On 1/25/23 7:23 PM, Richard Damon wrote:
On 1/25/23 6:51 PM, olcott wrote:
On 1/25/2023 5:29 PM, olcott wrote:
On 1/25/2023 5:15 PM, Richard Damon wrote:
On 1/25/23 11:51 AM, olcott wrote:
On 1/25/2023 5:46 AM, Richard Damon wrote:
On 1/25/23 12:13 AM, olcott wrote:That makes the Halting Problem ill-defined:
On 1/24/2023 10:09 PM, Richard Damon wrote:
On 1/24/23 10:51 PM, olcott wrote:
On 1/24/2023 9:40 PM, Richard Damon wrote:
On 1/24/23 10:10 PM, olcott wrote:
On 1/24/2023 8:03 PM, Richard Damon wrote:
On 1/24/23 7:26 PM, olcott wrote:
On 1/24/2023 5:41 PM, Richard Damon wrote:
On 1/24/23 10:41 AM, olcott wrote:
In computability theory, the halting problem is the >>>>>>>>>>>>>>>> problem ofRight, the Decider must decide if the actual running of >>>>>>>>>>>>>>> the program described by the input would halt when given >>>>>>>>>>>>>>> the input that is the rest of that input.
determining, from a description of an arbitrary computer >>>>>>>>>>>>>>>> program and an
input, whether the program will finish running, or >>>>>>>>>>>>>>>> continue to run
forever. https://en.wikipedia.org/wiki/Halting_problem >>>>>>>>>>>>>>>>
This definition of the halting problem measures >>>>>>>>>>>>>>>> correctness in a non-
pathological way, thus in the same way that ZFC >>>>>>>>>>>>>>>> (redefined set theory
and) eliminated Russell's Paradox the previously >>>>>>>>>>>>>>>> "impossible" input
ceases to be impossible.
In computability theory, the halting problem is the >>>>>>>>>>>>>>>> problem of defining
a machine that correctly determines from a description >>>>>>>>>>>>>>>> of an arbitrary
computer program and an input, whether or not its >>>>>>>>>>>>>>>> specified input pair
would terminate normally by reaching it own final state. >>>>>>>>>>>>>>>
It is NOT asking if the
The conventional proofs do not actually show that such a >>>>>>>>>>>>>>>> machine cannot
be defined. HH(PP, PP) does correctly determine that its >>>>>>>>>>>>>>>> correctly
simulated input cannot possibly reach the final state of >>>>>>>>>>>>>>>> PP and
terminate normally. (See pages 5-6 of this paper) >>>>>>>>>>>>>>>>
https://www.researchgate.net/publication/364657019_Simulating_Halt_Decider_Applied_to_the_Halting_Theorem
If the simulation is incorrect then there must a line of >>>>>>>>>>>>>>>> the simulation
that behaves differently than what its corresponding >>>>>>>>>>>>>>>> line of machine-
code specifies.
No, it is incorrect because it is incomplete.
You are just usong the INCORRECT definition of "Correct", >>>>>>>>>>>>>>> and since you have been told this in the past, it just >>>>>>>>>>>>>>> shows that you are a LIAR about this fact.
"Correct Simulation" is defined as a simulation that >>>>>>>>>>>>>>> exactly matches the actual behavior of the machine the >>>>>>>>>>>>>>> input describes.
It turns out that is incorrect. The ultimate measure of a >>>>>>>>>>>>>> correct
simulation is whether or not the simulated input exactly >>>>>>>>>>>>>> matches the
behavior specified by its machine code.
Nope, Sourcd for that claim.
Counter-examples cannot possibly exist.
Try and show any correct simulation where the simulator does >>>>>>>>>>>> not simulate what the machine language specifies.
This is the counter example.
Since the direct eecution of the machine language of PP(PP) >>>>>>>>>>> will Halt, the correct simulation of it must match.
That is merely a provably false assumption.
Then do so.
Remember, your HH has been admitted to return 0 from HH(PP,PP), >>>>>>>>> and to be a computation, it must do the same to EVERY call.
YOU have posted the execution trace of the direct execution of >>>>>>>>> the equivalent to PP, which shows it halts.
You can see on pages 5-6 that PP correctly simulated by HH
cannot possibly reach it own final state.
But that isn't the questipon, since that question, like the
Liar's paradox has no answer.
No, your RESTATEMENT is ill-defined. The behavior of P is well
defined given a proper definition of H
H(P,P) can do one of 4 things, and can't do anything else.
1) H(P,P) can return 0, in which case P(P) will halt, and H is
shown to be wrong. This is what you claim your H does when directly
called.
2) H(P,P) can retrun 1, in which case, P(P) will go into an
infinite loop, and H is shown to be wrong.
3) H(P,P) can just dies and halt and not return an answer, in which
case H fails to be the needed halt decider, and P(P) will be halting. >>>>>
4) H(P,P) can get stuck in an infinte loop, and never return an
answer, in which case H fails to be the needed halt decider, and
P(P) will be non-halting. This is what you seem to claim is what H
does when simulated inside P.
In the study of problem solving, any problem in which the initial
state or starting position, the allowable operations, and the goal >>>>>> state are clearly specified, *and a unique solution can be shown
to exist*
Right, and given an actual definition of the complete algorithm of
H (and 'Get the right answer' is NOT an complete algorithm) there
is a precise correct answer to the problem.
Unfortunately of H, H can never give that answer.
Thus making the halting problem ill-defined.
No machine can possibly be defined that divides all pairs of finite
strings into those that represent machines would halt on their input
when directly executed and those that do not in the same way and for the >>> same reason that there is no barber that shaves all and only those that
do not shave themselves. ZFC eliminated this problem by declaring it is
erroneous.
Right, which shows that the Haltng Theorm is CORRECT, that that the
Halting Function is not computable.
So, you agree with the Theorem that you have been arguing against for
2 decades?
Every decision problem where
*a unique solution CANNOT be shown to exist*
is erroneous.
Nope, just shows that the problem is not computable.
If you think uncompuatable problems are erroneous, then you can't
handle all of mathematics.
One simple comment that comes to mind that points out the error in your thinking:
The number of possible computing machines is a countable infinite,
because we can express every such machine as a finite string of a finite symbol set.
The number of possible deciders that can be defined is an UNCOUNTABLE infinite.
On 1/25/2023 6:33 PM, Richard Damon wrote:
One simple comment that comes to mind that points out the error in
your thinking:
The number of possible computing machines is a countable infinite,
because we can express every such machine as a finite string of a
finite symbol set.
The number of possible deciders that can be defined is an UNCOUNTABLE
infinite.
I had a very very long debate about this with a PhD computer scientist.
When we make one single universal halt decider that correctly determines
the halt status of any arbitrary input pair then the countability issue ceases to exist.
On 1/25/23 7:48 PM, olcott wrote:
On 1/25/2023 6:33 PM, Richard Damon wrote:
One simple comment that comes to mind that points out the error in
your thinking:
The number of possible computing machines is a countable infinite,
because we can express every such machine as a finite string of a
finite symbol set.
The number of possible deciders that can be defined is an UNCOUNTABLE
infinite.
I had a very very long debate about this with a PhD computer scientist.
When we make one single universal halt decider that correctly determines
the halt status of any arbitrary input pair then the countability issue
ceases to exist.
So, by assuming an impossible machine exists, you can prove a false statement.
On 1/25/2023 7:11 PM, Richard Damon wrote:
On 1/25/23 7:48 PM, olcott wrote:
On 1/25/2023 6:33 PM, Richard Damon wrote:
One simple comment that comes to mind that points out the error in
your thinking:
The number of possible computing machines is a countable infinite,
because we can express every such machine as a finite string of a
finite symbol set.
The number of possible deciders that can be defined is an
UNCOUNTABLE infinite.
I had a very very long debate about this with a PhD computer scientist.
When we make one single universal halt decider that correctly determines >>> the halt status of any arbitrary input pair then the countability issue
ceases to exist.
So, by assuming an impossible machine exists, you can prove a false
statement.
By hypothesizing a single halt decider for the infinite set of arbitrary finite string pairs countability ceases to be an issue.
