Path: news.eternal-september.org!eternal-september.org!.POSTED!not-for-mail
From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory
Subject: Re: DDD specifies recursive emulation to HHH and halting to HHH1
Date: Sun, 30 Mar 2025 21:17:14 -0500
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On 3/30/2025 7:35 PM, Richard Damon wrote:
> On 3/30/25 5:56 PM, olcott wrote:
>> On 3/30/2025 4:05 PM, Richard Damon wrote:
>>> On 3/30/25 4:32 PM, olcott wrote:
>>>> 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.
>>>>
>>>> THAT IS WHAT IT SAYS AND ANYONE THAT DISAGREES
>>>> IS A DAMNED LIAR OR STUPID.
>>>>
>>>
>>> How is that DDD correctly emulated beyond the call HHH instruction by 
>>> a program that is a pure function, and thus only looks at its input?
>>>
>>
>> *THE ENTIRE SCOPE IS*
>> DDD EMULATED BY HHH DOES SPECIFY THAT IT
>> CANNOT POSSIBLY REACH ITS OWN FINAL HALT STATE.
> 
>  From where? Remember, the Halting problem is SPECIFICALLY 

OFF F-CKING TOPIC. WE ABOUT ONE F-CKING STEP OF MY PROOF.
WE HAVE BEEN TALKING ABOUT ONE F-CKING STEP OF MY PROOF
FOR THREE F-CKING YEARS.

DDD correctly emulated by HHH DOES NOT F-CKING HALT !!!


-- 
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer