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BB(n) is roughly the maximum number of new states an n-state TM can run into without halting. So for a particular n, if we know BB(n), then we can find out if an arbitrary n-state TM halts by running it for BB(n) steps.

So given some n-state TM, can we construct an m-state TM, where m>n, to calculate BB(n) and thus decides if it halts?

Now some might say that this is equivalent to solving the halting problem, but I don't think so. Because given an arbitrary TM, we don't in fact know how many states it has. And if we don't know n, we certainly can't find m.

If the answer is in the negative, then how is it that we can in fact find out the values of BB(n) for small n? What's the connection between incomputability and unknowability?

Gilles 'SO- stop being evil'
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hollow7
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For a particular $n$, there are at most $2^{O(n)}$ different TMs with $n$ states, and we can have a list for all of them: 1 is the TM halts, and 0 otherwise. A TM that has this "list" in its memory can decide the problem you give in your question. Not only this language is decidable - it is even regular!

The "halting problem" states though, that there is no TM that works for all possible $n$'s.

Ran G.
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