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From: Mikko <mikko.levanto@iki.fi>
Newsgroups: comp.theory
Subject: Re: DDD specifies recursive emulation to HHH and halting to HHH1
Date: Tue, 1 Apr 2025 09:25:55 +0300
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On 2025-03-31 18:06:35 +0000, olcott said:

> On 3/31/2025 3:47 AM, Mikko wrote:
>> On 2025-03-30 20:32:07 +0000, olcott said:
>> 
>>> On 3/30/2025 1:52 PM, Richard Damon wrote:
>>>> On 3/30/25 2:27 PM, olcott wrote:
>>>>> On 3/30/2025 3:12 AM, joes wrote:
>>>>>> Am Sat, 29 Mar 2025 16:46:26 -0500 schrieb olcott:
>>>>>>> On 3/29/2025 3:14 PM, dbush wrote:
>>>>>>>> On 3/29/2025 4:01 PM, olcott wrote:
>>>>>> 
>>>>>>>>> We can know that when this adapted UTM simulates a finite number of
>>>>>>>>> steps of its input that this finite number of steps were simulated
>>>>>>>>> correctly.
>>>>>>>> And therefore does not do a correct UTM simulation that matches the
>>>>>>>> behavior of the direct execution as it is incomplete.
>>>>>>> It is dishonest to expect non-terminating inputs to complete.
>>>>>> A complete simulation of a nonterminating input doesn't halt.
>>>>>> 
>>>>>>>>>> 2) changing the input is not allowed
>>>>>>>>> The input is unchanged. There never was any indication that the input
>>>>>>>>> was in any way changed.
>>>>>>>> False, if the starting function calls UTM and UTM changes, you're
>>>>>>>> changing the input.
>>>>>>> When UTM1 is a UTM that has been adapted to only simulate a finite
>>>>>>> number of steps
>>>>>> So not an UTM.
>>>>>> 
>>>>>>> and input D calls UTM1 then the behavior of D simulated
>>>>>>> by UTM1 never reaches its final halt state.
>>>>>>> When D is simulated by ordinary UTM2 that D does not call Then D reaches
>>>>>>> its final halt state.
>>>>>> Doesn't matter if it calls it, but if the UTM halts.
>>>>>> 
>>>>>>>> Changing the input is not allowed.
>>>>>>> I never changed the input. D always calls UTM1.
>>>>>>> thus is the same input to UTM1 as it is to UTM2.
>>>>>> You changed UTM1, which is part of the input D.
>>>>>> 
>>>>> 
>>>>> UTM1 simulates D that calls UTM1
>>>>> simulated D NEVER reaches final halt state
>>>>> 
>>>>> UTM2 simulates D that calls UTM1
>>>>> simulated D ALWAYS reaches final halt state
>>>>> 
>>>> 
>>>> Only because UTM1 isn't actually a UTM, but a LIE since it only does a 
>>>> partial simulation, not a complete as REQURIED by the definition of a 
>>>> UTM.
>>>> 
>>> 
>>> _DDD()
>>> [00002172] 55         push ebp      ; housekeeping
>>> [00002173] 8bec       mov  ebp,esp  ; housekeeping
>>> [00002175] 6872210000 push 00002172 ; push DDD
>>> [0000217a] e853f4ffff call 000015d2 ; call HHH(DDD)
>>> [0000217f] 83c404     add  esp,+04
>>> [00002182] 5d         pop  ebp
>>> [00002183] c3         ret
>>> Size in bytes:(0018) [00002183]
>>> 
>>> DDD EMULATED BY HHH DOES SPECIFY THAT IT
>>> CANNOT POSSIBLY REACH ITS OWN FINAL HALT STATE.
>> 
>> No, it does not. HHH misintepretes, contrary to the semantics of x86,
>> the specification to mean that.
> 
> It is a truism that a correct x86 emulator
> would emulate itself emulating DDD whenever
> DDD calls this emulator with itself.

Irrelevant. You didn't say anything about a correct emulator or emulation.

-- 
Mikko