On 1/25/23 8:35 PM, olcott wrote:
On 1/25/2023 7:11 PM, Richard Damon wrote:
On 1/25/23 7:48 PM, olcott wrote:
On 1/25/2023 6:33 PM, Richard Damon wrote:
One simple comment that comes to mind that points out the error in
your thinking:
The number of possible computing machines is a countable infinite,
because we can express every such machine as a finite string of a
finite symbol set.
The number of possible deciders that can be defined is an
UNCOUNTABLE infinite.
I had a very very long debate about this with a PhD computer scientist. >>>> When we make one single universal halt decider that correctly
determines
the halt status of any arbitrary input pair then the countability issue >>>> ceases to exist.
So, by assuming an impossible machine exists, you can prove a false
statement.
By hypothesizing a single halt decider for the infinite set of arbitrary
finite string pairs countability ceases to be an issue.
So your answer to trying to do the impossible is to just assume you can
do something impossible.
On 1/25/2023 7:42 PM, Richard Damon wrote:
On 1/25/23 8:35 PM, olcott wrote:
On 1/25/2023 7:11 PM, Richard Damon wrote:
On 1/25/23 7:48 PM, olcott wrote:
On 1/25/2023 6:33 PM, Richard Damon wrote:
One simple comment that comes to mind that points out the error in >>>>>> your thinking:
The number of possible computing machines is a countable infinite, >>>>>> because we can express every such machine as a finite string of a
finite symbol set.
The number of possible deciders that can be defined is an
UNCOUNTABLE infinite.
I had a very very long debate about this with a PhD computer
scientist.
When we make one single universal halt decider that correctly
determines
the halt status of any arbitrary input pair then the countability
issue
ceases to exist.
So, by assuming an impossible machine exists, you can prove a false
statement.
By hypothesizing a single halt decider for the infinite set of arbitrary >>> finite string pairs countability ceases to be an issue.
So your answer to trying to do the impossible is to just assume you
can do something impossible.
*Not at all*
Under the conditions that I specified countability definitely does
become moot, thus conclusively proving that countability
is not always an issue.
On 1/25/23 9:07 PM, olcott wrote:
On 1/25/2023 7:42 PM, Richard Damon wrote:
On 1/25/23 8:35 PM, olcott wrote:
On 1/25/2023 7:11 PM, Richard Damon wrote:
On 1/25/23 7:48 PM, olcott wrote:
On 1/25/2023 6:33 PM, Richard Damon wrote:
One simple comment that comes to mind that points out the error
in your thinking:
The number of possible computing machines is a countable
infinite, because we can express every such machine as a finite
string of a finite symbol set.
The number of possible deciders that can be defined is an
UNCOUNTABLE infinite.
I had a very very long debate about this with a PhD computer
scientist.
When we make one single universal halt decider that correctly
determines
the halt status of any arbitrary input pair then the countability
issue
ceases to exist.
So, by assuming an impossible machine exists, you can prove a false
statement.
By hypothesizing a single halt decider for the infinite set of
arbitrary
finite string pairs countability ceases to be an issue.
So your answer to trying to do the impossible is to just assume you
can do something impossible.
*Not at all*
Under the conditions that I specified countability definitely does
become moot, thus conclusively proving that countability
is not always an issue.
So in a land of fairy dust power magical mythical unicorns, you think
you can count them.
On 1/25/2023 8:16 PM, Richard Damon wrote:
On 1/25/23 9:07 PM, olcott wrote:
On 1/25/2023 7:42 PM, Richard Damon wrote:
On 1/25/23 8:35 PM, olcott wrote:
On 1/25/2023 7:11 PM, Richard Damon wrote:
On 1/25/23 7:48 PM, olcott wrote:
On 1/25/2023 6:33 PM, Richard Damon wrote:
One simple comment that comes to mind that points out the error >>>>>>>> in your thinking:
The number of possible computing machines is a countable
infinite, because we can express every such machine as a finite >>>>>>>> string of a finite symbol set.
The number of possible deciders that can be defined is an
UNCOUNTABLE infinite.
I had a very very long debate about this with a PhD computer
scientist.
When we make one single universal halt decider that correctly
determines
the halt status of any arbitrary input pair then the countability >>>>>>> issue
ceases to exist.
So, by assuming an impossible machine exists, you can prove a
false statement.
