There is of course not currently any upper bound on BB(6).
I find the question about a probvious upper bound more interesting - There is also not a probvious upper bound on BB(6), as this would require at least some understanding of the high-level behavior of all remaining holdout machines. However, there may soon be a 'probvious' upper bound on the value of BB(3,3) (BB for 3-state, 3-symbol machines). Up to equivalence, there are four remaining machines to decide to find the value of BB(3,3). One is a 'probviously halting' machine which will be the new champion if it halts, and for which probabilistic models of its behavior predict with high probability an exact halting time. One is a 'probviously nonhalting' machine. The two other machines are not well-understood enough to say whether they have any probvious long term behavior, but some suspect that they both 'probviously nonhalt'. If this is true it could be said that a 'probvious' upper bound exists for BB(3,3).
I think finding an upper bound is basically just as difficult as finding the actual value itself, since both would require proving that all of the programs which run longer than that will run forever. That's why we can say BB(x) grows faster than any computable function. Being able to compute BB(x) algorithmically or any faster growing function would let you solve the halting problem
Not sure if this is a joke, but actually that is guaranteed to be true. It is proven that for all n: BB(n+1) >= BB(n) + 3. But it is not proven that BB(n+1) >= BB(n) + 4, haha.
The point stands: the hard part is proving that all the programs with longer runtime than your upper bound will never terminate, and once you've solved that, getting the exact value is just a little extra work
For arbitrary n, that proof is arbitrarily hard, even undecidable for large enough n. Again though, for the specific case n=6, that difficulty has not yet been demonstrated, especially if you're willing to accept probabilistic arguments instead of rigorous proofs. n-by-n checkers is PSPACE-complete but the specific case n=8 that people actually play, has been completely solved using computers.
There is no general procedure for computing upper bounds on busy beaver numbers (this can be proven). We haven't even come close to enumerating all of the interesting six-state Turing machines, so right now we don't even have a wild guess for an upper bound on BB(6).
There is of course not currently any upper bound on BB(6).
I find the question about a probvious upper bound more interesting - There is also not a probvious upper bound on BB(6), as this would require at least some understanding of the high-level behavior of all remaining holdout machines. However, there may soon be a 'probvious' upper bound on the value of BB(3,3) (BB for 3-state, 3-symbol machines). Up to equivalence, there are four remaining machines to decide to find the value of BB(3,3). One is a 'probviously halting' machine which will be the new champion if it halts, and for which probabilistic models of its behavior predict with high probability an exact halting time. One is a 'probviously nonhalting' machine. The two other machines are not well-understood enough to say whether they have any probvious long term behavior, but some suspect that they both 'probviously nonhalt'. If this is true it could be said that a 'probvious' upper bound exists for BB(3,3).
I think finding an upper bound is basically just as difficult as finding the actual value itself, since both would require proving that all of the programs which run longer than that will run forever. That's why we can say BB(x) grows faster than any computable function. Being able to compute BB(x) algorithmically or any faster growing function would let you solve the halting problem
Sure, but I only asked about the single case x=6.
If you want an unproven-but-almost-certainly-correct upper bound on BB(6), consider BB(12).
Not sure if this is a joke, but actually that is guaranteed to be true. It is proven that for all n: BB(n+1) >= BB(n) + 3. But it is not proven that BB(n+1) >= BB(n) + 4, haha.
The point stands: the hard part is proving that all the programs with longer runtime than your upper bound will never terminate, and once you've solved that, getting the exact value is just a little extra work
For arbitrary n, that proof is arbitrarily hard, even undecidable for large enough n. Again though, for the specific case n=6, that difficulty has not yet been demonstrated, especially if you're willing to accept probabilistic arguments instead of rigorous proofs. n-by-n checkers is PSPACE-complete but the specific case n=8 that people actually play, has been completely solved using computers.
There is no general procedure for computing upper bounds on busy beaver numbers (this can be proven). We haven't even come close to enumerating all of the interesting six-state Turing machines, so right now we don't even have a wild guess for an upper bound on BB(6).