An entry in Fortnightly Topic Challenge #47: "Wacky Sudokus"

A Sudoku with an oddly familiar shape. Here's a Penpa Link.

enter image description here


  • Normal Sudoku rules apply
  • (Special) Thermometers:
    • The digits in the thermometer shapes must be strictly increasing in each cell from the round bulb to the flat end (e.g., 12569).
    • Also, both ends of every thermometer must be an odd number.
    • Colors are only for visual purposes and do not influence the rules, sorry to disappoint you.
  • Between Line:
    • Numbers along a between line must be between the numbers on the circled ends of the line
  • $\begingroup$ @msh210 If one end is 9 and the other end is 6, all the values between would have to be 7 or 8, not less than 6. $\endgroup$ Jan 18, 2021 at 21:19
  • $\begingroup$ Oh! I misread that rule. $\endgroup$
    – msh210
    Jan 18, 2021 at 21:19
  • 2
    $\begingroup$ Cool puzzle! There were some really delicate deductions in there that I greatly enjoyed. Danke schön Lukas! $\endgroup$ Jan 18, 2021 at 22:37
  • $\begingroup$ @Jeremy Gern geschehen! :) Glad to hear you enjoyed it! $\endgroup$ Jan 18, 2021 at 22:48

1 Answer 1





First steps:

The left endpoint of the between line must be 9, since the 9 in the upper left square must be in the middle column, and a 9 cannot appear anywhere else on a between line. Then looking at the top 7 squares of the between line, they must be 7 distinct digits, none of which is either 1 or 9. Hence they are 2-7, and the right end of the between line must be 1. This places the 1 and 9 in the top row as well.

Now looking in the bottom middle square, the fact that every end of a thermometer must be odd implies that the bulb ends must be 1, 3, 5, and 7, since they must be distinct and 9 cannot be a bulb end. The 7 must be in the upper left corner, since the thermometer from the upper right is too long. The 5 cannot be in the upper right bulb since the thermometer would go past 9, and it cannot be the middle right bulb, since that would force the top of that thermometer to be either 7 or 9, which are both blocked. So the 5 must be in the middle left cell of this square, which forces the entire thermometer. Finally, R5C7 must be 9, since the 9 in this row is in the right square, R5C8 cannot be 9 because it is mid-thermometer, and R5C9 is blocked. The grid thus far:


Moving on:

In the upper middle square, the left middle cell must be 1, since the thermometer blocks the middle, and one of the two bulbs in R7-8C6 must be 1. R9C3 is 7 by Sudoku logic. This forces the 1 in column 3 to be in R8: R1-3 are blocked, R5-7 and 9 are filled, and R4C3 can't be 1 since it is greater than R5C2. This forces the bulbs in the lower middle square, and since R8C6 is 3, the other end of its thermometer must be 5 (it cannot be 7 or 9), forcing the whole thermometer.

Look at the thermometer starting from R7C6. It cannot have 5 anywhere on it: the spaces in the middle right square are blocked, the terminus must be 7 or 9, and R3C6 cannot be 5, because that would force R4C7 to be 4, which is blocked. This implies R2C5 cannot be 7, since if it were, we would again force R4C7 to be 4. This allows us to place all of the 9s in the grid, and forces R7C9 to be 3. Moreover, we find that R4C7 must be 7: it must be at least 4 and at most 7 given its position on the thermometer, and 4, 5, and 6 are all blocked. The grid thus far:

Some Sudoku forcing:

The 8 in the upper left square must be in either R1C2 or R1C3, so in particular neither R2C1 or R3C1 is 8. This forces R7C1 to be 8, and allows us to complete row 7.

About time to tackle the last thermometer. If its bulb in R5C2 were 3, then R4C6 would have to be at least 7, but both 7 and 8 are blocked; therefore R5C2 is 1. But this forces the 5 in column 3 to be in the upper left square, which again means neither R2C1 or R3C1 is 5, so R5C1 is 5.

In the bottom left square, R8-9C2 are 2 and 4 in some order, which means in the middle left square, R4C2 must be 3. The thermometer prevents R4C3 from being 4, so we can finish the square. In the upper left square, this same observation shows that R1-2C2 are 6 and 8 in some order, which the 8 in R2C9 forcing the positioning. The grid thus far:


Finishing up:

R4C6 cannot be 6, forcing the rest of the R5C2 thermometer. This then forces R3C3 to be 5. We have R3C9 is 7 from its column, allowing us to complete row 3 and column 1. We also find R9C9 is 5. This forces R9C7 to be 1.

We also find the 4 in row 8 must be in R8C2, since it is blocked elsewhere, finishing the bottom left square. This forces R9C4 and R5C6 to be 4 as well. The values remaining to be placed in C6 are 2, 6 and 7, but R2 already has a 6 and 7, so R2C6 is 2, forcing the rest of the column. Indeed, the rest falls with easy Sudoku fill-ins.


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.