On Tuesday, May 3, 2022 at 10:31:21 AM UTC-4, olcott wrote:
On 5/3/2022 7:12 AM, wij wij wrote:
Richard Damon 在 2022年5月3日 星期二上午8:48:57 [UTC+8] 的信中寫道：It is an easily verified fact that P(P) and the correct simulation of
On 5/2/22 8:35 PM, olcott wrote:
On 5/2/2022 5:47 PM, Mr Flibble wrote:Thus you have shown you don't even know what a "Definition" is, so it is >>>> impossible for you to reason by the meaning of the words.
On Mon, 2 May 2022 18:46:00 -0400
Richard Damon wrote:
On 5/2/22 6:38 PM, Mr Flibble wrote:
On Mon, 2 May 2022 18:32:16 -0400
Richard Damon wrote:
On 5/2/22 11:47 AM, Mr Flibble wrote:
Not all infinitely recursive definitions are invalid however >>>>>>>>>> infinitely recursive definitions that arise out of a category >>>>>>>>>> error (as is the case with the halting problem) are invalid. >>>>>>>>>>
The halting problem (as currently defined) is invalid due to the >>>>>>>>>> invalid "impossible program" [Strachey, 1965] that is actually >>>>>>>>>> impossible due to the category error present in its definition and >>>>>>>>>> *not* because of any function call-like recursion; confusion >>>>>>>>>> between these two types of recursion are why Olcott is having >>>>>>>>>> difficulty communicating his ideas with the rest of you shower. >>>>>>>>>>
The categories involved in the category error are the decider and >>>>>>>>>> that which is being decided. Currently extant attempts to
conflate the decider with that which is being decided are
infinitely recursive and thus invalid.
/Flibble
Except that the "impossible program" isn't part of the definition >>>>>>>>> of the Halting Problem.
It is according to [Wikipedia, 2022].
/Flibble
Nope, you comprehend worse that PO.
Note, and Encyclopedic entery, like Wikipedia, is NOT just a
definition but a full article explaining the subject.
Maybe if you look for a FORMAL source, that states what is the ACTUAL >>>>>>> definition, you would learn something.
If Wikipedia is wrong then correct it and have your corrections
reviewed; until then please shut the fuck up.
/Flibble
I think that the problem is that Richard has disagreeably as his highest >>>>> priority, thus doesn't really give a rat's ass for the truth. An
An impossible program C. Strachey
The Computer Journal, Volume 7, Issue 4, January 1965, Page 313,
Published: 01 January 1965
https://academic.oup.com/comjnl/article/7/4/313/354243
It is very common knowledge that the Wikipedia description is true and >>>>> this is affirmed in Sipser.
For any program f that might determine if programs halt, a
"pathological" program g, called with some input, can pass its own
source and its input to f and then specifically do the opposite of what >>>>> f predicts g will do. https://en.wikipedia.org/wiki/Halting_problem
Now we construct a new Turing machine D with H as a subroutine. This new >>>>> TM calls H to determine what M does when the input to M is its own
description ⟨M⟩. Once D has determined this information, it does the >>>>> opposite. https://www.liarparadox.org/Sipser_165_167.pdf
You have just proved yourself to be an IDIOT.
PO is incapable of logic reasoning (PO had shown he cannot even get the truth
table of logical implication/AND right). All he said is delusion including when
words from him happen to be correct to others (no real meaning).
IIRC, PO's revision that H(P,P) has no relation with P(P) is deliberately >>> fabricated this recent year after PO ran out his reasons to explain why HP is
wrong and he is correct. PO has no trouble to 'lie' to his bible (he can read
it his way), the HP thing is just piece of cake.
the input to H(P,P) specify different sequences of configurations, thus
have different halting behavior.
The easily verified fact is that the correct simulation to H(P,P) is performed by Hb(P,P) (which simulates for k more steps than H) which remains in UTM mode while simulating the same input to a final state.
Because H and Hb and both simulating halt deciders and are given the same input, they are deciding on the same sequence of configurations (namely starting with the first instruction of P). Because one answers false and one answers true, one must bewrong.
Since a simulating halt decider that simulates its input to a final state while remaining in UTM mode is necessarily correct, this proves that Hb(P,P) == true is correct and that H(P,P) == false is incorrect, and that H(P,P) does *not* in fact performa correct simulation of its input because it aborts too soon.
You've been asked several times what input must be given to H to determine if P(P) halts. It turns out that the input (P,P) can be given to Hb to determine exactly that, so the fact that H can't give the same result for the same input just shows thatit is wrong and that the halting problem is unsolvable as the existing proofs show.
