Re: Correcting the definition of the halting problem --- Computable functions ---HHH(DD)

Am Tue, 25 Mar 2025 13:57:18 -0500 schrieb olcott:

On 3/25/2025 1:37 PM, dbush wrote:

On 3/25/2025 2:27 PM, olcott wrote:

On 3/25/2025 12:24 PM, dbush wrote:

On 3/25/2025 1:18 PM, olcott wrote:

On 3/25/2025 12:04 PM, dbush wrote:

On 3/25/2025 12:56 PM, olcott wrote:

On 3/25/2025 11:46 AM, dbush wrote:

On 3/25/2025 12:40 PM, olcott wrote:

On 3/25/2025 10:17 AM, dbush wrote:

On 3/25/2025 11:13 AM, olcott wrote:

On 3/25/2025 10:02 AM, dbush wrote:

On 3/25/2025 10:53 AM, olcott wrote:

On 3/25/2025 9:45 AM, dbush wrote:

On 3/24/2025 11:29 PM, olcott wrote:

On 3/24/2025 10:12 PM, dbush wrote:

On 3/24/2025 10:07 PM, olcott wrote:

On 3/24/2025 8:46 PM, André G. Isaak wrote:

On 2025-03-24 19:33, olcott wrote:

On 3/24/2025 7:00 PM, André G. Isaak wrote:

It is impossible for an actual Turing machine to be input to
any other TM.

Therefore we encode it.

But a description of a turing machine can be, for example in
the form of source code or a binary.  And a turing machine by
definition *always* behaves the same for a given input when
executing directly.

IT IS COUNTER-FACTUAL THAT A MACHINE DESCRIPTION ALWAYS
SPECIFIES BEHAVIOR IDENTICAL TO THE DIRECTLY EXECUTED MACHINE.

WTF, a TM is specified by its description.

That is not the complete description.  The complete description
consists of the code of III

and the fact that EEE

Is called by III makes the code of EEE part of the fixed input,
as well as everything that EEE calls down to the OS level.

Which is not relevant to whether or not III emulated by EEE
reaches its own final halt state.

Which is why III emulated by EEE is not relevant.

Does III emulated by EEE reach its final halt state when III defines
a pathological relationship with its emulator?

I don't care what some simulator says, I want to know whether the
direct execution halts, and the simulator better give the same result.

But that's not the question.  The question is whether or not an H
exists that behaves as described below:

Turing machines are only capable operating on input finite strings.

And those finite strings can be a complete description of a turing
machine

The input to a Turing machine cannot possibly be the actual behavior of
any executing process.
A Turing machine can only port on the behavior that a finite string
input specifies.

And that is its direct execution.

Turing machine computable functions cannot compute anything that their
input doesn't specify.

Translation: algorithms only compute what they're programed to compute.

NO WRONG. Turing machine computable functions cannot compute any mapping
from anything that their input DOES NOT SAY.

Yes, exactly.

THEIR INPUT CANNOT POSSIBLY SAY THE ACTUAL BEHAVIOR OF ANY EXECUTING
PROCESS

Wrong, that is exactly what the description of a TM says.

And the algorithm your EEE is computing is not the mathematical halting
function, which has proven to not be computable:

When HHH rejects DD as specifying a computation that does not reach its
final halt state HHH IS CORRECT.

No, DD halts.

THEIR INPUT NEVER SPECIFIES THE ACTUAL BEHAVIOR OF ANY OTHER TURING
MACHINE

How else would you specify a TM?

What a particular turing machine is able to compute doesn't change
whether or not the input string fully describes another turing machine

--
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.