Sujet : Re: HHH(DDD)==0 is correct
De : polcott333 (at) *nospam* gmail.com (olcott)
Groupes : comp.theoryDate : 05. Jul 2025, 17:16:37
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <104bj55$1hqln$13@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
User-Agent : Mozilla Thunderbird
On 7/5/2025 8:01 AM, Richard Damon wrote:
On 7/4/25 6:17 PM, olcott wrote:
On 7/4/2025 3:33 PM, Richard Damon wrote:
On 7/4/25 2:24 PM, olcott wrote:
On 7/4/2025 12:48 PM, Richard Damon wrote:
On 7/4/25 9:32 AM, olcott wrote:
On 7/4/2025 8:22 AM, joes wrote:
Am Fri, 04 Jul 2025 07:50:23 -0500 schrieb olcott:
On 7/4/2025 2:35 AM, Mikko wrote:
On 2025-07-03 12:56:42 +0000, olcott said:
On 7/3/2025 3:57 AM, Mikko wrote:
On 2025-07-03 02:50:40 +0000, olcott said:
On 7/1/2025 11:37 AM, Mr Flibble wrote:
On Mon, 30 Jun 2025 21:12:48 -0400, Richard Damon wrote:
On 6/30/25 2:30 PM, Mr Flibble wrote:
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No. A simulator does not have to run a simulation to completion if
it can determine that the input, A PROGRAM, never halts.
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If. But here it confuses that with not being able to simulate past the
recursive call.
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It is the correct simulation of the input that
specifies the actual behavior of this input.
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Right, and that means correctly simulating EVERY instruction that makes up the PROGRAM, which must include the code of *THE* HHH, or you can't "correctly simulate" the call instruction.
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SIn
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If this simulation cannot simulate past the recursive
call then this correctly simulated input cannot possibly
reach its own correctly simulated final halt state.
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But it can, as the call goes into HHH, and you then just simulate the code of HHH.
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Do you mean like this? (as I have been telling you for three years)
https://liarparadox.org/HHH(DDD)_Full_Trace.pdf
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Yes, so where in that trace does HHH's "correct simulation" differ from the exact same simulation generated by HHH1 before HHH aborts?
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When we compare DDD emulated by HHH and DDD emulated
by HHH1 SIDE-BY-SIDE. (Mike didn't do it this way).
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*The difference is when*
HHH begins to simulate itself simulating DDD and
HHH1 NEVER begins to simulate itself simulating DDD.
And where does that show as a difference in the trace?
"itself" isn't something in the code/
Yes there aren't any machine addresses in x86 code.
x86 code always only works on the basis of its psychic ability.
We can't possibly see that DDD calls the machine address
of HHH because in x86 code there is no such thing as machine
addresses.
-- Copyright 2025 Olcott "Talent hits a target no one else can hit; Geniushits a target no one else can see." Arthur Schopenhauer
Re: ChatGPT agrees that HHH refutes the standard halting problem proof method
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