1. The solver stops here:
A13 must contain a 7 so A2=7, eliminating the other 7s in the row and column.
A79 cannot contain a 3 (if it did, A5 would be 4 and B79 could not contain the 4
necessary to make a Str8t with the 3). A79 must therefore be [456] solving A5 for 3
and B5 for 4. eliminating the other 4s from B59, and also eliminating the 6 from A3,
solving it for 9.
B13 must contain an 8 so B3=8 and 8 is eliminated from J3. 8 and 9 are thus
stranded in J24 and can be eliminated.
F2=1 (the only 1 in AJ2), eliminating the other 1s in F19.
2 can be eliminated from E3 (no 1 or 3 in F3).
J4 cannot be 7 (if it were, J24 would be [567], and no Str8t would be possible in J69).
Similarly, J4 cannot be 6 (if it were, J24 would be [456], and no Str8t would be
possible in J69. J4 is thus solved for 3, eliminating the 6s from J24 (not in sequence
with 3).
J24 must contain a 4 so J2=4, D2=6, B2=9 and E2=5. After eliminating the
inconsistencies with the solved cells, the naked pair 79 in row D eliminates the 7s
from D89.
J24 eliminates 3 and 4 (and the stranded 2) from J69.
AD9 must contain a 4, so 4 (plus the stranded 2) is eliminated from FJ9. FJ9 must
contain a 6, which is therefore eliminated from AD9.
H49 must be [456789] so H1 cannot be 4 and must be 2, solving F1 for 4 and
eliminating the 4 from F7. 2 can be eliminated from E7 (no 1 or 3 in F7).
1 is eliminated from D9 (no 2 in D8). 6 is eliminated from F3 (no 5 or 7 in E3).
4 is eliminated from H7 (if H7 were 4, H8 would be 5, and if 45 were eliminated from
AC7, the remaining candidates in that compartment could not make a Str8t). 5 is
eliminated from H7 and J7 (if either were 5, HJ7 would be [56] and EF7 would be

2. [23], and the remaining candidates in AC7 could not make a Str8t). 6 is eliminated
from H7 and J7 (if either were 6, HJ7 would be [67] and EF7 would be [23], and the
remaining candidates in AC7 could not make a Str8t).
1 is eliminated from AC7 (not in sequence with the 4 in A7).
Every column except 3 must contain a 4. Row G cannot contain a 4, so every other
row must contain a 4 (Setti). Hence D89 must be [34] and the 1 and 3 can be
eliminated from that compartment, and C8=1 (the only 1 in AJ8).
AD9 must contain a 2, so C9=2 (the only 2 in AD9), solving C5=4, C6=3, A7=4,
A9=5, A8=6, B9=3, B7=5, D9=4 and D8=3. After eliminating the inconsistencies with
the solved cells, the naked pair 67 in EF7 eliminates 7 from HJ7.
F3=2 (the only 2 in F19), solving E3=3 and J3=5. H8=4 (the only 4 in AJ8 and in
H49). The naked pair 79 in AJ8 solves G8=5 and thus G9=6 and G4=8.
J69 must contain a 6, so J5=6 (the only 6 in J69) solving B6 for 1 and eliminating all
9s from BJ6, thus solving G6=7, E6=2 and H6=5. The 2 in E6 eliminates the 9 from
E8 solving E8=7, E7=6 and F7=7. The 7 in E8 solves J8=9, J7=8, J9=7, H8=8 and
H9=8.
DH5 must contain a 5, so F5=5 (the only 5 in DH5), solving F9=9, F4=6, H4=7, D4=9,
E4=7, G4=9 and H4=6, solving the puzzle: