Normal Sudoku rules apply. In addition no L-shaped tetromino can contain four consecutive digits (for instance you can't have r2c9=5,r4c9=8). The little killer clue indicates cells on the indicated diagonal must sum to 40 and may contain repeated digits.

Source: Cracking The Cryptic, 10 December 2020. Obviously, one can solve the puzzle by watching the video but that’s kinda ruining the fun isn’t it? 😊

enter image description here

  • $\begingroup$ This sudoku looks awfully symmetrical. $\endgroup$
    – Anonymous
    Commented Dec 11, 2020 at 8:49
  • $\begingroup$ @Anonymous that doesn't mean it's easy $\endgroup$ Commented Dec 11, 2020 at 11:22
  • $\begingroup$ After the first half hour or so, I really struggled to find an obvious naked single. Watching the video, Simon missed the exact same one for several minutes. (After reaching that point a lot faster than me, of course.) Made me feel at least a little better. :-) $\endgroup$
    – Bass
    Commented Dec 11, 2020 at 16:09
  • 2
    $\begingroup$ "No L" - hoho, I get it! :) $\endgroup$
    – Stiv
    Commented Dec 11, 2020 at 19:25

1 Answer 1


OK, I finally found a solving path after being at this for hours.

I eventually found a breakthrough via a few observations:


Here, 4 and 5 must appear somewhere in the 3 middle orange squares. The green squares represent the positions that 6 can take and the yellow squares represent the positions that 3 can take. In fact, each of the 3 x 3 squares in each corner of the grid has a similar configuration.

The next thing I realised was that the 6 and 3 cannot be in the corner. If we try to place a 6 there, we will eventually run into a contradiction somewhere in the other 3 x 3 corner squares of the grid. Here is one example for 6:


Here, there is no way to place a 3 in the bottom-left corner. A similar contradiction arises for placing 3 in the bottom left corner.

This means the 6 must appear in either R1C2 or R2C1 and 3 must appear in R8C1 or R9C2. We can also extend this argument for all the corner 3 x 3 squares.

Lastly, 3 and 6 cannot appear in R9C2 and R1C2 together. This will force R4-6C8 to hold 4 numbers, as seen below:


Thus, 3 and 6 must appear in R8C1 and R1C2 respectively or R9C2 and R2C1 respectively. This also means that one of 3 or 6 must appear in the orange squares (R4-6C2).

Now, finally, for the breakthrough. If R4-6C2 were to be 4,5,6 in some order, then there is no room for a 7 in the middle left square. The 7 in R5C6 prevents placement in R5 and we cannot place 7 in any of the corners of the 3 x 3 square because it will form a L tetromino with consecutive digits. Therefore, the middle digits must be 3,4,5 instead. This lets us place the 3 and 6 and some other digits from chain deductions.


Since R4-6C2 is 3,4,5 in some order, 2 and 6 cannot appear in any of the corners of the 3 x 3 square. (otherwise, it will form an L-shape with consecutive digits). Similar arguments for the other squares.


Now, R2C7 cannot be an 8. If it was a 2, then it runs into a contradiction as both 1 and 9 are forced to R1C9. Therefore, it must be a 9. The only remaining numbers for R2 are 2 and 8. Since 2 already appears in C3, 2 must go in R2C1. This also means 2 must be in R9C2 of the bottom left 3 x 3 square. From there, we can make some more chain deductions:


Next, a few more deductions in several places:

R8C4-6 must be 4,5,6, so R8C7 and R8C9 must be 2 and 7. However, 7 cannot appear in R8C7 because it will form an L-shape.

Then, 5 must appear in R5C2. If it was 3 or 4, they will form an L-shape. Following that, 4 cannot appear in R4C2. So it must appear in R6C2 and 3 in R4C2.

Next, for R5, only 3, 8 and 9 are missing. 3 can only appear in R5C8 and after that, 9 can only appear in R5C5.


Next, it's time to finally make use of the 40. Counting along the diagonal, we have 6 + 8 + 7 + 7 = 28. Therefore, the sum of the remaining squares is 12. Currently, the bottom right square is missing a 1 and 8. If R7C8 was an 8 and therefore part of the sum for 40, we will have split 4 over the 3 remaining squares. That means there must be at least 2 '1's. However, that is impossible since 1 can only appear in one of those squares. Therefore, R7C8 is 1. From the corresponding deductions, we get:


Then, focus on the top middle 3 x 3 square. Note that 8 must be in either R1C6 or R3C4. Hence 6 cannot be in R2C4 since that prevents 8 from being placed in either of those squares. Similarly, 8 must appear in either R1C8 or R3C8. As a result, 6 cannot be in R2C6. Therefore, 6 must be in R2C5.

Now focus on the bottom middle 3 x 3 square. 6 cannot appear in R8C4 since that will cause 4 and 5 to form an L-shape with 2 and 3. Since 6 already appears in R2C5, this 6 must be in R8C6.


Next, note that the last column is currently missing a 2 and a 9. However, 2 cannot appear in R6C9 due to formation of an L-shape, so it must appear in R4C9. From there, there is a series of chain deductions involving the left-middle 3 x 3 square:


8 can now only go into R4C5. Since that diagonal must sum to 40, R3C4 must be 2. From the deductions that follow, we get:


Lastly, note that 6 cannot appear in R4C8 or it will form an L-shape otherwise. From there, it is routine to finish the grid from the chain deductions:



Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.