The following claim from p246 of Turing’s seminal paper On Computable Numbers is a fallacy:
/the problem of enumerating computable sequences is equivalent to the
problem of finding out whether a given number is the D.N of a circle-
free machine, and we have no general process for doing this in a finite number of steps/
Annoyingly, the linguists markup of a leading asterisk is also taken in >usenet formatting markup for a bullet mark:
The following claim from p246 of Turing’s seminal paper On Computable Numbers is a fallacy:You've missed something out. That something is something like "If I
/the problem of enumerating computable sequences is equivalent to the problem of finding out whether a given number is the D.N of a circle-What's an "infinite circle-free machine"? All the theory we are talking
free machine, and we have no general process for doing this in a finite number of steps/
For any given computable sequence, there are _infinite_ circle-free
machines ....
.... which compute that particular sequence. Not only can variousI think you'll need to prove that you can identify a machine that
machines differ significantly in the specific steps to produce the
same output, machines can be changed in superficial ways that do not meaningfully affect the steps of computation, akin to modern no-op
statements or unreachable code
The problem of enumerating computable sequences, however, only depends
on successfully identifying _one_ circle-free machine that computes any given computable sequences. While identifying more than one can
certainly be done, it is _not_ a requirement for enumerating computable sequences, as _one_ machine computing a sequence /suffices to output any
and all digits of that sequence/
The problem of enumerating computable sequences is therefore _not_Your given reason fails to rule out the equivalence. For a start, under
actually equivalent to a _general process_ of enumerating circle-free machines, as there is no need to identify all circle-free machines which compute any given computable sequence
Said problem is only equivalent to a _limited process_ of enumerating circle-free machines. The machine which identifies circle-free machinesAgain, can this machine exist? It seems to me that the "limited power"
only needs the limited power of determining _at least one_ circle-free machine for any given computable sequence, _not all_ machines for any
given computable sequence
Because of this fallacy, the proof found on the following p247, where an ill-defined machine 𝓗 (which attempts and fails to compute the direct diagonal β’) is found to be undecidable in respect to circle-freeIf this were a fallacy, you should write it up properly, get it
decider 𝓓; does not then prove an impossibility for enumerating computable sequences. As the problem of enumerating /all circle-free machines/ is _not_ equivalent to that of enumerating /just computable sequences/
----
arising us out of the computing dark ages,
please excuse my pseudo-pyscript,
~ the little crank that could
[ Followup-To: set ]
In comp.theory dart200 <[email protected]d> wrote:
The following claim from p246 of Turing’s seminal paper On Computable
Numbers is a fallacy:
You've missed something out. That something is something like "If I understand correctly" or "As far as I can see". Without such a qualification, your statement just looks like extreme hubris.
It is vanishingly unlikely that you have understood correctly or you
have seen far enough. If there were a flaw in Turing's 1936 paper, and
it were subtle enough to escape detection by the millions of specialists
who have verified it, it would certainly be beyond your powers as
somebody lacking education in the subject to spot it.
/the problem of enumerating computable sequences is equivalent to the
problem of finding out whether a given number is the D.N of a circle-
free machine, and we have no general process for doing this in a finite
number of steps/
For any given computable sequence, there are _infinite_ circle-free
machines ....
What's an "infinite circle-free machine"?
All the theory we are talking about deals only with finite machines.
.... which compute that particular sequence. Not only can various
machines differ significantly in the specific steps to produce the
same output, machines can be changed in superficial ways that do not
meaningfully affect the steps of computation, akin to modern no-op
statements or unreachable code
The problem of enumerating computable sequences, however, only depends
on successfully identifying _one_ circle-free machine that computes any
given computable sequences. While identifying more than one can
certainly be done, it is _not_ a requirement for enumerating computable
sequences, as _one_ machine computing a sequence /suffices to output any
and all digits of that sequence/
I think you'll need to prove that you can identify a machine that
computes a given computable sequence. By definition of "computable"
there is such a machine, but how do you "identify" it? Suppose you had
a computable number and a machine, how would you test whether or not
that machine generates the number?
The problem of enumerating computable sequences is therefore _not_
actually equivalent to a _general process_ of enumerating circle-free
machines, as there is no need to identify all circle-free machines which
compute any given computable sequence
Your given reason fails to rule out the equivalence. For a start, under
what relationship does/doesn't this equivalence hold?
Said problem is only equivalent to a _limited process_ of enumerating
circle-free machines. The machine which identifies circle-free machines
only needs the limited power of determining _at least one_ circle-free
machine for any given computable sequence, _not all_ machines for any
given computable sequence
Again, can this machine exist? It seems to me that the "limited power"
you describe is not at all limited. It is known that there is no
machine which can ascertain whether or not two turing machines produce
the same output. So the ability to discard machines redundant in this
sense will not exist either.
Because of this fallacy, the proof found on the following p247, where an
ill-defined machine 𝓗 (which attempts and fails to compute the direct
diagonal β’) is found to be undecidable in respect to circle-free
decider 𝓓; does not then prove an impossibility for enumerating
computable sequences. As the problem of enumerating /all circle-free
machines/ is _not_ equivalent to that of enumerating /just computable
sequences/
If this were a fallacy, you should write it up properly, get it
published and become famous. Far more likely, however, is that you have
just failed to understand what Turing's paper means.
--
arising us out of the computing dark ages,
please excuse my pseudo-pyscript,
~ the little crank that could
If there were a flaw in Turing's 1936 paper, and^^^^
it were subtle enough to escape detection by the millions of specialists^^^^^
who have verified it, it would certainly be beyond your powers as
somebody lacking education in the subject to spot it.
I think you'll need to prove that you can identify a machine that
computes a given computable sequence. By definition of "computable"
there is such a machine, but how do you "identify" it? Suppose you had
a computable number and a machine, how would you test whether or not
that machine generates the number?
On 10/03/2026 16:51, dart200 wrote:
The following claim from p246 of Turing’s seminal paper On Computable Numbers is a fallacy:
/the problem of enumerating computable sequences is equivalent to the problem of finding out whether a given number is the D.N of a circle-
free machine, and we have no general process for doing this in a finite number of steps/
I am delighted to see that your usage of forward-slashes is valid as an example of usenet text formatting markup but frustrated that it's so
close to being an example of the linguist's forward-slash markup to
indicate a wrong example.
Alas, the meaning of the sentence is what is wrong, although perhaps the syntax is technically wrong somehow too, and perhaps some intrinsic
semantic problem is present too so we can say they are the linguists
error markers too after all.
Annoyingly, the linguists markup of a leading asterisk is also taken in usenet formatting markup for a bullet mark:
* this nonsense sentence is
/too crap this is sentence a/
It is known that there is no
machine which can ascertain whether or not two turing machines produce
the same output.
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