No fixed left lookahead determines the boundary outcome

These are confined local rules for constructed configurations. They do not identify a halt from the blank tape. All words are written left to right. The starting state is C on cell 0, the first displayed one. The endpoint is at cell -1, and that endpoint cell is not read. The suffix v is arbitrary and unread; the exterior to the left is also arbitrary and unread.

For integers q,m >= 1, the first pair is:

C> 1^(3q) 00 v
    -> F at -1, right word 11 (011)^(q-1) 111 v
       8q+7 steps

C> 1^(3q) 0 1111 00 1^(3m) 00 v
    -> A at -1, right word (011)^q 110 (011)^(m+1) 111 v
       8q+8m+45 steps.

The inputs share the first 3q+1 bits. Because q is arbitrary, no bounded prefix of the right word determines the A/F exit. With a suitable suffix, both words have no internal 000 and no isolated one-runs.

A stronger pair uses 3q+1 initial ones:

C> 1^(3q+1) 00 v
    -> A at -1, right word (011)^q 111 v
       8q+9 steps

C> 1^(3q+1) 0 1111 00 1^(3m) 00 v
    -> D at -1, right word 0 (011)^q 110 (011)^(m+1) 111 v
       8q+8m+47 steps.

These share the first 3q+2 bits. Put a single one at cell -1 and zeros everywhere farther left. The first family then reaches E0 after two additional transitions and halts. The second family reaches F on a zero with an entirely blank left tail after one additional transition and never halts. Thus arbitrarily long shared left prefixes can coexist with opposite halting outcomes, even within these simple run constraints. This obstruction concerns candidate languages for arbitrary constructed core inputs, not a language restricted to the blank orbit.

Short finite bridges

The following bridges are confined to the displayed words, with endpoints unread:

1. A on the final one of 111 transforms it to 011, ends A immediately left, and takes five steps.
2. A on the first zero of 00 transforms it to 10, ends C immediately right, and takes two steps.
3. A on the final one of 11 leaves 11 unchanged, ends F immediately left, and takes four steps. Its primitive sequence is A1, D1, F0, F0.
4. A on the displayed 1 changes it to 0, ends D immediately left, and takes one step.

The existing C reflection rule sends C on the first bit of u00 to A on the penultimate bit of u, replaces the final zeros by ones, and takes |u|+4 zeros(u)+8 steps. Here u begins/ends in one and contains no 00.

The existing residue-one rule, for any j>=0, is:

A on final bit of 1 0 1^(3j+1)
    -> 110 (011)^j, A immediately left, 5j+3 steps.

The two-reflection bridge

Let N>=1, m>=0, and begin with

C> 1^N 0 1111 00 1^(3m) 00 v.

First reflect at the first 00. The scanned word before it is u=1^N 0 1111, which has length N+5 and one zero. After N+17 steps the tape is

1^N 0 1^(3m+6) 00 v,

with A at position N+3, the third one in the long run. Only the first pair of zeros was changed.

Apply the A111 bridge to the three ones at positions N+1..N+3. It changes the beginning of this region from 0 111 to 0 011, and leaves A at position N. The first two cells there are now 00. Apply A00. After seven additional steps the configuration is

1^(N+1) 0 C> 1^(3m+5) 00 v,

where C is at position N+2.

Reflect at the displayed second 00. This takes 3m+13 steps. The resulting configuration can be cut as

1^N [1 0 1^(3(m+1)+1)] 111 v,

with A on the final one of the bracketed block. Apply the residue-one rule with j=m+1, taking 5m+8 steps. The result is

1^N 110 (011)^(m+1) 111 v,

with A at position N-1, the last one of the initial 1^N. The total time so far is

N + 8m + 45.

Every read has remained in the original finite input before v. The bridge does not inspect any left exterior cell.

For N=3q, apply q copies of A111. They replace 1^N by (011)^q and end A at -1 in 5q more steps. This gives the long A family above.

For N=3q+1, q copies of A111 leave A on the first one, at position 0. Apply the one-step A1 bridge. This replaces that remaining one by zero and ends D at -1. The additional time is 5q+1, giving the long D family.

The short families

Reflect C on 1^N 00. It takes N+8 steps and leaves A on position N-2, with the three-bit suffix 111 on its right.

For N=3q, apply q-1 A111 bridges, then the four-step A11-to-F bridge. The right word is 11 (011)^(q-1)111, the endpoint is F at -1, and the total is 8q+7.

For N=3q+1, apply q A111 bridges. The endpoint is A at -1, the right word is (011)^q111, and the total is 8q+9.

Evidence and scope

unbounded-lookahead-obstruction.json records 875 paired instances of the A/F families, including arbitrary sample suffixes. halt-escape-common-prefix-families.json records 175 pairs of the A/D families, checked with a literal transition loop that asserts every read is inside the displayed finite input. The formulas and this derivation assert arbitrary parameters; the numerical checks by themselves are finite evidence. Whole-machine safety still needs the blank-orbit reserve argument.