The busy beaver function Σ

The busy beaver function quantifies the maximum score attainable by a busy beaver on a given measure. This is a noncomputable function. Also, a busy beaver function can be shown to grow faster asymptotically than any computable function.^[3]

The busy beaver function, ${\displaystyle \Sigma$ :\mathbb {N} \to \mathbb {N} } , is defined so that Σ(n) is the maximum attainable score (the maximum number of 1s finally on the tape) among all halting 2-symbol n-state Turing machines of the above-described type, when started on a blank tape.

It is clear that Σ is a well-defined function: for every n, there are at most finitely many n-state Turing machines as above, up to isomorphism, hence at most finitely many possible running times.

This infinite sequence Σ is the busy beaver function, and any n-state 2-symbol Turing machine M for which σ(M) = Σ(n) (i.e., which attains the maximum score) is called a busy beaver. Note that for each n, there exist at least four n-state busy beavers (because, given any n-state busy beaver, another is obtained by merely changing the shift direction in a halting transition, another by shifting all direction changes to their opposite, and the final by shifting the halt direction of the all-swapped busy beaver. Theoretically, there could be more than one kind of transition leading to the halting state, but in practice it would be wasteful, because there's only one sequence of state transitions producing the sought-after result).

Maximum shifts function S

In addition to the function Σ, Radó [1962] introduced another extreme function for Turing machines, the maximum shifts function, S, defined as follows:

• s(M) = the number of shifts M makes before halting, for any M ∈ E_n,
• S(n) = max{s(M) | M ∈ E_n} = the largest number of shifts made by any halting n-state 2-symbol Turing machine.

Because these Turing machines are required to have a shift in each and every transition or "step" (including any transition to a Halt state), the max-shifts function is at the same time a max-steps function.

Radó showed that S is noncomputable for the same reason that Σ is noncomputable — it grows faster than any computable function. He proved this simply by noting that for each n, S(n) ≥ Σ(n). Each shift may write a 0 or a 1 on the tape, while Σ counts a subset of the shifts that wrote a 1, namely the ones that hadn't been overwritten by the time the Turing machine halted; consequently, S grows at least as fast as Σ, which had already been proved to grow faster than any computable function.

The following connection between Σ and S was used by Lin & Radó [Computer Studies of Turing Machine Problems, 1965] to prove that Σ(3) = 6: For a given n, if S(n) is known then all n-state Turing machines can (in principle) be run for up to S(n) steps, at which point any machine that hasn't yet halted will never halt. At that point, by observing which machines have halted with the most 1s on the tape (i.e., the busy beavers), one obtains from their tapes the value of Σ(n). The approach used by Lin & Radó for the case of n = 3 was to conjecture that S(3) = 21, then to simulate all the essentially different 3-state machines for up to 21 steps. By analyzing the behavior of the machines that had not halted within 21 steps, they succeeded in showing that none of those machines would ever halt, thus proving the conjecture that S(3) = 21, and determining that Σ(3) = 6 by the procedure just described.

Inequalities relating Σ and S include the following (from [Ben-Amram, et al., 1996]), which are valid for all n ≥ 1:

${\displaystyle {\begin{aligned}S(n)&\geq \Sigma (n)\\S(n)&\leq (2n-1)\Sigma (3n+3)\\S(n)&<\Sigma (3n+6);\end{aligned}}}$

and an asymptotically improved bound (from [Ben-Amram, Petersen, 2002]): there exists a constant c, such that for all n ≥ 2,

${\displaystyle S(n)\leq \Sigma \left(n+\left\lceil {\frac {8n}{\log _{2}n}}\right\rceil +c\right).}$

${\displaystyle S(n)}$ tends to be close to the square of ${\displaystyle \Sigma (n)}$, and in fact many machines give ${\displaystyle S(n)}$ less than ${\displaystyle \Sigma ^{2}(n)}$.

