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Stuck with this for long time now, applied all the techniques i can think of

Blank: enter image description here

All possibilities:

missed 6 as possibility in R8C5

enter image description here

Verified that this has unique solution

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    $\begingroup$ As far as I see, if R7C2 is 7, then R7C8 is not 7, then R2C8 has to be 7, then R1C9 is 4, then R1C2 is 7, then R7C2 can't be 7, therefore R7C2 is 9. (I hope I haven't messed up the indexing here) $\endgroup$ – the default. May 30 at 10:26
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    $\begingroup$ @mypronounismonicareinstate that sounds correct that fixes 9 in R7C2 $\endgroup$ – Chetan Paliwal May 30 at 10:34
  • $\begingroup$ @mypronounismonicareinstate anything in particular that made you look into that area or just having 2 possible numbers was the cue? $\endgroup$ – Chetan Paliwal May 30 at 13:35
  • $\begingroup$ 2 possible numbers, some luck and also some of the relevant cells happened to be highlighted. $\endgroup$ – the default. May 30 at 13:44
  • $\begingroup$ Did that solve it? If not, can you update the question? $\endgroup$ – jonnybolton16 May 31 at 7:55
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I got the answer! Apologies for the graphics, this is a self-coded sudoku solver. The length of this answer is mostly screenshots.

Final solution:

sudoku solution

Explanation:

Okay, to start let's place the 9 at R7C2, per @mypronounismonicareinstate. When discussing the numbers placed in certain cells, I will use the syntax RXCY:Z, which means that the cell in row X and column Y, with top left as R1C1, has Z placed in it.step 1 of sudoku
Now several numbers are forced. Forced numbers: R7C3:6, R8C3:8, R1C3:2, R3C3:9, R3C8:2. Now the board looks like this:
step 2 of sudoku
There are no obvious numbers any more. So let's guess! The top-left 3-by-3 box looks good to guess at, since any one number will force two others automatically. Let's guess that R2C1:8. Forced numbers (sorry this is long, I just kept filling in any boxes had a forced number) R3C2:4 R1C1:7 R9C2:5 R5C2:8 R5C4:5 R5C1:9 R6C1:5 R1C9:4 R3C9:3 R5C8:1 R4C7:7 R7C8:7 R7C1:4 R3C9:3 R2C8:9 R2C7:1 R5C7:2
impossible sudoku
And here is a contradiction. The highlighted cell (middle of all blue cells) has no possible number. Since all numbers after our guess were forced, that means the guess was incorrect. I undo to the third image. The top-left box can only have 8 in R2C1 or R3C2, so it must be R3C2:8. After this one number is put in, the rest of the numbers are forced. I'll put some of the milestone images in here. The whole left block is filled in here, after R2C1:7 R1C2:4 R1C9:2 R5C2:5 R9C2:7 R7C1:4 R8C1:5 R5C4:8 R5C1:9 R6C1:8
left blocks filled in
The whole right block is filled in here, after R9C4:9 R8C4:6 R6C4:5 R5C8:1 R7C8:7 R2C8:9 R4C7:7 R4C6:1 R2C7:8 R2C9:1 R9C9:4 R3C9:3 R6C9:9 R8C9:2 R8C7:9 R7C7:1 R9C7:5 R6C7:3 R5C7:2 R3C7:4
right block filled in
And it should be pretty obvious by now, but to finish off the middle (refer to first picture for finished grid) R7C5:2 R9C5:1 R9C6:8 R6C6:7 R8C6:4 R8C5:7 R6C5:6 R5C6:3 R5C5:4 R2C5:3 R2C6:2

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  • $\begingroup$ looks good but wouldn't it be considered brute forcing what you did in top left 3x3 $\endgroup$ – Chetan Paliwal Jun 3 at 1:27
  • $\begingroup$ It depends on what you mean by brute force. I guessed, kept careful track of where I guessed, and then concluded that the guess was incorrect. It's proving that a certain number is impossible by writing numbers down instead of theorizing. $\endgroup$ – bobble Jun 3 at 15:57

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