On 1/22/26 3:56 PM, Tristan Wibberley wrote:
On 22/01/2026 23:21, dart200 wrote:
On 1/21/26 5:05 PM, Tristan Wibberley wrote:
On 21/01/2026 16:38, dart200 wrote:
a current thesis of mine is that given a total enumeration of TM (in >>>>> order of increasing complexity/number of states), no paradoxical
machine
produces an input->output mapping that is a first of it's kind - ie
there will always be some simpler machine that exists and produces an >>>>> equivalent input->output mapping.
Just for clarity, do you mean to say that for every tuple of tuples
{initial states {tape|umachine} |u final states {tape|umachine}} there >>>> is a
halting machine that has fewer states than any non-halting machine?
Yes, because none of the nonhalting machines has a final state so its
tuple can't be constructed and therefore is not first of anything, and >>>> is /not/, just /isn't/.
Do you mean there is a simpler machine with the same initial state for >>>> each of the nonhalting machine's reachable states?
let us refer to turing's solution for defining "output" in regards to
infinitely running machines, because they can still have output:
the tape is divided into alternating cells: F-squares (figure) and
E-squares (erasure). F-squares hold permanent output of the computed
sequence, and can only be written once. E-squares are for
temporary/scratch work necessary for the computation, but not actually
the output of the machine.
I think you need to restrict your quantification of machines called
"paradoxical" to those that only write such squares once yet have a
when i say paradoxical, i'm referring a machines that cannot be decided
upon from some specific interface because they query that interface and contradict it.
ur all fucking useless or i wouldn't have to be here posting...
that's because i'm exploring it in ways that previous have gone unexplored
On 24/01/2026 22:28, dart200 wrote:
that's because i'm exploring it in ways that previous have gone unexplored
I don't believe that at all. Do you mean ways whose previous
explorations are obscure to you and/or not set in front of you with an abstract that makes it clear to you that /that/ is what it is and that
it is worth your attention?
On 23/01/2026 05:51, dart200 wrote:
On 1/22/26 3:56 PM, Tristan Wibberley wrote:
On 22/01/2026 23:21, dart200 wrote:
On 1/21/26 5:05 PM, Tristan Wibberley wrote:
On 21/01/2026 16:38, dart200 wrote:
a current thesis of mine is that given a total enumeration of TM (in >>>>>> order of increasing complexity/number of states), no paradoxical
machine
produces an input->output mapping that is a first of it's kind - ie >>>>>> there will always be some simpler machine that exists and produces an >>>>>> equivalent input->output mapping.
Just for clarity, do you mean to say that for every tuple of tuples
{initial states {tape|umachine} |u final states {tape|umachine}} there >>>>> is a
halting machine that has fewer states than any non-halting machine?
Yes, because none of the nonhalting machines has a final state so its >>>>> tuple can't be constructed and therefore is not first of anything, and >>>>> is /not/, just /isn't/.
Do you mean there is a simpler machine with the same initial state for >>>>> each of the nonhalting machine's reachable states?
let us refer to turing's solution for defining "output" in regards to
infinitely running machines, because they can still have output:
the tape is divided into alternating cells: F-squares (figure) and
E-squares (erasure). F-squares hold permanent output of the computed
sequence, and can only be written once. E-squares are for
temporary/scratch work necessary for the computation, but not actually >>>> the output of the machine.
I think you need to restrict your quantification of machines called
"paradoxical" to those that only write such squares once yet have a
when i say paradoxical, i'm referring a machines that cannot be decided
upon from some specific interface because they query that interface and
contradict it.
You're talking about embodiments of fixed-points via machine state (what would be the finite-sized part of a machine, not relying on the tape and repeated nested simulation). Am I right?
That's a c-machine, not an a-machine, right?
ur all fucking useless or i wouldn't have to be here posting...
Do you mean to tell me you're posting to comp.theory under an obligation
to do so and because (some or all of) the readers of the group are useless?
I'm curious how you received the obligation to post for we useless few.
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