That several people here deny easily
verified facts is a little psychotic on their part.
You're projecting. Again. In fact you're *so* good a projecting that if you opened a movie theater I'll bet the picture quality would be second to none.
On Tuesday, May 3, 2022 at 11:19:33 AM UTC-4, olcott wrote:
On 5/3/2022 9:47 AM, Dennis Bush wrote:
On Tuesday, May 3, 2022 at 10:31:21 AM UTC-4, olcott wrote:I have no idea what you mean.
On 5/3/2022 7:12 AM, wij wij wrote:
Richard Damon 在 2022年5月3日 星期二上午8:48:57 [UTC+8] 的信中寫道：It is an easily verified fact that P(P) and the correct simulation of
On 5/2/22 8:35 PM, olcott wrote:
On 5/2/2022 5:47 PM, Mr Flibble wrote:Thus you have shown you don't even know what a "Definition" is, so it is >>>>>> impossible for you to reason by the meaning of the words.
On Mon, 2 May 2022 18:46:00 -0400
Richard Damon wrote:
On 5/2/22 6:38 PM, Mr Flibble wrote:
On Mon, 2 May 2022 18:32:16 -0400
Richard Damon wrote:
On 5/2/22 11:47 AM, Mr Flibble wrote:
Not all infinitely recursive definitions are invalid however >>>>>>>>>>>> infinitely recursive definitions that arise out of a category >>>>>>>>>>>> error (as is the case with the halting problem) are invalid. >>>>>>>>>>>>
The halting problem (as currently defined) is invalid due to the >>>>>>>>>>>> invalid "impossible program" [Strachey, 1965] that is actually >>>>>>>>>>>> impossible due to the category error present in its definition and >>>>>>>>>>>> *not* because of any function call-like recursion; confusion >>>>>>>>>>>> between these two types of recursion are why Olcott is having >>>>>>>>>>>> difficulty communicating his ideas with the rest of you shower. >>>>>>>>>>>>
The categories involved in the category error are the decider and >>>>>>>>>>>> that which is being decided. Currently extant attempts to >>>>>>>>>>>> conflate the decider with that which is being decided are >>>>>>>>>>>> infinitely recursive and thus invalid.
/Flibble
Except that the "impossible program" isn't part of the definition >>>>>>>>>>> of the Halting Problem.
It is according to [Wikipedia, 2022].
/Flibble
Nope, you comprehend worse that PO.
Note, and Encyclopedic entery, like Wikipedia, is NOT just a >>>>>>>>> definition but a full article explaining the subject.
Maybe if you look for a FORMAL source, that states what is the ACTUAL >>>>>>>>> definition, you would learn something.
If Wikipedia is wrong then correct it and have your corrections >>>>>>>> reviewed; until then please shut the fuck up.
/Flibble
I think that the problem is that Richard has disagreeably as his highest
priority, thus doesn't really give a rat's ass for the truth. An >>>>>>>
An impossible program C. Strachey
The Computer Journal, Volume 7, Issue 4, January 1965, Page 313, >>>>>>> Published: 01 January 1965
https://academic.oup.com/comjnl/article/7/4/313/354243
It is very common knowledge that the Wikipedia description is true and >>>>>>> this is affirmed in Sipser.
For any program f that might determine if programs halt, a
"pathological" program g, called with some input, can pass its own >>>>>>> source and its input to f and then specifically do the opposite of what >>>>>>> f predicts g will do. https://en.wikipedia.org/wiki/Halting_problem >>>>>>>
Now we construct a new Turing machine D with H as a subroutine. This new
TM calls H to determine what M does when the input to M is its own >>>>>>> description ⟨M⟩. Once D has determined this information, it does the
opposite. https://www.liarparadox.org/Sipser_165_167.pdf
You have just proved yourself to be an IDIOT.
PO is incapable of logic reasoning (PO had shown he cannot even get the truth
table of logical implication/AND right). All he said is delusion including when
words from him happen to be correct to others (no real meaning).
IIRC, PO's revision that H(P,P) has no relation with P(P) is deliberately >>>>> fabricated this recent year after PO ran out his reasons to explain why HP is
wrong and he is correct. PO has no trouble to 'lie' to his bible (he can read
it his way), the HP thing is just piece of cake.
the input to H(P,P) specify different sequences of configurations, thus >>>> have different halting behavior.
The easily verified fact is that the correct simulation to H(P,P) is performed by Hb(P,P) (which simulates for k more steps than H) which remains in UTM mode while simulating the same input to a final state.
