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

On 5/4/2025 10:49 PM, olcott wrote:

On 5/4/2025 7:21 PM, Richard Damon wrote:

On 5/4/25 6:15 PM, olcott wrote:

On 5/4/2025 2:21 PM, Richard Heathfield wrote:

On 04/05/2025 18:55, olcott wrote:

Changing my words then rebutting these changed
words is dishonest.
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Functions computed by Turing Machines require INPUTS
and produce OUTPUTS DERIVED FROM THESE INPUTS.

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Counter-example: a Turing Machine can calculate pi without any input whatsoever.
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As Mikko rightly said: a Turing machine does not need to require an input.
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IT IS NOT COMPUTING FUNCTION THEN
IT IS NOT COMPUTING FUNCTION THEN
IT IS NOT COMPUTING FUNCTION THEN
IT IS NOT COMPUTING FUNCTION THEN

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Right, not all Turing Machine compute Functions, they all do perform Computations.
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 Even those that know this pretend that they don't.
 

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Computable functions are the basic objects of study in computability theory. Computable functions are the formalized analogue of the intuitive notion of algorithms, in the sense that a function is computable if there exists an algorithm that can do the job of the function, i.e. given an input of the function domain it can return the corresponding output.
https://en.wikipedia.org/wiki/Computable_function
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given an input of the function domain it can return the corresponding output.

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Right, and the input to a Halt Decider is the representation of a Program,

 Not exactly. It is a 100% specific precise sequence of encoded steps.
Not at all the same as a mere description.

In other words, YES EXACTLY.  And those steps include the code of the function DD, the code of the function HHH, and the code of everything HHH calls down to the OS level.  As such, *none* of those may be modified, hypothetically or otherwise.

 

and the correct output is based on the behavior of that progrma when run.
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 It typically precisely coincides with the exact same
behavior as the direct execution.
 My idea of a simulating termination analyzer

Which as a termination analyzer is required to compute the following mapping:
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

conclusively proves to all those with great
expertise in programming that this is not
always the case.

It just seems that way because you change the input.
Changing the input is not allowed.

You seem to lack sufficient
expertise in programming
 int DD()
{
   int Halt_Status = HHH(DD);
   if (Halt_Status)
     HERE: goto HERE;
   return Halt_Status;
}
 Replacing the code of HHH with an unconditional simulator and subsequently running HHH(DD) cannot possibly
return anywhere. Most every expert C programmer
knows this.
 

Obviously, since you changed the input.
Changing the input is not allowed.