In my last post, I described the peculiar behavior of Skelet #34. After sharing that post on the bbchallenge.org Discord, Justin Blanchard and @Iijil shared some interesting machines with similar looking behavior. I will call these Shift Overflow Counters. They seem to characterized by having a completely orderly “Counter Phase” in which they implement basic double counter until one of the sides overflows (expands) at which point they shift the block offset leading to the other side counter needing to be “reparsed” (in Skelet #34 this shifted 1000 -> 0100).

## Known Examples

Name Machine BBC # Status
sk15 1RB---_1RC1LB_1LD1RE_1LB0LD_1RA0RC 2,204,428 Infinite (Proof Sketch)
sk26 1RB1LD_1RC0RB_1LA1RC_1LE0LA_1LC--- 13,134,219 Infinite (Proof Sketch)
sk33 1RB1LC_0RC0RB_1LD0LA_1LE---_1LA1RE 11,896,833 Infinite (Proof Sketch)
sk34 1RB1LC_0RC0RB_1LD0LA_1LE---_1LA1RA 11,896,832 Infinite (Proven)
sk35 1RB1LC_0RC0RB_1LD0LA_1LE---_1LA0LA 11,896,831 Infinite (Proven)
h19k 1RB1LD_1RC0RB_1LA0LE_1LC0LA_1RZ1RB Link Halt in 19283 steps
h24k 1RB1LD_1RC0RB_1LA1LE_1LC0LA_1RZ0RD Link Halt in 24567 steps

## Machines alike Skelet #34

All of the Skelet machines (sk#) have transition tables very closely related to that of Skelet #34. I provide rough analysis below for all of these here:

### Skelet #35 (sk35)

sk35 has the exact same behavior as sk34 (just slightly faster in one transition).

The only difference between these two machines is that the E1 transition:

• sk34 has E1 -> 1RA
• sk35 has E1 -> 0LA

In fact, the only way to reach state E in either of these machines is via the path: C0 -> 1LD, D0 -> 1LE. So, to reach the E1 transition, we must have tape config: 1 <E 11. Simulating directly from here we get:

• sk34: 1 <E 11 -> <A 011 in 3 step
• sk35: 1 <E 11 -> <A 011 in 1 step

### Skelet #33 (sk33)

sk33 differs from sk34 only in the E1 transition as well. However, in this case, this leads to somewhat different behavior. For example, the “Counter Phase” behavior is now $$D(n, m) \to D(n+1, m+2)$$ and the shift is to the left after overflow.

Some rules:

L(2k)   = L(k) 0000
L(2k+1) = L(k) 0001

G(n, m) = L(n)     <A 010 R(m)
D(n, m) = L(n) 000 <A 010 R(m)

G(n, m) -> G(n+1, m+1)  (if b(m) > 1)
L(n) <A -> L(n+1) B>
0001 <A -> <A 1111
0000 <A -> 0001 B>
E> R(n) -> <C R(n+1)
E> 11 -> 11 E>
E> 10 -> <C 11
11 <C -> <C 10
B> 010 -> 111 E>
111 <C -> <A 010

G(2n, 2^k - 1) -> D(n, 2^k)
0 <A 010 11^k 0 -> <A 010 10^k 11

D(n, m) -> D(n', 0, 2 m' + 1)  (if b(m) > 2n) (n' < n)
1000 0000^k <A 010 R(m) -> 1001 0001^k <A 010 R(m + 2^{k+1} - 1)
-> <A 010 11 10 R(m + 2^{k+1} - 1)

Start -> G(0, 0, 0, 13)  @ Step 83


Note: $$2 m^\prime + 1$$ is always odd, so each counter phase begins in a config $$G(0, 2m+1)$$ and so all end with $$n$$ even as well. The Reset Invariant $$b(m) > 2n$$ also seems to hold, but I have not thought deeply about why yet.

### Skelet #26 (sk26)

sk26 differs from sk33 only in the B0 transition. Specifically (using sk33 state names):

• sk33: B0 -> 0RC
• sk26: B0 -> 1RE

Non-normalized: 1RB1LC_1RE0RB_1LD0LA_1LE---_1LA1RE TNF Normalized: 1RB1LD_1RC0RB_1LA1RC_1LE0LA_1LC--- (C -> D -> E -> C)

sk26 appears to satisfy similar (although slightly more complex) rules to sk34. Specifically (Using the non-normalized state name):

L(2k)   = L(k) 0000
L(2k+1) = L(k) 0001

D(n, a, m) = L(n) 000 <C 101a R(m)
E(n, a, m) = L(n) 00  <C 101a R(m)
F(n, a, m) = L(n) 0   <C 101a R(m)
G(n, a, m) = L(n)     <C 101a R(m)

D(n, a, m) -> D(n+1, a, m+1)  (if b(m) > 1)
L(n) 000 <C -> L(n+1) 000 B>
1000 <C -> <C 1111
0000 <C -> 1000 B>
E> R(n) -> <C R(n+1)
E> 11 -> 11 E>
E> 10 -> <C 11
11 <C -> <C 10
B> 1010 -> 1111 E>
1111 <C -> <C 1010
B> 1011 -> 0111 E>
0111 <C -> <C 1011

D(n-1, 0, 2^k - 1) -> E(n, 1, 2^k)
0 1111 E> 11^k 0 -> <C 1011 10^k 11

E(n, a, m) -> E(n', 1, m')  (if b(m) > n) (n' < n)
0100 0000^k <C 101a R(m) -> 0100 1000^k <C 101a R(m + 2^k - 1)
0100 1000^k <C 101a R(m + 2^k - 1) -> <C 1011 10^2k 11 1a R(m + 2^k - 1)

D(n-1, 1, 2^k - 1) -> G(2n, 0, 2^{k-1})
1000 0000^n 0111 E> 11^m+1 0 -> 01 0000^n+1 <C 1010 10^m 11

G(n, a, m) -> E(n', a', m')  (if b(m) > n) (n' < n)
01 0000^k <C 101a R(m) -> <C 10 10^2k 11 1a R(m + 2^k - 1)

Start -> D(0, 0, 11)  @ Step 85


The tricky bit here appears to be preserving a “Reset Invariant” in this transition $$D(n-1, 1, 2^k - 1) \to G(2n, 0, 2^{k-1})$$ where $$n$$ doubles and $$m$$ is cut in half. For example, the trajectory of the TM actually reaches $$G(2, 0, 1)$$ where $$2 > b(1)$$ early on. In fact, this TM repeatedly hits $$G$$ configs which violate the strict invariant $$b(m) > n$$. However, this invariant is overly strict. In fact we only need $$b(m) > 2^k$$ where $$2^k$$ is the largest power of 2 which divides $$n$$. It appears that that is always true. In fact, experimentally, after an initial bit it looks like $$k = 1$$ every time we reach the $$G$$ config …

### Skelet #15 (sk15)

sk15 is the exact same machine as sk26 except starting on a different start state (sk33s state D or sk26s state E).

Non-normalized: 1RB1LC_1RE0RB_1LD0LA_1LE---_1LA1RE (Start state D) TNF Normalized: 1RB---_1RC1LB_1LD1RE_1LB0LD_1RA0RC (A -> C -> E -> B -> D -> A)

Analysis is the same as for sk26 except with initial condition: Start -> D(0, 0, 1) @ Step 21.