So to bring it back to cousin_it’s post, one could have a “busy beaver candidate” (a turing machines where we don’t know whether it goes on for ever or eventually halt or start repeating), and the encoding of a natural number n would be that the first bit is whether or not that candidate halts before n steps, followed by the unary encoding of n.
“Encoding” or decoding n would be easy (but long when n is big), but “breaking” that encoding for any usy beaver candidate would require a halting oracle.
Suppose I pass you the bit that says the candidate does not halt, followed by an infinite string of 1s. Then to decode this (by which I mean reject it as invalid) you would need to know whether the busy beaver candidate halts, which we’ve rejected as hard.
This is a problem with the Sphinxes, too, in retrospect. A Hollow Sphinx could just keep answering “yes” if it’s hard to check whether the Turing machine halts, making you do the hard work.
Agreed, but a non-oracle Sphinx wouldn’t be impossible to recognize any more, it would just be very hard to recognize (you would need the Sphinx to guess wrong, and to be patient enough to figure out that it did).
In summary, whoever has the halting oracle wins, and if nobody does, it’s a patience contest.
Huh, that looks like it could work, neat :)
So to bring it back to cousin_it’s post, one could have a “busy beaver candidate” (a turing machines where we don’t know whether it goes on for ever or eventually halt or start repeating), and the encoding of a natural number n would be that the first bit is whether or not that candidate halts before n steps, followed by the unary encoding of n.
“Encoding” or decoding n would be easy (but long when n is big), but “breaking” that encoding for any usy beaver candidate would require a halting oracle.
Suppose I pass you the bit that says the candidate does not halt, followed by an infinite string of 1s. Then to decode this (by which I mean reject it as invalid) you would need to know whether the busy beaver candidate halts, which we’ve rejected as hard.
This is a problem with the Sphinxes, too, in retrospect. A Hollow Sphinx could just keep answering “yes” if it’s hard to check whether the Turing machine halts, making you do the hard work.
Agreed, but a non-oracle Sphinx wouldn’t be impossible to recognize any more, it would just be very hard to recognize (you would need the Sphinx to guess wrong, and to be patient enough to figure out that it did).
In summary, whoever has the halting oracle wins, and if nobody does, it’s a patience contest.