• Re: Simulating halt deciders correct decider halting [ The only point t

    From olcott@21:1/5 to Ben Bacarisse on Wed Mar 16 20:51:57 2022
    XPost: comp.theory, sci.logic, sci.math

    On 3/16/2022 8:36 PM, Ben Bacarisse wrote:
    olcott <NoOne@NoWhere.com> writes:

    On 3/16/2022 8:16 PM, Ben Bacarisse wrote:
    olcott <NoOne@NoWhere.com> writes:

    On 3/12/2022 8:55 PM, Ben Bacarisse wrote:
    André G. Isaak <agisaak@gm.invalid> writes:

    On 3/12/2022 5:57 PM, André G. Isaak wrote:

    So what string, according to you, encodes the computation Ĥ applied >>>>>>>> to ⟨Ĥ⟩? If these two "different" computations don't have separate >>>>>>>> encodings as strings then they are not, in fact, different
    computations at all.

    No Comment?

    I know you've been asked this question before and have consistently >>>>>> ignored it. According to a recent post of yours that constitutes
    justification for a repetitive all-caps temper tantrum!
    I once tried to get a direct answer to this question. I asked 12 times >>>>> in consecutive posts but never got one.
    Later, on the related question of whether ⟨Ĥ⟩ ⟨Ĥ⟩ encodes a halting
    computation I got this dazzling display of equivocation:
    "When it is construed as input to H then ⟨Ĥ⟩ ⟨Ĥ⟩ encodes a halting
    computation.
    When it is construed as input to Ĥ.qx then ⟨Ĥ⟩ ⟨Ĥ⟩ DOES NOT encode a
    halting computation."
    Bear in mind that at time, PO's machines were magic: two identical state >>>>> transition functions could entail transitions to different states when >>>>> presented with identical inputs. He has since backed off from some of >>>>> these remarks, but it never exactly clear which previous claims he would >>>>> now accept were wrong.

    None-the-less...

    You mean you won't comment on the above but would rather present new
    junk about BASIC. Oh well... I can't stop you.

    None-the-less none of what you have ever said shows that I am
    incorrect.

    Ah! So from your point of view I did not point out an error in the post
    you replied to. That means you /still/ think that:

    "When it is construed as input to H then ⟨Ĥ⟩ ⟨Ĥ⟩ encodes a halting
    computation. When it is construed as input to Ĥ.qx then ⟨Ĥ⟩ ⟨Ĥ⟩ DOES
    NOT encode a halting computation."


    THIS IS THE ONLY POINT THAT MATTERS
    THIS IS THE ONLY POINT THAT MATTERS
    THIS IS THE ONLY POINT THAT MATTERS
    THIS IS THE ONLY POINT THAT MATTERS

    That the simulated input: ⟨Ĥ⟩ ⟨Ĥ⟩ to embedded_H would never reach the final state of this simulated input in any finite number of steps of
    correct simulation by embedded_H conclusively proves that a mapping from
    this input to the reject state of embedded_H is correct.

    That's useful to know. Either your strings or your TMs are still
    magic. Sadly, that means you have nothing to say about the halting
    problem for Turing machines. You need to fix that before anyone can
    take you seriously.



    --
    Copyright 2021 Pete Olcott

    Talent hits a target no one else can hit;
    Genius hits a target no one else can see.
    Arthur Schopenhauer

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  • From olcott@21:1/5 to Richard Damon on Wed Mar 16 21:22:26 2022
    XPost: comp.theory, sci.logic, sci.math

    On 3/16/2022 9:03 PM, Richard Damon wrote:
    On 3/16/22 9:51 PM, olcott wrote:
    On 3/16/2022 8:36 PM, Ben Bacarisse wrote:
    olcott <NoOne@NoWhere.com> writes:

    On 3/16/2022 8:16 PM, Ben Bacarisse wrote:
    olcott <NoOne@NoWhere.com> writes:

    On 3/12/2022 8:55 PM, Ben Bacarisse wrote:
    André G. Isaak <agisaak@gm.invalid> writes:

    On 3/12/2022 5:57 PM, André G. Isaak wrote:

    So what string, according to you, encodes the computation Ĥ >>>>>>>>>> applied
    to ⟨Ĥ⟩? If these two "different" computations don't have separate
    encodings as strings then they are not, in fact, different >>>>>>>>>> computations at all.

