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Fill each empty cell of the board on the left with a digit between 1 and 6. Each column and row must contain all digits.

The dots outside the board on the right indicate how many cells in the corresponding column or row of that board contain precisely the same digit as is to be found in the same cell on the board on the left.

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2 3 5 1 4 6
5 2 1 3 6 4
1 6 2 4 3 5
4 1 3 6 5 2
6 5 4 2 1 3
3 4 6 5 2 1

Logical steps:

Let $(a, b)$ denote the $a$-th row and $b$-th column. I filled in the cells in the following order:

(1,2) = 3: on row 1, the two 1's and two 4's can match at most 2, and the 6 already doesn't match. So the 3 is a match.
(1, 5) = 4, (3, 5) = 3, (5, 5) = 1: on column 5, there only remains these three possible matches.
(4, 5) = 5, (6, 5) = 2: only possibility for (4, 5).
(2, 2) = 2, (2, 5) = 5: since the only match on row 3 is (3, 5), we know (3, 2) is not a match. Thus (2, 2) and (2, 5) must be matches.
(2, 1) = 5, (2, 4) = 3: Now (2, 6) is not a match, these two are matches.
(5, 6) = 3: it's the only possible remaining match on row 5.
(1, 4) = 1: it's the only possible remaining match on row 1.

At this point we determined all the matches. After that it's easy (similar to a normal Sudoku).

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