropriate [and conciliatory].

...as I had already done, hence my (but see below) annotations in
that reply.

If you're looking for people to engage with you in topical
non-olcottian discussions, you're sure going a funny way about it.

--
Richard Heathfield
Email: rjh at cpax dot org dot uk
"Usenet is a strange place" - dmr 29 July 1999
Sig line 4 vacant - apply within

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