Eulero is a grid-based logic puzzle. Like Sudoku, it involves Latin squares; however, while Sudoku relies on a single Latin square, Eulero consists of two overlaid squares, one of numerals and one of letters. An additional constraint is that each letter-number combination has to appear exactly once.
This got me thinking: What are the fewest number of clues needed for a Eulero puzzle?
The only obvious constraint is on grid shape: It has to be square, but because the grid isn’t subdivided, it doesn’t have the limitation that Sudoku does that only certain square sizes work. The trivial case is 1×1, which requires no clues; the only 1×1 solution is a square with A1 written in it.
There are no 2×2 solutions. If the additional constraint is followed, the cells must contain A1, A2, B1, and B2. However, there’s no way to place those in a square configuration such that both the letters and the numbers make Latin squares. For instance, in the table below, the letters make a Latin square but the numbers don’t.
If both the numbers and the letters form Latin squares, as in the example below, then some of the combinations are repeated:
There’s no way to satisfy all three constraints at the same time.
There are 3×3 solutions. In fact, there are two basic shifts in a 3×3 Latin square (clockwise rotation and counterclockwise rotation), which means that all solutions are variations of the same solution:
That left me to wonder, then: What are the fewest clues needed to specify a unique solution? Two would seem to be the minimum for letters, since if there were only one letter, we couldn’t place the other two letters definitively; likewise, two would seem to be the minimum for numbers as well.
There is indeed a way to put in four clues that defines a single solution. It has the added beauty of being symmetrical:
The letters force the unfilled corners to be C, while the numbers force the unfilled corners to be 1:
The next step is to fill in the sides:
This leaves the remaining combination, C1, for the center:
Since a 3×3 Latin square will always have one diagonal that’s the same, and since this problem relies on the opposing corners being different, the clue set pattern only works with half the possible solutions. I’m not sure if there is a pattern of four clues that will work with all possible solutions, though. If you find one, please comment.