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computers / comp.ai.philosophy / Re: Concise refutation of halting problem proofs V27 [ input is not in domain ]

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* Concise refutation of halting problem proofs V27 [ finallyolcott
`* Re: Concise refutation of halting problem proofs V27 [ finallyolcott
 `* Re: Concise refutation of halting problem proofs V27 [ input is notolcott
  `- Re: Concise refutation of halting problem proofs V27 [ input is notolcott

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Concise refutation of halting problem proofs V27 [ finally mathematically precise ]

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Subject: Concise refutation of halting problem proofs V27 [ finally
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 by: olcott - Mon, 22 Nov 2021 20:32 UTC

#include <stdint.h>
#include <stdio.h>
typedef int (*ptr)();

int H(ptr x, ptr y)
{ x(y); // direct execution of P(P)
return 1;
}

// Minimal essence of Linz(1990) Ĥ
// and Strachey(1965) P
int P(ptr x)
{ H(x, x);
return 1; // Give P a last instruction at the "c" level
}

int main(void)
{ H(P, P);
}

Computation that halts
a computation is said to halt whenever it enters a final state.
(Linz:1990:234)

[PSR_set] Combinations of H/P having pathological self-reference
For every H of H(P,P) where P(P) calls this same H(P,P) and H simulates
or executes its input and aborts or does not abort its input.
∀H ∊ PSR_set ∀P ∊ PSR_set (Input_Never_Halts(H(P,P)))
[PSR_subset_A] ∃H ∊ PSR_set ∃P ∊ PSR_set (Halts( H(P,P) ))
[PSR_subset_B] ∃H ∊ PSR_set ∃P ∊ PSR_set (Halts( P(P) ))

[PSR_subset_C] The subset of the PSR_subset_A where H returns 0 on the
basis that H correctly detects that its input never halts (reaches its
final instruction). H could detect that its simulated P is calling
H(P,P) with the same parameters that it was called with, thus specifying
infinite recursion.

H is a computable function that accepts or rejects inputs in its domain
on the basis that these inputs specify a sequence of configurations that
reach their final state.
H is a correct decider and H has a correct halt deciding basis.

*(see page 3)*
Halting problem undecidability and infinitely nested simulation V2

https://www.researchgate.net/publication/356105750_Halting_problem_undecidability_and_infinitely_nested_simulation_V2)

--
Copyright 2021 Pete Olcott

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

Re: Concise refutation of halting problem proofs V27 [ finally mathematically precise ]

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 by: olcott - Tue, 23 Nov 2021 03:59 UTC

On 11/22/2021 7:44 PM, André G. Isaak wrote:
> On 2021-11-22 18:16, olcott wrote:
>> On 11/22/2021 7:07 PM, André G. Isaak wrote:
>>> On 2021-11-22 17:13, olcott wrote:
>>>> On 11/22/2021 5:16 PM, André G. Isaak wrote:
>>>
>>>>> And this bit is particularly mystifying:
>>>>>
>>>>>  > PSR_set: for n = 0 to ∞
>>>>>  > {
>>>>>  >    (Input_Never_Halts(  Hn(Pn,Pn)  ))
>>>>>  >    (Sometimes_Halts(  Hn(Pn,Pn)  ))
>>>>>  >    (Sometimes_Halts(   Pn(Pn)  ))
>>>>>  > }
>>>>>
>>>>> André
>>>>>
>>>>>
>>>>
>>>> Numbered elements of the infinite set of finite string C encodings
>>>> of H and P. The source-code that I provided is the (H0, P0) element.
>>>
>>> That wasn't the mystifying part despite it being a strange abuse of
>>> syntax. I assumed this was just a deranged way of writing ∀n rather
>>> than being a syntactically ill-formed C program, though why you
>>> continuously insist on inventing your own undefined notation is
>>> beyond me.
>>>
>>> How can the input of Hn(Pn, Pn) never halt when Pn(Pn) sometimes halt?
>>
>> The input to Hn(Pn,Pn) never reaches its final state.
>> For some of these exact same Pn(Pn) P does reach its final state.
>
> Which goes back to Richard's question which you refused to answer: What
> does Input_Never_Halts mean?
>
> The input to a halt decider is a *string*. The input can't halt. It also
> can't not halt. Only the computation which it describes has a halting
> status, and in the case of Hn(Pn, Pn) the input describes the
> computation Pn(Pn).
>
> You keep talking about what the 'input' does, but unless by 'input' you
> mean 'the computation described by the input' this is a completely
> undefined and, as far as I can tell, meaningless notion. Halting is a
> property of *computations*, not of strings.
>
> So what does Input_Never_Halts mean?
>
> What does it mean for an input to halt or not halt?
>

It turns out that the independent execution of P(P) actually means an
input to H(P,P) that *is not* an input to H(P,P).

It turns out that the independent execution of P(P) means that main()
directly executes P(P) without H being involved.

No deciders can actually work that way. All deciders either accept or
reject finite string inputs.

H is a computable function that accepts or rejects inputs in its domain
on the basis that these inputs specify a sequence of configurations that
reach their final state.

