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From: Mikko <mikko.levanto@iki.fi>
Newsgroups: comp.theory
Subject: Re: Anyone that disagrees with this is not telling the truth --- V5 --- Professor Sipser
Date: Wed, 21 Aug 2024 10:54:04 +0300
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On 2024-08-21 01:17:44 +0000, olcott said:

> On 8/20/2024 7:50 PM, Richard Damon wrote:
>> On 8/20/24 7:28 PM, olcott wrote:
>>> On 8/20/2024 6:18 PM, Richard Damon wrote:
>>>> On 8/20/24 9:09 AM, olcott wrote:
>>>>> On 8/19/2024 11:02 PM, Richard Damon wrote:
>>>>>> On 8/19/24 11:50 PM, olcott wrote:
>>>>>>> On 8/19/2024 10:32 PM, Richard Damon wrote:
>>>>>>>> On 8/19/24 10:47 PM, olcott wrote:
>>>>>>>>> *Everything that is not expressly stated below is*
>>>>>>>>> *specified as unspecified*
>>>>>>>> 
>>>>>>>> Looks like you still have this same condition.
>>>>>>>> 
>>>>>>>> I thought you said you removed it.
>>>>>>>> 
>>>>>>>>> 
>>>>>>>>> void DDD()
>>>>>>>>> {
>>>>>>>>>    HHH(DDD);
>>>>>>>>>    return;
>>>>>>>>> }
>>>>>>>>> 
>>>>>>>>> _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]
>>>>>>>>> 
>>>>>>>>> *It is a basic fact that DDD emulated by HHH according to*
>>>>>>>>> *the semantics of the x86 language cannot possibly stop*
>>>>>>>>> *running unless aborted* (out of memory error excluded)
>>>>>>>> 
>>>>>>>> But it can't emulate DDD correctly past 4 instructions, since the 5th 
>>>>>>>> instruciton to emulate doesn't exist.
>>>>>>>> 
>>>>>>>> And, you can't include the memory that holds HHH, as you mention HHHn 
>>>>>>>> below, so that changes, but DDD, so the input doesn't and thus is CAN'T 
>>>>>>>> be part of the input.
>>>>>>>> 
>>>>>>>> 
>>>>>>>>> 
>>>>>>>>> X = DDD emulated by HHH∞ according to the semantics of the x86 language
>>>>>>>>> Y = HHH∞ never aborts its emulation of DDD
>>>>>>>>> Z = DDD never stops running
>>>>>>>>> 
>>>>>>>>> The above claim boils down to this: (X ∧ Y) ↔ Z
>>>>>>>> 
>>>>>>>> And neither X or Y are possible.
>>>>>>>> 
>>>>>>>>> 
>>>>>>>>> x86utm takes the compiled Halt7.obj file of this c program
>>>>>>>>> https://github.com/plolcott/x86utm/blob/master/Halt7.c
>>>>>>>>> Thus making all of the code of HHH directly available to
>>>>>>>>> DDD and itself. HHH emulates itself emulating DDD.
>>>>>>>> 
>>>>>>>> Which is irrelevent and a LIE as if HHHn is part of the input, that 
>>>>>>>> input needs to be DDDn
>>>>>>>> 
>>>>>>>> And, in fact,
>>>>>>>> 
>>>>>>>> Since, you have just explicitly introduced that all of HHHn is 
>>>>>>>> available to HHHn when it emulates its input, that DDD must actually be 
>>>>>>>> DDDn as it changes.
>>>>>>>> 
>>>>>>>> Thus, your ACTUAL claim needs to be more like:
>>>>>>>> 
>>>>>>>> X = DDD∞ emulated by HHH∞ according to the semantics of the x86 language
>>>>>>>> Y = HHH∞ never aborts its emulation of DDD∞
>>>>>>>> Z = DDD∞ never stops running
>>>>>>>> 
>>>>>>>> The above claim boils down to this: (X ∧ Y) ↔ Z
>>>>>>>> 
>>>>>>> 
>>>>>>> Yes that is correct.
>>>>>> 
>>>>>> So, you only prove that the DDD∞ that calls the HHH∞ is non-halting.
>>>>>> 
>>>>>> 
>>>>>> Not any of the other DDDn
>>>>>> 
>>>>>>> 
>>>>>>>> Your problem is that for any other DDDn / HHHn, you don't have Y so you 
>>>>>>>> don't have Z.
>>>>>>>> 
>>>>>>>>> 
>>>>>>>>> void EEE()
>>>>>>>>> {
