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

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   Message 242,446 of 243,242   
   Mikko to olcott   
   Re: Very simple first principles showing   
   15 Dec 25 11:20:33   
   
   XPost: comp.theory, sci.logic, comp.lang.c++   
   From: mikko.levanto@iki.fi   
      
   On 15/12/2025 02:15, olcott wrote:   
   > On 12/14/2025 4:46 AM, Mikko wrote:   
   >> On 11/12/2025 16:38, olcott wrote:   
   >>> On 12/11/2025 2:53 AM, Mikko wrote:   
   >>>> olcott kirjoitti 10.12.2025 klo 18.27:   
   >>>>>   
   >>>>> DD() executed from main() calls HHH(DD) thus is   
   >>>>> not one-and-the-same-thing as an argument to HHH.   
   >>>>   
   >>>> If the last sentence is true then this is not the counter exmaple   
   >>>> mentioned in certain proofs of noncomputability of halting and   
   >>>> therefore not relevant in that context. The halting problem reuqires   
   >>>> that HHH can determine whether the counter example halts. That is,   
   >>>> you must be able to replace "???" in   
   >>>>   
   >>>>    #include  // or your replacement   
   >>>>    int main (void)   
   >>>>    {   
   >>>>      int Halt_Status = HHH(???); // put the correct argument here   
   >>>>      printf("HHH says: %s\n", Halt_Status ? "halts" : "does not   
   halt");   
   >>>>      return Halt_Status;   
   >>>>    }   
   >>>>   
   >>>> with whatever specifies the behaviour of DD to HHH. If you can't   
   >>>> do this then HHH is not a halt decider nor a partial halt decider.   
   >>   
   >>> When the halting problem requires a halt decider   
   >>> to report on the behavior of a Turing machine this   
   >>> is always a category error.   
   >>   
   >> That you don't know what "category error" means does not justify your   
   >> claim. Apparently you can't apply definitions.   
   >   
   > Turing machines only compute functions from finite   
   > strings they never compute functions from Turing   
   > machines.   
      
   True, but irrelevant to questions about category errors.   
      
   > A halt decider can at best compute the behavior of   
   > a Turing machine through the proxy of a finite   
   > string machine description it never computes it   
   > directly from another Turing machine.   
   >   
   > Whenever any textbook says that a halt decider   
   > must compute halting for machine M on input w   
   > is it wrong.   
      
   Which textbook actually says "must"? It is not wrong to say "must" in   
   the sense that any decider that does not compute whether machine M   
   halts on input w is not a halt decider. But using "must" is not the   
   clearest way to say it because the word "must" other meanings.   
      
    > It actually computes halting that this input pair specifies (⟨M⟩, w).   
      
   There is an unbalanced parenthesis above.   
      
   --   
   Mikko   
      
   --- SoupGate-Win32 v1.05   
    * Origin: you cannot sedate... all the things you hate (1:229/2)   

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