Extreme str8ts #124: steps that the solver does not find.

1. The solver takes us to this position:
Compartment CE8 must contain 678 or 789, and thus 78 are certain candidates for
the compartment and can be eliminated from A8.
AD9 is a naked quad 6789, and those numbers can be eliminated from the rest of
column 9, solving E9 for 5 which can thus also be eliminated from the rest of column
9. E6 is solved for 6, G6 for 8, H6 for 5, H5 for 6 and J5 for 5. 5 can be eliminated
from the rest of row J and 8 from the rest of column 8. 9 can thus be eliminated from
C9 (no sequential 8 in C8).
Compartment EG2 must contain 234 or 345 so 4 can be eliminated from A2.
If C1 were 9 then C2 would have to be 8 to be in sequence, but it cannot be (B2 is 8)
so 9 can be eliminated from C1.
The X-wing formation at E1, E2, F1, F2 eliminates 3 from G1, G2 and H1. Now 2
can be eliminated from G1 (no sequential 1 or 3 in G2) and 4 can be eliminated from
G2 (no sequential 3 or 5 in G1). E2 is thus solved for 4 (the only 4 in the
compartment EG2), E1 for 3, F1 for 2, F2 for 3 and H1 for 9, eliminating 29 from the
rest of column 1.
That brings us to the following position:

2. Counting 9s, there are 6 rows that must have a 9 (ACFGHJ) 2 that do not (DE) and
one row which is unknown (B). There are 4 columns that must have a 9 (1479) 3 that
do not (356) and two that are unknown (28). From that we can deduce (by Setti's
rule) that 6 rows and 6 columns each contain a 9, and 3 rows and 3 columns do not
contain a 9. It follows that row B does not contain a 9 and 9 can be eliminated from
B9, solving A9 for 9.
The naked pair 67 in B9 and D9 eliminate 67 from C9 solving it for 8. 89 can be
eliminated from C1 and C2. The naked pair 67 in C12 solves C8 for 9 and thus D8
for 7, D9 for 6 and B9 for 7.
The rest is easily solved.