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enter image description here

Rules of MIXXOX:

  • Put either X or O in each empty cell.
  • A border between two adjacent cells may have a mirror, denoted by a bold line. Some mirrors are already given. Note that the outer border of the grid does not count.
  • A border has a mirror if and only if the border separates two lines of cells that are mirrored to each other, without taking the part of the reflection outside the grid into account. The two lines of cells are the cells perpendicular to the border.
  • Each border or mirror should be treated independently.
  • The numbers outside the grid tell you how many mirrors each row and column have.
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  • $\begingroup$ (Long overdue comment but anyway:) The instructions didn't seem to be explicit about this so I want to clarify something: it seems that in a row and column, the length of the shortest line of cells, sandwiched by a mirror and the outer border (regardless of any mirrors in between), must also be the length of its reflection. Is this correct? $\endgroup$
    – oAlt
    May 11 at 3:09
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    $\begingroup$ @oAlt yes, that's correct; basically the mirror reflects the whole row or column (and may assume everything outside the grid can be any symbols.) $\endgroup$
    – athin
    May 11 at 4:12
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I think this is the solution:

enter image description here

Explanation (red):

1. The right column contains five mirrors, i.e. on all positions. This is only possible if the six cells all contain the same symbol. That symbol can't be an X; otherwise, the bottom row would have a mirror between columns 5 and 6, so it must be an O.
2. For row 1, we can fill in an X in column 1 and an in column 5 because of the existing mirrors. Since those are the only mirrors, the central cells must have different symbols (either OX or XO).
3. Since row 5 doesn't have a mirror, the cell in column 5 must be an X.
4. That means in column 5 the only mirror is at the bottommost position, and therefore row 2 is an X, and therefore row 3 is an O.
5. The only mirror in both row 3 and row 4 is at the rightmost position. For row 3, that means the cell in column 1 must be an O.

Continued (blue):

6. Because of the mirrors in columns 3 and 4, row 1 is a copy of row 6. Step 2. left two possibilities: OX and XO, but OX would produce a mirror in row 6, so it must be XO.
7. Since there is no other mirror in column 4, row 2 and 5 must be Xes.
8. Since there is no mirror in row 5, column 3 must be an X, and therefore row 2, column 3 as well.
9. If row 3 and 4 in column 3 are Xes, it would contain 5 mirrors; therefore, they must both be Os.
10. Since there are no other mirrors in row 3 and 4, column 4 must be Xes.

Final part (green):

11. The only way to get two mirrors in row 2 is if the first two cells are Xes.
12. Row 5, column 2 must be an O; an X causes 1 or 2 mirrors in that row while there are none.
13. The only remaining way to get 3 mirrors in column 2 is if row 4 is an X and row 6 is an O.
14. The remaining cells in column 1 must be different from column 2 in the same row, otherwise there would be mirrors in the leftmost positions.

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    $\begingroup$ Greatly solved! Very well done! :) $\endgroup$
    – athin
    Apr 13 '20 at 14:01

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