What all you guys have been asking for is for H to have an input that
*is not* its input.

The x86 equivalent of a TM description and its input is an x86 machine
language string and some other finite string or references to these.

>>
>>> And if Hn is a halt decider, how can it only sometimes halt?
>>>
>>
>> I am using categorically exhaustive reasoning examining the behavior
>> of every possible Hn(Pn,Pn) by examining the categories of possible
>> behavior.
>>
>> Only a subset of Hn(Pn,Pn) is H a decider and in only a subset of
>> these is H a correct halt decider.
>>
>>> And what does it even mean for something to sometimes halt? For any
>>> given value of n it either halts or it doesn't.
>
> Not going to answer this, I take it?
>
> André
>

--
Copyright 2021 Pete Olcott

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

Re: Concise refutation of halting problem proofs V27 [ input is not in domain ]

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 by: olcott - Wed, 24 Nov 2021 03:24 UTC

On 11/23/2021 8:22 PM, André G. Isaak wrote:
> On 2021-11-23 18:53, olcott wrote:
>> On 11/23/2021 7:34 PM, André G. Isaak wrote:

> Crucially, the halting function exists *independently* and *prior* to
> any Turing Machine which attempts to compute this function. The halting
> function is *not* associated with any sort of algorithm; it is simply a
> mapping. In other words it is simply a denumerably infinite list of all
> possible computations and their halting status expressed as ordered pairs.
>

You don't realize it but you are asking a decider to make its decision
on the basis of a hypothetical string that does not actually exist.

int H(ptr x, ptr y) { x(y); }
int P(ptr x) { H(x, x); }
int H1(ptr x, ptr y) { x(y); }

The input to H requires that P call this very same H.
The input to H1 has no such pathological relationship to P.

By simply ignoring the pathological relationship between H and P you are
answering the wrong question.

You are answering the question:
What would the behavior of P be if P called H1 instead of calling H?

That makes the "input" to H hypothetical rather than an actual string.

> The computation P(P) will occur in this list exactly once, as will be
> true for all other computation. Since your P(P) halts, it will map to
> 'halts'.
>
> A halt decider, given <P> <P> as its input, is tasked with determining
> what the entry for that computation in the list described above actually
> is in a finite number of steps. If <P> <P> maps to 'halts' in the
> halting function, then a halt decider must accept it. If it maps to
> 'doesn't halt', it must reject it.
>
> That means that the halting value of P(P) is *not* specific to a given
> halt decider. Your H and H1 must both provide the same answer if they
> are both halt deciders. It means that it is non-sensical to talk about
> the halting value of P(P) 'inside' some H, since H's job is to determine
> what value this computation maps to in the halting FUNCTION which is
> independent of any halt decider.
>
> André
>

--
Copyright 2021 Pete Olcott

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

Re: Concise refutation of halting problem proofs V27 [ input is not in domain ]

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 by: olcott - Thu, 25 Nov 2021 04:54 UTC

