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Solve this puzzle.(Handmade)

Rules:

  1. Every rows and columns should have unique digits(1~9) once.
  2. Regions(Regions): Yellow-colored cells should not have same digits.
  3. Max((Ddu Ddu Ddudu )Max( Verstappen)): Bigger than adjacent cells.
  4. Min(Min): Smaller than adjacent cells.
  5. Even(Even): Cells with gray square (except Max/Min) should have even digits (2, 4, 6, 8).
  6. Odd(Odd): Cells with gray circle (except Thermometer) should have odd digits (1, 3, 5, 7, 9).
  7. Thermometer(Thermometer): Digits should increase from the gray circle, following the thermometer line.

Now, here is the puzzle:

Puzzle
Click the puzzle to play in Sudokupad.(Sudokupad may say your answer wrong, but nevermind if that’s valid.)

When I posted the Min-Max Circuit, the rates were “Nice Puzzle”, “Enjoyable.”

This time, the rates may be “Evil Puzzle”. :) Good Luck!

Criteria for Upvote / Acception

  • Upvote: Correct answer

  • Accept: Logical Step (No need to be fully logical - some skips are allowed.)

[Edited]

Since Hint 2 is provided, I will accept the answer using additional clues in Hint 2.

[Partial answers are welcome.]

Hints will be evil too.

Hint 1.

No 6 on the yellows.

Hint 2.

Puzzle with more clues
Click the puzzle to play in Sudokupad.(Sudokupad may say your answer wrong, but nevermind if that’s valid.)

Hint 3.

Puzzle with more and more clues
Click the puzzle to play in Sudokupad.(Sudokupad may say your answer wrong, but nevermind if that’s valid.)

Solver says logical steps are found.

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  • $\begingroup$ Do thermometer bulbs count as grey circles (odd digit)? $\endgroup$ Commented Aug 3 at 13:26
  • 3
    $\begingroup$ @FlorianF No, it doesn’t count. See Rule 6. $\endgroup$ Commented Aug 3 at 14:15
  • 1
    $\begingroup$ This one's stupid hard. Two hours in, I've only got one new digit in addition to the "given" thermometers. $\endgroup$ Commented Aug 4 at 15:53
  • $\begingroup$ @Bass Brute force leads to that result ;) I think this puzzle is no more than a lucky box… $\endgroup$ Commented Aug 4 at 17:03
  • 2
    $\begingroup$ I verified with a computer that there is a unique solution. Good luck! $\endgroup$ Commented Aug 4 at 20:16

2 Answers 2

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Solution (to the board in hint 3):

completed sudoku

Steps:

First, complete the top and right thermos and mark the other outer thermos. sudoku after the first step

Next, mark the middle row and column. Since only two values are greater than 6/7, the two max cells must have 8/9 and the middle sections must have 1-5. Since the cells between the min cells must be greater than two values in each row/column, they cannot be 1/2. Using the odd/even constraints, the center cell must have a 2. sudoku after the second step

Next, the remaining cells in r1 and c9 must be 7-9. Since r1c3 must be odd and c3 already contains a 9, r1c3 must be a 7. sudoku after the third step

Now, (using hint 2), the only candidates for cell r6c3 are 2/3/4. Since cells r6c3-6 contain 1/2/3/4, the remaining cells must have 6/7/9. The even cell must be a 6, and we can use the thermo to place 7/9. sudoku after the fourth step

Due to the odd constraint, there is only one cell in c1 that can contain an 8, and that 8 can be used to place 8/9 in c5. sudoku after the fifth step

Now, add pencil marks to the remaining odd/even cells and the thermos. The even cells contain a 2/6 pair, so remove them as candidates from the other cells in c7. Now, the only cell in c7 that can contain a 9 is r9c7. sudoku after the sixth step

The only cells in c3 that can contain a 6/8 are r3c3 and r7c3. Due to an X-wing on 6s in c3/c7, we can remove 6 as a candidate from r7c8. Now, the only cell that can contain a 6 in c8 is r9c8. Placing the 6 also lets us fill out r9. sudoku after the seventh step Note: I missed removing some pencil marks in this step; I notice and remove them in a later step.

