Green's Machines
In 1964 (only 2 years after Tibor Radó first described the Busy Beaver game), Milton W. Green of the Stanford Research Institute handcrafted a family of fastgrowing Turing Machines with \(n\) states, 2 symbols for all \(n \ge 4\).^{1} At the time of publication, Green’s Machines were the Busy Beaver champions for \(BB(n)\) for all \(n \ge 6\). This family grows roughly as fast as the Ackermann function
\[BB_{2n} \approx 3 \uparrow^{n2} 3\]and (as far as I can tell) some of them have not been beaten to this day.
Green’s Machines are still referenced often in discussion about Busy Beavers. However, I have not seen much discussion about the machines themselves. Even reading the original paper is a bit of a challenge. In it Green defines 4 different families of Turing Machines (\(M_N, G_N, B_N, BB_N\)) and I find the subtleties of how each is defined and extended to the other families to be quite confusing every time I read it. In order to support more widespread understanding of the details of Green’s constructions, this post will describe Green’s Machines and their scores in modern notation.
Note: I use names in this article that are adapted from Green’s names, but are not quite identical. The difference is somewhat subtle.
The Machines
The Green Machines are made up of two components, split up by states. These components will be a setup component \(S_k\) and a compute component \(G_n\).
G Component
The main workhorse is the \(G_n\) component. For every odd \(n\), \(G_n\) defined by the transition table:
State  0  1 

\(Q_n\)  \(1 L T_n\)  \(1 R Q_{n+2}\) 
\(T_n\)  \(0 L Q_{n2}\)  \(0 L T_n\) 
…  …  … 
\(Q_k\)  \(1 L T_k\)  \(1 R Q_{k+2}\) 
\(T_k\)  \(0 L Q_{k2}\)  \(0 L T_k\) 
…  …  … 
\(Q_1\)  \(1 R Q_1\)  \(1 R Q_3\) 
Where the \(Q_k, T_k\) are defined this way for all odd \(3 \le k \le n\).
This component has a single entry state (\(Q_n\)) which the setup component should transition into when it wants to run the \(G_n\) component and a single exit state (\(Q_{n+2}\)) which the component will transition out to when it is done with it’s computation.
These compute a series of fastgrowing functions \(G_n(k)\) that I will describe in detail below.
Setup Components
There are two different setup components \(S_3\) and \(S_4\) (with 3 and 4 states respectively). Either can be added to any \(G_n\) component in order to make a Green Machine of either even or odd number of states (respectively).
ThreeState Setup
The \(S_3\) component is defined by:
State  0  1 

\(A = Q_{n+2}\)  \(1LB\)  \(1RZ\) 
\(B\)  \(0LT\)  \(1LT\) 
\(T\)  \(0 L Q_n\)  \(0LT\) 
where \(Q_n\) is the entry state and \(A = Q_{n+2}\) is the exit state of the \(G_n\) component.
When attached to \(G_{2k+1}\), this produces an even state \(BB_{2k+4}\) Green Machine (with start state A
and halt state Z
).
FourState Setup
The \(S_4\) component is defined by:
State  0  1 

