Re: Anyone with sufficient knowledge of C knows that DD specifies non-terminating behavior to HHH

Am Tue, 25 Feb 2025 20:13:43 -0600 schrieb olcott:

On 2/25/2025 5:41 PM, Richard Damon wrote:

On 2/25/25 4:12 PM, olcott wrote:

On 2/25/2025 8:59 AM, Mikko wrote:

On 2025-02-24 23:36:04 +0000, olcott said:

On 2/24/2025 2:47 AM, Mikko wrote:

On 2025-02-23 17:44:25 +0000, olcott said:

On 2/23/2025 4:59 AM, Mikko wrote:

On 2025-02-22 16:11:31 +0000, olcott said:

On 2/22/2025 3:04 AM, Mikko wrote:

On 2025-02-21 22:35:16 +0000, olcott said:

On 2/21/2025 2:18 AM, Mikko wrote:

On 2025-02-20 21:31:44 +0000, olcott said:

On 2/20/2025 2:38 AM, Mikko wrote:

On 2025-02-20 00:31:33 +0000, olcott said:

On 2/19/2025 3:01 AM, Mikko wrote:

On 2025-02-18 11:26:25 +0000, olcott said:

On 2/18/2025 3:24 AM, Mikko wrote:

On 2025-02-17 09:05:42 +0000, Fred. Zwarts said:

Op 16.feb.2025 om 23:51 schreef olcott:

On 2/16/2025 4:30 PM, joes wrote:

What’s confusing about „halts”? I find it clearer as
it does not imply an ambiguous „abnormal
termination”. How does HHH simulate DD terminating
abnormally, then? Why doesn’t it terminate
abnormally itself?
You can substitute the term: the input DD to HHH
does not need to be aborted, because the simulated
decider terminates.
>

Every simulated input that must be aborted to prevent
the non-termination of HHH is stipulated to be
correctly rejected by HHH as non-terminating.
>

A very strange and invalid stipulation.

It merely means that the words do not have their
ordinary meaning.

Unless HHH(DD) aborts its simulation of DD itself cannot
possibly terminate normally.

That cannot be determined without examination of HHH,
which is not in the scope of OP.

I have given everyone here all of the complete source code
for a few years

True but irrelevant. OP did not specify that HHH means that
particular code.

OP had a pointer of that code but didn's state that that code
is a part of the problem. OP did not spacify any range for
variation.

There are at least two algorithms the current one that was also
the original one is easiest to understand. This algorithm
essentially spots the equivalent of infinite recursion. The code
provides all of the details.

HHH is exactly as specified. Assuming otherwise is silly.

>
The words "as specified" when nothing is specified are not a good
use of the language.
>

When DD is correctly simulated by HHH according to the behavior
that the above machine code specifies then the call from DD to
HHH(DD) cannot possibly return making it impossible for DD
emulated by HHH to terminate normally.

>
That code does not specify whether HHH ever returns or what value
HHH returns if it does.
>

When HHH is known to emulate the above code with an x86 emulator
THEN

>
Although OP says that HHH simulates it does not specify what is
simulated.
>

the above code
>

Which isn't all of the program.
 

The behavior of DD emulated by HHH only refers to DD and the fact that
HHH emulates this DD.

On on hand, the simulator can have no influence on the execution.
On the other, that same simulator is part of the program.
You don't understand this simple entanglement.

--
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.