by **Mel-o-rama** » Thu Apr 17, 2008 5:20 am

I believe I have a solution:

1 0 8 7 2 6 4 5 3 9

3 6 5 0 4 9 2 1 7 8

2 7 4 3 1 5 0 8 9 6

8 1 9 2 6 4 7 3 0 5

0 2 3 1 5 7 8 9 6 4

6 5 7 9 8 3 1 4 2 0

5 3 2 4 0 8 9 6 1 7

9 8 0 6 3 2 5 7 4 1

7 4 6 5 9 1 3 0 8 2

4 9 1 8 7 0 6 2 5 3

This solution includes the 10th row and column I used to help solve the puzzle.

This puzzle was harder than expected, but it was interesting. I learned a new solving technique. (Perhaps this is described in general somewhere else.)

Let's say you have two rows where a number can go into only one cell. Since the number can be missing only once, it has to be in at least one of the cells. Thus, you can eliminate any instances of that number in the intersection between those two cells. What is this - a modified swordfish?

I got stuck twice, and had to use trial & error, but I think this is the only solution.

Hints:

#1 The diagonals are very easy.

#2 Fill in all the pencil markings, get as far as you can and do the modified swordfish.

#3 If you did everything right, you should be able to place all the 3's (though at this point you don't know which ones might be "missing".)

Now, why did I get stuck? I'm not sure what to do after #3.

Conjecture/question: It turns out that 3 is the number that appears 9 times. If you can place a number into 9 cells, does this mean that the number is the 9-times number? I don't think so - but if it's true, it could be a useful solving tool.

Last hint: 0 is the under-killer. I still don't know of a good way to identify the under-killer number until you've already determined some of the missing numbers.

Thanks for the fun puzzle, HATMAN!