Known values for Σ and S

As of 2016 the function values for Σ(n) and S(n) are only known exactly for n < 5.^[6]

The current (as of 2023) 5-state busy beaver champion produces 4098 1s, using 47176870 steps (discovered by Heiner Marxen and Jürgen Buntrock in 1989), but there remain many machines with non-regular behavior which are believed to never halt, but which have not been proven to run infinitely. Various sources list different numbers of these holdouts. Skelet lists 42 or 43 unproven machines.^[13]

At the moment the record 6-state champion produces over 10⇈15^[14] (found by Pavel Kropitz in 2022). As noted above, these are 2-symbol Turing machines.

Milton Green, in his 1964 paper "A Lower Bound on Rado's Sigma Function for Binary Turing Machines", constructed a set of Turing machines demonstrating that

${\displaystyle \Sigma (2k)>3\uparrow ^{k-2}3>A(k-2,k-2)\qquad {\text{for }}k\geq 2,}$

where ↑ is Knuth up-arrow notation and A is Ackermann's function.

Thus

${\displaystyle \Sigma (10)>3\uparrow \uparrow \uparrow 3=3\uparrow \uparrow 3^{3^{3}}=3^{3^{3^{.^{.^{.^{3}}}}}}}$

(with 3³³ = 7625597484987 terms in the exponential tower), and

${\displaystyle \Sigma (12)>3\uparrow \uparrow \uparrow \uparrow 3=g_{1},}$

where the number g₁ is the enormous starting value in the sequence that defines Graham's number.

In 1964 Milton Green developed a lower bound for the Busy Beaver function that was published in the proceedings of the 1964 IEEE symposium on switching circuit theory and logical design. Heiner Marxen and Jürgen Buntrock described it as "a non-trivial (not primitive recursive) lower bound". This lower bound can be calculated but is too complex to state as a single expression in terms of n. When n=8 the method gives

Σ(8) ≥ 3 × (7 × 3⁹² - 1) / 2 ≈ 8.248×10⁴⁴.

It can be derived from current lower bounds that:

${\displaystyle \Sigma (2k+1)>3\uparrow ^{k-2}31>A(k-2,k-2)\qquad {\text{for }}k\geq 2,}$

In contrast, the best current (as of 2023) lower bound on Σ(6) is 10⇈15, which is greater than the lower bound given by Green's formula, 3³ = 27 (which is tiny in comparison). In fact, it is much greater than the lower bound: 3⇈3 = 3³³ = 7625597484987, which is Green's first lower bound for Σ(8), and also much greater than the second lower bound: 3×(7×3⁹²−1)/2.

Proof for uncomputability of S(n) and Σ(n)

Suppose that S(n) is a computable function and let EvalS denote a TM, evaluating S(n). Given a tape with n 1s it will produce S(n) 1s on the tape and then halt. Let Clean denote a Turing machine cleaning the sequence of 1s initially written on the tape. Let Double denote a Turing machine evaluating function n + n. Given a tape with n 1s it will produce 2n 1s on the tape and then halt. Let us create the composition Double | EvalS | Clean and let n₀ be the number of states of this machine. Let Create_n₀ denote a Turing machine creating n₀ 1s on an initially blank tape. This machine may be constructed in a trivial manner to have n₀ states (the state i writes 1, moves the head right and switches to state i + 1, except the state n₀, which halts). Let N denote the sum n₀ + n₀.

Let BadS denote the composition Create_n₀ | Double | EvalS | Clean. Notice that this machine has N states. Starting with an initially blank tape it first creates a sequence of n₀ 1s and then doubles it, producing a sequence of N 1s. Then BadS will produce S(N) 1s on tape, and at last it will clear all 1s and then halt. But the phase of cleaning will continue at least S(N) steps, so the time of working of BadS is strictly greater than S(N), which contradicts to the definition of the function S(n).