On 11/24/2021 9:50 AM, André G. Isaak wrote:
> On 2021-11-24 08:02, olcott wrote:
>> On 11/24/2021 12:27 AM, André G. Isaak wrote:
>>> On 2021-11-23 22:48, olcott wrote:
>>>> On 11/23/2021 10:18 PM, André G. Isaak wrote:
>>>>> On 2021-11-23 21:03, olcott wrote:
>>>>>> On 11/23/2021 9:44 PM, André G. Isaak wrote:
>>>>>>> On 2021-11-23 20:24, olcott wrote:
>>>>>>>> On 11/23/2021 8:22 PM, André G. Isaak wrote:
>>>>>>>>> On 2021-11-23 18:53, olcott wrote:
>>>>>>>>>> On 11/23/2021 7:34 PM, André G. Isaak wrote:
>>>>>>>>
>>>>>>>>> Crucially, the halting function exists *independently* and
>>>>>>>>> *prior* to any Turing Machine which attempts to compute this
>>>>>>>>> function. The halting function is *not* associated with any
>>>>>>>>> sort of algorithm; it is simply a mapping. In other words it is
>>>>>>>>> simply a denumerably infinite list of all possible computations
>>>>>>>>> and their halting status expressed as ordered pairs.
>>>>>>>>>
>>>>>>>>
>>>>>>>>
>>>>>>>> You don't realize it but you are asking a decider to make its
>>>>>>>> decision on the basis of a hypothetical string that does not
>>>>>>>> actually exist.
>>>>>>>
>>>>>>> ???
>>>>>>>
>>>>>>> So now you claim the STRING <P> <P> can't exist? But at the same
>>>>>>> time you claim that your H1 returns the correct answer for this
>>>>>>> allegedly impossible string? How on earth does that work.
>>>>>>>
>>>>>>
>>>>>> int H(ptr x, ptr y) { x(y); }
>>>>>> int H1(ptr x, ptr y) { x(y); int P(ptr x) { H(x, x); }
>>>>>> int P1(ptr x) { H1(x, x); }
>>>>>
>>>>>
>>>>> What does any of this have to do with your ridiculous claim that
>>>>> the string <P> <P> does not actually exist?
>>>>
>>>> If you want the actual behavior of P(P) run independently
>>>> then you have to change P or H into something that they are not
>>>> because they are defined with hardwired inter-dependency.
>>>
>>> If you want the actual behaviour of P(P) run independently, you
>>> simply run P(P) independently.
>>>
>>
>> So in other words H must report on the behavior of a finite string
>> that is not is input.
>
> H doesn't report on the behaviour of finite strings at all. It reports
> on the behaviour of computations.
>
> P(P) is a computation which halts.
>
> Your H needs to be able to report that P(P) halts (or any other
> arbitrary computation).
>
> Determining exactly how that particular computation must be encoded as a
> finite string so it can be passed as an input to H is part of your job
> when designing H.
>
> Maybe you don't grasp this. Consider two UTMs, X and Y (and by UTM I
> mean actual UTMs). Both of these should be able to emulate P(P) if we
> give them two copies of a string which describes the Turing Machine P.
>
> But the strings we pass to X and the string we pass to Y are *not*
> necessarily the same string. X and Y may use different alphabets and may
> use different encoding schemes. To constitute a UTM, it must be the
> case, though, that it is possible to represent every possible
> computation as a string that can be passed to that UTM.
>
> If you've designed an H such that there is no possible way to represent
> the independent computation P(P) as a string which can be passed to H as
> an input, then there is something wrong with your design.
>
>>> I think what you mean to say is if you want H to give you the correct
>>> answer about the actual behaviour of P(P) run independently, which it
>>> can't. The interdependency prevents this.
>>>
>>
>> It is incorrect to think of H having the hypothetical input where P
>> does not call H.All deciders must have finite string inputs. Because P
>> foes
>
> Where did I (or anyone) mention a P that does not simulate/call H?
>
> H is required to describe the halting behaviour of the independent
> computation P(P). That means it must describe how P(P) behaves when
> called directly from main rather than from within your halt decider.
>
>> call this same H changing P so that it does not call this same H is
>> incorrect.
>>
>> It is equally incorrect to have H report on the behavior of P as if H
>> was H1 where P does not call H1.
>>
>>> But that does not change the fact that the computation P(P) halts, and
>>
>> The fact there there are no cats in the kitchen does not have one damn
>> thing to do with dogs in the living room. I can't imagine how people
>> can so persistently make this strawman error even after being warned
>> dozens of times.
>>
>> #include <stdint.h>
>> #include <stdio.h>
>> typedef int (*ptr)();
>>
>> int H(ptr x, ptr y)
>> {
>>    x(y); // direct execution of P(P)
>>    return 1;
>> }
>>
>> // Minimal essence of Linz(1990) Ĥ
>> // and Strachey(1965) P
>> int P(ptr x)
>> {
>>    H(x, x);
>>    return 1; // Give P a last instruction at the "c" level
>> }
>>
>> int main(void)
>> {
>>    H(P, P);
>> }
>>
>> Combinations of (H,P) having pathological self-reference (PSR)
>>    H(P,P) simulates or executes its input and aborts or does
>>    not abort its input and P(P) calls this same H(P,P) with itself.
>>
>> for all Hn(Pn,Pn) in the above PSR set Pn never reaches its final
>> instruction.
>>
>>
>>> that this therefore is the only correct answer to the question which
>>> a halt decider is required to answer.
>>
>> That is not the freaking way that deciders work.
>
> A decider accepts or rejects all possible inputs based on some criteria
> determined by the problem at hand.
>
> For the halting problem, the criterion is that the decider must accept
> all and only those independent computations which HALT.
>
>
>> If H(P,P) reports on the behavior of H1(P,P) or H(P1,P1) it freaking
>> answered the wrong freaking question.
>>
>> Those are the only pair of combinations that show P(P) as an
>> independent computation and they are both answers to the wrong question.
>
> NEITHER of those show P(P) as an independent computation.
>
> H(P,P) must report on the behaviour of P(P). That's the *only* behaviour
> it is being asked about.
>
>>
>> The one key verified fact that everyone simply assumes away even
>> though it is a verified fact is that there are elements of the PSR set
>> such that the input to Hn(Pn,Pn) never reaches its final state and
>> Pn(Pn) does reach its final state in this exact same sequence of
>> configurations.
>>
>> People assume that the placement in the execution trace can't make a
>> difference and yet it is a verified fact that this placement does make
>> a difference. The one way dependency relationship that Pn(Pn) has on
>> Hn(Pn,Pn) causes the outer Pn to behave differently than the inner Pn.
>>
>> Simply assuming this away is flat out dishonest.
>
> None of this is relevant (or even coherent). H(P, P) by the definition
> of the problem must report the behaviour of the independent computation
> P(P). Whether the "placement in the execution trace can make a
> difference" has no bearing on the behaviour of P(P). It might have some
> bearing on the behaviour of the simulation of P(P) inside of H, but
> that's not what the halt decider is being asked about.

The sequence of 1 to N configurations specified by the input to H(X, Y)
cannot be correctly construed as anything other than the sequence of 1
to N steps of the (direct execution, x86 emulation or UTM simulation of
this input by H.

--
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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