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Is it possible to fill a rectangular board with 12 squares completely with the numbers 1 to 12, in such a way that there exists a square on the board, where if we start there and traverse the board according to the value of the number in the square we're in and below diagram (as in the puzzle sands of time) that we will visit all 12 squares.

enter image description here

As a much simpler example, in a 2x2 square we could place the numbers

3  6
12 9

And we could start on 3, move according the diagram above to the 6, move down to 9, then left to the 12, visiting each square.

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    $\begingroup$ If such a rectangle existed, it would have to be 4x3 (or 3x4, same thing), because a 1x12 or 2x6 could not contain both 1 and 2, for example. $\endgroup$ – dpwilson Feb 5 '16 at 18:38
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 6  4  5  7
2 3 8 12
1 11 9 10

is one of 12 solutions, but there are fundamentally only two.

The one given above traces a path which has horizontal symmetry; it can be flipped vertically and traced in either direction, giving four equivalent solutions. The other was posted by hvd; it has no symmetries, meaning that it can be flipped horizontally and/or vertically and traced in either direction, giving eight equivalent solutions.

The paths traced by the two fundamental solutions are:

Graphs showing the two paths

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Sure, it's possible.

 3   7   5   6
 4   9   12  8
 2   11  1   10
 
Start anywhere you like.

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  • $\begingroup$ Sorry for swapping the accepted answer away, when you answered, i doubted anyone else would take the time to go into more detail. Turns out someone did, and Peter Taylors answer is more complete $\endgroup$ – DrunkWolf Feb 5 '16 at 20:23
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    $\begingroup$ @DrunkWolf Not a problem, that's only fair when a better answer comes along. $\endgroup$ – hvd Feb 5 '16 at 20:38

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