The uncomputability of Σ(n) may be proved in a similar way. In the above proof, one must exchange the machine EvalS with EvalΣ and Clean with Increment — a simple TM, searching for a first 0 on the tape and replacing it with 1.

The uncomputability of S(n) can also be established by reference to the blank tape halting problem. The blank tape halting problem is the problem of deciding for any Turing machine whether or not it will halt when started on an empty tape. The blank tape halting problem is equivalent to the standard halting problem and so it is also uncomputable. If S(n) was computable, then we could solve the blank tape halting problem simply by running any given Turing machine with n states for S(n) steps; if it has still not halted, it never will. So, since the blank tape halting problem is not computable, it follows that S(n) must likewise be uncomputable.

Generalizations

For any model of computation there exist simple analogs of the busy beaver. For example, the generalization to Turing machines with n states and m symbols defines the following generalized busy beaver functions:

1. Σ(n, m): the largest number of non-zeros printable by an n-state, m-symbol machine started on an initially blank tape before halting, and
2. S(n, m): the largest number of steps taken by an n-state, m-symbol machine started on an initially blank tape before halting.

For example, the longest-running 3-state 3-symbol machine found so far runs 119112334170342540 steps before halting.^[15] The longest running 6-state, 2-symbol machine which has the additional property of reversing the tape value at each step produces 6147 1s after 47339970 steps. So for the Reversal Turing Machine (RTM) class,^[16] S_RTM(6) ≥ 47339970 and Σ_RTM(6) ≥ 6147.

It is possible to further generalize the busy beaver function by extending to more than one dimension.

Likewise we could define an analog to the Σ function for register machines as the largest number which can be present in any register on halting, for a given number of instructions.

Exact values and lower bounds

The following table lists the exact values and some known lower bounds for S(n, m) and Σ(n, m) for the generalized busy beaver problems. Note: entries listed as "?" are bounded from below by the maximum of all entries to left and above. These machines either haven't been investigated or were subsequently surpassed by a smaller machine.

Values of S(n, m)
n m	2-state	3-state	4-state	5-state	6-state
2-symbol	6	21	107	≥ 47176870	> 10⇈15
3-symbol	38	≥ 119112334170342540	> 1.0×1014072	?	?
4-symbol	≥ 3932964	> 10⇈2048	?	?	?
5-symbol	> 1.9×10704	?	?	?	?
6-symbol	> 10⇈10⇈1010115	?	?	?	?
Values of Σ(n, m)
n m	2-state	3-state	4-state	5-state	6-state
2-symbol	4	6	13	≥ 4098	> 10⇈15
3-symbol	9	≥ 374676383	> 1.3×107036	?	?
4-symbol	≥ 2050	> 10⇈2048	?	?	?
5-symbol	> 1.7×10352	?	?	?	?
6-symbol	> 10⇈10⇈1010115	?	?	?	?

Nondeterministic Turing machines

The problem can be extended to Nondeterministic Turing machines by looking for the system with the most states across all branches or the branch with the longest number of steps.^[17] The question of whether a given NDTM will halt is still computationally irreducible, and the computation required to find an NDTM busy beaver is significantly greater than the deterministic case, since there are multiple branches that need to be considered. For a 2-state, 2-color system with p cases or rules, the table to the right gives the maximum number of steps before halting and maximum number of unique states created by the NDTM.

Notable instances

A 745-state binary Turing machine has been constructed that halts if and only if ZFC is inconsistent.^[10][11] A 744-state Turing machine has been constructed that halts if, and only if, the Riemann hypothesis is false.^[7][20] A 43-state Turing machine has been constructed that halts if, and only if, Goldbach's conjecture is false, and a 27-state machine for that conjecture has been proposed but not yet verified.^[7][20]

A 15-state Turing machine has been constructed that halts if and only if the following conjecture formulated by Paul Erdős in 1979 is false: for all n > 8 there is at least one digit 2 in the base 3 representation of 2ⁿ.^[21][22]