XPost: comp.theory, comp.lang.c, comp.lang.c++   
   From: polcott333@gmail.com   
      
   On 8/11/2025 2:04 PM, dbush wrote:   
   > On 8/11/2025 3:00 PM, olcott wrote:   
   >> On 8/11/2025 1:51 PM, dbush wrote:   
   >>> On 8/11/2025 2:50 PM, olcott wrote:   
   >>>> On 8/11/2025 1:45 PM, dbush wrote:   
   >>>>> On 8/11/2025 2:40 PM, olcott wrote:   
   >>>>>> On 8/11/2025 1:33 PM, Richard Heathfield wrote:   
   >>>>>>> On 11/08/2025 18:48, olcott wrote:   
   >>>>>>>> On 8/11/2025 11:42 AM, Richard Heathfield wrote:   
   >>>>>>>>> On 11/08/2025 17:29, olcott wrote:   
   >>>>>>>>>   
   >>>>>>>>>    
   >>>>>>>>>   
   >>>>>>>>>>   
   >>>>>>>>>> I have proven that DD correctly simulated by HHH   
   >>>>>>>>>   
   >>>>>>>>> No, you haven't. You have asserted that the simulation is   
   >>>>>>>>> correct, but we all know that it derives a different result   
   >>>>>>>>> from that produced by direct execution, and therefore we all   
   >>>>>>>>> know that the simulation is not correct.   
   >>>>>>>>   
   >>>>>>>> You can only fully know that your assumption is incorrect   
   >>>>>>>> when you notice that no specific error exists.   
   >>>>>>>   
   >>>>>>> The specific error is that you get the wrong answer.   
   >>>>>> It is incorrect to say it *is* a wrong answer until after   
   >>>>>> a specific error in the basis of this answer is found.   
   >>>>>   
   >>>>> It's the wrong answer because it doesn't meet the required   
   >>>>> specification:   
   >>>>>   
   >>>>   
   >>>> That merely presumes that the required specification   
   >>>> is correct.   
   >>>   
   >>>   
   >>> It's correct because I want to know if any arbitrary algorithm X with   
   >>> input Y will halt when executed directly.   
   >>   
   >> Likewise I want to know the radius of a square circle   
   >> that has a length of 2.0 of one of its equal length   
   >> four sides.   
   >   
   > The difference is that every algorithm X with input Y either halts or   
   > does not halt when executed directly, so there is a correct answer in   
   > all cases.   
   >   
      
   Because no Turing machine ever takes another   
   Turing machine as an input no Turing machine can   
   directly compute the mapping from an input that   
   is not in its domain.   
      
   In all but one case input ⟨M⟩ to simulating halt   
   decider H has the exact same behavior machine M.   
      
   M.q0 ⟨M⟩ ⊢* M.H ⟨M⟩ ⟨M⟩ ⊢* M.qy   
   M.q0 ⟨M⟩ ⊢* M.H ⟨M⟩ ⟨M⟩ ⊢* M.qn   
   (a) M copies its input ⟨M⟩   
   (b) M invokes M.H ⟨M⟩ ⟨M⟩   
   (c) M.H simulates ⟨M⟩ ⟨M⟩   
      
   When H is embedded within M then the behavior of   
   ⟨M⟩ ⟨M⟩ simulated by M.H cannot possibly reach a final   
   halt state, even if there is no loop trick at the end.   
      
   The halting problem definition assumes that this case   
   does not exist. Thus they used the short-hand that   
   H applied to ⟨M⟩ ⟨M⟩ is reporting on M applied to ⟨M⟩.   
      
      
   --   
   Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius   
   hits a target no one else can see." Arthur Schopenhauer   
      
   --- SoupGate-Win32 v1.05   
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