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I present a *deep breath* Even-Odd TetroThermoDoku. From the back of the name to the front:

Rules

  • Sudoku: Fill each cell with a digit 1-9 such that no number repeats in any row, column, or 3x3 heavy-outlined region.
  • Thermometer: Digits within a thermometer (gray-shaded shapes) must strictly increase from the bulb (circle) to the flat end. For thermometers with multiple ends, digits increase along any contiguous path. For example, R2C2 (R1C1 is the top left) must be smaller than both R2C1 and R2C3, but the latter two may have any values relative to each other.
  • Tetromino: Shade cells such that each heavy-outlined region contains a tetromino. This matters because...
  • Even-Odd: Shaded cells contain even digits.

Puzzle

The puzzle looks vaguely like a crab if you squint (click for full-size version):

puzzle

Here is a playable Penpa version (no automatic answer checking).

This is a logic puzzle, and is intended to be solved logically. Please include description in your answer as to how the puzzle was solved. This could be as little as explaining a few key deductions, or as much detail as you would like. Answers with just the final solution will not be accepted.

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  • $\begingroup$ Sorry all, forgot the L. Also removed the I because it's cause more trouble than it's worth $\endgroup$
    – bobble
    Commented Apr 15 at 13:48

2 Answers 2

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The solved grid:

Final grid

Starting in the middle:

The long thermo in the middle box is a good place to start. Being 7-long, initially the bulb could be 1/2/3, the next cell 2/3/4, etc. However, the tetromino rule means that we cannot isolate an even digit in a corner. For example, this implies R6C5 cannot be 7, since we would have 789 along thermo in the lower right corner, which would not allow the 8 to join any other even digits. Similarly R6C5 cannot be 5, since that would isolate 4 in the lower left corner, and this cell is forced to be 6.
Since it is in a tetromino, the 6 must be adjacent to an even digit. If R6C4 is even (necessarily 4), then R5C4 is odd, and if R6C6 is even, then R5C6 is odd. Thus the only way a tetromino of even digits in the middle box can fit is if R5C5 is even. Now, R5C4 cannot be 3: if it were, then 2 would be forced in the upper left corner and could not join the tetromino. This places the 4 and 5 in the box. Finally, we find R3C6 must be odd, since it cannot join with the other three even digits in the box. The grid thus far (blue denotes even cells, red odd cells):

Initial progress

Pencilmark the thermos:

For the longer thermos, including the Fs in box 4 and box 8, we can pencilmark using the thermo rule and sudoku, and specifically we reveal in row 6 that the only places 1 and 2 can go is in columns 3 and 7, so these form a 1/2 pair. That's nice, I guess, but...

A general observation:

No box can have its middle row or column be entirely odd digits. If so, then there must be digits on either side that cannot join up. A corollary to this is that no box can have its middle row or column be entirely even digits either, since that would force another box to have only even digits. Specifically in column 5, R4C5 cannot be even, since if it were, either box 2 or box 8 would have to have a middle column of all odd digits. Hence we can conclude R4C5 is 1. Now this 1 looks over into box 4, removing 1 as a candidate for the thermo bulb. This then leaves only one location for the 3 in row 6, namely R6C9, which forces a 2 and 1 below it on the thermo. This 2 must now be in a tetromino, which forces R7C8 to be even. The grid thus far:

Progress 1

Back to box 5 real quick:

The general observation also shows that R5C6 must be odd, and hence 9.

Then back on the track:

The 1 in box 9 looks left, removing 1 as a candidate for the bulb of the thermo in box 8. This again eliminates the smallest candidate in each cell on the thermo. That didn't help nearly as much yet as I hoped. Let's go back to box 9. We know the location of 2 even digits in the box, and no tetromino can contain those two cells and reach R9C7 or R9C9. Moreover, a tetromino reaching R9C8 would have a middle column of all even digits. So we can conclude that all of R9C7-9 are odd. Pencilmarking candidates, we can remove 3 as a candidate from R9C8 (since the bulb can be neither 1 nor 2).

Let's look back in row 6:

Another consequence of the tetromino rule is that the middle column or row of a box cannot be even/odd/even, since no tetromino can reach around and contain both even digits. So in box 4, this eliminates 4 as a candidate for R6C2, since it would yield a 2/3/4 column down the middle. This leaves a 7/8 pair in the row, and places the 4 and the 9. The grid thus far:

Progress 2

Some small deductions:

In box 6, R4C8 must be odd. If it were even, then the middle column of this box would be even/odd/even, whose evens can't fit in a tetromino, or even/even/even, which would force an all odd middle column in another box. So in fact, this is general: if we have an even digit on the edge of a box, then the opposite edge cell is odd. Not sure we can use that yet, but good to have in the toolkit.
Staying in box 6, note that R5C6 cannot be even. We can have at most (in fact exactly) one even in the middle row of box 6, and putting it in R5C6 would isolate it from the 4. This now isolates R4C9, which must be odd, and in fact gives us a 579 triple in the box. Now, R5C7 cannot be odd, since it would force R4C7 to be an isolated even digit. But as noted, there can be at most one even in the middle row of box 6, so R5C8 must be odd (and in fact 1). This allows us to place the 2 as well, which looks left into box 4. Note that we have also placed all of the odds in column 9, forcing R123C9 to all be even, which forces all of R1-3C7 to be odd. As C8 has only one even digit in box 3, we find R8C8 must be even, and sudoku combined with the thermo resolves the 6 and 8 in this box. The grid thus far:

