Re: Turing computable functions

On 3/25/25 5:46 PM, olcott wrote:

On 3/25/2025 4:12 PM, joes wrote:

Am Tue, 25 Mar 2025 15:50:33 -0500 schrieb olcott:

On 3/25/2025 3:05 PM, dbush wrote:

On 3/25/2025 3:47 PM, olcott wrote:

On 3/25/2025 2:32 PM, dbush wrote:

On 3/25/2025 3:24 PM, olcott wrote:

>

Cannot possibly derive any outputs not computed from their inputs.

Correct, algorithms can only compute computable mathematical
function.
>

A Turing machine halt decider

Does not exist because the required mapping is not computable:
Given any algorithm (i.e. a fixed immutable sequence of instructions)
X described as <X> with input Y:
A solution to the halting problem is an algorithm H that computes the
following mapping:
(<X>,Y) maps to 1 if and only if X(Y) halts when executed directly
(<X>,Y) maps to 0 if and only if X(Y) does not halt when executed
directly
>

cannot possibly report on the behavior of any directly executing
process. No Turing machine can every do this. This has always been
beyond what any Turing machine can ever do.

Strawman: reporting on an executing process is not a requirement.

YOU JUST SAID THAT IT WAS YOU KEEP MINDLESSLY REPEATING THAT IT IS

I never said it had to actually watch an executing process, only report
what would happen if it did run.

*It has been conclusively proven as a verified*
*fact many hundreds of times over several years*
That the behavior that the finite string input specifies is not perfect
proxy for the behavior of the underlying directly executed machine.

 

A TM can be completely specified in a finite string.
>

 That has different behavior when it is simulated by a UTM
that defines a pathological relationship to this UTM than
a UTM where no such pathological relationship exists.
 

Really, you mean a REAL UTM, not yor fake UTM?
How can an input have pathology to a UTM, since the only goal of a UTM is to accurately recreate the behavior of the input, so an infinite recursion loop is not "pathological" but the behavior of the machine.
Of course, when the machine being called isn't actually a UTM, but a program trying to play one on TV and then getting an answer to a question, it gets tripped up by believing that it actually was a UTM when it wasn't.
Of course your problem is you don't understand what a UTM actually is, think a partial emulator can be a UTM, when it isn't.
You are just proving your utter stupidity and ignorance.