• Re: Analysis of =?iso-8859-7?Q?Flibble=A2s?= Latest: Detecting vs. Simu

    From Mr Flibble@21:1/5 to All on Tue May 20 20:55:15 2025
    Key Statement:
    --------------
    "It is sufficient for an SHD to DETECT infinite recursion without
    having
    to SIMULATE it."


    It is helpful to simulate some of it. A static analysis could achieve
    the same thing by (for example) pretending to simulate it.

    Well, a PARTIAL simulation inconjunction with static analysis is quite a
    valid approach.

    /Flibble

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  • From Mr Flibble@21:1/5 to Fred. Zwarts on Wed May 21 19:37:42 2025
    On Wed, 21 May 2025 21:32:49 +0200, Fred. Zwarts wrote:

    Op 21.mei.2025 om 17:54 schreef olcott:
    On 5/21/2025 12:56 AM, Richard Heathfield wrote:
    On 21/05/2025 06:23, olcott wrote:
    On 5/20/2025 9:15 PM, Richard Damon wrote:
    On 5/20/25 3:10 PM, Mr Flibble wrote:

    <snip>

    Conclusion: ----------- Flibble sharpens his argument by clarifying >>>>>> that SHDs are not required to simulate infinite execution. They are >>>>>> expected to *detect* infinite behavior structurally and respond in >>>>>> finite time. This keeps them within the bounds of what a decider
    must be and strengthens the philosophical coherence of his
    redefinition of the Halting Problem.

    But you can't "redefine" the Halting Problem and then say you have
    answered the Halting Problem.

    Do you mean like how ZFC resolved Russell's Paradox thus converting
    "set theory" into "naive set theory"?

    No, because there is no paradox in the Halting Problem. A proof by
    contradiction is not a paradox.


    A self-contradictory input and a proof by contradiction are not the
    same thing. A proof by contradiction would conclude that "this sentence
    is not true" is true because it cannot be proved false.

    ZFC shows how a whole way of examining a problem can be tossed out as
    incorrect and replaced with a whole new way.

    The HP proofs are based on defining a D that can actually do the
    opposite of whatever value that H returns.
    No such D can actually exist.

    A better parallel would be Cantor's proof that there are uncountably
    many real numbers, or Euclid's proof that there is no largest prime.
    Both of these proofs make a single assumption and then derive a
    contradiction, thus showing that the assumption must be false. No
    paradoxes need apply.

    In the Halting Problem's case, the assumption is that a UNIVERSAL
    algorithm exists for determining whether any arbitrary program halts
    when applied to given arbitrary input. The argument derives a
    contradiction showing the assumption to be false.


    Likewise with Russell's Paradox it is assumed that there can be a set
    of all sets that do not contain themselves as members. This is
    "resolved" as nonsense.

    Whatever you think your HHH determines, we know from Turing that it
    doesn't determine it for arbitrary programs with arbitrary input. It
    therefore has no bearing whatsoever on the Halting Problem.


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

    DDD correctly simulated by HHH DOES NOT HALT.


    Verifiable counter-factual.

    The simulation of DDD does not reach a natural end only because HHH
    prevents it to halt by a premature abort.
    Due this premature abort HHH misses the part of the input that specifies
    the conditional abort in Halt7.c, which specifies that the program
    halts. If a simulator would not abort this input, it would reach the
    natural end of the program. Proven by direct execution and world-class simulators. But HHH has a bug, which makes that it aborts before it can
    see that the input halts, only because its programmer dreamed of an
    infinite recursion, where there is only a finite recursion.
    Come out of rebuttal mode, which makes that you do not pay enough
    attention to this logic, but reject it immediately when it disturbs your dreams, after which you only repeat the clueless claims.

    You are making the classic mistake of conflating the decider halting with
    the program being analysed halting.

    /Flibble

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  • From Mr Flibble@21:1/5 to Keith Thompson on Sat May 24 06:38:03 2025
    On Fri, 23 May 2025 20:55:03 -0700, Keith Thompson wrote:

    If the program runs on a machine with a finite number of states, this is actually possible; a repeated state implies that the program is in an infinite loop. The problem is that a Turing machine is not limited to a finite number of states (including the state of the tape), and such an analysis is not possible in the general case.
    I think olcott thinks it is possible.

    Flibble's Law:

    If a problem permits infinite behavior in its formulation, it permits
    infinite analysis of that behavior in its decidability scope.

