Let S denote the squares that are shot only once, and let T denote the set of squares that are shot at least twice. Then S forms an independent set (in the graph theoretic set), and the length of the underlying shooting sequence is at least |S|+2|T|. On the other hand, for an arbitrary independent set S the seqnence that first shoots all vertices in T, then shoots all vertices in S, and finally shoots again all vertices in T is a feasible shooting sequence. (If the tank hides in T, it is hit in the first phase and once again hit in the second or third phase; if the tank hides in S, then it is hit in the sceond phase and once again hit in the third phase).

All this implies that the length of the shortest feasible shooting sequence of any graph equals twice the number of vertices minus the size of the largest independent set. For the 41×41 chessboard, the largest independent set consists of the 21×21 black squares in a standard chessboard coloring.

Hence the answer to the 2015 All-Russian Math Olympiad puzzle is 2*41*41-21*21=2921 shots.

]]>I can come up with a two-coloring of the board in the first case and with a four-coloring in the second.

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