1RB0RA_1LC1LF_1RD0LB_1RA1LE_1RZ0LC_1RG1LD_0RG0RF

1RB0RA_1LC1LF_1RD0LB_1RA1LE_1RZ0LC_1RG1LD_0RG0RF (bbch) is the current BB(7) champion. It runs for over $2{\uparrow }^{11}2{\uparrow }^{11}3$ steps before it halts. It was discovered by Pavel Kropitz on 10 May 2025 (Discord link) and analyzed by Shawn Ligocki (here) on 13 May 2025.

Analysis by Shawn Ligocki

Consider general configurations matching the regex:

$$0^{\mathrm{\infty }}11(1(01{)}^{*}{)}^{*}0011100\text{<mrow data-mjx-texclass="ORD">A<mo>></mo></mrow>}{0}^{\mathrm{\infty }}$$

Low level rules

              01 1 01^n 0011100 A> 00   -->                 1 01^n+2 0011100 A>
           01^3 11 01^n 0011100 A> 0^6  -->            1 01^n+5 1 01 0011100 A>
01^3 (1 01)^k+1 11 01^n 0011100 A> 0^6  -->  1 01^n+6 (1 01)^k 11 01 0011100 A>
 011 (1 01)^k   11 01^n 0011100 A> 0^2  -->  1 Z> 111 01^n+1 00 101^k+2

Mid level rules

Let

$$B(a;b,c,...,z)=0^{\mathrm{\infty }}111(01{)}^{3z+1}1\cdots 1(01{)}^{3c+1}1(01{)}^{3b+1}1(01{)}^{0}1(01{)}^{3a+1}0011100\text{<mrow data-mjx-texclass="ORD">A<mo>></mo></mrow>}{0}^{\mathrm{\infty }}$$

and let B(a; [x]*k, y, ...) = $B(a;\underset{k}{\underbrace{x,\cdots ,x}},y,...)$ (In other words, [x]*k represents k repeats of the value x in a config).

then

B(a; b+1, ...) -> B(2a+4; b, ...)
B(a; [0]*k, 0, n+1, ...) -> B(0; [0]*k, a+2, n, ...)
B(a; [0]*k) -> Halt(3a + 2k + 9)

Start at step 8178: B(2, [1]*12)

High level rule

Let

$$\begin{array}{lll}{g}_0(x)& =& 2x+4\\ g_{k+1}(x)& =& g_k^{x+2}(0)\\ \end{array}$$

then

$$B(a;\underset{k}{\underbrace{0,\cdots ,0}},n,...)\to B(g_k^n(a);\underset{k}{\underbrace{0,\cdots ,0}},0,...)$$

Bound

Let

$$\begin{array}{lll}{a}_0& =& 2\\ a_{k+1}& =& g_k(a_k)\\ \end{array}$$

then

$$B(a_0;\underset{k}{\underbrace{1,\cdots ,1}})\to B(a_k,\underset{k}{\underbrace{0,\cdots ,0}})\to \text{Halt}(3a_k+2k+9)$$and

$$\text{<mrow data-mjx-texclass="ORD">Start</mrow>}\to B(a_0;\underset{12}{\underbrace{1,\cdots ,1}})$$

and so this TM halts with a sigma score of $\sigma =3a_{12}+33$

Note that $g_k(x)=(2{\uparrow }^k(x+4))-4$ and so for $k\ge 2$,

$$a_{k+1}+4>2{\uparrow }^{k}2{\uparrow }^{k}3$$

and so this TM halts with sigma score $\sigma >2{\uparrow }^{11}2{\uparrow }^{11}3$.

This bound is pretty tight: $\sigma <2{\uparrow }^{11}2{\uparrow }^{11}4=2{\uparrow }^{12}4$.