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From: dbush <dbush.mobile@gmail.com>
Newsgroups: comp.theory
Subject: =?UTF-8?Q?Re=3A_Analysis_of_Flibble=E2=80=99s_Latest=3A_Detecting_v?=
=?UTF-8?Q?s=2E_Simulating_Infinite_Recursion_ZFC?=
Date: Sat, 24 May 2025 12:33:11 -0400
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On 5/24/2025 12:13 PM, 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.
>
Nope:
On 3/24/2025 10:07 PM, olcott wrote:
> A halt decider cannot exist
On 4/28/2025 2:47 PM, olcott wrote:
> On 4/28/2025 11:54 AM, dbush wrote:
>> And the halting function below is not a computable function:
>>
>
> It is NEVER a computable function
>
>> 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
On 3/14/2025 1:19 PM, olcott wrote:
> When we define the HP as having H return a value
> corresponding to the halting behavior of input D
> and input D can actually does the opposite of whatever
> value that H returns, then we have boxed ourselves
> in to a problem having no solution.
On 6/21/2024 1:22 PM, olcott wrote:
> the logical impossibility of specifying a halt decider H
> that correctly reports the halt status of input D that is
> defined to do the opposite of whatever value that H reports.
> Of course this is impossible.
On 7/4/2023 12:57 AM, olcott wrote:
> If you frame the problem in that a halt decider must divide up finite
> strings pairs into those that halt when directly executed and those that
> do not, then no single program can do this.
On 5/5/2025 5:39 PM, olcott wrote:
> On 5/5/2025 4:31 PM, dbush wrote:
>> Strawman. The square root of a dead rabbit does not exist, but the
>> question of whether any arbitrary algorithm X with input Y halts when
>> executed directly has a correct answer in all cases.
>>
>
> It has a correct answer that cannot ever be computed
On 5/13/2025 5:16 PM, olcott wrote:
> There is no time that we are ever going to directly
> encode omniscience into a computer program. The
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