dailysudoku.com

[Beginner]

I found this little gem at Boldt's:

[chobans]

r4c2 can be 3,4,8
but r5c1 and r5c3 both share two same numbers: 1,4. So 1 and 4 can be eliminated from box 4 (r4c1-r6c3). So r4c2 can only be 3,8.

r9c2 can be 2,4,5,9
but in box 8 (r7c4-r9c6), 4 can ONLY go in row 9. So 4 can be eliminated from other cells in row 9. So r9c2 can only be 2,5,9.

In column 2, after the elimination I mentioned above, 4 only shows up in r1c2 and r2c2. So 4 can be eliminated from the other cells in box 1 (r1c1-r3c3). Which means, eventhough r3c3 had 4 and 9, 4 can be eliminated, leaving only the 9.

[squirrel]

There are hidden pairs {2,5} at r5c7 and r5c9. That means a 4 must appear in either r5c1 or r5c3 and thus cannot appear in r4c2 or r6c2.

Also, since there's already a 4 in both r8 and c5, there must be a 4 in either r9c4 or r9c6.

We have now eliminated 4 as a candidate from every row in c2 except for r1c2 and r2c2. There must be a 4 in one of these, and thus 4 cannot appear in r3c3.

[squirrel]

Note to self: Next time, type faster.

[Beginner]

Thanks for your efforts. This time I found the solution (4 only possible in c2 in box 1) myself, but unfortunately only after I left my computer access, so I could not stop you!