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

Am Thu, 27 Feb 2025 17:28:58 -0600 schrieb olcott:

On 2/27/2025 2:45 PM, joes wrote:

Am Thu, 27 Feb 2025 14:13:13 -0600 schrieb olcott:

On 2/27/2025 3:50 AM, Fred. Zwarts wrote:

Op 27.feb.2025 om 06:00 schreef olcott:

On 2/26/2025 5:50 PM, Richard Damon wrote:

On 2/26/25 9:49 AM, olcott wrote:

On 2/26/2025 3:48 AM, Fred. Zwarts wrote:

Op 26.feb.2025 om 05:50 schreef olcott:

On 2/25/2025 10:21 PM, Richard Damon wrote:

>

You already know that you are stupidly wrong about the emulation
being incorrect or you would have provided the correct emulation
sequence long ago.

 

What are the first 15 lines of DD correctly emulated by HHH?

The error in the simulation occurs already at the 5th
instruction,
the 'call 000015c3'. Instead of simulating this instruction,

You dodged the actual question.
What are the correct first 15 lines of DD emulated by HHH.

And you are missing the point that HHH can not correctly emulate 15
line of this DD, as it is missing information and thus the task is
improperly defined.

If we assume that HHH emulates the above machine code then there is
NO MORE relevant information needed and we know that machine address
0000213c of DD is followed by machine address 00002133 of DD.

No. Relevant information includes the code at address 0x213c and
everything it points to.

THE ONLY FREAKING THING THAT NEED BE KNOWN ABOUT THE CODE AT THAT
ADDRESS (TO DETERMINE THE SEQUENCE OF MACHINE INSTRUCTIONS OF DD) IS
THAT IT EMULATES THE MACHINE INSTRUCTIONS OF DD.

Well, it doesn't *call* D.

 From this base assumption we know that the first
five instructions of DD are repeated three times when the first 15
instructions of DD are emulated.

No, there is no loop. The repetition only happens in the simulation.
There are only a single instance of HHH and DD actually running, the
rest is just "software", if you will, just like the encoding of a
TM on the tape of an UTM (the hardware is the state table of the UTM,
which it has no reflective access to). The processor running HHH
has no clue what is being simulated; as far as it is concerned, it
is just executing HHH.

If assuming that following the call would lead to infinite recursion
(which is not true, because DD has been proven to halt),

If it was true that the above instance of DD reaches its "ret"
instruction then you could show how it does this.

Easy. It calls a program that returns *by definition* .

^

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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.