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Path: ...!weretis.net!feeder8.news.weretis.net!newsfeed.bofh.team!paganini.bofh.team!not-for-mail 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: Tue, 27 Aug 2024 10:55:24 +0300 Organization: To protect and to server Message-ID: <vak0pc$2ldi6$1@paganini.bofh.team> References: <va104l$376ed$4@dont-email.me> <cd375f68f97a737988bab8c1332b7802509ff6ea@i2pn2.org> <va13po$376ed$7@dont-email.me> <d42e5d30ea5f1c067283cb04d8a7293e2117188e@i2pn2.org> <va24hl$3cvgv$1@dont-email.me> <431deaa157cdae1cae73a1b24268a61cf8ec2c1c@i2pn2.org> <va38qh$3ia79$1@dont-email.me> <7a1c569a699e79bfa146affbbae3eac7b91cd263@i2pn2.org> <va3f7o$3ipp3$1@dont-email.me> <729cc551062c13875686d266a5453a488058e81c@i2pn2.org> <va3kac$3nd5c$1@dont-email.me> <148bf4dd91f32379a6d81a621fb7ec3fc1e00db0@i2pn2.org> <va3lai$3nd5c$2@dont-email.me> Mime-Version: 1.0 Content-Type: text/plain; charset=utf-8; format=flowed Content-Transfer-Encoding: 8bit Injection-Info: paganini.bofh.team; logging-data="2799174"; posting-host="ArmERdYYIOOJVi41tgCxGQ.user.paganini.bofh.team"; mail-complaints-to="usenet@bofh.team"; posting-account="9dIQLXBM7WM9KzA+yjdR4A"; User-Agent: Unison/2.2 X-Notice: Filtered by postfilter v. 0.9.3 Bytes: 11116 Lines: 237 On 2024-08-21 03:01:38 +0000, olcott said: > On 8/20/2024 9:53 PM, Richard Damon wrote: >> On 8/20/24 10:44 PM, olcott wrote: >>> On 8/20/2024 8:56 PM, Richard Damon wrote: >>>> On 8/20/24 9:17 PM, olcott wrote: >>>>> 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. >>>> >>>> >>>> Right, so HHH must do just one of them, and the DDD that calls it will >>>> act differently based on which one it is. >>>> >>>>> >>>>> Professor Sipser would not have construed that I am referring >>>>> to self-modifying code. >>>> >>>> Nope, just giving the exact same input to two different version of the >>>> decider, the one that doesn't abort as the hypothetical, and the one >>>> that does as the actual (if the hypothical one doesn't halt). >>>> >>>> The key is that both get the EXACT SAME input, the DDD that calls the >>>> HHH that does abort. >>>> >>>> The point is that your "reasoning" can't actually be done just by >>>> experimentation, (as to experimentally show that the input when ========== REMAINDER OF ARTICLE TRUNCATED ==========