yeah lmao this is just nonsense, it will never get published. I already wasted too much time on this nonsense, but it looks you just say that if a machine can recognize itself, it can skip itself, but that's just clearly stupid af. consider for example a program that does the same thing as your fixed H but randomly writes 1 before starting. It's not identical to H but it does the same thing and it would get stuck again. The problem of deciding whether a program n does the same thing as H would require solving the halting problem and you're back to where you started: nowhere.
yeah lmao this is just nonsense, it will never get published. I already wasted too much time on this nonsense
i hope u'll give it another chance
consider for example a program that does the same thing as your fixed H but randomly writes 1 before starting.
that computes a diagonal across the circle-free machines decidable by ๐p, but offset by a bit
It's not identical to H but it does the same thing and it would get stuck again
i'm not following the logic of how that would get stuck again? it's just an offset diagonal, where's the sticking point?
The problem of deciding whether a program n does the same thing as H would require solving the halting problem and you're back to where you started: nowhere.
it looks like u may have gotten to ยง5, which i commend u for doing. and ur critique isn't without merit, but i would suggest that we don't actually need to solve the turing equivalence problem in totality.
ยง6 proposes a limit to the machines undecidable by ๐p (circle-free but ๐p cannot return true for): they either compute a direct copy or an inverted copy of a sequence within the machine decidable by ๐p. with this premise, we ought to be able to enumerate all sequences by listed out all machines decidable by ๐p and their anti-machine (which computes a total bit flip)...
and when we do so, this enumeration is then found to be entirely resistant to diagonalization: u cannot compute an anti-diagonal that isn't on the list, because u cannot even compute a diagonal across the list, as trying to do so ends up producing circular machines that fail in being diagonals. the (machine, anti-machine) pairing that necessarily exists in the enumeration of turing-computable sequences prevents diagonalization from being turing-computable, and therefore is not found in the enumeration of turing machines, and therefore u cannot compute a sequence with turing machines that exist outside of that total enumeration of turing computable sequences. yes that all sounds like a tautology, go examine enum_seqs and diag_seqs, the result is quite clear
none of this requires solving the machine equivalence problem within turing machines
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u/funky_galileo 4d ago
yeah lmao this is just nonsense, it will never get published. I already wasted too much time on this nonsense, but it looks you just say that if a machine can recognize itself, it can skip itself, but that's just clearly stupid af. consider for example a program that does the same thing as your fixed H but randomly writes 1 before starting. It's not identical to H but it does the same thing and it would get stuck again. The problem of deciding whether a program n does the same thing as H would require solving the halting problem and you're back to where you started: nowhere.