For a given n-state busy beaver game, is the busy beaver function unique, or might there be multiple functions with the same maximum score? Perhaps it has not been proven either way?
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2A very interesting question, but perhaps better suited for http://cstheory.stackexchange.com/ ? – Aasmund Eldhuset Sep 06 '11 at 14:29
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Yes, it is.
The busy beaver function is defined so that
\Sigma(n) = max { \sigma(M) | M is a halting n-state 2-symbol Turing machine}
The maximum is unique if it exists, which it does (Rado proved this). This is just a number.
Therefore \Sigma(n) is also unique, and so the discrete function \Sigma: N --> N is also unique. There may be multiple ways to extend \Sigma to a continuous function, but why someone would want to do this is beyond me.
It's possible to compute small values of \Sigma; check out the OEIS entry for the largest known values.
PengOne
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As @PengOne pointed out, the function is indeed unique. It is a completely defined N -> N discrete function.
However, from your formulation ("or might there be multiple functions with the same maximum score") it can also be understood that you want to know whether there are multiple busy-beavers that give the same maximum. If that is the case, then yes, there are at least 2 busy-beavers given an N, one is constructed from the other by simply reversing the shifts.
Luchian Grigore
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Never mind. I understand. Because the mapping is the same, just the intermediate steps are different. Thank you. – Daniel Sep 06 '11 at 21:39
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@Daniel, I think you are using a non-standard terminology. The standard terminology (for example, used in the Wikipedia article you link) is that the busy beaver function is a function which tells you the maximum score for all n-state Turing machines. There is only one function. However, there are multiple Turing machines which attain this maximum (as Luchian Grigore mentioned). These machines can be thought of as programs in the Turing machine language; you appear to be calling them functions which has caused confusion here. – sligocki Sep 12 '11 at 16:55
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This has been asked a long time ago, but I found this interesting: http://www.win.tue.nl/~wijers/shallit.pdf
Also, I coded an algorithm that brute forces the 3-state busy-beaver problem, and it gave me about 22 non-symetrical configurations that produced 6 symbols (consecutive or not). This means there are perhaps 60-some configurations if you consider you can swap state 1 and state 2, as well as inverse the first transition.
But that's only for the amount of symbols produced, not the 'longest execution' one.
Tama Yoshi
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