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Failed to connect to MySQL: (1203) User howardkn already has more than 'max_user_connections' active connectionsPath: ...!eternal-september.org!feeder3.eternal-september.org!news.eternal-september.org!eternal-september.org!.POSTED!not-for-mail From: dbush Newsgroups: comp.theory Subject: Re: Incorrect requirements --- Computing the mapping from the input to HHH(DD) Date: Sat, 10 May 2025 11:01:13 -0400 Organization: A noiseless patient Spider Lines: 131 Message-ID: References: <87r00xchn5.fsf@nosuchdomain.example.com> <23a27379d226b7b3b9f8c303a492f66edc9019ff.camel@gmail.com> <1020d30c2c5b5a7cce584777131d5ce414b480ea.camel@gmail.com> <0323d5ca6d757a1e35d7e4cf5eb4fc8f41bc866a.camel@gmail.com> <9f5774bfb493325652f97d72f760ad98442c333d.camel@gmail.com> <84b09a0d53d77e2a8fddf567226d05c0d65e60c0.camel@gmail.com> <235107421488220fb79fa83cbe8bf44709f6445a.camel@gmail.com> MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8; format=flowed Content-Transfer-Encoding: 8bit Injection-Date: Sat, 10 May 2025 17:01:13 +0200 (CEST) Injection-Info: dont-email.me; posting-host="ceabab99a32dd3b60042c0bcdabf6faa"; logging-data="3777206"; mail-complaints-to="abuse@eternal-september.org"; posting-account="U2FsdGVkX1+dNUzLdrMMnMLKnVMA10F9" User-Agent: Mozilla Thunderbird Cancel-Lock: sha1:ZM4EQKXOqdThGHqnY1TRO6EuqV4= Content-Language: en-US In-Reply-To: Bytes: 6550 On 5/10/2025 10:51 AM, olcott wrote: > On 5/10/2025 1:19 AM, wij wrote: >> On Sat, 2025-05-10 at 01:06 -0500, olcott wrote: >>> On 5/10/2025 1:00 AM, wij wrote: >>>> On Sat, 2025-05-10 at 00:41 -0500, olcott wrote: >>>>> On 5/10/2025 12:27 AM, wij wrote: >>>>>> On Sat, 2025-05-10 at 00:19 -0500, olcott wrote: >>>>>>> On 5/10/2025 12:13 AM, wij wrote: >>>>>>>> On Sat, 2025-05-10 at 00:06 -0500, olcott wrote:>> >>>>>>>>> When mathematical mapping is properly understood >>>>>>>>> it will be known that functions computed by models >>>>>>>>> of computation must transform their input into >>>>>>>>> outputs according to the specific steps of an >>>>>>>>> algorithm. >>>>>>>>> >>>>>>>>> _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] >>>>>>>>> >>>>>>>>> For example HHH(DDD) only correctly map to the >>>>>>>>> behavior that its input actually specifies by correctly >>>>>>>>> emulating DDD according to the rules of the x86 language. >>>>>>>>> >>>>>>>>> This causes the first four instructions of DDD >>>>>>>>> to be emulated followed by HHH emulating itself >>>>>>>>> emulating the first three instructions of DDD. >>>>>>>>> >>>>>>>>> It is right at this recursive simulation just >>>>>>>>> before HHH(DDD) is called again that HHH recognizes >>>>>>>>> the repeating pattern and rejects DDD. >>>>>>>> >>>>>>>> Yes, but you still did not answer the question: Is POOH exactly >>>>>>>> about HP? >>>>>>>> >>>>>>> >>>>>>>     >>>>> H(D)=1 if D() halt. >>>>>>>     >>>>> H(D)=0 if D() not halt. >>>>>>> >>>>>>> Right now it is mostly about proving the >>>>>>> above requirements are is mistaken. >>>>>>> >>>>>> >>>>>> Why is the requirement invalid? >>>>>> >>>>>> H(D)=1 if D() halt. >>>>>> H(D)=0 if D() not halt. >>>>>> >>>>> >>>> >>>>> The notion that the behavior specified by the finite >>>>> string input to a simulating termination analyzer >>>> >>>> POOH reads(takes) its input as a function, not 'finite string'. >>>> Are you talking about POOH now? There is no POOH that takes >>>> 'finite string'. >>>> >>> >>> It a finite string of x86 bytes. >> >> Disagree. >> The D in Halt7.c (I just saw once) does not treat H as 'finite string', >> D calls H. H also does not treat D as 'finite string'. >> > > HHH and DDD and DD are the most recent functions. > HHH does emulate its finite strings of x86 machine code > according to the rules of the x86 language. > >>>>> does sometimes differ from the behavior of its direct >>>>> execution. It is a provably different sequence of steps. >>>> >>> >>> This is a verified fact. >>> The pathological relationship that inputs can have >>> with their simulating termination analyzer changes >>> the behavior of these inputs relative to their direct >>> execution. >> >> So, you redefined the halting problem should be about the behavior of D >> decided by POOH, not the 'direct' behavior of D? >> > > Not at all. HHH(DDD) reports on the behavior that DDD > actually specifies. Which is the behavior of the algorithm described by the input, which halts. > All my critics require HHH to report > on behavior that DDD does not actually specify. > > The actual behavior that DDD specifies is measured > by Executing DDD directly, as per the requirements: Given any algorithm (i.e. a fixed immutable sequence of instructions) X described as with input Y: A solution to the halting problem is an algorithm H that computes the following mapping: (,Y) maps to 1 if and only if X(Y) halts when executed directly (,Y) maps to 0 if and only if X(Y) does not halt when executed directly > int sum(int x, int y) { return x + y ;} > sum must apply the rules of arithmetic to > its inputs thus sum(3,2) cannot correctly > derive the sum of 5 + 7. > > HHH must apply the rules of the x86 language > to its input DDD. And those rules say that when DDD is actually run on an actual x86 processor that it will halt. > > When HHH1(DDD) applies these same rules to its input > there is no recursive emulation because DDD does > not call HHH1(DDD). >