Busy Beavers 15-0 Busy Beavers Consider Turing machines - - PDF document

✬ ✫ ✩ ✪ Griffith University 3130CIT Theory of Computation (Based on slides by Harald Søndergaard of The University of Melbourne)

Busy Beavers

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Busy Beavers

Consider Turing machines with tape alphabet { , 1}. Starting with an empty tape, a machine may write a string of 1s and halt. We say that its productivity is the number of 1s. Define Σ(n) =    the productivity of most productive n-state machines An optimally productive machine is a busy beaver.

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Busy Beavers (cont.)

Here is a trivial machine T3 which has productivity 3:

• →1,L
• →1,L
• →1,L
• Of course there are more productive 4-state

machines. (The Turing machine model used here is the two-way tape machine with anonymous halting state.) Note that there are (n + 1)-state machines Tn with productivity n.

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A Busy Beaver with 3 States

Here is a proof that Σ(3) ≥ 6:

• →1,R
• 1,L
• 1,R
• →1,L
• →1,L
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About Busy Beavers

It is known that Σ(1) = 1 (trivial) Σ(2) = 4 (easy) Σ(3) = 6 (fairly hard) Σ(4) = 13 (hard) In early 2003, all that is known about Σ(5) is that it is at least 4098. Out of about 88 million 5-state machines, more than 99.9% have been checked! Σ(6) > 10865. About 30% of the 6-state machines have been checked. Clearly Σ is a very fast-growing function.

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Σ Is Not Computable

Here is a 10-state Turing machine D to double a string of 1s:

• ,R
• ,L

1,R

• ,L
• 1,L
• ,L
• 1,R
• 1→

,L

• ,L
• 1,L
• →1,L

1,L

• →1,R
• 1,R
• ,R
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Σ Is Not Computable (cont.)

If we run D after Tn, we have a machine of productivity 2n. − → Tn − → D − → We have Σ(n + 10) ≥ 2n (1) To show that Σ is not computable, assume that we have a machine BB which computes Σ: n → Σ(n) Let k be the number of states in BB. The following machine: − → Tn − → BB − → BB − → shows that Σ(n + 2k − 1) ≥ Σ(Σ(n)) (2)

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Σ Is Not Computable (cont.)

In summary: Σ(n + 10) ≥ 2n (1) Σ(n + 2k − 1) ≥ Σ(Σ(n)) (2) Note that Σ is total. Σ is also monotonic and strictly increasing, so Σ(j) ≥ Σ(i) ⇒ j ≥ i. Hence for all n, n + 2k − 1 ≥ Σ(n). In particular, n + 2k + 9 ≥ Σ(n + 10) ≥ 2n by (1). So 2k + 9 ≥ n for all n. But this is clearly false—take n = 2k + 10. We conclude that BB does not exist.

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Undecidability Again

We can think of the busy beaver problem as a decidability problem. If we can decide Σ(n) ≤ k for all n and k, then we can compute Σ, and vice versa. So in showing Σ uncomputable we have produced an undecidable problem, without relying on any

• ther results about undecidability.

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TM Halting Revisited

This can also give us an easy alternative proof of the undecidability of TM halting. Namely, we can reduce the problem of computing Σ to the halting problem. Here is why Σ would be computable if the halting problem was decidable: On input n:

• 1. Generate all n-state Turing machines.
• 2. Filter out all the non-terminating machines.
• 3. Run the rest until all have halted.
• 4. Pick the most productive machine.
• 5. Print its productivity.

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