# How can I solve this sudoku?

I've almost finished this hard sudoku, but I cannot finish it without guessing or making multiple blind steps to see if I'm wrong or not. Even https://www.sudoku-solutions.com/ gives up and says "there is no hint available". I've tried to apply as many solving techniques as possible, but to no avail. Can anyone help?

• It is often a mistake, but some sudoku puzzles have multiple solutions... I'll have a look though. Jul 31 '19 at 20:42
• Thank you, I've also just found this: puzzling.stackexchange.com/questions/252/…, from which I understood it's acceptable to make multiple assumptions on multiple cells (what I called "blind steps forward") and see if you come to a dead end or not... Jul 31 '19 at 20:50
• It's not enough to see if you come to a dead end though; you'll need to see if the solution you found was unique; otherwise you'll end up claiming to have solved a sudoku that wasn't actually a sudoku.
– Bass
Jul 31 '19 at 20:58
• There is the following nice trick: If every open cell in the sudoku has exactly two possibilities left except for one cell which has three, this cell has one possible number that is a possibility an odd number of times. This is the correct digit for that cell. In this puzzle, 2 is still possible in eleven cells, 4 is possible in six and 8 is possible in 10. Hence, E7 has to be a 2. Aug 1 '19 at 13:16
• Look up the solving technique called Sashimi-X-Wing. If F1 does not contain a 2, then you have a basic X-Wing on E1, G1, E9 and G9 which eliminates the 2 in E2. Of course if F1 is a 2, then the 2 in E2 is also eliminated. So in no case can E2 contain a 2. Aug 1 '19 at 14:13

If F1 is a 2, then

G1 is "not 2", G9 must be 2, and E9 cannot be a 2.

On the other hand, if F1 is not a 2, then

E2 must be a 2, so E9 cannot be 2.

Therefore,

E9 cannot be 2.

• Genius, thank you. Is there a name for this logic, such as swordfish? Or just logic? :) Jul 31 '19 at 21:27
• @Andrejs sudokusnake.com/coloring.php Aug 1 '19 at 7:39
• @Andrejs I tried to colour them here E9 cannot be be both red and blue, if we consider number 2. Aug 1 '19 at 7:45
• @Andrejs I think it might be called "universal elimination" in formal logic, but sudoku terms are not my strong suit :-)
– Bass
Aug 1 '19 at 7:48
• @Andrejs To add to this answer, look up "Bivalue Universal Grave + 1" or BUG+1. If every cell has only 2 options except for one cell that has 3 options, you call that a BUG+1. Aug 1 '19 at 13:26

Solution method that uses value propagation and digit elimination, but not explicit case-by-case analysis:

G1 = E9. (That's because of how they interact with G9.) That digit is not 2 because that would squeeze out digit 2 from the DEF123 square altogether. Therefore it is 8.