By hypothesizing a single halt decider for the infinite set of
arbitrary
finite string pairs countability ceases to be an issue.
So your answer to trying to do the impossible is to just assume you
can do something impossible.
*Not at all*
Under the conditions that I specified countability definitely does
become moot, thus conclusively proving that countability
is not always an issue.
So in a land of fairy dust power magical mythical unicorns, you think
you can count them.
Anyone can count to one. Do you disagree?
One TM and an arbitrary finite string pair also applies to deriving the
sum of natural numbers as pairs of finite strings. We need not count
these strings we only need to compute the sum of any arbitrary pair.
On 1/25/2023 8:45 PM, Richard Damon wrote:
On 1/25/23 9:26 PM, olcott wrote:
On 1/25/2023 8:16 PM, Richard Damon wrote:
On 1/25/23 9:07 PM, olcott wrote:
On 1/25/2023 7:42 PM, Richard Damon wrote:
On 1/25/23 8:35 PM, olcott wrote:
On 1/25/2023 7:11 PM, Richard Damon wrote:
On 1/25/23 7:48 PM, olcott wrote:
On 1/25/2023 6:33 PM, Richard Damon wrote:
One simple comment that comes to mind that points out the
error in your thinking:
The number of possible computing machines is a countable
infinite, because we can express every such machine as a
finite string of a finite symbol set.
The number of possible deciders that can be defined is an
UNCOUNTABLE infinite.
I had a very very long debate about this with a PhD computer >>>>>>>>> scientist.
When we make one single universal halt decider that correctly >>>>>>>>> determines
the halt status of any arbitrary input pair then the
countability issue
ceases to exist.
So, by assuming an impossible machine exists, you can prove a
false statement.
By hypothesizing a single halt decider for the infinite set of
arbitrary
finite string pairs countability ceases to be an issue.
So your answer to trying to do the impossible is to just assume
you can do something impossible.
*Not at all*
Under the conditions that I specified countability definitely does
become moot, thus conclusively proving that countability
is not always an issue.
So in a land of fairy dust power magical mythical unicorns, you
think you can count them.
Anyone can count to one. Do you disagree?
One TM and an arbitrary finite string pair also applies to deriving the
sum of natural numbers as pairs of finite strings. We need not count
these strings we only need to compute the sum of any arbitrary pair.
Nope, not the "sum" of an arbtrary pair. Yes, "Summing" is computable.
The number of possible functions to compute goes exponentially to the
number of input values.
Yes of course everyone knows that we must have a separate TM to compute
the sum of each distinct input pair. It is utterly impossible to define
a single machine that could compute the sum of 3+4 and also compute the
sum of 2+1.
Even if by some magic this would be possible everyone knows that it is utterly impossible to chain these calls together to derive the sum of
more elements than a pair.
sum(sum(2,3), sum(4,5)) is utterly unimaginable total nonsense.
Recursive calls are a form of psychosis.
Did you notice that was sarcasm?
On 1/25/23 9:26 PM, olcott wrote:
On 1/25/2023 8:16 PM, Richard Damon wrote:
On 1/25/23 9:07 PM, olcott wrote:
On 1/25/2023 7:42 PM, Richard Damon wrote:
On 1/25/23 8:35 PM, olcott wrote:
On 1/25/2023 7:11 PM, Richard Damon wrote:
On 1/25/23 7:48 PM, olcott wrote:
On 1/25/2023 6:33 PM, Richard Damon wrote:
One simple comment that comes to mind that points out the error >>>>>>>>> in your thinking:
The number of possible computing machines is a countable
infinite, because we can express every such machine as a finite >>>>>>>>> string of a finite symbol set.
The number of possible deciders that can be defined is an
UNCOUNTABLE infinite.
I had a very very long debate about this with a PhD computer
scientist.
When we make one single universal halt decider that correctly
determines
the halt status of any arbitrary input pair then the
countability issue
ceases to exist.
So, by assuming an impossible machine exists, you can prove a
false statement.
By hypothesizing a single halt decider for the infinite set of
arbitrary
finite string pairs countability ceases to be an issue.
So your answer to trying to do the impossible is to just assume you
can do something impossible.
