Re: Turing computable function for sum of two integers

On 2025-04-30 19:57:00 +0000, dbush said:

On 4/30/2025 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 an algorithm for the sum of two integers.
 int sum(int x, int y) { x + y; }
Is an algorithm for the sum of two integers.

 Obviously, but that has nothing to do with a solution to the halting function:
 Given any algorithm (i.e. a fixed immutable sequence of instructions) X described as <X> with input Y:

It does. The first "sum" is analogous to Olcott's halting deciders that are
like a real halting decider except that they compute another function.
--
Mikko