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On 10/15/2024 8:43 PM, Mike Terry wrote:Is there any difference between 100% correct and necessarily correct?On 14/10/2024 23:17, Chris M. Thomasson wrote:DDD emulated by HHH according to the semantics of the x86On 10/13/2024 7:57 PM, olcott wrote:>On 10/13/2024 11:34 AM, Kaz Kylheku wrote:[...]On 2024-10-12, Chris M. Thomasson <chris.m.thomasson.1@gmail.com> wrote:>On 10/12/2024 11:28 AM, Janis Papanagnou wrote:>On 12.10.2024 11:32, Jan van den Broek wrote:>2024-10-12, Chris M. Thomasson <chris.m.thomasson.1@gmail.com> schrieb:>On 10/11/2024 7:50 PM, olcott wrote:>
[Schnipp]
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As I see it, the main Halting Problem is Olcott not halting.
LOL! - A very nice one. Thanks for that. :-)
I second that. :^)
You're likely thousand-seconding that. The Olcott not halting joke
is many years old now, and will likely come up again.
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My cancer has gotten worse.
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*ChatGPT explains why rebuttals of my work are incorrect*
https://chatgpt.com/share/6709e046-4794-8011-98b7-27066fb49f3e
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I had to dumb this down from the original halting problem
input so that reviewers can verify that HHH is correct
without hardly paying any attention at all:
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void DDD()
{
HHH(DDD);
return;
}
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When HHH is an x86 emulation based termination analyzer
then each DDD emulated by any HHH that it calls never returns.
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Isn't that similar to:
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void foobar()
{
foobar();
}
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Similar, but different because HHH only performs a /partial/ step by step emulation of DDD - it stops emulating after a while and returns, so DDD() halts.
language cannot possibly reach its own return instruction
no matter what HHH does.
When HHH rejects DDD on this basis it is necessarily correct.
https://chatgpt.com/share/6709e046-4794-8011-98b7-27066fb49f3e
ChatGPT finally totally explains away all of the quibbling
about whether HHH is correct to reject DDD as non-halting.
foobar() will never halt (ignoring physical resource constraints like running out of stack). foobar() undergoes infinite recursive call. DDD() exhibits /finite/ recursive emulation, then halts.
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Mike.
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