Re: HHH(DDD) computes the mapping from its input to HHH emulating itself emulating DDD --- anyone that says otherwise is a liar

Am Fri, 22 Nov 2024 08:50:33 -0600 schrieb olcott:

On 11/22/2024 6:20 AM, joes wrote:

Am Thu, 21 Nov 2024 15:19:43 -0600 schrieb olcott:

On 11/21/2024 3:11 PM, joes wrote:

Am Thu, 21 Nov 2024 09:19:03 -0600 schrieb olcott:

On 11/20/2024 10:00 PM, Richard Damon wrote:

On 11/20/24 9:57 PM, olcott wrote:

On 11/20/2024 5:51 PM, Richard Damon wrote:

On 11/20/24 5:03 PM, olcott wrote:

On 11/20/2024 3:53 AM, Mikko wrote:

On 2024-11-20 03:23:12 +0000, olcott said:

On 11/19/2024 4:12 AM, Mikko wrote:

On 2024-11-18 20:42:02 +0000, olcott said:

On 11/18/2024 3:41 AM, Mikko wrote:

The "the mapping" on the subject line is not correct. The
subject line does not specify which mapping and there is no
larger context that could specify that. Therefore it should
be "a mapping".
On 2024-11-17 18:36:17 +0000, olcott said:

>

My code is one example of the infinite set of every possible HHH
that emulates DDD according to the semantics of the x86 language.

Like all of them, it is unable to simulate DDD to its undeniable
halting state.

>

But it gets the wrong answer for the halting problem, as DDD dpes
halt.

DDD emulated by HHH does not halt.

Whatever. DDD halts and HHH should return that.

IT IS NOT THE SAME INSTANCE OF DDD.

All instances of DDD behave the same (if it is a pure function and the
HHH called from it doesn't switch behaviour by a static variable).

Only HHH is required to be a pure function, DDD is expressly allowed to
be any damn thing.

TMs don't have side effects, such as reading a static Root variable.

The behavior of DDD emulated by HHH is
different than the actual behavior of DDD emulated by HHH1.

Yes. HHH simulates it incorrectly.

Note we have been on this one point about the behavior of DDD emulated
by HHH for many months and have not yet even begun to talk about how HHH
would report this behavior.

Yes, you seem hellbent on dying before you can make your point.

The question does DDD halt?
Is answered by Can DDD emulated by any HHH reach its own "return"
instruction final state?

Not in the case of an erroneous simulation.
If DDD halts, you can simply run it to find out.

The question: How could HHH determine whether or not DDD emulated by HHH
can possibly reach its own final state?
is another entirely different question.

It seems very relevant. Since you have an answer to the former question,
maybe you can proceed to this one?

--
Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
It is not guaranteed that n+1 exists for every n.