On Two By Two

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The Baron's most recent wager with Sir R----- set him the challenge of being the last to remove a horizontally, vertically or diagonally adjacent pair of draughts from a five by five square of them, with the Baron first taking a single draught and Sir R----- and he thereafter taking turns to remove such pairs.

When I heard these rules I was reminded of the game of Cram and could see that, just like it, the key to figuring the outcome is to recognise that the Baron could always have kept the remaining draughts in a state of symmetry, thereby ensuring that however Sir R----- had chosen he shall subsequently have been free to make a symmetrically opposing choice. Specifically, if we give the places upon the table the coordinates
-2, 2-1, 20, 21, 22, 2
-2, 1-1, 10, 11, 12, 1
-2, 0-1, 00, 01, 02, 0
-2, -1-1, -10, -11, -12, -1
-2, -2-1, -20, -21, -22, -2
then the symmetry of interest is that between \((x, y)\) and \((-x, -y)\). I told the Baron of this, but I am not entirely sure that he was following my line of reasoning.
Now, the only way that Sir R----- may have prevented the Baron from picking a pair that were, in this sense, opposed to his would have been to remove a pair that included both a draught and its opposite so that the latter could no longer be chosen. The difference between such opposites is trivially
\[ (x, y) - (-x, -y) = (2x, 2y) \]
and so they will be neighbours if, and only if, the magnitudes of both \(2x\) and \(2y\) are no greater than one, which is clearly only possible if both \(x\) and \(y\) equal zero.
To ensure victory the Baron should have simply removed the centre draught upon his first turn and subsequently removed the pair at the coordinates that were the negations of those chosen by Sir R-----, and I should consequently have advised him to refuse the Baron's wager.
\(\Box\)

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