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Friday 3-Jul-2015

... by: JC Van Hay

Corrected presentation of the solution :

Thanks to the exotic patterns, this puzzle is solvable:
1. G4=5
2. Double Exocet : (3467)H45,I89 => +2H7, +9I6, HP(29)G12
3. Rank 0 Logic : (3467)Columns367 belongs to {(34)RowA, (46)RowB, (67)RowD, (37)RowE, the 4 target cells of the double exocet G67,HI3} => -(34)A159, -(46)B248, -(67)D159, -(37)E2489
4. Optional : The solutions of (3467)Columns367 also exclude 1B6 and 5E7.
5. Andrew's solver is now able to solve the puzzle using strategies 1 to 34.

Thursday 2-Jul-2015

... by: JC Van Hay

A bridge too far ... :(

After checking again with Andrew's solver , -1B6 and -5E7 are optional, even though the "solver path" option only gives the first nine steps !?

Thursday 2-Jul-2015

... by: JC Van Hay

Typo ... read 1.G4=5

Thursday 2-Jul-2015

... by: GoldFish

G4 is 5. 1 is locked in G89, so G12 cannot be 1. So far it's trivial. The next step is where it needs ideas - however, if searching for back-doors in the Sudoku, you can find that A3<>3, D3<>6, C9<>7 and J8<>6. With all those considered, the puzzle collapses.
[C8=7, J8=3, H3=3, J3=6, E7=3, then locked 4 in block 7 makes BC2<>4, locked 7 gives EF2<>7, naked triple in F128 => leaving only 7 in F9, J9=4, locked 4 in C removes that in the rest of block 2, G6=4, H4=6, H5=7 etc, with 1x skyscraper and 1x remote pair on path, but that's no pain.]

Tuesday 30-Jun-2015

... by: JC Van Hay

Note : Besides the obvious exclusions due to the solutions of (3467)Column367, 1B6 [and 5E7] are also excluded. Otherwise ROOKS !

Tuesday 30-Jun-2015

... by: JC Van Hay

Thanks to the exotic patterns, this puzzle is solvable:
1. F4=5
2. Double Exocet : (3467)H45,I89 => +2G7, +9I6, HP(29)G12
3. Rank 0 Logic : (3467)Columns367 belongs to {(34)RowA, (46)RowB, (67)RowD, (37)RowE, the 4 target cells of the double exocet G67,HI3} => -(34)A159, -(46)B248, -(67)D159, -(37)E2489; HP(46)B67; B2=1=H1
4. Andrew's solver is now able to solve the puzzle using strategies 1 to 34.

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