Re: Indirect Reference Changes the Behavior of DDD() relative to DDD emulated by HHH

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Sujet : Re: Indirect Reference Changes the Behavior of DDD() relative to DDD emulated by HHH
De : F.Zwarts (at) *nospam* HetNet.nl (Fred. Zwarts)
Groupes : comp.theory
Date : 08. Sep 2024, 11:31:38
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <vbjuec$1sml7$3@dont-email.me>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29
User-Agent : Mozilla Thunderbird
Op 07.sep.2024 om 16:00 schreef olcott:
On 9/7/2024 5:19 AM, Fred. Zwarts wrote:
Op 06.sep.2024 om 13:31 schreef olcott:
On 9/6/2024 4:36 AM, Fred. Zwarts wrote:
Op 05.sep.2024 om 15:48 schreef olcott:
>
HHH MUST ABORT AFTER SOME FIXED NUMBER OF RECURSIVE EMULATIONS
AND THE OUTERMOST HHH ALWAYS SEE ONE MORE THAN THE NEXT INNER ONE.
>
And the outer one, when aborting after two cycles , misses the behaviour of the inner one in the next cycle, where the inner one would see the 'special condition', abort, return to DDD, which would halt as well.
That HHH misses the last part of the behaviour of the program, does not change the fact that this is the behaviour that was coded in the program
>
>
If we have an infinite chain of people each waiting for
the next one down the line to do something then that thing
is never done.
>
The infinite chain exists only in your dream. In fact there are only two recursions, so never more that a chain of three HHH in the simulation.
HHH is incorrect in assuming the there is an infinite chain, but this incorrect assumption makes that it aborts and halts. This applies both to the simulating and the simulated HHH.
>
The way it is encoded now there are only two recursions.
>
If we encode it as you suggest the outermost directly
executed HHH would wait for the first emulated HHH which
would wait for the second which would wait for third
on and on...
>
>
What is olcott's problem with English?
If one way is incorrect, he thinks that it suggests that another way must be correct.
I never suggested to change HHH, because there is *no* correct way to do it. Every HHH that simulates itself is incorrect. No matter what clever code it includes.
 You must be a brain dead moron.
Ad hominem attack without evidence. Does olcott really think that it will convince anyone?

As long as HHH emulates the sequence of instructions
it was provided then HHH is correct even if it catches
your computer on fire.
HHH only made a correct start of the simulation, but failed to reach the end of the halting program.
There is no need to set the computer in fire to halt a program, if the program halts of its own.
No matter how clever HHH is coded, HHH cannot possibly simulate itself correctly up to the end and there is no way to fix it.
Even olcott himself has repeatedly proven that HHH cannot reach the end of its own simulation, but he is unable to accept his own proofs.

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