This is one of the hardest Sudokus (and one of the best Sudokus) I have ever seen, with only 4 given digits. There are some more rules too:

  1. First, normal Sudoku rules apply here.

  2. Both the diagonals also should contain 1 to 9 in some order.

  3. You cannot put the same number in cells which are at a knight's move (in chess) apart.

  4. The center box will be a magic square.

Sudoku, row 4 columns 1 to 3 are 3, 8, 4 in order, and row 9 column 9 is 2

Now comes the real part, can you solve it?

Bonus: Is the solution unique, or can you find any other solutions?

This is not a puzzle of my own devising. However, to avoid spoiling the solve for others I will wait to declare its source until the puzzle is solved. Credit goes to Mark Goodliffe and Simon Anthony for solving this puzzle, and also to Aad Van de Wetering for making such a wonderful puzzle !

Edit: This puzzle got solved here - https://www.youtube.com/watch?v=hAyZ9K2EBF0

  • $\begingroup$ Very nice puzzle! I can place like 11 numbers than I am stuck. Looking forward to other's solutions. $\endgroup$
    – daw
    Oct 9 '20 at 11:03
  • $\begingroup$ @Ananymous. Is it Simon Anthony? Not Timothy $\endgroup$
    – DrD
    Oct 9 '20 at 12:56



A tough one. Spent much time on it. I can also confirm that the solution is unique (assuming I didn't make any mistake).

For the beginning, I only give some snapshot of my progress. Some of the reasoning are a bit complicated. But everything is logically deduceable until the last step.

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At this point I got stuck and cannot see a possible logical solution. So I looked at the circled grid below. It must be either 7 or 8.
Assuming that it is 7, I arrived at the following situation, where the two grids with question marks can neither be 9 now.
Thus the circled grid must be 8. enter image description here

And it easily leads to the final answer. ans

  • $\begingroup$ You got it, nice ! I can see that those digits you got from the snapshot are actually correct, and maybe others can think of how they came using logic. $\endgroup$
    – Anonymous
    Oct 10 '20 at 6:05

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