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Am Wed, 04 Jun 2025 21:15:52 -0500 schrieb olcott:HHH emulates itself emulating DDDOn 6/4/2025 8:48 PM, Richard Damon wrote:On 6/4/25 10:52 AM, olcott wrote:On 6/4/2025 1:54 AM, Mikko wrote:On 2025-06-03 19:57:09 +0000, olcott said:On 6/3/2025 2:37 AM, Mikko wrote:On 2025-06-02 15:52:53 +0000, olcott said:Their code is the same and has the same meaning. DDD always calls HHH.The DDD emulated (correctly or otherwise) by HHH is the same DDD as>
the one emulated (correctly or otherwise) so both specify the same
behaviour.
No they do not. When DDD calls its own emulator its behavior is
different than when DDD calls another different emulator.
If the input string does not unambiguously specify one and only oneThe code proves what it proves.
behaviour it is incorrectly encoded and not a valid input string. The
halting problem of Truing machines requires that every pair of a
Turing macnine and input is descibed so that the behaviour to be
decided about is the only behaviour that meets to the description.
>HHH1 also does that, and more, because it doesn't abort.So what "simulation" is the above? It seems that you are showing aWhat I am showing is DDD emulated by HHH1 side-by-side with DDD emulated
trace from x86, not what HHH is doing.
>
by HHH
>
*They initially match up*
DDD emulated by HHH1 DDD emulated by HHH [00002183] push
ebp [00002183] push ebp [00002184] mov ebp,esp
[00002184] mov ebp,esp [00002186] push 00002183 ; DDD [00002186] push
00002183 ; DDD [0000218b] call 000015c3 ; HHH [0000218b] call
000015c3 ; HHH *The matching is now all used up*
>
*Then DDD emulated by HHH does something*
*that DDD emulated by HHH1 never does*
*it emulates DDD all over again*
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