Sujet : Re: HHH(DD) --- COMPUTE ACTUAL MAPPING FROM INPUT TO OUTPUT --- Using Finite String Transformations
De : F.Zwarts (at) *nospam* HetNet.nl (Fred. Zwarts)
Groupes : comp.theoryDate : 24. Apr 2025, 21:05:16
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <vue5ht$27hl3$5@dont-email.me>
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User-Agent : Mozilla Thunderbird
Op 24.apr.2025 om 21:47 schreef olcott:
On 4/24/2025 2:15 PM, Fred. Zwarts wrote:
Op 24.apr.2025 om 19:46 schreef olcott:
On 4/24/2025 3:11 AM, Fred. Zwarts wrote:
Op 24.apr.2025 om 05:34 schreef olcott:
On 4/23/2025 7:31 PM, Mike Terry wrote:
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...because HHH stops simulating before reaching that step in the computation. Note that I said
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MT: Both traces were of course /identical/,
*up to the point where HHH stops simulating*
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So I was factually correct.
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Mike.
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It *is not* up to the point where HHH stops simulating.
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It is up to the point where the simulated versus directly
executed calls HHH(DD).
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That is exactly the same point. If not, show the difference in the traces before that point.
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As soon as the directly executed DD calls HHH(DD) this
call immediately returns.
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When DD emulated by HHH calls HHH(DD) then HHH emulates
DD and also emulates itself emulating DD. This is one
whole recursive emulation than the directly executed
DD can possibly get to.
Again a lot of words, which hide that you cannot show where the traces differ up to that point.
THEY DIFFER BY THE EMULATED DD REACHES RECURSIVE EMULATION
AND THE DIRECTLY EXECUTED DD NEVER DOES.
No, they both reach a finite recursion, but the HHH aborts and the direct execution does not abort. That difference, however, is not visible in the trace up to that point.
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