>>>>>>>>>    HERE: goto HERE;
>>>>>>>>> }
>>>>>>>>> 
>>>>>>>>> HHHn correctly predicts the behavior of DDD the same
>>>>>>>>> way that HHHn correctly predicts the behavior of EEE.
>>>>>>>>> 
>>>>>>>> 
>>>>>>>> Nope, HHHn can form a valid inductive proof of the input.
>>>>>>>> 
>>>>>>> 
>>>>>>>> It can't for DDDn, since when we move to HHHn+1 we no longer have DDDn 
>>>>>>>> but DDDn+1, which is a different input.
>>>>>>>> 
>>>>>>> 
>>>>>>> You already agreed that (X ∧ Y) ↔ Z is correct.
>>>>>>> Did you do an infinite trace in your mind?
>>>>>> 
>>>>>> But only for DDD∞, not any of the other ones.
>>>>>> 
>>>>>>> 
>>>>>>> If you can do it and I can do it then HHH can
>>>>>>> do this same sort of thing. Computations are
>>>>>>> not inherently dumber than human minds.
>>>>>>> 
>>>>>> 
>>>>>> But HHHn isn't given DDD∞ as its input, so that doesn't matter.
>>>>>> 
>>>>>> HHHn is given DDDn as its input,
>>>>>> 
>>>>>> Remeber, since you said that the input to HHH includes all the memory, 
>>>>>> if that differs, it is a DIFFERENT input, and needs to be so marked.
>>>>>> 
>>>>>> You are just admittig that you are just stupid and think two things 
>>>>>> that are different are the same.
>>>>>> 
>>>>>> 
>>>>> 
>>>>> *attempts to use misdirection to weasel word around this are dismissed*
>>>>> *attempts to use misdirection to weasel word around this are dismissed*
>>>>> *attempts to use misdirection to weasel word around this are dismissed*
>>>>> 
>>>>> <MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
>>>>>      If simulating halt decider H correctly simulates its input D
>>>>>      until H correctly determines that its simulated D would never
>>>>>      stop running unless aborted then
>>>>> 
>>>>> 
>>>> 
>>>> Right, so the decider needs top be able to show that its exact input 
>>>> will not halt.
>>> 
>>> No it cannot possibly mean that or professor Sipser
>>> would not agreed to the second half:
>>> 
>>>      H can abort its simulation of D and correctly report that D
>>>      specifies a non-halting sequence of configurations.
>>> </MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
>>> 
>>> 
>> 
>> Of course it means that, because Professoer Sipser would have presumed 
>> that you built the machines PROPERLY, so that you COULD think of 
>> changing THIS H to be non-aborting, while the input still used the 
>> final version that it always uses,
>> 
> 
> A machine cannot both abort and fail to abort an input
> unless it modifies its own code dynamically.

Even if it does it cannot do both. If it aborts it does not fail to abort.
However, at one point of execution it may decide to continue and then
later it may consider again.

> Professor Sipser would not have construed that I am referring
> to self-modifying code.

The agreed text does not exclude self-modifying code.

> This means that he must have understood that HHHn(DDD)
> is predicting the behavior of HHH∞.

No, that is clearly outside of the agreement. Of course, if the
behaviour HHH∞ is the same as the behavour of D then as a
byproduct that becames determined, too.

> You continue to use the screwy reasoning that because
> you are no longer hungry after you have eaten this
> proves that you never had to eat.
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