I hardly ever bifurcate, especially in sudoku. Bifurcation just leads to errors for me. Showcased in philadelphia when the last round you were basically told to bifurcate and 2 of my 4 solutions had errors. I understand the early harder steps will more avoid bifurcation, but which isn't bad for the higher point puzzles but maybe isn't necessary for all puzzles. It comes over a bit weird to me when my fastest solve is the 85 point puzzle.

For the argyle I missed a hidden single in row 7, after I checked to see what I missed. I was just thinking it would have been nicer for there to be something easier. The easiest puzzle being 4-5 minutes for the faster solvers seems very long for me. But maybe I'm just more used to puzzle competitions which generally have more puzzles than Sudoku competitions and the easier puzzles take less time. It's just a personal thing.

Some little explanations on some grids:

1. Diagonal sudoku

Preliminary remark: The 1rst glance suggests that the opening has something to do with the first diagonal (R1C1 to R9C9), with lot of clues in boxes 1, 5 and 9.

There are a couple of easy placements: 7 in R4C6 and 3 in R6C7.

One can view the clues 134 in boxes 5 and 9, which leads to have a triplet 134 on the diagonal in box 1. Then you can place the 7 and 9 in box 1. Then you can place some other 7's, and then some 9's, which leads to a pair 89 in the central box.

After this point you can solve box 9, and then, etc... finish to solve the grid





Attachments
----------------
diago.png (33KB - 0 downloads)

2. Creasing sudoku

Preliminary remark: This sudoku has a hard opening, because you have to think "globally" about the configuration of marked diagonals, especially 5-cells diagonals !

There are a couple of easy placements: 8 in R2C7 (can't be in C8 because no greater digit can be in this marked diagonal) and 6 in R8C8.

Let's analyse a bit the 5-cells diagonals. If you follow the path of these marked diagonal, it has to alternate between increasing and decreasing diagonal (If you had 2 increasing diagonals following each other, you should have digits 1-9, and then it would be impossible to have 2 decreasing diagonals from 9 to 1). So digits in R1C5 and R9C5 have to be a) both greater than 5 or b) smaller than 5. Ditto for cells R5C1 and R5C9.
We can see that only one digit greater than 5, the 7 can be placed in R1C5 and R9C5. So these cells must contain digits smaller than 5. Thus Cells R5C1 and R5C9 must contain digits greater than 5. We have found the direction of increasing and decreasing diagonals !

Knowing that, you can place the 8 in R5C1, then in R4C8, then the 9 in R5C9 (must be greater than the 8), etc...





Edited by Fred76 2012-03-16 8:51 PM








Attachments
----------------
creasinga.png (24KB - 0 downloads)
creasingb.png (25KB - 0 downloads)

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).









Attachments
----------------
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





Attachments
----------------
diagotwin2.png (18KB - 0 downloads)