Re: Correcting the definition of the halting problem --- Computable functions

On 3/24/2025 12:35 PM, dbush wrote:

On 3/24/2025 12:44 PM, olcott wrote:

On 3/24/2025 10:14 AM, dbush wrote:

On 3/24/2025 11:03 AM, olcott wrote:

On 3/24/2025 6:23 AM, Richard Damon wrote:

On 3/23/25 11:09 PM, olcott wrote:

It is impossible for HHH compute the function from the direct
execution of DDD because DDD is not the finite string input
basis from which all computations must begin.
https://en.wikipedia.org/wiki/Computable_function

>
WHy isn't DDD made into the correct finite string?i
>

>
DDD is a semantically and syntactically correct finite
stirng of the x86 machine language.

>
Which includes the machine code of DDD, the machine code of HHH, and the machine code of everything it calls down to the OS level.
>

>

That seems to be your own fault.
>
The problem has always been that you want to use the wrong string for DDD by excluding the code for HHH from it.
>

>
DDD emulated by HHH directly causes recursive emulation
because it calls HHH(DDD) to emulate itself again. HHH
complies until HHH determines that this cycle cannot
possibly reach the final halt state of DDD.
>

>
Which is another way of saying that HHH can't determine that DDD halts when executed directly.
>

>
given an input of the function domain it can
return the corresponding output.
https://en.wikipedia.org/wiki/Computable_function
>
Computable functions are only allowed to compute the
mapping from their input finite strings to an output.
>

  The HHH you implemented is computing *a* computable function, but it's not computing the halting function:
 

The whole point of this post is to prove that
no Turing machine ever reports on the behavior
of the direct execution of another Turing machine.

 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

Cannot possibly be a computable function because computable
functions cannot possibly have directly executing Turing
machines as their inputs.

(<X>,Y) maps to 0 if and only if X(Y) does not halt when executed directly
 

--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer