On 12/13/25 5:15 AM, polcott wrote:
On 12/12/2025 11:22 PM, dart200 wrote:
On 12/11/25 1:20 PM, polcott wrote:
On 12/11/2025 3:02 PM, dart200 wrote:
On 12/11/25 12:45 PM, polcott wrote:
On 12/11/2025 1:35 PM, dart200 wrote:
On 12/9/25 8:02 PM, Richard Damon wrote:
On 12/9/25 1:55 PM, dart200 wrote:
On 12/9/25 4:42 AM, Richard Damon wrote:
On 12/9/25 12:23 AM, dart200 wrote:
On 12/8/25 8:12 PM, Richard Damon wrote:
???
Given Machine H is chosen as one partial decider then the >>>>>>>>>>>> machine:
H^(d): if H(d, d) returns halting, loop forever
else halt.
i'm sorry now ur claiming H(d,d) actually returns an answer??? >>>>>>>>>>>
when did this happen, and what does it return buddy???
what ever its programs says it will.
Do you not understand the concept of a parameter to an arguement? >>>>>>>>>>
My claim is if *YOU* give me a machine H, I can prove it wrong. >>>>>>>>>>
YOU need to provide some machine that my arguement will label >>>>>>>>>> as H.
Then H^(H^) will show that H was wrong for H(H^, H^)
How is that not showing the machine which that machine can >>>>>>>>>>>> not decider.
partial decidable does not fly it loses to BB
Nope, because "partial deciability" means the machine is
allowed to not answer.
so what ur saying is H won't answer, so H^ will have an answer? >>>>>>>>> i did explore that paradigm in one of my papers, a believe it's >>>>>>>>> possible to create a program that seeks out an contradicts any >>>>>>>>> and all deciders that try to decide on it:
H^ must have a behavior, so there is a correct answer.
One semi-useful class of partial decider, which are also called >>>>>>>> recognizer, are machines that never give a wrong answer, but
sometimes
yeah that's what i explored in the paper i posted
don't answer. This class is more useful if they always
eventually answer for one side of the decision, and only not- >>>>>>>> answer sometimes for the
no, there's always going to be some machine which they cannot
answer for both sides
please do read §2 of that paper
other. Halting is partially decidable by this criteria, with a >>>>>>>> decider that eventually answer halting for all halting machines, >>>>>>>> and non- halting for a large class of non-halting machines. I >>>>>>>> looked at machines of this type in the late 70's in school.
Also, "beleive" is not proof, and doesn't mean you framework is >>>>>>>> useful.
It is easy to create a system where Halting can be decided, it >>>>>>>> just needs making the system less than Turing Complete, so if >>>>>>>> you idea is based on that, so what.
https://www.academia.edu/136521323/
how_to_resolve_a_halting_paradox
(partial decidability also wouldn't work in Turing's
"satisfactory" problem from the og paper /on computable
numbers/, but we'll get there later)
The Abstract talks about changing the meaning of basics of
Conputation theory and the defintion of Halting (I haven't read >>>>>>>> the full paper).
All that is doing is admitting that by the definitions accepted >>>>>>>> for defining a computation and what halting means, the author is >>>>>>>> conceeding that Halting is uncomputable.
The paper than says:
This paper does not attempt to present at depth arguments or
reasons for why we should accept either of these proposals vs a >>>>>>>> more conventional perspective,
because the implications are so broad my interest was to just
focus on the idea of the technique vs why
But, what good is an alternate formulation if you aren't going >>>>>>>> to discuss why the alternative is going to be useful.
i cannot condense meaning into the abstract and conclusions, u'd >>>>>>> actually have to read it 🤷
It seems this is just another person not understand the
reasoning behind how computations were defined, and why
"Halting" was an important question, but just wants to create a >>>>>>>> somewhat similar solvable problem, even if such an alternative >>>>>>>> problem has no use.
if BB has some limit L (which is must if u believe halting >>>>>>>>>>> problem), then there must be some specifically L-state
machine which *no* machine could decide upon, for if that >>>>>>>>>>> machine was decidable by anything, then BB could find that >>>>>>>>>>> anything and subvert the limit L
WHy does BB need to have a limit L?
my my richard, u don't know that in ur theory BB must have a >>>>>>>>> limit?
You seem to be using undefined terms.
BB is apparently the Busy Beaver problem, which since it is
uncomputable, can't actually be a machine.
yeah but it's certainly computable up until a limit, as we've
already computed it up to 5, there cannot be any machines <6
states that are not decidable
BB(n) is the maximum length tape that a Turing Machine of n
states can create and then halt.
technically it's steps: https://en.wikipedia.org/wiki/Busy_beaver >>>>>>>
but for the purposes of this discussion it doesn't really matter >>>>>>> whether it's space or steps we're talking about
BB(n) is, by definitiion a "finite" number. Talking about the >>>>>>>> "limit" of a finite number is a misuse of the term.
i mean the natural number limit L >5 at which point BB(L) becomes >>>>>>> fundamentally *unknowable* due to some L-state machine being
fundamentally undecidable.
if L doesn't exist, that would make halting generally decidable, >>>>>>> so therefore L must exist
if L does exist, then there must be some L-state machine U which >>>>>>> cannot be decided on *by any partial* decider, because the BB
computation would find it and use it
We can sometimes establish upper and lower bounds on the value >>>>>>>> of BB(n), is that what you mean by "a limit L"?
if you believe the halting problem, then BB must have a limit >>>>>>>>> L, or else halting becomes generally solvable using the BB
function. see, if you can compute the BB number for any N-state >>>>>>>>> machines, then for any N-state machine u can just run the N- >>>>>>>>> state machine until BB number of steps. any machine that halts >>>>>>>>> on or before BB(N) steps halts, any that run past must be
nonhalting
No, if we could establish an upper limit for BB(n) for all n, >>>>>>>> then we could solve the hatling problem, as we have an upper
limit for the number of steps we need to simulate the machine. >>>>>>>>
BB(n) has a value, but for sufficiently large values of n, we >>>>>>>> don't have an upper bound for BB(n).
and the problem with allowing for partial decidability is that >>>>>>>>> BB can run continually run more and more deciders in parallel, >>>>>>>>> on every N- state machine, until one comes back with an halting >>>>>>>>> answer, for every N-state machine, which then it can the use to >>>>>>>>> decide what the BB number is for any N ...
So, what BB are you running? Or are you misusing "running" to >>>>>>>> try to mean somehow trying to calculate?
contradicting the concept it must have a limit L, where some L- >>>>>>>>> state machine cannot be decidable by *any* partial decider on >>>>>>>>> the matter,
No, it can have a limit, just not a KNOWN limit.
consensus is there can a known limit L to the BB function, and
proofs have been put out in regards to this
so no richard, partial decidability does not work if BB is to >>>>>>>>> have a limit
You only have the problem is BB has a KNOWN limit. Again, you >>>>>>>> trip up on assuming you can know any answer you want.
That some things are not knowable breaks your logic.
I just glanced at your paper and skipped to the conclusion.
Why do we care about the undecidability of the halting problem?
Because undecidability in general (if it is correct) shows
that truth itself is broken. Truth itself cannot be broken.
This is the only reason why I have worked on these things
for 28 years.
because it makes us suck as developing and maintaining software,
and as a 35 year old burnt out SWE, i'm tired of living in a world
running off sucky software. it really is limiting our potential,
and i want my soon to be born son to have a far better experience
with this shit than i did.
a consequence of accepting the halting problem is then necessarily
accepting proof against *all* semantic deciders, barring us from
agreeing on what such general deciders might be
Exactly: Tarski even "proved" that we can't even directly
compute what is true. This lets dangerous liars get away
with their dangerous lies.
this has lead to not only an unnecessary explosion in complexity of >>>>> software engineering, because we can't generally compute semantic
(turing) equivalence,
but the general trend in deploying software that doesn't have
computed semantic proofs guaranteeing they actually do what we want >>>>> them to do.
Yes without computing halting total proof of
correctness is impossible.
"testing" is poor substitute for doing so, but that's the most we
can agree upon due to the current theory of computing.
i think my ideas might contribute to dealing with incompleteness in >>>>> fundamental math more generally ... like producing more refined
limits to it's philosophical impact. tho idk if it can be gotten
rid of completely, anymore than we can get rid of the words "this
statement is false"
I don't think that there actually are any limits
except for expressions requiring infinite proofs.
but i am currently focused on the theory of computing and not
anything more generally. the fundamental objects comprising the
theory of computing (machines) are far more constrained in their
definitions than what set theory needs to encompass, and within
those constraints i think i can twist the consensus into some
contradiction that are just entirely ignorant of atm
I have explored all of the key areas. None of them
can be made as 100% perfectly concrete and unequivocal
as computing.
that's the slam dunk left that i need. i have a means to rectify
whatever contradiction we find thru the use of RTMs, but i'm still
teasing out the contradiction that will *force* others to notice
I do have my refutation of the halting problem itself
boiled down to a rough draft of two first principles.
When the halting problem requires a halt decider
to report on the behavior of a Turing machine this
is always a category error because Turing machines
only take finite string inputs.
The corrected halting problem requires a Turing
machine decider to report in the behavior that its
actual finite string input actually specifies.
polcott, i'm working on making the halting problem complete and
consistent in regards to a subset of the improved "reflective turing
machines" that encompasses all useful computations
i'm sorry, but not about trying to reaffirm the halting function as
still uncomputable by calling it a category error
I do compute the halting function correctly.
the halting *function* is an abstract mathematical object that maps a machine description to whether the machine described halts or not, not
the associated machine description that attempts to compute this
I have been doing this for more than three years.
We probably should not be spamming alt.buddha.short.fat.guy
i hang out there tho, i have my own reasons for posting this there
int sum(int x, int y){return x + y;}
sum(3,2) should return 5 and it is incorrect
to require sum(3,2) to return the sum of 5+6.
u can argue about what computing machines actually exist all u want, and whether anything actually computes the halting *function*, i'm not going
to argue over what the halting *function* itself is
On 12/13/2025 10:01 AM, dart200 wrote:
On 12/13/25 5:15 AM, polcott wrote:
On 12/12/2025 11:22 PM, dart200 wrote:
On 12/11/25 1:20 PM, polcott wrote:
On 12/11/2025 3:02 PM, dart200 wrote:
On 12/11/25 12:45 PM, polcott wrote:
On 12/11/2025 1:35 PM, dart200 wrote:
On 12/9/25 8:02 PM, Richard Damon wrote:
On 12/9/25 1:55 PM, dart200 wrote:
On 12/9/25 4:42 AM, Richard Damon wrote:
On 12/9/25 12:23 AM, dart200 wrote:
On 12/8/25 8:12 PM, Richard Damon wrote:
???
Given Machine H is chosen as one partial decider then the >>>>>>>>>>>>> machine:
H^(d): if H(d, d) returns halting, loop forever
else halt.
i'm sorry now ur claiming H(d,d) actually returns an answer??? >>>>>>>>>>>>
when did this happen, and what does it return buddy???
what ever its programs says it will.
Do you not understand the concept of a parameter to an
arguement?
My claim is if *YOU* give me a machine H, I can prove it wrong. >>>>>>>>>>>
YOU need to provide some machine that my arguement will label >>>>>>>>>>> as H.
Then H^(H^) will show that H was wrong for H(H^, H^) >>>>>>>>>>>>>
How is that not showing the machine which that machine can >>>>>>>>>>>>> not decider.
partial decidable does not fly it loses to BB
Nope, because "partial deciability" means the machine is >>>>>>>>>>> allowed to not answer.
so what ur saying is H won't answer, so H^ will have an
answer? i did explore that paradigm in one of my papers, a >>>>>>>>>> believe it's possible to create a program that seeks out an >>>>>>>>>> contradicts any and all deciders that try to decide on it:
H^ must have a behavior, so there is a correct answer.
One semi-useful class of partial decider, which are also called >>>>>>>>> recognizer, are machines that never give a wrong answer, but >>>>>>>>> sometimes
yeah that's what i explored in the paper i posted
don't answer. This class is more useful if they always
eventually answer for one side of the decision, and only not- >>>>>>>>> answer sometimes for the
no, there's always going to be some machine which they cannot >>>>>>>> answer for both sides
please do read §2 of that paper
other. Halting is partially decidable by this criteria, with a >>>>>>>>> decider that eventually answer halting for all halting
machines, and non- halting for a large class of non-halting >>>>>>>>> machines. I looked at machines of this type in the late 70's in >>>>>>>>> school.
Also, "beleive" is not proof, and doesn't mean you framework is >>>>>>>>> useful.
It is easy to create a system where Halting can be decided, it >>>>>>>>> just needs making the system less than Turing Complete, so if >>>>>>>>> you idea is based on that, so what.
https://www.academia.edu/136521323/
how_to_resolve_a_halting_paradox
(partial decidability also wouldn't work in Turing's
"satisfactory" problem from the og paper /on computable
numbers/, but we'll get there later)
The Abstract talks about changing the meaning of basics of
Conputation theory and the defintion of Halting (I haven't read >>>>>>>>> the full paper).
All that is doing is admitting that by the definitions accepted >>>>>>>>> for defining a computation and what halting means, the author >>>>>>>>> is conceeding that Halting is uncomputable.
The paper than says:
This paper does not attempt to present at depth arguments or >>>>>>>>> reasons for why we should accept either of these proposals vs a >>>>>>>>> more conventional perspective,
because the implications are so broad my interest was to just >>>>>>>> focus on the idea of the technique vs why
But, what good is an alternate formulation if you aren't going >>>>>>>>> to discuss why the alternative is going to be useful.
i cannot condense meaning into the abstract and conclusions, u'd >>>>>>>> actually have to read it 🤷
It seems this is just another person not understand the
reasoning behind how computations were defined, and why
"Halting" was an important question, but just wants to create a >>>>>>>>> somewhat similar solvable problem, even if such an alternative >>>>>>>>> problem has no use.
if BB has some limit L (which is must if u believe halting >>>>>>>>>>>> problem), then there must be some specifically L-state >>>>>>>>>>>> machine which *no* machine could decide upon, for if that >>>>>>>>>>>> machine was decidable by anything, then BB could find that >>>>>>>>>>>> anything and subvert the limit L
WHy does BB need to have a limit L?
my my richard, u don't know that in ur theory BB must have a >>>>>>>>>> limit?
You seem to be using undefined terms.
BB is apparently the Busy Beaver problem, which since it is >>>>>>>>> uncomputable, can't actually be a machine.
yeah but it's certainly computable up until a limit, as we've >>>>>>>> already computed it up to 5, there cannot be any machines <6
states that are not decidable
BB(n) is the maximum length tape that a Turing Machine of n >>>>>>>>> states can create and then halt.
technically it's steps: https://en.wikipedia.org/wiki/Busy_beaver >>>>>>>>
but for the purposes of this discussion it doesn't really matter >>>>>>>> whether it's space or steps we're talking about
BB(n) is, by definitiion a "finite" number. Talking about the >>>>>>>>> "limit" of a finite number is a misuse of the term.
i mean the natural number limit L >5 at which point BB(L)
becomes fundamentally *unknowable* due to some L-state machine >>>>>>>> being fundamentally undecidable.
if L doesn't exist, that would make halting generally decidable, >>>>>>>> so therefore L must exist
if L does exist, then there must be some L-state machine U which >>>>>>>> cannot be decided on *by any partial* decider, because the BB >>>>>>>> computation would find it and use it
We can sometimes establish upper and lower bounds on the value >>>>>>>>> of BB(n), is that what you mean by "a limit L"?
if you believe the halting problem, then BB must have a limit >>>>>>>>>> L, or else halting becomes generally solvable using the BB >>>>>>>>>> function. see, if you can compute the BB number for any N- >>>>>>>>>> state machines, then for any N-state machine u can just run >>>>>>>>>> the N- state machine until BB number of steps. any machine >>>>>>>>>> that halts on or before BB(N) steps halts, any that run past >>>>>>>>>> must be nonhalting
No, if we could establish an upper limit for BB(n) for all n, >>>>>>>>> then we could solve the hatling problem, as we have an upper >>>>>>>>> limit for the number of steps we need to simulate the machine. >>>>>>>>>
BB(n) has a value, but for sufficiently large values of n, we >>>>>>>>> don't have an upper bound for BB(n).
and the problem with allowing for partial decidability is that >>>>>>>>>> BB can run continually run more and more deciders in parallel, >>>>>>>>>> on every N- state machine, until one comes back with an
halting answer, for every N-state machine, which then it can >>>>>>>>>> the use to decide what the BB number is for any N ...
So, what BB are you running? Or are you misusing "running" to >>>>>>>>> try to mean somehow trying to calculate?
contradicting the concept it must have a limit L, where some >>>>>>>>>> L- state machine cannot be decidable by *any* partial decider >>>>>>>>>> on the matter,
No, it can have a limit, just not a KNOWN limit.
consensus is there can a known limit L to the BB function, and >>>>>>>> proofs have been put out in regards to this
so no richard, partial decidability does not work if BB is to >>>>>>>>>> have a limit
You only have the problem is BB has a KNOWN limit. Again, you >>>>>>>>> trip up on assuming you can know any answer you want.
That some things are not knowable breaks your logic.
I just glanced at your paper and skipped to the conclusion.
Why do we care about the undecidability of the halting problem?
Because undecidability in general (if it is correct) shows
that truth itself is broken. Truth itself cannot be broken.
This is the only reason why I have worked on these things
for 28 years.
because it makes us suck as developing and maintaining software,
and as a 35 year old burnt out SWE, i'm tired of living in a world >>>>>> running off sucky software. it really is limiting our potential,
and i want my soon to be born son to have a far better experience >>>>>> with this shit than i did.
a consequence of accepting the halting problem is then necessarily >>>>>> accepting proof against *all* semantic deciders, barring us from
agreeing on what such general deciders might be
Exactly: Tarski even "proved" that we can't even directly
compute what is true. This lets dangerous liars get away
with their dangerous lies.
this has lead to not only an unnecessary explosion in complexity
of software engineering, because we can't generally compute
semantic (turing) equivalence,
but the general trend in deploying software that doesn't have
computed semantic proofs guaranteeing they actually do what we
want them to do.
Yes without computing halting total proof of
correctness is impossible.
"testing" is poor substitute for doing so, but that's the most we >>>>>> can agree upon due to the current theory of computing.
i think my ideas might contribute to dealing with incompleteness
in fundamental math more generally ... like producing more refined >>>>>> limits to it's philosophical impact. tho idk if it can be gotten
rid of completely, anymore than we can get rid of the words "this >>>>>> statement is false"
I don't think that there actually are any limits
except for expressions requiring infinite proofs.
but i am currently focused on the theory of computing and not
anything more generally. the fundamental objects comprising the
theory of computing (machines) are far more constrained in their
definitions than what set theory needs to encompass, and within
those constraints i think i can twist the consensus into some
contradiction that are just entirely ignorant of atm
I have explored all of the key areas. None of them
can be made as 100% perfectly concrete and unequivocal
as computing.
that's the slam dunk left that i need. i have a means to rectify
whatever contradiction we find thru the use of RTMs, but i'm still >>>>>> teasing out the contradiction that will *force* others to notice
I do have my refutation of the halting problem itself
boiled down to a rough draft of two first principles.
When the halting problem requires a halt decider
to report on the behavior of a Turing machine this
is always a category error because Turing machines
only take finite string inputs.
The corrected halting problem requires a Turing
machine decider to report in the behavior that its
actual finite string input actually specifies.
polcott, i'm working on making the halting problem complete and
consistent in regards to a subset of the improved "reflective turing
machines" that encompasses all useful computations
i'm sorry, but not about trying to reaffirm the halting function as
still uncomputable by calling it a category error
I do compute the halting function correctly.
the halting *function* is an abstract mathematical object that maps a
machine description to whether the machine described halts or not, not
the associated machine description that attempts to compute this
All Turing machines only compute the mapping
from input finite strings to some value.
On this basis I do compute halting correctly.
I have been doing this for more than three years.
We probably should not be spamming alt.buddha.short.fat.guy
i hang out there tho, i have my own reasons for posting this there
Theory of computation issues are disrespectful
spam to that group that violate Buddhist compassion.
I was a long time poster to alt.zen. I still
have 8969 messages posted there since 2005.
