This solution makes some deductions without comment. First, if a square is identified as a ship piece, then we know that none of its diagonally adjacent pieces are ship pieces. Also, if identifying a ship square accounts for the complete sum in its row (or column), then it is assumed that all of the remaining squares in the row (or column) are ignored from further consideration.
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The first thing that we can do is to ignore any squares that have values that are greater than the sum for their row or column. |
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Column S will not use the 3 at ES, so EQ is the only ship square in row E. BP is the only ship square in column P. (Recall that DP is not a ship square because it is diagonally adjacent to EQ.) The only way to legally get a sum of 16 from column Q is to have BQ and DQ be ships. |
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The only place with "room" for the four-unit ship is BN-BQ. Row C can only be satisfied with ships at CK, CS, and CT. BK must be a ship square, as must AT, FL, and JL. FM must be a ship square to satisfy column M, so FN is the last ship square in row F and column N. GT is the last ship square in row G or column T, and JO must be filled in to satisfy column O. The only squares unaccounted for are HR-JR, which must all be filled in to satisfy column R. |
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The submarines are at AT, GT, JL, JO. |
Notes: I didn't attempt this problem until I only had about twenty minutes left in the test. I thought that it would be a good puzzle to finish off with, but it actually only took about five minutes to solve.