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Am Fri, 06 Jun 2025 12:35:40 -0500 schrieb olcott:HHH1(DDD) simulates DDD that eventually halts.On 6/6/2025 12:26 PM, joes wrote:Am Wed, 04 Jun 2025 21:15:52 -0500 schrieb olcott:On 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: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.Their code is the same and has the same meaning. DDD always calls HHH.
You overlooked this.
I didn't say that it did. Like HHH it simulates HHH simulating DDD.HHH emulates itself emulating DDD>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.
>>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
trace from x86, not what HHH is doing.
>
emulated 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*
HHH1 also does that, and more, because it doesn't abort.
>
HHH1 NEVER emulates itself
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