Many solvers made a comment on the **Killer Sudoku** – either it was unreasonably hard or it was one of the most beautiful puzzles they have ever solved. Couldn’t tell whether you did the right thing, huh?

I met up with Sinchai at the annual Thailand Puzzle Open last month and we went through the Killer Sudoku together at McDonalds. Here are some of his notes on his own puzzle:

**A: **If 9 is placed in the 13-cage, it will need to be grouped with 1 and 3 (to get 13). We know 1 has to be in one of the top three cells (since 1s will be in the 7-cage and the 8-cage). Therefore, the 9 has to be in R9C9.

**B: **1 cannot be in the 15-cage because it will need to be grouped with either: [9 and 5] and we can’t use the 9 or [6 and 8] and we can’t use them since the 22-cage will either need [9,7,6] or [9,8,5]. Lastly, the 7 can be placed by adding the regions to get 45.

**C: **1 can be placed in the 8-cage.

**D: **Take a look at the bottom row: the two uncaged squares and one cell in the 13-cage has to contain 6, 7 and 8 since they can’t be anywhere else in that bottom row. Note that 6 and 8 can’t both be in the 13-cage. Since we see the 23-cage needing 9, 8 and 6 – R9C3 has to be a 7.

**E: **1 is forced to be in the 15-cage and has to be grouped with either [9 and 5] or [6 and 8]. If [6 and 8], we can see that the first column will have no place for both 6 and 8. Therefore the 15-cage has to be comprised of 1, 5 and 9.

**F: **The three blank squares of the top left 3×3 box must add up to 24, namely; 9, 8 and 7. Using simple slice and dice we can fill in those 3 cells.

**G: **After you fill in 9, 8 and 7. Relay the information down to the 23-cage at the bottom. You can see the third column can be broken down into unique pairs; which leaves a 2 at R4C3. **(H)**

I: The 8-cage houses 1, 2 and 5 so it will share 1s and 2s with the 7-cage at the bottom. The 5 and 4 must be in the same column, therefore you can now place the 3 and 5 of the bottom left 3×3 box.

**J:** That leaves 3 and 4 in the 2^{nd} column marked as shown with 6, 7 and 8 remaining in the 1^{st} column.

**K:** Only possible location for the 9 at the left-middle 3×3 box.

**L:** Still keeping track of the 9s, you can see that the three uncaged cells of the upper middle 3×3 box adds to 22, which needs a 9. The only place for the 9 in the 6^{th} column is therefore, R4C6.

**M:** Careful here. The 3 has to belong in either the 13 or 15-cage. If the latter, there is no way to get 15 using [3,?,?]. There are no more big numbers available. So the 3 has to go to the 13-cage.

**N:** Similarly, the 5 can’t also go to the 13-cage as [3,5,?] needs another 5 to get 13. Therefore, the 5 goes to the 15-cage.

**O:** As we’ve pointed out earlier (at step D) that R8C4 has to be 6, 7, or 8. The 5 has to go in the 8-cage, making [1,2,5]. Now, several cells can be filled in.**P: **2 is forced by the 7 and 8-cage to be in the 15-cage making [2,5,8]. The 13-cage is therefore [3,4,6].

**Q:** 8 can be filled in R9C4 leaving [6,7,9] for the 22-cage. 3 and 4 can be placed in their uncaged cells.

We’ll leave it here for now.

Those were Sinchai’s initial steps and from here on he believes simple notations should get you through the whole puzzle. A very tough Killer Sudoku indeed.

The solutions to all the puzzles have been combinsudoked into this PDF