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International Journal of Computational Engineering Research(IJCER) ijceronline
nternational Journal of Computational Engineering Research (IJCER) is dedicated to protecting personal information and will make every reasonable effort to handle collected information appropriately. All information collected, as well as related requests, will be handled as carefully and efficiently as possible in accordance with IJCER standards for integrity and objectivity.
Solution to Diabolic Str8ts #8 puzzle (http://is.gd/slowthinker_diabolic_str8ts_8c) and its tough variation Diabolic #8b (http://is.gd/slowthinker_diabolic_str8ts_8d)
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: