Cec said

Bigtone53 wrote:

"...Look at box 1. r3c3 has to be a 4. Consider boxes 1,4,7, adding up to 135. r9c3 has to be a 2, so r9c4 has to be a 7. Consider rows 6,7,8.9. These 4 rows add up to 180 so r5c9 is a 4

and so on...."

I'm sure I am missing something - yes, I know what you're thinking when I say that - but I still don't get this.

I can see why r9c3 has to be a 2 but I don't understand why r3c3 has to be a 4. As the numbers to be placed in each of the three cells in column 3 (r345c3) must total 17 then why is cell r3c3 restricted to 4? With a 2 later placed in r9c3 following Bigtone's above comments then there are alternative choices for different numbers between 1 to 9 (excluding 2 of course) which could be placed in r345c3 and still give a total of 17.

On the matter of posting this puzzle in the wrong forum, I too agree with Ronk's above post that the puzzle should belong in the Sudoku Variants Forum. That said, Bigtone's comments are also relevant as to how this "problem" can be avoided.

Cec

Cec,

The rules of Killer Sudoku as used in The Times are

Fill the grid so that every column, every row and every 3x3 box contains the digits 1 to 9. The digits within the cells joined by dotted lines add up to the printed top left-hand figure. Within each dotted-line "shape", a digit cannot be repeated

Look at box 1. 6 of the cells add up to 29 and two of the remainder add up to 12. As this box contains 1-9, the total for box 1 (as every box, line and column) is 45 so the last cell r3c3 is 45-29-12 = 4

The 1 in c3 cannot be in the two-cell 12 or (given the 4) in the 3-cell 17 in c3. It has to be part of the 14 in c3. Knowing this, the 9 in c3 cannot be in the 14 as the third number would be 4, making two 4s in c3, not allowed. Similarly, the 9 is not in the 17 in c3 as again, this would involve another 4 in c3. The 9 is therefore in the 12 in c3, together with the 3, in some order.

In box 7, the 9 is not in c3 (there is a 9 already in c3). It is not in the two cell 11 as the other number would be 2 and there is already a 2 in box 7. It is obviously not in the two cell 9 so it is in either r7 c1 or r7c2. Because of this, either r6c1 is a 1 or r6c2 is a 2.

Going back to c3, the other two numbers in the three cell 17 add up to 13. They cannot be 9/4 (as we already have both a 9 and a 4 in c3) so they are either 8/5 or 7/6 in some order. If they were 7/6, then the 2 two cell 10s in box 4 can only be 1/9 and 8/2 in some order (the 7/3 and 6/4 possibilities causing duplications in box 4 of a 7 or a 6). But we have deduced above that either the 1 or the 2 in box 4 is in r6c1/2. So the remaining 13 in the three cell 17 in c3 cannot be 7/6 and is 8/5 in some order.

and so on.

It may be a bit adventurous to try and learn on a Deadly killer. Here is yesterday's Gentle one.

I hope that the link works