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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: Wed, 2 Apr 2025 12:09:19 +0300
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On 2025-04-01 23:31:23 +0000, olcott said:

> On 4/1/2025 1:25 AM, Mikko wrote:
>> 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.
> 
> Sure all trolls would agree that when-so-ever a statement
> is made many dozens of time this proves that this statement
> was never said.

Trolls don't care what was said. But I do. My comment was about your words
I quoted. Your response was not about my or your quoted words. Instead you
talked obout something else as trolls typically do.

-- 
Mikko