Riddler Classic for 27 July 2018 (part 2)

Here’s a way to think about those time symmetric solutions.

Color the squares like this:There’s a 2 by 3 block in each corner, each containing one square of each of the colors red, orange, yellow, green, blue, purple. Now starting from the top left go East, South, West, Northeast, South:You’ve visited one square of each color. Obviously if you started in a different corner and followed a similar path (rotated 90°, 180°, or 270°) you’d visit a different square of each color. Do that from all four corners and you visit 24 of the 25 squares, all but the center one. Of course that’s four paths, not one. But!

The path ends on a purple square. Notice, though, from a purple square you have a legal move to another purple square. From there you can follow the same path (rotated) backwards, ending up on a red square. Now you’ve visited twelve squares.

But now from that red square you can go to the center square, and from there to the red square diagonally opposite.

From there you can follow the path a third time (forward, rotated) to reach a third purple square; from there you can move to the fourth purple square; and from there you can follow the path a fourth time (backwards, rotated) to reach the final red square. And that gives you a time symmetric path visiting all 25 squares. Not only is it time symmetric, but it’s doubly so: the first 12 steps also are time symmetric with a rotation (as are, necessarily, the last 12).

Specifically it gives you the first two time symmetric solutions of the four I mentioned — one if you go Northwest from the first purple square to the second, the other if you go Northeast.

The other two time symmetric solutions are a little more complicated, because they begin with a path that starts on a yellow square and ends on purple:From there you could go to another purple square and then follow a backwards, rotated path that would land you on a yellow square— but then what? You could go to another yellow square, do another forward path and another reversed path, and you’d have a doubly time symmetric solution… but for 24 squares, not 25. But instead, after the jump to the second purple square, you can follow a different path, one that hits the same squares but in a different order, ending on red:And now you can go to the center square, then to the opposite red square, and then follow a path that time reverses the first steps to finish.

What these solutions rely upon is that there’s a color (purple) from which you can move to another square of the same color, and a central square from and to which you can move to another color (red). With a 6 by 6 square, using the same idea with 3 by 3 blocks in each corner, there is no center square to use as a stepping stone. So to get a similar solution you’d need two colors from which you can move to squares of the same colors. Start on (say) red, hit all the colors ending on (say) purple, jump to another purple, reverse the path to get to red, jump to another red, and then reverse everything to finish. But if you check you’ll find there’s only one color from which you can move to another square of the same color. So doubly time symmetric solutions of this sort don’t exist.

In principle there could still be a singly time symmetric solution, where you use 18 colors to color two 3 by 6 blocks. Then follow a path that hits every color once, ending with (say) purple from which you can go to the other purple, then follow the reversed path to finish. But it seems there’s no such path.

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