In each cell of a 5 × 5 square there is a token that is black on one side and white on the other. Initially all tokens are placed with the white side facing up. On each move three tokens in consecutive cells in a row or in a column are flipped over. What is the smallest number of moves needed in order to obtain the chessboard colouring shown below ?
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3$\begingroup$ The move order doesn't really matter, does it? $\endgroup$– JafeJun 25, 2019 at 14:11
1$\begingroup$ That was quick! I think you can prove this is optimal by looking at the group of three tokens along an edge of the square. At least two moves are needed to get them correct. The same goes for the other edges, and no move can affect two or more such groups. $\endgroup$ Jun 25, 2019 at 14:13
$\begingroup$ @jafe you are right! $\endgroup$– MattiJun 25, 2019 at 14:21
$\begingroup$ @JaapScherphuis yeah, good idea. $\endgroup$– MattiJun 25, 2019 at 14:22