*Not at all*
Under the conditions that I specified countability definitely does
become moot, thus conclusively proving that countability
is not always an issue.
So in a land of fairy dust power magical mythical unicorns, you think
you can count them.
Anyone can count to one. Do you disagree?
One TM and an arbitrary finite string pair also applies to deriving the
sum of natural numbers as pairs of finite strings. We need not count
these strings we only need to compute the sum of any arbitrary pair.
Nope, not the "sum" of an arbtrary pair. Yes, "Summing" is computable.
The number of possible functions to compute goes exponentially to the
number of input values.
On 1/25/23 10:26 PM, olcott wrote:
On 1/25/2023 8:45 PM, Richard Damon wrote:
On 1/25/23 9:26 PM, olcott wrote:
On 1/25/2023 8:16 PM, Richard Damon wrote:
On 1/25/23 9:07 PM, olcott wrote:
On 1/25/2023 7:42 PM, Richard Damon wrote:
On 1/25/23 8:35 PM, olcott wrote:
On 1/25/2023 7:11 PM, Richard Damon wrote:
On 1/25/23 7:48 PM, olcott wrote:
On 1/25/2023 6:33 PM, Richard Damon wrote:
One simple comment that comes to mind that points out the >>>>>>>>>>> error in your thinking:
The number of possible computing machines is a countable >>>>>>>>>>> infinite, because we can express every such machine as a >>>>>>>>>>> finite string of a finite symbol set.
The number of possible deciders that can be defined is an >>>>>>>>>>> UNCOUNTABLE infinite.
I had a very very long debate about this with a PhD computer >>>>>>>>>> scientist.
When we make one single universal halt decider that correctly >>>>>>>>>> determines
the halt status of any arbitrary input pair then the
countability issue
ceases to exist.
So, by assuming an impossible machine exists, you can prove a >>>>>>>>> false statement.
By hypothesizing a single halt decider for the infinite set of >>>>>>>> arbitrary
finite string pairs countability ceases to be an issue.
So your answer to trying to do the impossible is to just assume
you can do something impossible.
*Not at all*
Under the conditions that I specified countability definitely does >>>>>> become moot, thus conclusively proving that countability
is not always an issue.
So in a land of fairy dust power magical mythical unicorns, you
think you can count them.
Anyone can count to one. Do you disagree?
One TM and an arbitrary finite string pair also applies to deriving the >>>> sum of natural numbers as pairs of finite strings. We need not count
these strings we only need to compute the sum of any arbitrary pair.
Nope, not the "sum" of an arbtrary pair. Yes, "Summing" is computable.
The number of possible functions to compute goes exponentially to the
number of input values.
Yes of course everyone knows that we must have a separate TM to compute
the sum of each distinct input pair. It is utterly impossible to define
a single machine that could compute the sum of 3+4 and also compute the
sum of 2+1.
????? That is a standard first year exercise.
Even if by some magic this would be possible everyone knows that it is
utterly impossible to chain these calls together to derive the sum of
more elements than a pair.
So, "Sum" is a computable function.
You falling into the proof by example problem agian.
sum(sum(2,3), sum(4,5)) is utterly unimaginable total nonsense.
Recursive calls are a form of psychosis.
Did you notice that was sarcasm?
So, rather than try to actually come up with an answewr, you deflect
with garbage.
On 1/25/2023 9:39 PM, Richard Damon wrote:
On 1/25/23 10:26 PM, olcott wrote:
On 1/25/2023 8:45 PM, Richard Damon wrote:
On 1/25/23 9:26 PM, olcott wrote:
On 1/25/2023 8:16 PM, Richard Damon wrote:
On 1/25/23 9:07 PM, olcott wrote:
On 1/25/2023 7:42 PM, Richard Damon wrote:
On 1/25/23 8:35 PM, olcott wrote:
On 1/25/2023 7:11 PM, Richard Damon wrote:
On 1/25/23 7:48 PM, olcott wrote:
On 1/25/2023 6:33 PM, Richard Damon wrote:
One simple comment that comes to mind that points out the >>>>>>>>>>>> error in your thinking:
The number of possible computing machines is a countable >>>>>>>>>>>> infinite, because we can express every such machine as a >>>>>>>>>>>> finite string of a finite symbol set.
