Puzzle time - the forth corner
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wrote on 8 Nov 2020, 19:02 last edited by
Pegs occupy three corners of a square. At any time, a peg can jump over another peg, landing an equal distance on the other side. Jumped pegs are not removed. Can you get a peg onto the fourth corner of the square?
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wrote on 8 Nov 2020, 19:12 last edited by
Not 100% confident, but I would say no off the top of my head.
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wrote on 8 Nov 2020, 20:10 last edited by
I assume the pegs can land outside of the square, right? Otherwise it doesn't seem to make sense.
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wrote on 8 Nov 2020, 20:22 last edited by
Yeah that was the first thing I had to come to terms with. Otherwise the problem is incoherent.
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wrote on 8 Nov 2020, 20:29 last edited by
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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wrote on 8 Nov 2020, 20:29 last edited by Klaus 11 Aug 2020, 20:32
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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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wrote on 8 Nov 2020, 22:57 last edited by jon-nyc 11 Aug 2020, 22:58
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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wrote on 8 Nov 2020, 23:07 last edited by
Now consider the situation that the square is on a Möbius strip...
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wrote on 9 Nov 2020, 02:09 last edited by
I have to say no also
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wrote on 14 Nov 2020, 21:53 last edited by
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.