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   comp.lang.c      Meh, in C you gotta define EVERYTHING      243,242 messages   

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   Message 241,995 of 243,242   
   Kaz Kylheku to olcott   
   Re: D simulated by H cannot possibly rea   
   19 Nov 25 04:30:52   
   
   XPost: comp.theory, sci.logic, comp.lang.c++   
   From: 643-408-1753@kylheku.com   
      
   On 2025-11-19, olcott  wrote:   
   > On 11/18/2025 8:53 PM, Kaz Kylheku wrote:   
   >> On 2025-11-19, olcott  wrote:   
   >>> On 11/18/2025 7:01 PM, Kaz Kylheku wrote:   
   >>>> On 2025-11-18, olcott  wrote:   
   >>>>> On 11/18/2025 3:21 PM, Kaz Kylheku wrote:   
   >>>>>> On 2025-11-18, olcott  wrote:   
   >>>>>>> If you ask a decider to determine if my   
   >>>>>>> sister's name is "Sally" and I don't tell   
   >>>>>>> it who I am then the information contained   
   >>>>>>> in the input is insufficient. This does not   
   >>>>>>> in any way limit computation itself.   
   >>>>>>   
   >>>>>> The problem is that UTM(D) can work out the fact that   
   >>>>>> D halts. Why is it that UTM knows that D's sister's   
   >>>>>> name is Sally, but H does not?   
   >>>>>>   
   >>>>>   
   >>>>> UTM(D) is answering a different question.   
   >>>>> (a) It is not providing any answer at all.   
   >>>>   
   >>>> Well, of course, by "UTM" we mean a /decider/ that purely simulates:   
   >>>>   
   >>>>    bool UTM(ptr P) {   
   >>>>      sim S = sim_create(P);   
   >>>>      sim_step_exhaustively(S);   
   >>>>      return true;   
   >>>>    }   
   >>>>   
   >>>> All deciders applied to D are tasked with answering exactly the same   
   >>>> question.   
   >>>>   
   >>>> Pretending that a different question was asked is nonproductive;   
   >>>> the answer will be interpreted to the original question.   
   >>>>   
   >>>> All the information needed to answer is positively contained in D.   
   >>>>   
   >>>> It is just too complex relative to H.   
   >>>>   
   >>>   
   >>> What The F does UTM decide when DD calls UTM(DD)?   
   >>   
   >> That doesn't happen; DD calls HHH(DD).   
   >>   
   >> A diagonal functon set against UTM, call it DDUTM,   
   >> cannot be decided by UTM(DDUTM).   
   >>   
   >> That call simply does not return.   
   >>   
   >   
   > Yes, and the other one does return proving the   
   > whole point that I have been making for three   
   > years that everyone (besides Ben) was too damned   
   > dishonest to acknowledge has been true all along.   
      
   What "other one"? Is that referring to HHH(DD)?   
      
   HHH(DD) returns; UTM(DDUTM) does not return.   
      
   That's four functions; HHH isn't UTM; DD isn't DDUTM.   
      
   HHH and DDUTM are unrelated; UTM and DD are unrelated.   
      
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