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On 5/7/2025 8:19 PM, olcott wrote:Does there exist an HHH such that DDD emulated byOn 5/7/2025 7:15 PM, dbush wrote:Category error. There is no "can" as algorithm HHH is fixed and immutable, as is algorithm DDD.On 5/7/2025 7:40 PM, olcott wrote:>On 5/7/2025 6:31 PM, dbush wrote:>On 5/7/2025 7:26 PM, olcott wrote:>>>
When N instructions of DD are emulated by HHH
according to the rules of the x86 language then
The subject was "DD emulated by HHH", not "N instructions of DD emulated by HHH".
>
Changing the subject is the dishonest dodge of the strawman deception.
>
That you and Richard construe anything less than an
infinite number of steps of DD emulated by HHH
(according to the rules of the x86 language)
as an incorrect emulation IS MORONICALLY STUPID.
>
The fixed immutable code of HHH simulates a fixed number X of instructions of DD, the last of which was simulated incorrectly. Any number other than X is not what HHH simulates and is therefore irrelevant to HHH.
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UTM simulates X+Y instruction of DD correctly and reaches a final state.
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I will make it easier to understand.
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void DDD()
{
HHH(DDD);
return;
}
>
Can DDD simulated by HHH reach its own "return" instruction?
>
Algorithm HHH *does not* simulate algorithm DDD to the end but instead aborts in violation of the x86 language.there is no end to reach.
Algorithm UTM *does* simulate algorithm DDD to the end.--
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