Re: Turing computable function for sum of two integers

On 4/30/25 8:24 PM, olcott wrote:

On 4/30/2025 6:55 PM, Richard Damon wrote:

On 4/30/25 1:36 PM, olcott wrote:

On 4/29/2025 4:50 AM, Mikko wrote:

On 2025-04-28 19:55:35 +0000, olcott said:
>

On 4/28/2025 11:01 AM, dbush wrote:

On 4/28/2025 11:52 AM, olcott wrote:

On 4/28/2025 4:01 AM, Mikko wrote:

On 2025-04-16 17:36:31 +0000, olcott said:
>

On 4/16/2025 7:29 AM, Richard Heathfield wrote:

On 16/04/2025 12:40, olcott wrote:

sum(3,2) IS NOT THE SAME AS sum(5,2).
IT IS EITHER STUPID OR DISHONEST FOR YOU TO TRY TO
GET AWAY FOR CLAIMING THIS USING THE STRAW DECEPTION
INTENTIONALLY INCORRECT PARAPHRASE OF MY WORDS.

>
Whether sum(3,2) is or is not the same as sum(5,2) is not the question. The question is whether a universal termination analyser can be constructed, and the answer is that it can't.
>
This has been rigorously proved. If you want to overturn the proof you've got your work cut out to persuade anyone to listen, not least because anyone who tries to enter into a dialogue with you is met with contempt and scorn.
>
The proof stands.
>

>
*corresponding output to the input*
*corresponding output to the input*
*corresponding output to the input*
*corresponding output to the input*
*corresponding output to the input*
>
Not freaking allowed to look at any damn thing
else besides the freaking input. Must compute whatever
mapping ACTUALLY EXISTS FROM THIS INPUT.

>
A halt decider is is not allowed to compute "whatever" mapping. It is
required to compute one specific mapping: to "no" if the computation
described by the input can be continesd forever without halting, to
"no" otherwise.
>

>
It must do this by applying the finite string transformation
rules specified by the x86 language to the input to HHH(DD).
>
This DOES NOT DERIVE THE BEHAVIOR OF THE DIRECTLY EXECUTED DD.
It DOES DERIVE DD EMULATED BY HHH AND ALSO DERIVES THE RECURSIVE
EMULATION OF HHH EMULATING ITSELF EMULATING DD.
>

>
>
In other words, no H exists that satisfies the following requirements,

>
BECAUSE THOSE REQUIREMENTS HAVE ALWAYS BEEN WRONG AND NO ONE NOTICED.
BECAUSE THOSE REQUIREMENTS HAVE ALWAYS BEEN WRONG AND NO ONE NOTICED.
BECAUSE THOSE REQUIREMENTS HAVE ALWAYS BEEN WRONG AND NO ONE NOTICED.
BECAUSE THOSE REQUIREMENTS HAVE ALWAYS BEEN WRONG AND NO ONE NOTICED.
BECAUSE THOSE REQUIREMENTS HAVE ALWAYS BEEN WRONG AND NO ONE NOTICED.

>
You have not proven that the requirements are wrong in any sense.
>

>
int sum(int x, int y) { return 5; }
Is NOT a Turing Computable function for the sum of two integers.

>
But it *
>

>
int sum(int x, int y) { x + y; }
Is a Turing Computable function for the sum of two integers.
>

>
No, it is an algorithm in the C language to compute the Computable Function for addition.
>
No algorithm/program is a "Function" in the computation theory sense of the word, that is just a category error.
>
algorithms/programs COMPUTE Functions, which are just the complete mapping of inputs to outputs that the Function Defines.
>

 YES that means that HHH(DD) is not allowed to report
on the direct execution of DD(DD) because the input
to HHH(DD) maps to something else.

No. The Halting Function maps the input DD to Halting.
The fact that the algorithm of HHH maps DD to Non-Halting just says that HHH doesn't correctly compute the required mapping.

 When one maps an input to an output one must apply
very specific finite string transformation rules.
One cannot simply guess what one wants to end up
with and simply assume that this mapping is correct.

The mapping generated by an algorithm are limited by that.
The mapping required of an algorithm are not, but are specified by the mapping of the function that is being asked to compute.
That mapping is fully and completely specified by the mapping rule of the halting function:
Halt(P, D) maps (P,D) to Halting if the program P when given the input D will halt in a finite number of steps, and to Non-Halting, if the program P when givne the input D will never halt, even after doing an unbonded number of steps.
Since DD *WILL* reach a final state when run, it maps to Halting.
The fact that HHH can't emulate it to that point is irrelevent, it just is part of what makes it non-computable.
Note, input DD, to be a valid input, must be a program, and thus include the code of the specific HHH that it calls, which by your problem stipulations is the one that aborts and returns 0 as defined in Halt7.c

 *DD correctly emulated by HHH DOES NOT HALT*

That is a NULL statement, as it is based on a lie.
HHH doesn't do a correct emulation, and thus your house of cards has been blown to smithereens by being based on errors.

 

Being Computable just means that an algorithm/program exists that computes it.
>
Perhaps you need to go back to Freshman Programming to learn the meanings of the words you are misusing.

 No level of computer science education ever gets
around to specifying the details of how the mapping
from inputs to outputs must occur.
 I came up with this on my own. Turing Machine algorithms
that compute functions must apply finite string transformations
to inputs to derive outputs.
 That means that DD correctly emulated by HHH DOES NOT HALT!