Re: How Partial Simulations correctly determine non-halting -- TM as finite string

Am Sun, 09 Jun 2024 11:09:23 -0500 schrieb olcott:

On 6/9/2024 10:11 AM, Mikko wrote:

On 2024-06-09 13:59:57 +0000, olcott said:

On 6/9/2024 3:47 AM, Mikko wrote:

On 2024-06-08 13:04:14 +0000, olcott said:

On 6/8/2024 1:42 AM, Mikko wrote:

On 2024-06-07 22:26:05 +0000, olcott said:

On 6/7/2024 4:00 PM, joes wrote:

Am Fri, 07 Jun 2024 09:47:35 -0500 schrieb olcott:

On 6/7/2024 1:30 AM, Mikko wrote:

On 2024-06-06 15:31:36 +0000, olcott said:

On 6/6/2024 10:14 AM, Mikko wrote:

On 2024-06-06 13:53:58 +0000, olcott said:

On 6/6/2024 5:15 AM, Mikko wrote:

On 2024-06-05 13:29:28 +0000, olcott said:

On 6/5/2024 2:37 AM, Mikko wrote:

On 2024-06-04 18:02:03 +0000, olcott said:

Any finite string can be an input to some Turing machine. Can you
prove that a Turing machine is not a finite string?
>

By definition Turing Machines are not finite strings in the
conventional model. In my x86utm model of computation x86 machine
language <is> the input to another function written in the x86
language.

It is common knowledge (that you lack) that a Turing machine <is not> a
finite string. It is also common knowledge that finite strings can
describe Turing machines.

I know Turing machines as tuples, consisting of an alphabet, a tape, a
state table, a final state... what did I forget?
They can be described in a finite string. What's your point?

--
joes