In other words you don't want to admit that this proves you are wrong.
wrong.Because H and Hb and both simulating halt deciders and are given the same input, they are deciding on the same sequence of configurations (namely starting with the first instruction of P). Because one answers false and one answers true, one must be
It is ridiculously stupid to assume that an input having pathological
self-reference to its decider would have the same behavior as an input
NOT having pathological to its decider.
Which is another way of saying that H can't give a correct answer for (P,P).
perform a correct simulation of its input because it aborts too soon.Since a simulating halt decider that simulates its input to a final state while remaining in UTM mode is necessarily correct, this proves that Hb(P,P) == true is correct and that H(P,P) == false is incorrect, and that H(P,P) does *not* in fact
It is very easy to verify the fact that the simulated input to H(P,P)
would never stop unless aborted. It is pretty psychotic that many of my
reviewers deny easily verified facts.
There is no "unless". The fixed algorithm of H, which will henceforth be referred to as Ha and similarly P will be referred to as Pa, *does* abort.
Because of this, Hb(Pa,Pa) explicitly shows that the simulated input to Ha(Pa,Pa) *does* stop. The fact that Pn(Pn) does not halt and that Hn(Pn,Pn) does not halt is irrelevant.
it is wrong and that the halting problem is unsolvable as the existing proofs show.You've been asked several times what input must be given to H to determine if P(P) halts. It turns out that the input (P,P) can be given to Hb to determine exactly that, so the fact that H can't give the same result for the same input just shows that
It is ridiculously stupid to assume that an input having pathological
self-reference to its decider would have the same behavior as an input
NOT having pathological to its decider.
Which is another way of saying that Ha can't give a correct answer for (Pa,Pa).
It is an easily verified fact that H does correctly reject its input
Ha does not correctly reject its input as easily verified by Hb.
and that deciders only compute the mapping from their inputs.
And all halt deciders must compute the same mapping from the same input. Ha(Pa,Pa) and Hb(Pa,Pa) do not perform the same mapping from the same input so one must be wrong.
Since a simulating halt decider that simulates its input to a final state while remaining in UTM mode is necessarily correct, this proves that Hb(Pa,Pa) == true is correct and that Ha(Pa,Pa) == false is incorrect
Anyone that denies easily verified facts has (by definition) a breakThat several people here deny easily
verified facts is a little psychotic on their part.
You're projecting. Again. In fact you're *so* good a projecting that if you opened a movie theater I'll bet the picture quality would be second to none.
from reality.
I can't *wait* to see your movie theater. Such a expert at projection must have a great picture.
On 2022-05-03 14:38:57 +0000, olcott said:
On 5/3/2022 4:36 AM, Mikko wrote:
On 2022-05-02 16:18:36 +0000, olcott said:
It seems to me that all infinitely recursive definitions are invalid
and I am having an excellent dialogue with some Prolog folks about
this in comp.lang.prolog.
One of the rules that define Prolog language is
arguments ::= argument | argument "," arguments
which is infinitely recursive. Is it invalid? Is Prolog invalid because
of this and other infinitely recursive rules?
Mikko
If would have to be invalid because it can never be resolved.
What would be invalid? Prolog? Definition of Prolog?
Why "would be" and not "is"?
Mikko
On 2022-05-03 14:42:32 +0000, olcott said:
On 5/3/2022 4:31 AM, Mikko wrote:
On 2022-05-02 15:47:32 +0000, Mr Flibble said:
Not all infinitely recursive definitions are invalid however infinitely >>>> recursive definitions that arise out of a category error (as is the
case with the halting problem) are invalid.
An infinite recursion cannot arise out of a category error as the
recursion
stops at the category error.
Mikko
The category error is that an expression of language X is construed as
a logic sentence / truth bearer that is true or false. It is because
of the infinitely recursive definition that X is neither of these.
Only if the recursive expression is used as if it were a truth bearer. Definitions usually don't use expression that way.
Mikko
On Tuesday, May 3, 2022 at 12:39:49 PM UTC-4, olcott wrote:
On 5/3/2022 10:36 AM, Dennis Bush wrote:
On Tuesday, May 3, 2022 at 11:19:33 AM UTC-4, olcott wrote:No I can't understand what you mean.
On 5/3/2022 9:47 AM, Dennis Bush wrote:
On Tuesday, May 3, 2022 at 10:31:21 AM UTC-4, olcott wrote:I have no idea what you mean.
On 5/3/2022 7:12 AM, wij wij wrote:
Richard Damon 在 2022年5月3日 星期二上午8:48:57 [UTC+8] 的信中寫道：It is an easily verified fact that P(P) and the correct simulation of >>>>>> the input to H(P,P) specify different sequences of configurations, thus >>>>>> have different halting behavior.
