Re: The execution trace of HHH1(DDD) shows the divergence

Am Sat, 07 Jun 2025 13:59:54 -0500 schrieb olcott:

On 6/7/2025 1:51 PM, dbush wrote:

On 6/7/2025 2:49 PM, olcott wrote:

On 6/7/2025 1:43 PM, dbush wrote:

On 6/7/2025 2:36 PM, olcott wrote:

On 6/7/2025 1:26 PM, dbush wrote:

On 6/7/2025 1:10 PM, olcott wrote:

On 6/7/2025 11:37 AM, dbush wrote:

On 6/7/2025 12:29 PM, olcott wrote:

On 6/7/2025 11:20 AM, dbush wrote:

On 6/7/2025 12:17 PM, olcott wrote:

On 6/7/2025 11:14 AM, dbush wrote:

On 6/7/2025 11:33 AM, olcott wrote:

On 6/7/2025 10:17 AM, dbush wrote:

On 6/7/2025 11:12 AM, olcott wrote:

On 6/7/2025 10:08 AM, dbush wrote:

On 6/7/2025 11:06 AM, olcott wrote:

On 6/7/2025 10:01 AM, dbush wrote:

On 6/7/2025 10:58 AM, olcott wrote:

On 6/7/2025 9:56 AM, dbush wrote:

On 6/7/2025 10:54 AM, olcott wrote:

On 6/7/2025 9:51 AM, dbush wrote:

On 6/7/2025 10:32 AM, olcott wrote:

The execution trace of HHH1(DDD) shows the
divergence of DDD emulated by HHH from DDD
emulated by HHH1.
Shows that DDD emulated by HHH and DDD emulated by
HHH1 diverges as soon as HHH begins emulating
itself emulating DDD.

The next instruction of DDD that HHH emulates is
at the machine address of 00002183.
The next instruction of DDD that HHH1 emulates is
at the machine address of 00002190.

>
False.
The next instruction of DDD that both HHH and HHH1
emulates is at the machine address of 000015c3,

>
*That is not an instruction of DDD*

>
In other words, you're not operating on algorithms.

I didn't say that you didn't specify an algorithm.  I said your HHH
doesn't work with algorithms,
 

Termination analyzers are encoded algorithms that work with finite
strings that specify a precise sequence of configurations.

Yes, they should. Thus HHH is not a termination analyser if it doesn't
consider itself being part of the input. This is how computation works.

--
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.