Re: HHH(DD) --- COMPUTE ACTUAL MAPPING FROM INPUT TO OUTPUT --- Ignoramus !!!

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.
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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.
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The proof stands.
>

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*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, as Linz and others have proved:
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