On 5/2/22 8:35 PM, olcott wrote:
On 5/2/2022 5:47 PM, Mr Flibble wrote:Thus you have shown you don't even know what a "Definition" is, so it is
On Mon, 2 May 2022 18:46:00 -0400
Richard Damon wrote:
On 5/2/22 6:38 PM, Mr Flibble wrote:
On Mon, 2 May 2022 18:32:16 -0400
Richard Damon wrote:
On 5/2/22 11:47 AM, Mr Flibble wrote:
Not all infinitely recursive definitions are invalid however >>>>>>>>>>>>>> infinitely recursive definitions that arise out of a category >>>>>>>>>>>>>> error (as is the case with the halting problem) are invalid. >>>>>>>>>>>>>>
The halting problem (as currently defined) is invalid due to the >>>>>>>>>>>>>> invalid "impossible program" [Strachey, 1965] that is actually >>>>>>>>>>>>>> impossible due to the category error present in its definition and
*not* because of any function call-like recursion; confusion >>>>>>>>>>>>>> between these two types of recursion are why Olcott is having >>>>>>>>>>>>>> difficulty communicating his ideas with the rest of you shower. >>>>>>>>>>>>>>
The categories involved in the category error are the decider and
that which is being decided. Currently extant attempts to >>>>>>>>>>>>>> conflate the decider with that which is being decided are >>>>>>>>>>>>>> infinitely recursive and thus invalid.
/Flibble
Except that the "impossible program" isn't part of the definition >>>>>>>>>>>>> of the Halting Problem.
It is according to [Wikipedia, 2022].
/Flibble
Nope, you comprehend worse that PO.
Note, and Encyclopedic entery, like Wikipedia, is NOT just a >>>>>>>>>>> definition but a full article explaining the subject.
Maybe if you look for a FORMAL source, that states what is the ACTUAL
definition, you would learn something.
If Wikipedia is wrong then correct it and have your corrections >>>>>>>>>> reviewed; until then please shut the fuck up.
/Flibble
I think that the problem is that Richard has disagreeably as his highest
priority, thus doesn't really give a rat's ass for the truth. An >>>>>>>>>
An impossible program C. Strachey
The Computer Journal, Volume 7, Issue 4, January 1965, Page 313, >>>>>>>>> Published: 01 January 1965
https://academic.oup.com/comjnl/article/7/4/313/354243
It is very common knowledge that the Wikipedia description is true and
this is affirmed in Sipser.
For any program f that might determine if programs halt, a
"pathological" program g, called with some input, can pass its own >>>>>>>>> source and its input to f and then specifically do the opposite of what
f predicts g will do. https://en.wikipedia.org/wiki/Halting_problem >>>>>>>>>
Now we construct a new Turing machine D with H as a subroutine. This new
TM calls H to determine what M does when the input to M is its own >>>>>>>>> description ⟨M⟩. Once D has determined this information, it does the
opposite. https://www.liarparadox.org/Sipser_165_167.pdf
impossible for you to reason by the meaning of the words.
You have just proved yourself to be an IDIOT.
PO is incapable of logic reasoning (PO had shown he cannot even get the truth
table of logical implication/AND right). All he said is delusion including when
words from him happen to be correct to others (no real meaning). >>>>>>>
IIRC, PO's revision that H(P,P) has no relation with P(P) is deliberately
fabricated this recent year after PO ran out his reasons to explain why HP is
wrong and he is correct. PO has no trouble to 'lie' to his bible (he can read
it his way), the HP thing is just piece of cake.
The easily verified fact is that the correct simulation to H(P,P) is performed by Hb(P,P) (which simulates for k more steps than H) which remains in UTM mode while simulating the same input to a final state.
In other words you don't want to admit that this proves you are wrong.
I think that I see it now, I had forgotten the notation.
An input having a pathological self-reference relationship to its
decider H would necessarily derive a different halt status than an input
not having a pathological self-reference relationship to its decider Hb.
The P having a pathological self-reference relationship to H is not the
same as the Px NOT having a pathological self-reference relationship to
Hb. Because P.H calls itself and Px.Hb does not call itself P is not the
same input as Px.
The P we're talking about is a *specific* P, namely Pa which is built from Ha, and Ha is a *specific* H. So Pa and Px are the *same*.
So just because Pa contains an embedded copy of Ha but not an embedded copy of Hb doesn't means that it's not the same.
Ha(Pa,Pa) and Hb(Pa,Pa) have the *exact* same input.
