• Re: Every sufficiently competent C programmer knows --- Paraphrase ofSipser's agreement

    From olcott@polcott333@gmail.com to comp.theory,comp.lang.c,comp.lang.c++ on Sat Mar 15 16:27:00 2025
    From Newsgroup: comp.lang.c++

    On 3/15/2025 5:12 AM, Mikko wrote:
    On 2025-03-14 14:39:30 +0000, olcott said:

    On 3/14/2025 4:03 AM, Mikko wrote:
    On 2025-03-13 20:56:22 +0000, olcott said:

    On 3/13/2025 4:22 AM, Mikko wrote:
    On 2025-03-13 00:36:04 +0000, olcott said:


    void DDD()
    {
       HHH(DDD);
       return;
    }

    int DD()
    {
       int Halt_Status = HHH(DD);
       if (Halt_Status)
         HERE: goto HERE;
       return Halt_Status;
    }

    When HHH correctly emulates N steps of the
    above functions none of them can possibly reach
    their own "return" instruction and terminate normally.

    Nevertheless, assuming HHH is a decider, Infinite_Loop and
    Infinite_Recursion
    specify a non-terminating behaviour, DDD specifies a terminating
    behaviour

    _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]

    What is the sequence of machine language
    instructions of DDD emulated by HHH such that DDD
    reaches its machine address 00002183?

    Irrelevant off-topic distraction.

    Proving that you don't have a clue that Rice's Theorem
    is anchored in the behavior that its finite string input
    specifies. The depth of your knowledge is memorizing
    quotes from textbooks.

    Another irrelevant off-topic distraction, this time involving
    a false claim.

    One can be a competent C programmer without knowing anyting about Rice's Theorem.


    YES.

    Rice's Theorem is about semantic properties in general, not just
    behaviours.
    The unsolvability of the halting problem is just a special case.


    A property about Turing machines can be represented as the language of
    all Turing machines, encoded as strings, that satisfy that property. http://kilby.stanford.edu/~rvg/154/handouts/Rice.html

    Does THE INPUT TO simulating termination analyzer
    HHH encode a C function that reaches its "return"
    instruction [WHEN SIMULATED BY HHH] (The definition
    of simulating termination analyzer) ???

    <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

    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>

    <Accurate Paraphrase>
    If emulating termination analyzer H emulates its input
    finite string D of x86 machine language instructions
    according to the semantics of the x86 programming language

    until H correctly determines that this emulated D cannot
    possibly reach its own "ret" instruction in any finite
    number of correctly emulated steps then

    H can abort its emulation of input D and correctly report
    that D specifies a non-halting sequence of configurations.
    </Accurate Paraphrase>

    Memorizing quotes from textbooks is useful for practical purposes but
    if it is too hard for you there are other ways.


    The whole X represents TM X that halts on its input is inaccurate.
    If you did not merely learn-by-rote you would already know this.

    *Input Y to TM Z specifies a TM that halts when directly measured by Z*
    --
    Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
    hits a target no one else can see." Arthur Schopenhauer
    --- Synchronet 3.20c-Linux NewsLink 1.2
  • From olcott@polcott333@gmail.com to comp.theory,comp.lang.c,comp.lang.c++ on Sat Mar 15 20:28:27 2025
    From Newsgroup: comp.lang.c++

    On 3/15/2025 5:35 PM, dbush wrote:
    On 3/15/2025 5:27 PM, olcott wrote:

    <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

         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>


    But he didn't agree to what you think he did:



    THIS IS EXACTLY AND PRECISELY WHAT I THINK HE DID AGREE TO
    (It took me years to form this precise paraphrase)

    <Accurate Paraphrase>
    If emulating termination analyzer H emulates its input
    finite string D of x86 machine language instructions
    according to the semantics of the x86 programming language
    until H correctly determines that this emulated D cannot
    possibly reach its own "ret" instruction in any finite
    number of correctly emulated steps then

    H can abort its emulation of input D and correctly report
    that D specifies a non-halting sequence of configurations.
    </Accurate Paraphrase>
    --
    Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
    hits a target no one else can see." Arthur Schopenhauer
    --- Synchronet 3.20c-Linux NewsLink 1.2