Also my great grand uncle Henry Steel Olcott
was a very famous Buddhist.
int sum(int x, int y){return x + y;}
sum(3,2) should return 5 and it is incorrect
to require sum(3,2) to return the sum of 5+6.
u can argue about what computing machines actually exist all u want,
and whether anything actually computes the halting *function*, i'm not
going to argue over what the halting *function* itself is
Then you get the wrong answer.
In computability theory and computational complexity
theory, a decision problem is a computational problem
that can be posed as a yes–no question on a set of
input values. https://en.wikipedia.org/wiki/Decision_problem
Most people have no idea that there is such a thing
as an incorrect question. Because of this they misclassify
incorrect yes/no questions as undecidable decision problem
instances.
On 13/12/2025 18:22, olcott wrote:
On 12/13/2025 10:01 AM, dart200 wrote:
On 12/13/25 5:15 AM, polcott wrote:
On 12/12/2025 11:22 PM, dart200 wrote:
On 12/11/25 1:20 PM, polcott wrote:
On 12/11/2025 3:02 PM, dart200 wrote:
On 12/11/25 12:45 PM, polcott wrote:
On 12/11/2025 1:35 PM, dart200 wrote:
On 12/9/25 8:02 PM, Richard Damon wrote:
On 12/9/25 1:55 PM, dart200 wrote:
On 12/9/25 4:42 AM, Richard Damon wrote:H^ must have a behavior, so there is a correct answer.
On 12/9/25 12:23 AM, dart200 wrote:
On 12/8/25 8:12 PM, Richard Damon wrote:what ever its programs says it will.
???
Given Machine H is chosen as one partial decider then the >>>>>>>>>>>>>> machine:
H^(d): if H(d, d) returns halting, loop forever
else halt.
i'm sorry now ur claiming H(d,d) actually returns an answer??? >>>>>>>>>>>>>
when did this happen, and what does it return buddy??? >>>>>>>>>>>>
Do you not understand the concept of a parameter to an >>>>>>>>>>>> arguement?
My claim is if *YOU* give me a machine H, I can prove it wrong. >>>>>>>>>>>>
YOU need to provide some machine that my arguement will >>>>>>>>>>>> label as H.
Then H^(H^) will show that H was wrong for H(H^, H^) >>>>>>>>>>>>>>
How is that not showing the machine which that machine can >>>>>>>>>>>>>> not decider.
partial decidable does not fly it loses to BB
Nope, because "partial deciability" means the machine is >>>>>>>>>>>> allowed to not answer.
so what ur saying is H won't answer, so H^ will have an >>>>>>>>>>> answer? i did explore that paradigm in one of my papers, a >>>>>>>>>>> believe it's possible to create a program that seeks out an >>>>>>>>>>> contradicts any and all deciders that try to decide on it: >>>>>>>>>>
One semi-useful class of partial decider, which are also
called recognizer, are machines that never give a wrong
answer, but sometimes
yeah that's what i explored in the paper i posted
don't answer. This class is more useful if they always
eventually answer for one side of the decision, and only not- >>>>>>>>>> answer sometimes for the
no, there's always going to be some machine which they cannot >>>>>>>>> answer for both sides
please do read §2 of that paper
other. Halting is partially decidable by this criteria, with a >>>>>>>>>> decider that eventually answer halting for all halting
machines, and non- halting for a large class of non-halting >>>>>>>>>> machines. I looked at machines of this type in the late 70's >>>>>>>>>> in school.
Also, "beleive" is not proof, and doesn't mean you framework >>>>>>>>>> is useful.
It is easy to create a system where Halting can be decided, it >>>>>>>>>> just needs making the system less than Turing Complete, so if >>>>>>>>>> you idea is based on that, so what.
https://www.academia.edu/136521323/
how_to_resolve_a_halting_paradox
(partial decidability also wouldn't work in Turing's
"satisfactory" problem from the og paper /on computable >>>>>>>>>>> numbers/, but we'll get there later)
The Abstract talks about changing the meaning of basics of >>>>>>>>>> Conputation theory and the defintion of Halting (I haven't >>>>>>>>>> read the full paper).
All that is doing is admitting that by the definitions
accepted for defining a computation and what halting means, >>>>>>>>>> the author is conceeding that Halting is uncomputable.
