Once again, 9 out of 29 sudokus were left unfinished, unsolved or with errors, but 7 out of 8 from the 1st round I have later solved with either minor or major difficulties. Unfortunately, the unfinished one from the 2nd round, or sprint, was lost because I forgot the competition location on my sprint sheet. But there remains 1 classic sudoku, or 70-point, which I can't seem to solve. I have tried different methods, but to no avail. There are quite a few squares with only 2 options, so I could solve it using 50-50 (by experimenting), but I assume that it will still be solved logically, and let's be honest, solving it with 50-50 is still not right. So I'll add a picture of this sudoku, maybe some pro can help me?
PS The grid is not even and is crooked because I didn't have a ruler handy when I drew it, but I hope you understand.
Help with Estonian Sudoku
Lu par 43 utilisateurs
Hmm, can't anyone else do it? 😕
It is solvable. The 70p problem was easier for me than the 55p one. The ones you have now are correct.
It may be solvable, but how exactly? I still only see the option of using 50-50, but if there is a logical solution, maybe you can help? Or another pro can...
The key lies here..
Okay, I looked. If you number the squares from left to right and top to bottom, if 2.4 was 3, then 1.8 must be 3 and consequently 8.7 must be 3 and 7.4 must also be 3, which means that there would be two 3s in row 4, so 2.4 cannot be 3 and must be 6.
Did you mean that?
Did you mean that?
Yes, I thought the other way around, that if you put 3 in Row 4 Column 2 , then R8T7 must be 3 and therefore there could be no more places in the bottom left 3x3 box ( R7T2, R8T1, R8T2 ) where 3 could come in, so R4T2 can only be 6), but it works the same way.
Thanks, I finally got it sorted! But after this point, there's still one more place to get stuck, I'll add a picture of that.
But I still managed to move on from there by also using 3, namely if 8.6 was 3, then 8.7 should be 7, 9.8 therefore 6 and 9.7 8, which would mean that nothing could go into square 3.7, because both 7 and 8 would already be in the 7th row.
So 8.6 must be 1.
Anyway, you have to experiment in both places (3 or 6, 3 or 1) and see what results you get in three or four squares. I always try to avoid this and come up with a logical solution, so there is one possible option (untested), but in this sudoku it is still not possible, which is not good. And at least I generally can't find a good place to experiment, but maybe there is a trick for that.
But I still managed to move on from there by also using 3, namely if 8.6 was 3, then 8.7 should be 7, 9.8 therefore 6 and 9.7 8, which would mean that nothing could go into square 3.7, because both 7 and 8 would already be in the 7th row.
So 8.6 must be 1.
Anyway, you have to experiment in both places (3 or 6, 3 or 1) and see what results you get in three or four squares. I always try to avoid this and come up with a logical solution, so there is one possible option (untested), but in this sudoku it is still not possible, which is not good. And at least I generally can't find a good place to experiment, but maybe there is a trick for that.
This is where I got stuck in the meantime:
