Re: Turing Machine computable functions apply finite string transformations to inputs VERIFIED FACT

On 4/30/2025 6:08 PM, olcott wrote:

On 4/30/2025 2:57 PM, dbush wrote:

On 4/30/2025 1:38 PM, olcott wrote:

On 4/29/2025 5:00 AM, joes wrote:

Am Mon, 28 Apr 2025 21:50:03 -0500 schrieb olcott:

On 4/28/2025 3:13 PM, Richard Heathfield wrote:

>

What matters is whether a TM can be constructed that can accept an
arbitrary TM tape P and an arbitrary input tape D and correctly
calculate whether, given D as input, P would halt. Turing proved that
such a TM cannot be constructed.
This is what we call the Halting Problem.
>

Yet it is H(P,D) and NOT P(D) that must be measured. Computer science
has been wrong about this all of these years. When I provide the 100%
concrete example of the x86 language there is zero vagueness to slip
through the cracks of understanding.

>

No, H gets P(D) as input, not itself. H is the "measurer", not being
measured.
>

>
H NEVER gets P(D) as input.
H always gets finite strings P and D as inputs.

>
Which is stipulated to be a complete description of algorithm P with input D

 It <is> a complete specification

An algorithm that halts when executed directly

 int sum(int x, int y) { return 5; }
Does not apply transformations to its inputs
to derive its outputs thus is no kind of algorithm
not even for sum(2,3).
 

Sure it is, it's algorithm that computes this computable function:
For all integers X and Y:
(X,Y) maps to 5