Puzzle time - the forth corner
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Then you realize you can go backwards too. Not just the peg opposite to the blank, jumping its neighbor. Maybe it's a peg adjacent to the blank, jumping its neighbor, the one opposite to the blank. But somehow the system seems to exclude any peg ever reaching the blank.
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I'd say no.If we think of the points as being on a grid whose cell size is the square, then the pegs can always move only an even number of positions (since its two times the distance to another peg), but the distance from the fourth corner is odd (namely 1) for all three pegs. You cannot sum even numbers up to yield an odd number.
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Yep. Think of the points as pairs of parity flags or numbers mod 2. Your starting set {(0,0), (0,1), (1,0)} maps to itself under any peg jump.
You can almost tell from the question that the answer is no. Otherwise it would probably be ‘what is the minimum number of peg jumps....’
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I have to say no also
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The official solution came out today:
SOLUTION: The first thing to do here is think of the initial square as a cell of the plane grid, e.g., the points (0,0), (0,1), (1,0) and (1,1) on the X-Y plane. Then the pegs will always be on grid points.
Grid points, however, have four possible parities: each coordinate can be even or odd. When a peg jumps, its parity is preserved; its X-coordinate goes up or down by an even number, and likewise its Y-coordinate.
The points of the unit cell above have all four parities, so the corner that starts without a peg can never be occupied.