The number of possible deciders that can be defined is an >>>>>>>>>>>> UNCOUNTABLE infinite.
I had a very very long debate about this with a PhD computer >>>>>>>>>>> scientist.
When we make one single universal halt decider that correctly >>>>>>>>>>> determines
the halt status of any arbitrary input pair then the
countability issue
ceases to exist.
So, by assuming an impossible machine exists, you can prove a >>>>>>>>>> false statement.
By hypothesizing a single halt decider for the infinite set of >>>>>>>>> arbitrary
finite string pairs countability ceases to be an issue.
So your answer to trying to do the impossible is to just assume >>>>>>>> you can do something impossible.
*Not at all*
Under the conditions that I specified countability definitely
does become moot, thus conclusively proving that countability
is not always an issue.
So in a land of fairy dust power magical mythical unicorns, you
think you can count them.
Anyone can count to one. Do you disagree?
One TM and an arbitrary finite string pair also applies to deriving
the
sum of natural numbers as pairs of finite strings. We need not count >>>>> these strings we only need to compute the sum of any arbitrary pair. >>>>>
Nope, not the "sum" of an arbtrary pair. Yes, "Summing" is computable. >>>>
The number of possible functions to compute goes exponentially to
the number of input values.
Yes of course everyone knows that we must have a separate TM to compute
the sum of each distinct input pair. It is utterly impossible to define
a single machine that could compute the sum of 3+4 and also compute the
sum of 2+1.
????? That is a standard first year exercise.
Even if by some magic this would be possible everyone knows that it is
utterly impossible to chain these calls together to derive the sum of
more elements than a pair.
So, "Sum" is a computable function.
You falling into the proof by example problem agian.
sum(sum(2,3), sum(4,5)) is utterly unimaginable total nonsense.
Recursive calls are a form of psychosis.
Did you notice that was sarcasm?
So, rather than try to actually come up with an answewr, you deflect
with garbage.
I have proven that countability is not an issue.
That you fail to understand that is not my problem.
On 1/25/23 9:26 PM, olcott wrote:
On 1/25/2023 8:16 PM, Richard Damon wrote:
On 1/25/23 9:07 PM, olcott wrote:
On 1/25/2023 7:42 PM, Richard Damon wrote:
On 1/25/23 8:35 PM, olcott wrote:
On 1/25/2023 7:11 PM, Richard Damon wrote:
On 1/25/23 7:48 PM, olcott wrote:
On 1/25/2023 6:33 PM, Richard Damon wrote:
One simple comment that comes to mind that points out the error >>>>>>>>> in your thinking:
The number of possible computing machines is a countable
infinite, because we can express every such machine as a finite >>>>>>>>> string of a finite symbol set.
The number of possible deciders that can be defined is an
UNCOUNTABLE infinite.
I had a very very long debate about this with a PhD computer
scientist.
When we make one single universal halt decider that correctly
determines
the halt status of any arbitrary input pair then the
countability issue
ceases to exist.
So, by assuming an impossible machine exists, you can prove a
false statement.
By hypothesizing a single halt decider for the infinite set of
arbitrary
finite string pairs countability ceases to be an issue.
So your answer to trying to do the impossible is to just assume you
can do something impossible.
*Not at all*
Under the conditions that I specified countability definitely does
become moot, thus conclusively proving that countability
is not always an issue.
So in a land of fairy dust power magical mythical unicorns, you think
you can count them.
Anyone can count to one. Do you disagree?
One TM and an arbitrary finite string pair also applies to deriving the
sum of natural numbers as pairs of finite strings. We need not count
these strings we only need to compute the sum of any arbitrary pair.
Nope, not the "sum" of an arbtrary pair. Yes, "Summing" is computable.
The number of possible functions to compute goes exponentially to the
number of input values.
8 inputs to decide, 256 functions
9 inputs to decide, 512 functions
10 inputs to decide, 1024 functions
On 1/25/2023 8:45 PM, Richard Damon wrote:
On 1/25/23 9:26 PM, olcott wrote:
Nope, not the "sum" of an arbtrary pair. Yes, "Summing" is computable.
The number of possible functions to compute goes exponentially to the
number of input values.
