If the BB function is computable, does that mean we know how to compute {i | program i eventually halts when run with input 0}, which is a clear contradiction with halting problem.
Does this proof work? Am I missing anything?
Thanks
If the BB function is computable, does that mean we know how to compute {i | program i eventually halts when run with input 0}, which is a clear contradiction with halting problem.
Does this proof work? Am I missing anything?
Thanks
Since the question has yet to receive an answer, I'll add this comment as an answer for future visitors.
You ask "If the BB function is computable [...]", well, the Busy Beaver function is not computable, if by BB function you mean given a size $s$ find $BB(s)$.
Now, some terminology: Busy beaver is a game where you want to find, given an integer $n$, the Turing machine $BB(n)$ that has
Now, as you correctly guessed,
Determining whether an arbitrary Turing machine is a busy beaver is undecidable.