Just because it appears from a glance that Ha is starting its simulation of Pa "in the middle" doesn't mean that's what's actually happening. That's just how the incorrect simulation is manifesting itself. It's kind of like undefined behavior in a Cprogram.
wrong.
Because H and Hb and both simulating halt deciders and are given the same input, they are deciding on the same sequence of configurations (namely starting with the first instruction of P). Because one answers false and one answers true, one must be
Different computations must give different answers.It is ridiculously stupid to assume that an input having pathological
self-reference to its decider would have the same behavior as an input >>>> NOT having pathological to its decider.
Which is another way of saying that H can't give a correct answer for (P,P).
That you don't fully understand all of the nuances of how this applies
to H/P and Hb/Px is OK, it is difficult to understand.
Just because Pa contains an embedded copy of Ha but not an embedded copy of Hb doesn't means that it's not the same.
perform a correct simulation of its input because it aborts too soon.Since a simulating halt decider that simulates its input to a final state while remaining in UTM mode is necessarily correct, this proves that Hb(P,P) == true is correct and that H(P,P) == false is incorrect, and that H(P,P) does *not* in fact
Which is *NOT* halting. A halting input must reach its own final state.It is very easy to verify the fact that the simulated input to H(P,P)
would never stop unless aborted. It is pretty psychotic that many of my >>>> reviewers deny easily verified facts.
There is no "unless". The fixed algorithm of H, which will henceforth be referred to as Ha and similarly P will be referred to as Pa, *does* abort.
Because of this, Hb(Pa,Pa) explicitly shows that the simulated input to Ha(Pa,Pa) *does* stop. The fact that Pn(Pn) does not halt and that Hn(Pn,Pn) does not halt is irrelevant.It it not Hb(Pa,Pa) it is Hb(Px,Px). That P calls H makes it an entirely
different input than Px that does not call Hb.
No it is *exactly* Hb(Pa,Pa). The same encoding passed to Ha is passed to Hb.
Peter Olcott wrote:
On 5/3/2022 5:46 PM, Python wrote:
Peter Olcott wrote:
On 5/3/2022 5:21 PM, Python wrote:
Peter Olcott wrote:
On 5/3/2022 1:59 PM, Mr Flibble wrote:...
On Tue, 3 May 2022 11:57:39 -0500
olcott <polcott2@gmail.com> wrote:
...I don't buy into the whole imaginary numbers game.
We could imagine that 2 + 3 = 17 and call that an imaginary sum.
Nice to know, thanks. Thus your rebuttal seems complete it is not
an infinite anything. Imagining the square root of a negative
number or that parallel lines meet seems a little nuts to me.
Before jumping to such outrageously uninformed conclusions you may
want
to learn how complex numbers are actually defined nowadays.
They are defined to directly contradict the verified facts.
I really hate anything that directly contradicts the verified facts
because this can result in:
(a) Climate change making humans extinct quite soon,
(b) Nazi "big lie" propaganda about election fraud is making very
significant inroads to transforming Democracy ion the USA to Fascism.
(c) It directly resulted in many covid-19 deaths
You should definitely call a doctor and ask for help, your mental
state is utterly ill. What the f* are complex numbers related to
your rants on politics?
It is true that, at first, it was used without any proper definition >>>>> better than "let's assume we can deal with sqrt(-1) as usual". The
surprising point at that time is it works pretty well.
We can see what happens when we hypothesize (against the facts) that
square roots of negative numbers and parallel lines that meet exist
simply to see where this leads. I am sure that this is the intent.
Not quite you're wrong. The intent was to find *real* roots of degree 3
polynomials even if square root of negative quantities appears at
intermediary steps.
None-the-less as I just said this whole think relies on accepting a
known false premise.
Let me guess, you are some kind of "information engineer" and consider yourself also as "one of the greatest logicians Humanity ever had",
right? There is a demented guy of this kind on sci.physics.relativity,
his name is Maciej Wozniak. You guys should definitely mate (NOT).
You know *nothing* about history of Science, Peter. Guessing is quite
absurd when it comes to History.
*Then*, in the XIXth Century, Gallois showed how to define complex
numbers rigorously.
You've never heard of that, Peter, really?
[for the record: C is the set of equivalence classes of polynomials
on R by the relation p ~ q iff p - q = 0 [mod x^2+1], compatibility
of + and * on R and C can be proven easily, R is naturally injected
into C as a set of constant polynomials, i is the equivalence class of >>>>> the polynomial x]
No reaction? Well... Not a big surprise, your eyes cancel out any
sensible arguments proving you wrong, as usual.
Die in Hell, idiotic annoying crank. You deserve it.
(bis)
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