The paper than says:
This paper does not attempt to present at depth arguments or >>>>>>>>>> reasons for why we should accept either of these proposals vs >>>>>>>>>> a more conventional perspective,
because the implications are so broad my interest was to just >>>>>>>>> focus on the idea of the technique vs why
But, what good is an alternate formulation if you aren't going >>>>>>>>>> to discuss why the alternative is going to be useful.
i cannot condense meaning into the abstract and conclusions, >>>>>>>>> u'd actually have to read it 🤷
It seems this is just another person not understand the
reasoning behind how computations were defined, and why
"Halting" was an important question, but just wants to create >>>>>>>>>> a somewhat similar solvable problem, even if such an
alternative problem has no use.
if BB has some limit L (which is must if u believe halting >>>>>>>>>>>>> problem), then there must be some specifically L-state >>>>>>>>>>>>> machine which *no* machine could decide upon, for if that >>>>>>>>>>>>> machine was decidable by anything, then BB could find that >>>>>>>>>>>>> anything and subvert the limit L
WHy does BB need to have a limit L?
my my richard, u don't know that in ur theory BB must have a >>>>>>>>>>> limit?
You seem to be using undefined terms.
BB is apparently the Busy Beaver problem, which since it is >>>>>>>>>> uncomputable, can't actually be a machine.
yeah but it's certainly computable up until a limit, as we've >>>>>>>>> already computed it up to 5, there cannot be any machines <6 >>>>>>>>> states that are not decidable
BB(n) is the maximum length tape that a Turing Machine of n >>>>>>>>>> states can create and then halt.
technically it's steps: https://en.wikipedia.org/wiki/Busy_beaver >>>>>>>>>
but for the purposes of this discussion it doesn't really
matter whether it's space or steps we're talking about
BB(n) is, by definitiion a "finite" number. Talking about the >>>>>>>>>> "limit" of a finite number is a misuse of the term.
i mean the natural number limit L >5 at which point BB(L)
becomes fundamentally *unknowable* due to some L-state machine >>>>>>>>> being fundamentally undecidable.
if L doesn't exist, that would make halting generally
decidable, so therefore L must exist
if L does exist, then there must be some L-state machine U
which cannot be decided on *by any partial* decider, because >>>>>>>>> the BB computation would find it and use it
We can sometimes establish upper and lower bounds on the value >>>>>>>>>> of BB(n), is that what you mean by "a limit L"?
if you believe the halting problem, then BB must have a limit >>>>>>>>>>> L, or else halting becomes generally solvable using the BB >>>>>>>>>>> function. see, if you can compute the BB number for any N- >>>>>>>>>>> state machines, then for any N-state machine u can just run >>>>>>>>>>> the N- state machine until BB number of steps. any machine >>>>>>>>>>> that halts on or before BB(N) steps halts, any that run past >>>>>>>>>>> must be nonhalting
No, if we could establish an upper limit for BB(n) for all n, >>>>>>>>>> then we could solve the hatling problem, as we have an upper >>>>>>>>>> limit for the number of steps we need to simulate the machine. >>>>>>>>>>
BB(n) has a value, but for sufficiently large values of n, we >>>>>>>>>> don't have an upper bound for BB(n).
So, what BB are you running? Or are you misusing "running" to >>>>>>>>>> try to mean somehow trying to calculate?
and the problem with allowing for partial decidability is >>>>>>>>>>> that BB can run continually run more and more deciders in >>>>>>>>>>> parallel, on every N- state machine, until one comes back >>>>>>>>>>> with an halting answer, for every N-state machine, which then >>>>>>>>>>> it can the use to decide what the BB number is for any N ... >>>>>>>>>>
contradicting the concept it must have a limit L, where some >>>>>>>>>>> L- state machine cannot be decidable by *any* partial decider >>>>>>>>>>> on the matter,
No, it can have a limit, just not a KNOWN limit.
consensus is there can a known limit L to the BB function, and >>>>>>>>> proofs have been put out in regards to this
so no richard, partial decidability does not work if BB is to >>>>>>>>>>> have a limit
You only have the problem is BB has a KNOWN limit. Again, you >>>>>>>>>> trip up on assuming you can know any answer you want.