\(A\)  \(1LT\)  \(1LB\) 
\(B\)  \(1LZ\)  \(1LA\) 
\(C = Q_{n+2}\)  \(0LB\)  \(1LT\) 
\(T\)  \(0 L Q_n\)  \(0LT\) 
where \(Q_n\) is the entry state and \(C = Q_{n+2}\) is the exit state of the \(G_n\) component.
which, when attached to \(G_{2k+1}\), this produces an odd state \(BB_{2k+5}\) Green Machine (with start state A
and halt state Z
).
Example Machines
Putting this all together and converting into Standard Text format we get for example:
 \(BB_7\):
1LD1LB_1LZ1LA_0LB1LD_0LE0LD_1LF1RC_0LG0LF_1RG1RE
(on bbchallenge)  \(BB_8\):
1LB1RZ_0LC1LC_0LD0LC_1LE1RA_0LF0LE_1LG1RD_0LH0LG_1RH1RF
(on bbchallenge)  \(BB_9\):
1LD1LB_1LZ1LA_0LB1LD_0LE0LD_1LF1RC_0LG0LF_1LH1RE_0LI0LH_1RI1RG
(on bbchallenge)  \(BB_{10}\):
1LB1RZ_0LC1LC_0LD0LC_1LE1RA_0LF0LE_1LG1RD_0LH0LG_1LI1RF_0LJ0LI_1RJ1RH
(on bbchallenge)
Note:
 I have not normalized these to TNF. Instead I’ve left them with the same directions and state order as Green described them for ease of comparison.
 I am using state
Z
in all of these as the halt state.H
is not a halt state, but instead the 8th state.
Class G Machines
In order to motivate the \(G_n\) component, Green defines a general specification for TMs: Class G Machines. A TM, M
, as a Class G Machine (with entry state \(M_{in}\) and exit state \(M_{out}\)) if it conforms to the following behavioral specification:
In words:
 When started in its entry state (\(M_{in}\)) on a
0
 with a string on \(k \ge 0\)
1
s to its left (and nothing else),  it performs some computation without ever traveling to the right of the initial location
 until it eventually leaves to the right in the same state (\(M_{in}\))
 with a string
1
s to its left (and nothing else).
 with a string on \(k \ge 0\)
 When started in its entry state (\(M_{in}\)) on a
1
, it moves immediately to the right and enters the exit state (\(M_{out}\)).
We will say that M
computes the function \(M(k) = m\).
Recurrence
Note: An immediate consequence of the above specification is that:
\[\begin{array}{l} 0^\infty & 1^k & M_{in} > & 0^r & 1 & \to & 0^\infty & 1^{M^r(k) + 1} & M_{out} > \\ \end{array}\]G Components
Each \(G_n\) component is a Class G Machine (with entry state \(Q_n\) and exit state \(Q_{n+2}\)). We can see this via induction.
Base Case
We can see directly that \(G_1\)
State  0  1 

\(Q_1\)  \(1 R Q_1\)  \(1 R Q_3\) 
is a Class G Machine (with entry state \(Q_1\), exit state \(Q_3\)) which computes the function:
\[G_1(k) = k + 1\]Inductive Step
We can build \(G_{n}\) by simply adding two states to \(G_{n2}\):
State  0  1 

\(Q_n\)  \(1 L T_n\)  \(1 R Q_{n+2}\) 
\(T_n\)  \(0 L Q_{n2}\)  \(0 L T_n\) 
Now, assuming that \(G_{n2}\) is a Class G Machine (with entry state \(Q_{n2}\), exit state \(Q_{n}\)), we will prove that \(G_{n}\) is a Class G Machine (with entry state \(Q_{n}\), exit state \(Q_{n+2}\)):
First, the easy part:
\[\begin{array}{l} Q_n > & 1 & \to & 1 & Q_{n+2} > \\ \end{array}\]in a single step.
Second, we will use the inductive assumption which tells us that
\[\begin{array}{l} 0^\infty & 1^k & Q_{n2} > & 0^r & 1 & \to & 0^\infty & 1^{G_{n2}^r(k) + 1} & Q_n > \\ \end{array}\]in order to prove:
\[\begin{array}{l} 0^\infty & 1^k & Q_n > & 0 & \to & 0^\infty & 1^k & < T_n & & 1 \\ & & & & \to & 0^\infty & & < T_n & 0^k & 1 \\ & & & & \to & 0^\infty & & < Q_{n2} & 0^{k+1} & 1 \\ & & & & = & 0^\infty & & Q_{n2} > & 0^{k+2} & 1 \\ & & & & \to & 0^\infty & 1^{G_{n2}^{k+2}(0) + 1} & Q_n > \\ \end{array}\]thus proving that \(G_n\) is a Class G Machine and also that
\[G_n(k) = G_{n2}^{k+2}(0) + 1\]Fastgrowing Hierarchy
Thus we can see that \(G_n\) are all Class G Machines which compute a fastgrowing hierarchy of functions. The sequence begins with:
# States  Function  

\(G_1\)  1  \(G_1(k) = k + 1\) 
\(G_3\)  3  \(G_3(k) = G_1^{k+2}(0) + 1 = k + 3\) 
\(G_5\)  5  \(G_5(k) = G_3^{k+2}(0) + 1 = 3k + 7\) 
\(G_7\)  7  \(G_7(k) = G_5^{k+2}(0) + 1 = \frac{7 \cdot 3^{k+2}  5}{2} > 3^k\) 
\(G_9\)  9  \(G_9(k) = G_7^{k+2}(0) + 1 > 3 \uparrow\uparrow k\) 
Setup Components
While the sequence of \(G_n\) machines above compute impressively large functions, they need a little help when starting from a blank tape. This is where the Setup Components \(S_3\) and \(S_4\) come in. These setup components are very clever, they find a way to efficiently use the \(G_n\) components to initialize the tape for themselves.
ThreeState Setup
Starting from any \(G_n\), we can produce the \(n+3\) state Green Machine \(BB_{n+3}\) by adding the \(S_3\) component:
State  0  1 