Progress 3

Look in box 4:

R6C2 cannot be 8. If it were, then R5C2 would have to be even for the tetromino, which would force both R5C1 and R5C3 to be odd by sudoku. The tetromino rule would then force R4C2 to be even, which is forbidden. So we must have R6C2 odd, and hence 7. This resolves all of box 5, and also sets 7 in R5C9. Moreover, it places all odds in row 4, forcing all of R4C1-3 to be even. Moreover, in column 4, we now have only two locations which can be 1 and 2, and they resolve from the 2 in R7C9 looking over. We can also place the 6 in R1C4 (only location in the column). Again the tetromino rule forces R2C4 to be even, and thus 8. We can then resolve the row, and the resulting 9 in R9C4 looks right and resolves the bottom row of box 9. The grid thus far:

Progress 4

Hopefully finishing up:

The tetromino rule forces all of R1-3C6 to be odd, which forces all of R7-9C6 to be even. Sudoku and the thermo in box 8 force R9C5 to be 8. In box 9, the remaining cells are placed by the thermo, and the 5 looks up to resolve the odds in box 6. In row 7 all odds are now placed, forcing R7C1-3 to all odd. The 9 cannot be on the bulb of the thermo and is blocked in C1, so R7C2 is 9. The 9 in C3 must go in box 1, and R2 is the only possibility.

In box 7, sudoku forces R9C3 to be even. We also know that the middle row has exactly two evens, which cannot be even/odd/even, so we have R8C2 is even.

In box 1, R1C3 cannot be 2, since that would force R1C2 to be 1 and isolate the 2. Let's look at R2C2, the thermo bulb. It clearly cannot be 7, since neither 8 nor 9 can be in R2C1. But it also cannot be 6, since it would force the thermo to be 6/7, forcing both R2C5 and R2C9 to be 4. Not too sure how this helps yet...so let's pencilmark the rest of the thermo cells in box 1.

I left box 7 too soon. R9C1 cannot be even. If it were, R9C2 would be odd and no tetromino could reach around. Hence this cell is odd, and necessarily 1, and R9C2 is even. Now, box 1 must have an even in its middle column, so R5C2 must be odd. Staying in box 1, R1C3 must be odd. If it is even it is 4 and the tetromino rule forces R1C2 to be 2; sudoku forces R1C1 to be odd, and the elimination of 2 and 4 as candidates forces R2C2 to be odd, leaving the 2/4 pair isolated. Now note that R1C2 cannot be even. If it were, then R2C2 would be odd by sudoku, and thus all of R1-3C1 would be even, forcing R3C2 and R3C3 to be odd. The only place for 7 would be R3C3, but the biggest R3C2 could be is 5. So R1C2 is odd, and in particular one of 1 and 3. This gives a 1/3/5 triple in R1, forcing R1C7 to be 7 and resolving C7 in box 3. Pencilmarking R1C1 shows it can only be 2/4/8, so the tetromino rule forces R2C1 to be even as well. Only one of R2C2 and R3C2 can be even, so we must have R3C1 even and R3C3 odd. This sets all the odds in column 3 and evens in column 1. This breaks open the puzzle, with straightforward sudoku finishing the grid, the only wrinkles being that the 5 in R2C2 resolves the 4/6 pair via the thermo, and then 3 in R5C2 resolves the 2/4 pair in box 4.

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Partial answer. I found an interesting deduction early on.

After finding the 456 in the corner of box five, notice that the center cell has exactly two options, which each have a consequence. If the center cell is 2, then there is a 789 triple in the right column of the box. Otherwise, the center cell is 8, and there is a 123 triple in the top row of the box. That 123 triple is suspicious because it points at a thermometer bulb in box four, which can be no higher than 5. If the 123 triple was the case, that bulb must be at least 4, and thus all of the other cells on the thermo must also be at least 4. This means digits 123 are eliminated from seven cells in box four (highlighted red here), leaving only two cells where 123 must go. This is impossible, so the 123 triple is not the case and the center cell of box five is a 2. Places where 123 cannot go in box four, if box five has 123 across the top row

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    $\begingroup$ Please don't post a partial answer with a single deduction after someone has posted a full answer. Honestly even if there wasn't yet a full answer, this is too little and too soon for a partial answer. $\endgroup$
    – bobble
    Commented Apr 15 at 20:23
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    $\begingroup$ The rest of the solution follows Jeremy's path closely. I didn't post the whole thing because it seemed superfluous. I just thought this early deduction was neat. $\endgroup$ Commented Apr 15 at 20:49

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