8 inputs to decide, 256 functions
9 inputs to decide, 512 functions
10 inputs to decide, 1024 functions
The above is pure Cockamamie bullshit.
One TM has an arbitrary number of space delimited finite strings of
ASCII digits that it computes the sum of.
On 1/25/23 11:52 PM, olcott wrote:
On 1/25/2023 8:45 PM, Richard Damon wrote:
On 1/25/23 9:26 PM, olcott wrote:
Nope, not the "sum" of an arbtrary pair. Yes, "Summing" is computable.
The number of possible functions to compute goes exponentially to the
number of input values.
8 inputs to decide, 256 functions
9 inputs to decide, 512 functions
10 inputs to decide, 1024 functions
The above is pure Cockamamie bullshit.
One TM has an arbitrary number of space delimited finite strings of
ASCII digits that it computes the sum of.
So, you have no idea of what you are talking, and think all computations
are mearly sumations.
On 1/26/2023 5:47 AM, Richard Damon wrote:
On 1/25/23 11:52 PM, olcott wrote:
On 1/25/2023 8:45 PM, Richard Damon wrote:
On 1/25/23 9:26 PM, olcott wrote:
Nope, not the "sum" of an arbtrary pair. Yes, "Summing" is computable. >>>>
The number of possible functions to compute goes exponentially to
the number of input values.
8 inputs to decide, 256 functions
9 inputs to decide, 512 functions
10 inputs to decide, 1024 functions
The above is pure Cockamamie bullshit.
One TM has an arbitrary number of space delimited finite strings of
ASCII digits that it computes the sum of.
So, you have no idea of what you are talking, and think all
computations are mearly sumations.
If a TM that computes the sum of a finite number of arbitrary finite
strings of ASCII digits is computable then a TM that computes anything
else on a finite number of arbitrary finite strings is not uncomputable
for any countability reason.
On 1/26/23 11:46 AM, olcott wrote:
On 1/26/2023 5:47 AM, Richard Damon wrote:
On 1/25/23 11:52 PM, olcott wrote:
On 1/25/2023 8:45 PM, Richard Damon wrote:
On 1/25/23 9:26 PM, olcott wrote:
Nope, not the "sum" of an arbtrary pair. Yes, "Summing" is computable. >>>>>
The number of possible functions to compute goes exponentially to
the number of input values.
8 inputs to decide, 256 functions
9 inputs to decide, 512 functions
10 inputs to decide, 1024 functions
The above is pure Cockamamie bullshit.
One TM has an arbitrary number of space delimited finite strings of
ASCII digits that it computes the sum of.
So, you have no idea of what you are talking, and think all
computations are mearly sumations.
If a TM that computes the sum of a finite number of arbitrary finite
strings of ASCII digits is computable then a TM that computes anything
else on a finite number of arbitrary finite strings is not uncomputable
for any countability reason.
WHy? Summing is a very simple operation, that we can do that doesn't say anthing about more complicated operations.
On 1/26/2023 5:37 PM, Richard Damon wrote:
On 1/26/23 11:46 AM, olcott wrote:It does completely prove that countability is not an issue for the
On 1/26/2023 5:47 AM, Richard Damon wrote:
On 1/25/23 11:52 PM, olcott wrote:
On 1/25/2023 8:45 PM, Richard Damon wrote:
On 1/25/23 9:26 PM, olcott wrote:
Nope, not the "sum" of an arbtrary pair. Yes, "Summing" is
computable.
The number of possible functions to compute goes exponentially to
the number of input values.
8 inputs to decide, 256 functions
9 inputs to decide, 512 functions
10 inputs to decide, 1024 functions
The above is pure Cockamamie bullshit.
One TM has an arbitrary number of space delimited finite strings of
ASCII digits that it computes the sum of.
So, you have no idea of what you are talking, and think all
computations are mearly sumations.
If a TM that computes the sum of a finite number of arbitrary finite
strings of ASCII digits is computable then a TM that computes anything
else on a finite number of arbitrary finite strings is not uncomputable
for any countability reason.
WHy? Summing is a very simple operation, that we can do that doesn't
say anthing about more complicated operations.
halting problem thus all proofs to the contrary have been refuted.
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