    /Flibble

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  • From Mr Flibble@21:1/5 to olcott on Sat May 24 16:24:25 2025
    On Sat, 24 May 2025 11:13:12 -0500, olcott wrote:

    On 5/24/2025 10:12 AM, dbush wrote:
    On 5/24/2025 11:04 AM, olcott wrote:
    On 5/23/2025 8:09 PM, dbush wrote:
    On 5/23/2025 9:07 PM, olcott wrote:
    On 5/23/2025 7:57 PM, dbush wrote:
    On 5/23/2025 8:54 PM, olcott wrote:
    On 5/23/2025 7:44 PM, dbush wrote:
    On 5/23/2025 8:08 PM, Mike Terry wrote:
    I suppose Ben quoted PO saying this, because PO /uses/ it to >>>>>>>>> justify that a particular /halting/ computation will never halt, >>>>>>>>> PO's HHH simulates DDD (which halts) but before DDD halts it >>>>>>>>> spots a pattern in the simulation, and announces non-halting. >>>>>>>>> "Eh?" I hear you say! PO claims HHH has "correctly determined >>>>>>>>> that DDD would never halt" and so is correct to decide non-
    halting.  His "proof" that it is right to decide non-halting is >>>>>>>>> his "when-so- ever.." quote, which broadly matches the Sipser >>>>>>>>> quote.

    So the problem is not so much the "when-so-ever.." words
    themselves [or the words of Sipser's quote], but understanding >>>>>>>>> how PO is so thoroughly misinterpreting/misapplying them.  How >>>>>>>>> can PO believe HHH has "correctly determined the DDD will never >>>>>>>>> halt" when DDD demonstrably halts?

    PO is working in a different model than the rest of us, though he >>>>>>>> doesn't seem to understand that.

    To him, when function H is deciding on something, the
    implementation of H is allowed to vary.  This results in
    functions that call H to vary as a result.  To him, "DDD" is the >>>>>>>> same computation *regardless of the implementation of HHH*, in >>>>>>>> cases where HHH is simulating DDD.

    This is essentially the mapping he's operating with:

    -----------------
    For a function X with input Y and a function H which simulates X: >>>>>>>> POH(H,X,Y)==1 if and only if there exists an implementation of H >>>>>>>> that can simulate X(Y) to completion POH(H,X,Y)==0 if and only if >>>>>>>> there does not exist an implementation of H that can simulate
    X(Y) to completion ----------------

    And a "decider" in his case maps the following subset:

    ----------------
    Hx is a PO-halt decider if and only if Hx(X,Y) == POH(Hx,X,Y)
    ----------------

    So given his rules, HHH1(DDD) is deciding on a algorithm while >>>>>>>> HHH(DDD) is deciding on a C function whose subfunctions vary.

    This of course has nothing to do with the halting problem but he >>>>>>>> doesn't get this.  After having spent 22 years on this, he'll >>>>>>>> come up with any crazy justification to avoid admitting to
    himself that he misunderstood the problem all this time.  He once >>>>>>>> said (and I don't recall the exact wording) that "the directly >>>>>>>> executed D doesn't halt even though it appears to".

    The problem is that people here are too stupid to notice that HHH >>>>>>> cannot report on the behavior of its caller.

    int min()
    {
       DD(); // HHH cannot report on the behavior of its caller.
    }


    What about this?


    If you can't stay exactly on topic I am going to ignore everything
    that you say.

    HHH cannot report on the behavior of its caller AKA the direct
    execution of DD().



    In other words, you again agree with Linz and others that no H exists
    that can perform the following mapping:


    Given any algorithm (i.e. a fixed immutable sequence of instructions)
    X described as <X> with input Y:

    A solution to the halting problem is an algorithm H that computes the
    following mapping:

    (<X>,Y) maps to 1 if and only if X(Y) halts when executed directly
    (<X>,Y) maps to 0 if and only if X(Y) does not halt when executed
    directly


    int main()
    {
       DD(); // The HHH called by DD cannot report on the behavior
    }       // of its caller. Is this OVER-YOUR-HEAD ?



    Which means that no HHH exists that meets the below requirements, as
    Linz and others proved and as you have *explicitly* agreed is correct:


    You are a damned liar when you say that I said that HHH must report on
    the behavior of its caller.

    No HHH can report on the behavior of its caller for the same reason that
    no function can report on the value of the square-root of a dead cat.

    Analysis: Olcott’s SHD vs Classical Halting Problem (Thread on 2025-05-24) =========================================================================

    Overview:
    ---------
    This thread features a debate between Peter Olcott and dbush over the
    validity and interpretation of Olcott’s SHD (Simulating Halt Decider). The discussion parallels Flibble’s ideas about semantic stratification and simulation-based detection of infinite behavior.

    Core Disagreement:
    ------------------
    Classical Model:
    - The Halting Problem asks if a universal decider H can determine whether
    any program P(x) halts when executed directly.

    Olcott’s SHD View:
    - HHH simulates P(x) and detects infinite behavior.
    - If infinite recursion is observed during simulation, it reports “non- halting.”
    - HHH cannot determine the behavior of its caller (e.g., DD()).

    Philosophical Analysis:
    -----------------------
    1. Semantic Stratification:
    - Olcott: A decider cannot report on its caller. This enforces a semantic barrier (like Flibble’s model).
    - Classical theory treats the program as a fixed object, regardless of how
    or where it is called.