\(A = Q_{n+2}\)  \(1LB\)  \(1RZ\) 
\(B\)  \(0LT\)  \(1LT\) 
\(T\)  \(0 L Q_n\)  \(0LT\) 
Then we can see, that, starting in state A
on a blank tape we get the trajectory:
Which halts with exactly \(BB_{n+3} = G_n^{G_n^3(0) + 2}(0) + 2\) 1
s on the tape (this TMs “score”).
Note that:
\[G_{n+2}(k) = G_n^{k+2}(0) + 1\]so we can rewrite these scores as
\[BB_{n+3} = G_n^{G_n^3(0) + 2}(0) + 2 = G_n^{G_{n+2}(1) + 1}(0) + 2 = G_{n+2}(G_{n+2}(1)  1) + 1\]FourState Setup
Starting from any \(G_n\), we can produce the \(n+4\) state Green Machine \(BB_{n+4}\) by adding the \(S_4\) component:
State  0  1 

\(A\)  \(1LT\)  \(1LB\) 
\(B\)  \(1LZ\)  \(1LA\) 
\(C = Q_{n+2}\)  \(0LB\)  \(1LT\) 
\(T\)  \(0 L Q_n\)  \(0LT\) 
Prerequisites
This setup is more subtle to prove. It actually depends upon the fact that \(G_n(k)\) is a “parity swapping function”, in other words that:
\[\begin{array}{l} G_n(2k) & \equiv & 1 & \pmod{2} \\ G_n(2k+1) & \equiv & 0 & \pmod{2} \\ \end{array}\]This turns out to be true for all \(G_n(k)\) (which can be proven easily by induction).
Specifically, we use this fact below in the following transitions:
\[\begin{array}{l} 1^{G_{n+2}(0)} & <B & \to & <A & 1^{G_{n+2}(0)} \\ 1^{G_{n+2}^{G_{n+2}(0)}(G_{n+2}(0))} & <B & \to & <B & 1^{G_{n+2}^{G_{n+2}(0)}(G_{n+2}(0))} \\ \end{array}\]Since \(G_{n+2}(0)\) is odd and \(G_{n+2}^{G_{n+2}(0)}(G_{n+2}(0))\) is even.
We also need to prove a general rule first:
\[\begin{array}{l} 0^\infty & 1^k & C> & 1 & \to & 0^\infty & 1^k & <T & 1 \\ & & & & \to & 0^\infty & <T & 0^k & 1 \\ & & & & \to & 0^\infty & < Q_n & 0^{k+1} & 1 \\ & & & & = & 0^\infty & Q_n > & 0^{k+2} & 1 \\ & & & & \to & 0^\infty & 1^{G_n^{k+2}(0) + 1} & Q_{n+2} > \\ & & & & = & 0^\infty & 1^{G_{n+2}(k)} & C> \\ \end{array}\]Applying this recursively we get:
\[\begin{array}{l} 0^\infty & 1^k & C> & 1^r & \to & 0^\infty & 1^{G_{n+2}^r(k)} & C> \\ \end{array}\]Behavior
Finally, we can see, that, starting this TM in state A
on a blank tape we get the trajectory:
Which halts with exactly \(BB_{n+4} = G_{n+2}^{G_{n+2}(0)}(G_{n+2}(0)) + 1\) 1
s on the tape (this TMs “score”).
Once again, we can simplify:
\[BB_{n+4} = G_{n+2}^{G_{n+2}(0)}(G_{n+2}(0)) + 1 = G_{n+2}^{G_{n+2}(0) + 1}(0) + 1 = G_{n+4}(G_{n+2}(0)  1)\]Summary
Putting this all together, Green defined components \(G_n\) which compute a fast growing hierarchy of functions and two different setup components \(S_3\) and \(S_4\) which can be added on top in order to produce a sequence of TMs \(BB_n\) which start on a blank tape and leave large numbers of 1
s upon halting. Precisely,
Values and Bounds
We can compute the following exact values and (reasonably tight) bounds using Knuth uparrow notation:
\[\begin{array}{l} G_1(k) & = & k + 1 && G_1(0) = 1 & G_1(1) = 2 \\ G_3(k) & = & k + 3 && G_3(0) = 3 & G_3(1) = 4 \\ G_5(k) & = & 3k + 7 && G_5(0) = 7 & G_5(1) = 10 \\ G_7(k) & = & \frac{7 \cdot 3^{k+2}  5}{2} > 3^{k+3} && G_7(0) = 29 & G_7(1) = 92 \\ G_9(k) & > & 3 \uparrow\uparrow (k + 3) && G_9(0) = \frac{7 \cdot 3^{31}  3}{2} \\ G_{11}(k) & > & 3 \uparrow\uparrow\uparrow (k + 3) \\ & \vdots \\ G_{2n+5}(k) & > & 3 \uparrow^n (k + 3) \\ \end{array}\]And similarly for \(BB_n\) values:
\[\begin{array}{lcrcl} BB_{ 4} & = & 7 \\ BB_{ 5} & = & 13 \\ BB_{ 6} & = & 35 \\ BB_{ 7} & = & 22\,961 & > & 3^{ 9} & = & 3^{3^2} \\ BB_{ 8} & = & \frac{7 \cdot 3^{93}  3}{2} & > & 3^{94} & > & 3^{3^4} \\ BB_{ 9} & = & G_9(28) & > & 3 \uparrow\uparrow 31 & > & 3 \uparrow\uparrow 3 \uparrow\uparrow 2 \\ BB_{10} & > & G_9(3 \uparrow\uparrow 4) && & > & 3 \uparrow\uparrow 3 \uparrow\uparrow 4 \\ & \vdots \\ BB_{2n+5} & > & G_{2n+5}(3 \uparrow^{n1} 3) & > & 3 \uparrow^n (3 \uparrow^{n1} 3) & = & 3 \uparrow^n 3 \uparrow^n 2 \\ BB_{2n+6} & > & G_{2n+5}(3 \uparrow^n 4) && & > & 3 \uparrow^n 3 \uparrow^n 4 \\ \end{array}\]Green’s Champions
When Green published his paper in 1964, all \(BB_n\) for \(n \ge 6\) were Busy Beaver champions. However, over the years several have been slowly chipped away. According to Pascal Michel’s Historical survey of Busy Beavers:
 \(BB_6\) was surpassed in 1972 by Donald Lynn: \(\Sigma(6) \ge 42\)
 \(BB_7\) was surpassed in 1990 by Marxen and Buntrock: \(\Sigma(6) \ge 136\,612\) and
 \(BB_8\) was surpassed in 2000 by Marxen and Buntrock: \(\Sigma(6) > 1.4 \times 10^{60}\)
Yet, beyond this I do not know of any TMs beating Green’s Machines until finally, they are eclipsed by Daniel Nagaj’s massive BB(16) > Graham champion (since \(BB_{16} \ll g_2 \ll G\)).
Green Machine  Busy Beaver Bound 