That some things are not knowable breaks your logic.
I just glanced at your paper and skipped to the conclusion.
Why do we care about the undecidability of the halting problem? >>>>>>>> Because undecidability in general (if it is correct) shows
that truth itself is broken. Truth itself cannot be broken.
This is the only reason why I have worked on these things
for 28 years.
because it makes us suck as developing and maintaining software, >>>>>>> and as a 35 year old burnt out SWE, i'm tired of living in a
world running off sucky software. it really is limiting our
potential, and i want my soon to be born son to have a far better >>>>>>> experience with this shit than i did.
a consequence of accepting the halting problem is then
necessarily accepting proof against *all* semantic deciders,
barring us from agreeing on what such general deciders might be
Exactly: Tarski even "proved" that we can't even directly
compute what is true. This lets dangerous liars get away
with their dangerous lies.
this has lead to not only an unnecessary explosion in complexity >>>>>>> of software engineering, because we can't generally compute
semantic (turing) equivalence,
but the general trend in deploying software that doesn't have
computed semantic proofs guaranteeing they actually do what we
want them to do.
Yes without computing halting total proof of
correctness is impossible.
"testing" is poor substitute for doing so, but that's the most we >>>>>>> can agree upon due to the current theory of computing.
i think my ideas might contribute to dealing with incompleteness >>>>>>> in fundamental math more generally ... like producing more
refined limits to it's philosophical impact. tho idk if it can be >>>>>>> gotten rid of completely, anymore than we can get rid of the
words "this statement is false"
I don't think that there actually are any limits
except for expressions requiring infinite proofs.
but i am currently focused on the theory of computing and not
anything more generally. the fundamental objects comprising the >>>>>>> theory of computing (machines) are far more constrained in their >>>>>>> definitions than what set theory needs to encompass, and within >>>>>>> those constraints i think i can twist the consensus into some
contradiction that are just entirely ignorant of atm
I have explored all of the key areas. None of them
can be made as 100% perfectly concrete and unequivocal
as computing.
that's the slam dunk left that i need. i have a means to rectify >>>>>>> whatever contradiction we find thru the use of RTMs, but i'm
still teasing out the contradiction that will *force* others to >>>>>>> notice
I do have my refutation of the halting problem itself
boiled down to a rough draft of two first principles.
When the halting problem requires a halt decider
to report on the behavior of a Turing machine this
is always a category error because Turing machines
only take finite string inputs.
The corrected halting problem requires a Turing
machine decider to report in the behavior that its
actual finite string input actually specifies.
polcott, i'm working on making the halting problem complete and
consistent in regards to a subset of the improved "reflective
turing machines" that encompasses all useful computations
i'm sorry, but not about trying to reaffirm the halting function as >>>>> still uncomputable by calling it a category error
I do compute the halting function correctly.
the halting *function* is an abstract mathematical object that maps a
machine description to whether the machine described halts or not,
not the associated machine description that attempts to compute this
All Turing machines only compute the mapping
from input finite strings to some value.
On this basis I do compute halting correctly.
I have been doing this for more than three years.
We probably should not be spamming alt.buddha.short.fat.guy
i hang out there tho, i have my own reasons for posting this there
Theory of computation issues are disrespectful
spam to that group that violate Buddhist compassion.
I was a long time poster to alt.zen. I still
have 8969 messages posted there since 2005.
Also my great grand uncle Henry Steel Olcott
was a very famous Buddhist.
int sum(int x, int y){return x + y;}
sum(3,2) should return 5 and it is incorrect
to require sum(3,2) to return the sum of 5+6.
u can argue about what computing machines actually exist all u want,
and whether anything actually computes the halting *function*, i'm
not going to argue over what the halting *function* itself is
Then you get the wrong answer.
In computability theory and computational complexity
theory, a decision problem is a computational problem
that can be posed as a yes–no question on a set of
input values. https://en.wikipedia.org/wiki/Decision_problem
Most people have no idea that there is such a thing
as an incorrect question. Because of this they misclassify
incorrect yes/no questions as undecidable decision problem
instances.
Most people understand that a question is incorrect if it does not ask.
They also understand that a question can be correct ever when you don't
know the answer.
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