Let's speak a bit about little killer:

Preliminary remark: The grid is constructed on the 6-cells diagonals !

Of course you can place digits on the 4 corners of the grid (I try to delete these clues, but I think it was hard enough with them ).

One can see the 6-cells diagonal with clue 48. That's the maximum possible. Cells have to contain 2 triplet 789. With these triplets, both 6-cells diagonal with clue 16 are then the minimum possible. So you can place pairs 12 and 7 in one box and a triplet 123 in the other box.
And finally, with these triplets placed, you can see that the last 6-cells diagonal, with clue 20 is now the minimum possible, with the sum of two 145 triplet which can be placed.
One fun placement now is the 2 in the middle of the grid (fun because it's rare to be able to place a digit in the center of a little killer in the beginning of the resolution).









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littlekillera.png (31KB - 0 downloads)
littlekillerb.png (34KB - 0 downloads)
littlekillerc.png (35KB - 0 downloads)

Finally (then I stop and let you concentrate on the tapa contest ), I want to speak about the diagonal twin sudoku.

There is (at least) 2 openings for these grids. You can begin with the corners. With grid A you can place 8 in R1C9, 4 in R9C9, and Grid B let you know the two other corners.

But there is another opening, which I find more elegant: You can start with the center of the grids: box 5. Grid A let you know that digits in white cells are 3,5,7 and 8. Knowing that individually they can't be placed in the same cell let you place the 5 in R5C6, the 3 in R5C4, then the 8 in R6C5 and 7 in R4C5 in grid B.
Then in both grids, 6 can be placed in R5C5, you have a pair 14 in cells R6C46, then 2 in R4C4 and 9 in R4C6.

Of course, you'll need still lot of work to solve these grids, but box 5 is almost filled





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diagotwin2.png (18KB - 0 downloads)