The New Sudoku Players' Forum

[Loren Pechtel]

And this one's driving me nuts. I've gotten this far:

* * * * * * * 2 *
1 3 * 4 * 2 * 8 9
* * 2 6 * 8 3 * *

2 * * 5 * 9 * * 4
9 7 * * 1 * 2 5 *
3 * * 2 * 7 9 * 8

6 * 9 7 * 5 8 * *
8 4 * * * * * 9 2
* * * * * * * * *

I'm not so much interested in a solution as how to attack it.

[simes]

Look for "disjoint subsets" in row 8, block 7 and block 8. (The numbers are 1,3,6 and 5,7 and 1,3,6 respectively.)

(If you need to know what these are, here's my attempt at an explanation www.sadmansoftware.com/sudoku/nakedsubset.htm)

[george-no1]

Yes, if you look at r8c4, r8c5 and r8c6, you should be able to see that together they must take up the 1, 3 and 6 leaving r8c3 and r8c7 as 5 and 7. You can use pairings in box 7 to work out where the 3 goes in that box.

Hope this helps!!!

George

[Loren Pechtel]

Look for "disjoint subsets" in row 8, block 7 and block 8. (The numbers are 1,3,6 and 5,7 and 1,3,6 respectively.)

(If you need to know what these are, here's my attempt at an explanation http://www.simes.clara.co.uk/programs/sudokutechnique5.htm)

Simes
http://www.simes.clara.co.uk/programs/sudoku.htm

ARGH!!

I should have seen that one. When I initially pencilled in the 5's I screwed up and had a slew of extra 5's from missing the 5 in block 8. In going over things I caught the extra 5's but apparently only after evaluating that row for disjoint sets.

Yes, if you look at r8c4, r8c5 and r8c6, you should be able to see that together they must take up the 1, 3 and 6 leaving r8c3 and r8c7 as 5 and 7. You can use pairings in box 7 to work out where the 3 goes in that box.

Hope this helps!!!

George

Here I'm puzzled. I found the disjoint set I was overlooking and it was useful but so far it hasn't revealed where the 3 goes. With the 1, 3 and 6 known to be on row 8 that forced r9c6 to be 4 and some other numbers fell into place with that. I don't have time to hunt the whole thing right now, though.

[abailes]

To me, an easier way of solving this puzzle is to notice the 8,9 pair in the cells (4,9) and (5,9).

I haven't tried the other methods of solving this, but it looks at first glance as though they involve more complexity.

I don't know about other people, but when I am solving Sudoku, one of the things I immediately note in pencil (in the corner of the cell) are those numbers that are restricted to only two cells in any row/column/box. This makes finding pairs (and x-wings) a lot easier. It also immediatlely leads you to another answer if you fill one of the two cells with a different number.

If you find two numbers that are both restricted to two numbers in a row/column/box then you know immediately that there are no other candidates for that cell.

After spotting the 8,9 pair, the puzzle snowballs.

[george-no1]

Here I'm puzzled. I found the disjoint set I was overlooking and it was useful but so far it hasn't revealed where the 3 goes.

OK well I might be wrong. but because r9c1 and r8c3 can only be 5 and 7, you can assume that the 3 must go in either r7c2, r9c2 or r9c3. However, there is already a 3 in r2 so that only leaves one candidate for a 3 in box 7.

Have fun!

George

[Loren Pechtel]

Here I'm puzzled. I found the disjoint set I was overlooking and it was useful but so far it hasn't revealed where the 3 goes.

OK well I might be wrong. but because r9c1 and r8c3 can only be 5 and 7, you can assume that the 3 must go in either r7c2, r9c2 or r9c3. However, there is already a 3 in r2 so that only leaves one candidate for a 3 in box 7.

Have fun!

George

Oh, I understand now--I was looking at the stuff on row 8.

[george-no1]

Have you solved it now?

George

[Loren Pechtel]

Have you solved it now?

George

Yup.