\(BB_{ 9}\)  \(\Sigma(9) > 3 \uparrow\uparrow 31 > 10 \uparrow\uparrow 30\) 
\(BB_{10}\)  \(\Sigma(10) > 3 \uparrow\uparrow 3 \uparrow\uparrow 4 > 10 \uparrow\uparrow 10^{10^{12}}\) 
\(BB_{11}\)  \(\Sigma(11) > 3 \uparrow\uparrow\uparrow 3 \uparrow\uparrow\uparrow 2 > 10 \uparrow\uparrow\uparrow 10^{12}\) 
\(BB_{12}\)  \(\Sigma(12) > 3 \uparrow\uparrow\uparrow 3 \uparrow\uparrow\uparrow 4\) 
\(BB_{13}\)  \(\Sigma(13) > 3 \uparrow^4 3 \uparrow^4 2\) 
\(BB_{14}\)  \(\Sigma(14) > 3 \uparrow^4 3 \uparrow^4 4\) 
\(BB_{15}\)  \(\Sigma(15) > 3 \uparrow^5 3 \uparrow^5 2\) 
Do any of you know of any better results in this range? Let me know in the new comments system below (just set up today!).
Footnotes

Milton W. Green. “A Lower Bound on Rado’s Sigma Function for Binary Turing Machines”, Preceedings of the IEEE Fifth Annual Symposium on Switching Circuits Theory and Logical Design, 1964, pages 91–94, doi: 10.1109/SWCT.1964.3. ↩