Since the yellow squares do not contain a 6, then they must contain 1-5 and 7-9. The only yellow squares that can contain an 8 are in c3, so we can fill out the even squares in c3/c7. sudoku after the eighth step

r8c12/7/8 only contain 1/5/7, so we can remove them from the rest of r8 and place a 2 in r8c3. This lets us place a 3/4 in r6. sudoku after the ninth step

r2c3, r4c7, and r8c7 only contain 1/5/7, so we can remove them from the rest of the yellow cells. There is a 2/9 pair in r7, so we can mark the remaining yellow squares. The 4/8 pair in r3 places a 5 in r3c1. sudoku after the tenth step

I missed removing some pencil marks and fixed them, which placed the remaining cells in c1 and placed r3c2 and r7c4.
sudoku after the eleventh step

Pairs in c2 and c5 place 1 and 2 in r7, so we can now finish r7 and place a 1 in r8.
sudoku after the twelfth step

r2c8 must contain a 9, and we can place the rest of the 8/9s. sudoku after the thirteenth step

A 5/7 pair in c2 places a 4 in r4c2 sudoku after the fourteenth step

And that lets us fill out everything except for pairs in c4 and c6 sudoku after the fifteenth step

The thermo puts a 6 in r2c6, which lets us fill in the remaining cells.
completed sudoku

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  • $\begingroup$ Nice logical deduction!(+1) $\endgroup$ Commented Aug 14 at 5:21
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This is not solution

Here is the app I created to play the game online:

you can load the game by using the code below:

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

This will load the game along with its rules, allowing you to play online if you wish. The code isn't perfect, but it should be clear and easy to understand even not the same symbols are used.

Note: I haven’t tried to solve it yet; even I can solve it with computer, I just wrote the code for you. I also created a game creator and its solver, but as this will be one time probably, I don't believe you would need it.

Here is the very partial answer so far:

![enter image description here

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  • $\begingroup$ Wow, +1 for the app! I appreciate it. However maybe checking if there are some errors, such as in the right-downside thermometer or something else, will be helpful. :) $\endgroup$ Commented Aug 4 at 22:48
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    $\begingroup$ Pretty sure you can't put a 4 in the middle, that would place the two 2s in column 5 and row 5 so that there can be only one 2 in the area (call it "area A") you get by combining the red and yellow squares and removing the centre. If you then fill in the solved thermo on the right, you'll note 4 can also only appear once. The 6 and 8 can each appear at most twice in A, for a maximum of 6 even digits in the area. But the yellow has 8 distinct digits in is, so it must have at least three even digits, and adding in the 4 reds in A, we actually need 7 evens. (I may be mistaken, of course.) $\endgroup$ Commented Aug 5 at 0:53
  • $\begingroup$ And since that's quite a mouthful, here's a picture. The squares with the red outline must have (at least) 7 even digits in them, and that will be impossible if there are two 2s in the pink squares. $\endgroup$ Commented Aug 5 at 1:07
  • $\begingroup$ (Incidentally, this logic seems to make the gray square in r5c5 redundant, and thus a gift to us solvers from the creator, so at least the evilness isn't maximal :-) ) $\endgroup$ Commented Aug 5 at 1:23
  • $\begingroup$ @Bass If you watch this video (at this timestamp), it talks about a parity rule for a checkerboard pattern. If you apply the parity rule to the given puzzle, you'll see that 1 and 2 can only appear on "even" squares in either the thermometer bulbs or the yellow squares. That means that if 1 appears in a yellow square once, it must appear in a bulb 1 or 3 times (same goes for 2). So far that hasn't helped me, but maybe that will help someone else. $\endgroup$ Commented Aug 5 at 19:07

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