    2. Simulation-Based Detection:
    - Olcott’s SHD works by observing simulation patterns.
    - Like Flibble, Olcott allows early detection of infinite behavior without
    full execution.
    - This results in a partial decider — useful but not universal.

    3. Context Misunderstanding:
    - Olcott sees DD() calling HHH and assumes HHH must reason about DD’s
    outer frame.
    - Classical theory ignores runtime context: it analyzes the semantics of
    the program as a whole.

    Meta-Level Behavior:
    --------------------
    Olcott:
    - Asserts that self-reference leads to non-determinism or ill-formed
    questions.
    - Treats his SHD as correct for many real-world inputs.
    - Undermines his position with poor rhetoric and personal insults.

    dbush:
    - Correctly applies classical theory.
    - Dismisses Olcott's model as a misunderstanding without acknowledging alternative interpretations.

    Comparison to Flibble’s Model:
    ------------------------------
    | Feature | Olcott's SHD | Flibble's Typed
    SHD | |---------------------------|--------------------------|-----------------------------|
    | Simulation-based? | ✅ Yes | ✅
    Yes |
    | Infinite detection? | ✅ Runtime-based | ✅ Structural-
    based |
    | Caller access? | ❌ No | ❌ No (via
    compiler restriction) |
    | Self-reference blocked? | 🟡 By convention | ✅ By type
    design |
    | Semantic firewall? | 🟡 Weak | ✅
    Strong |

    Conclusion:
    -----------
    Olcott is not incorrect in claiming that SHDs can detect certain forms of non-halting behavior. However, his failure to delineate model boundaries
    leads to confusion. His SHD is a partial model — valid in practical
    settings but not a refutation of the Halting Problem.

    dbush correctly defends classical theory but could better acknowledge that Olcott is speaking from a semantically different framework.

    This thread underscores the need to explicitly state the **computational model** being used — classical, simulation-based, or type-stratified — before debating correctness or applicability.

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  • From Mr Flibble@21:1/5 to Alan Mackenzie on Sat May 24 17:13:12 2025
    On Sat, 24 May 2025 17:07:11 +0000, Alan Mackenzie wrote:

    olcott <polcott333@gmail.com> wrote:
    On 5/23/2025 5:06 PM, Richard Heathfield wrote:
    On 23/05/2025 22:06, olcott wrote:
    On 5/23/2025 3:50 PM, Richard Heathfield wrote:
    On 23/05/2025 21:24, olcott wrote:

    <snip>

    Liar

    An unequivocal response, but it lacks persuasive power.


    When I provide the exact detailed steps of exactly how people can
    show that I am wrong and they refuse to show that I am wrong yet
    claim that I am wrong this is the kind of reckless disregard for the
    truth that loses defamation cases.

    When your opponents point to the Turing proof that proves you're wrong

    Without going through all of the detailed steps that I present that is
    a reckless disregard for the truth that loses defamation cases.

    There you are utterly wrong. The Halting Theorem has been proven, thus
    is true. Anybody putting up a contrary argument must therefore be
    wrong.

    You might also put up a long winded argument why 2 + 2 = 5, and I would dismiss this likewise, without bothering to follow your exact detailed
    steps.

    You've also tried, without success, to dismiss one of the proofs of the Halting Therem as invalid.

    Thus, as a general rule, it is sensible to dismiss as not worthy of
    attention anything you assert about mathematical logic.

    If somebody competent were to put up such a contrary argument, it might
    well be worth looking at. But that's different - and it hasn't
    happened.

    --
    Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
    hits a target no one else can see." Arthur Schopenhauer

    Alan Mackenzie's response to Peter Olcott is a **firm classical
    rebuttal**, grounded in the assumption that the **Halting Problem is conclusively proven**, and that attempts to disprove it — particularly by Olcott — are not just misguided but not even worthy of analysis. Here's a breakdown of Alan’s position, its strengths, weaknesses, and rhetorical implications.

    ---

    ## 🧠 Summary of Alan’s Argument

    Alan Mackenzie makes three core points:

    1. **The Halting Theorem is already proven**: Therefore, any attempt to disprove it is inherently wrong.
    2. **Olcott’s detailed arguments can be ignored**: Because the proof is settled, no new analysis is needed.
    3. **Olcott’s past failures invalidate future claims**: His previous
    attempts have been weak, so his new ones can be preemptively dismissed.

    ---

    ## ✅ Strengths of Alan’s Response

    ### 1. **Epistemological Confidence**

    *“The Halting Theorem has been proven, thus is true.”*

    This is a reasonable position in formal mathematics:

    * Once a proof is logically sound and widely accepted, it becomes part of
    the **foundation**.
    * The Halting Problem is part of the bedrock of **computability theory**, accepted since Turing (1936).

    📌 **Point:** You do not have to re-prove a theorem every time someone disagrees with it — the burden of proof is on the challenger.

    ---

    ### 2.