1RB1RF_1LC1LF_0RE1LD_0LB1LD_---1RC_1RA0RD

1RB1RF_1LC1LF_0RE1LD_0LB1LD_---1RC_1RA0RD (bbch) is an unverified probviously non-halting BB(6) machine found by Racheline on 24 October 2025.

Argument by Racheline:

this one is probviously non-halting i think
basically, the tape length has a parity, which changes every time the bouncer hits both ends at the same time. after the parity changes, the iterated bouncer starts from the left end, roughly doubling in size with each iteration, and when it hits the right end, it moves the right end to fix the parity, *except* if it hits both ends at the same time again, in which case it halts.
assuming that the starting position of the bouncer behaves pseudorandomly and with uniform distribution, the probability of a parity change at a given length should decay linearly with the length, which means the lengths at which the parity changes happen should grow exponentially. for the machine to halt, the length would have to be one of the lengths of the iterated bouncer, and those grow exponentially too. so we need two exponentially growing sequences to intersect, which i think makes it probviously non-halting (unless it gets unlucky and halts early)

Analysis by Opus 4.7 / DrDisentangle

This machine only has ONE clean rule (DBL), with the rest being irregular state-dependent transitions that can't be captured by parametric rules.

start: C(2, 1)       [reached from blank tape in 5 steps]

C(z, m) := 0^inf C> (01)^z 1^m 0^inf    (state C on the leftmost 0)

Rule DBL:   C(z, m) -> C(2z, m+2-2z)            in exactly 6*z^2 micro-steps. Requires z >= 2 and m+2 >= 2z.

Rule P123:  C(z, m) -> {D, 1, [false]++ones 4++(zebra(z-2)).reverse, ones(m-2)++zeros p}
                                                in 2z+9 steps. (Intermediate — not a C-macro state; used as a stepping stone inside tm_DBL.)
                                                Requires z >= 2 and m >= 2.

When the DBL hypothesis $m+2>=2z$ FAILS (i.e. $m$ is "too small"), the next macro transition is irregular. Different $(z,m)$ states with the same shape can evolve to very different $(z^{\prime },m^{\prime })$.

  Step      State     DBL applies?                                                                           
  ─────────────────────────────────        
     5     C(2,1)     no  → 21 irregular steps                                                               
    26     C(2,4)     yes (DBL chain begins)                                                                 
    50     C(4,2)     no  → parity flip (C(4,2)→C(2,11))                                                     
   102     C(2,11)    yes (chain: 2→4→8)                                                                     
   126     C(4,9)     yes                                                                                    
   222     C(8,3)     no  (m+2=5 < 16) → 194 irregular steps                                                 
   416     C(10,1)    no  → 196 irregular                                                                    
   612     C(8,7)     no  → 186 irregular                                                                    
   798     C(6,13)    yes (chain: 6→12)                                                                      
  1014     C(12,3)    no  → 292 irregular                                                                    
  …                                                                                                          

The only TM-level halting transition is E,0 -> ---. At the macro level this would require a zebra-sweep to run off the end of the zebra into a right-side 0 — which never happens in the DBL regime (the zebra is bounded by the ones region). It COULD happen in the irregular regime if the bouncer "hits both ends simultaneously".

Proofs are formalized in Lean