This puzzle is part of the Puzzling Stack Exchange Advent Calendar 2023. The accepted answer to this question will be awarded a bounty worth 50 reputation.

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This puzzle is a variant of Thermometers, but with peppermint sticks! The goal is to shade some (or no) cells on each peppermint stick so that the shaded cells in each peppermint stick form a single contiguous strand starting at the large end. The twist (see what I did there!) is that there are two colors for shading, red and green, which must alternate along each peppermint stick. The first cell on each peppermint stick may be either red or green. Numbers on the top/left of the grid define the number of red cells in each column/row, and numbers on the bottom/right of the grid define the number of green cells in each column/row.

Puzzle Grid

Penpa link!

I hope you enjoy! This puzzle can be solved with constraint programming, but there is a logical path through, so no computer solutions please. The accepted answer will have the final grid, as well as the logical steps describing the solution. Please try to avoid bifurcating; contradictions are OK, but should not require more than a single sentence to describe.

  • $\begingroup$ I might be misunderstanding something but should the top right green 3 be a 4 or is it correct? $\endgroup$ Commented Dec 9, 2023 at 11:47
  • 1
    $\begingroup$ @Prim3numbah The top right green 3 is correct. $\endgroup$ Commented Dec 9, 2023 at 12:26
  • 1
    $\begingroup$ @Prim3numbah think of it like an actual thermometer, it start at the bulb and gets filled up, but not necessarily the whole way $\endgroup$ Commented Dec 9, 2023 at 13:34
  • $\begingroup$ Regarding "The first cell on each peppermint stick may be either red or green" - can it be unshaded as well? Or it HAS to be either red or green? $\endgroup$
    – Alaiko
    Commented Dec 9, 2023 at 15:16
  • $\begingroup$ @Alaiko having made a decent amount of progress I can confirm it can also be unshaded $\endgroup$ Commented Dec 9, 2023 at 16:45

2 Answers 2


The answer is:


Solving path:

Step 1:

Start with the top row. Given that red and green must alternate and to satisfy the red '5' clue, both R1C1 and R1C10 must start with red. 2 green cells must also appear in the peppermint stick on the left in Row 1.(Black here represents not filled with either red or green).


Then, in column 5, there must be 6 green cells. Even if R2C5 and R12C5 are green cells, 4 of the green cells must appear in the long peppermint stick in Col 5. The red '4' at the top of column 5 needs 3 more red cells, which have to be used in the alternating pattern in the peppermint stick. We can then fully fill up Column 5. Then, in R12, at least 3 red cells must appear in the long peppermint stick so we can fill it up as shown below.


Step 2:

Then, in column 9, there are only a few areas to satisfy the red '4' clue at the top as R9 and R12 cannot be red due to the alternating pattern. Hence, no matter what, at least one of the red cells must lie in R6 to R8. This means that the peppermint stick in the bottom right in Col 11 must be filled. However, the top clue for that column indicates that there can only be three red cells in that column. Therefore, that peppermint stick must start with green. Additionally, for R12, we can resolve where two more red cells will lie.


Then, look at Col 7. For the green '5' clue at the bottom, there is currently only 1 shaded green cell. R1C7 cannot be green. Even if there was one green cell in R2C7 and two green cells in Rows 5 to 9 for Col 7, there would at least need to be one green cell at R11C7. Since there can only be 4 green cells in Col 6, we now know how to fill up that long peppermint stick in Col. 6. We can also fully resolve Row 1.


Step 3:

Then, look at Col 9 again. R6 to R8 only had one red cell; 1 must lie in either R2 or R3; 1 must lie in R11 (the green '3' clue for that row is already satisfied, so the red cell cannot be in R10). There are no other spaces for the last red cell so it must lie in R5C9. Meanwhile, R6C7 must also be a green cell at the very least to satisfy the green '5' clue in Col 7, which places a total of 2 green cells for Col 7. R8C7 and R9C7 cannot both be green cells since it would otherwise violate the red '2' clue for col 7. So R2C7 must be a green cell. Additionally, because the '3' clue in R11 is already satisfied, there cannot be any more green cells in R11, C2 to 4. So, we can mark R11C3 onwards with black (R11C2 might still be red)


Looking at Row 5, for both the '4' green and red clues, we have three cells each that are already filled in and need one more of each. Now, note that the green '4' in Row 6 is already satisfied. So, for the green '3' clue in Col 4, there must be at least one cell in R2 to R4, C4. This means that the long peppermint stick that appears in R6 to R11 in Col 1 must be empty. However, if that is empty, then the peppermint stick in R2 to R5, C1 must be filled to fulfill the red '4' clue in Col 1. Therefore, the remaining green and red cell for Row 5 must be in R5C1 and R5C4 in some order.


We can fully resolve Col 12 after this. Then, looking at Col 2, to satisfy the green '6' clue at the bottom, at least two must appear in R6 to R11. R6C2 cannot be green, so it must be red. Doing so place green cells at R7C2 adn R9C2, which satisfies the green '2' clue in R9. Doing so blacks out R9C7, so we now know that R7C7 must be red and R8C7 must be green to satisfy the clue for the column. (I also belatedly realised that I could have blacked R11C10 a bit earlier, but I did so here).


Step 4:

After that, for the red '4' clue in Row 11, the last red cell cannot be in R11C2, so it must be in R11C8, which forces R10C8 to be green. Then, for Row 10, R10C3 cannot be red as it would force R9C3 to be green, but there are already enough greens in Row 9. Additionally, R10C3 also cannot be green since it would cause R6C3 to be green, but R6 is already satisfied.


Focus on Col 2 after that. For the green '6' clue, there are already 4 green cells. 1 of them must be in R3 to R4 and the other one must be in R2C2. This means that R5C1 is also green, R5C4 is red, and we can fully resolve peppermint stick beginning at R5C4. After that, the red '3' clue in row 4 needs two more red cells, so one of them is in R4C3 and the other is in R4C10. The green '2' clue in Row 3 is also resolved.


After that, R6C3 must be red because that column is still missing one red cell. Similarly, R7C8 must be green to satisfy the green '3' clue in col 8. Doing so satisfies the green '3' clue for R7. Since R7C3 can neither be red nor green, that peppermint stick is black from that cell onwards. Doing so allows us to fill the remaining cells rather easily and we get the final answer:


  • 1
    $\begingroup$ @BenReiniger Right, thanks. I accidentally uploaded the wrong one. The image before that was correct. $\endgroup$
    – Alaiko
    Commented Dec 9, 2023 at 19:50
  • $\begingroup$ That's it! Hope you enjoyed! $\endgroup$ Commented Dec 9, 2023 at 22:29

Final grid:

final solution

Solution milestones:

The top row gets us started: both of the two thermometers need to start and end red. That puts a red in the fifth column, increasing the imbalance enough to fill in some there. Purple below means known to be filled, but color not yet known: first steps

The red-labeled columns on the right side together give us the next breakthrough: a little more...

and now the middle row has all its greens, which gives a bit more almost there and the rest is straightforward compared to what came before.

  • $\begingroup$ Nice solution, but sniped by a well-written answer from Alaiko...that's quite the rite of passage :-) Hope you enjoyed! $\endgroup$ Commented Dec 9, 2023 at 22:30

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