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

On 5/4/25 2:21 PM, olcott wrote:

On 5/3/2025 2:55 AM, Mikko wrote:

On 2025-05-03 04:14:27 +0000, olcott said:
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On 5/2/2025 4:03 AM, Mikko wrote:

On 2025-04-30 15:28:33 +0000, olcott said:
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On 4/29/2025 4:49 AM, Mikko wrote:

On 2025-04-28 15:52:13 +0000, olcott said:
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On 4/28/2025 4:01 AM, Mikko wrote:

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

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

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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.

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It must do this by applying the finite string transformation
rules specified by the x86 language to the input to HHH(DD).

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No, it needn't. A halt decider cannot do other than certain finite string
operations. No relation to x86 language is required.
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This DOES NOT DERIVE THE BEHAVIOR OF THE DIRECTLY EXECUTED DD.

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Whether the execution is "direct" or otherwise is irrelevant. A computation
either halts or not. A halt decider must just tell whether the somputation
halts. It is true that no Turing machine can determine this about every
computation, i.e., no Turing machine is a halt decider.
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It DOES DERIVE DD EMULATED BY HHH AND ALSO DERIVES THE RECURSIVE
EMULATION OF HHH EMULATING ITSELF EMULATING DD.

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Which are not mentioned in the halting problem.

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When understand rather than simply ignore the HHH/DD
example it can be seen that every conventional halting
problem proof suffers the same fate.

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That you (or some other people) don't understand the proof is not fatal.
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The contradictory part of the "impossible" input IS NEVER REACHABLE.
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int DD()
{
   int Halt_Status = HHH(DD);
   if (Halt_Status)
     HERE: goto HERE;
   return Halt_Status;
}

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It is unless HHH never returns.

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HHH cannot possibly return to any DD correctly
emulated by HHH.

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In that case what I said is true. Another way to say the same is that
if you can determine that the above DD does not execute the "if" line
you can infer that HHH(DD) does not return and therefore HHH is not a
halt decider (nor any other decder).
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 That would be true IFF (if and only if) the directly
executed HHH did not see that DD has caused its emulated
self to get stuck in recursive emulation.

No, it can't see that, as that isn't what happens, as HHH is, and can onoy be, the one and only HHH that is defined in Halt7.c, which you have stipulated as part of the problem.

 HHH(DD) does see this and rejects its input on that basis.
 

And is wrong, as that isn't the HHH that DD calls, that HHH deosn't get stuck but always aborts and returns.
This means your HHH(DD) is just like your sum(2,3) thinking it was given 5 and 7 and returning 12. Those weren't the inputs. The input DD calls an HHH that WILL abort and return 0, so HHH is just incorrect to not consider that.
The fact that you made the logical error when you designed HHH, just makes HHH incorect, and you are bad programmer that doesn't understand logic.