    No Comment?

    I know you've been asked this question before and have consistently >>>>>>>> ignored it. According to a recent post of yours that constitutes >>>>>>>> justification for a repetitive all-caps temper tantrum!
    I once tried to get a direct answer to this question.  I asked 12 >>>>>>> times
    in consecutive posts but never got one.
    Later, on the related question of whether ⟨Ĥ⟩ ⟨Ĥ⟩ encodes a halting
    computation I got this dazzling display of equivocation:
         "When it is construed as input to H then ⟨Ĥ⟩ ⟨Ĥ⟩ encodes a
    halting
          computation.
          When it is construed as input to Ĥ.qx then ⟨Ĥ⟩ ⟨Ĥ⟩ DOES NOT
    encode a
          halting computation."
    Bear in mind that at time, PO's machines were magic: two
    identical state
    transition functions could entail transitions to different states >>>>>>> when
    presented with identical inputs.  He has since backed off from
    some of
    these remarks, but it never exactly clear which previous claims
    he would
    now accept were wrong.

    None-the-less...

    You mean you won't comment on the above but would rather present new >>>>> junk about BASIC.  Oh well...  I can't stop you.

    None-the-less none of what you have ever said shows that I am
    incorrect.

    Ah!  So from your point of view I did not point out an error in the post >>> you replied to.  That means you /still/ think that:

       "When it is construed as input to H then ⟨Ĥ⟩ ⟨Ĥ⟩ encodes a halting
       computation.  When it is construed as input to Ĥ.qx then ⟨Ĥ⟩ ⟨Ĥ⟩ DOES
       NOT encode a halting computation."


    THIS IS THE ONLY POINT THAT MATTERS
    THIS IS THE ONLY POINT THAT MATTERS
    THIS IS THE ONLY POINT THAT MATTERS
    THIS IS THE ONLY POINT THAT MATTERS

    That the simulated input: ⟨Ĥ⟩ ⟨Ĥ⟩ to embedded_H would never reach the
    final state of this simulated input in any finite number of steps of
    correct simulation by embedded_H conclusively proves that a mapping
    from this input to the reject state of embedded_H is correct.

    Except that 'correct simulation by embedded_H' only has meaning if
    embedded_H is actually
    embedded_H has all of the functionality of a UTM and is able to
    (a) Perfectly simulate its input as if it was a UTM.
    (b) Watch the behavior of this simulated input.
    (c) Match the behavior of this simulated input with infinite behavior
    patterns.

    As soon as an infinite behavior pattern is correctly matched embedded_H
    has complete proof that its input would never reach the final state of
    this input, thus the input never halts even if aborted.

    Since we can see that there is an infinite behavior pattern we can see
    that a transition to the embedded_H reject state would be correct.



    This by itself refutes the Linz proof because the Linz proof concludes
    that Ĥ applied to ⟨Ĥ⟩ results in a contradiction. (see direct quote below) and there is no actual contradiction.

    </Linz:1990:320>
    Now Ĥ is a Turing machine, so that it will have some description in Σ*,
    say ŵ . This string, in addition to being the description of Ĥ can also
    be used as input string. We can therefore legitimately ask what would
    happen if Ĥ is applied to ŵ .

    The contradiction tells us that our assumption of the existence of H,
    and hence the assumption of the decidability of the halting problem,
    must be false.
    </Linz:1990:320>



    --
    Copyright 2021 Pete 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
    * Origin: fsxNet Usenet Gateway (21:1/5)