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Let's place a cube on a table and flip it around a bit. In fact, flip it according to the following instructions:

  1. Flip forward twice.
  2. Flip left twice.
  3. Flip backward twice.
  4. Flip right twice.

Assuming no sliding has occurred (e.g. your flips were perfect along the edge), the cube will have ended its traversal with the same position and orientation it started with. Unfortunately, the perfectionist in me wants to press this further, so repeat the same process, but this time track the right face of the cube. Notice that it never touches the table?


What is the minimum number of flips required to ensure that every face touches the table at least once before returning the cube to its original position and orientation?

This puzzle is fun, and it inspired this one; be sure to show it some love!

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1 Answer 1

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This can be done in

10 flips.

Since this puzzle

relaxes the other puzzle's rule that a move may not undo the previous move.

The most straightforward flips are

forward twice, back twice, left, right, right, left, back, forward.

But @Penguino has found a series of flips that

adheres to the other puzzle's rule and still takes only 10 flips: F, F, F, F, R, B, B, B, B, L.

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    $\begingroup$ Fa la la la la? Try it! $\endgroup$ Jun 14 at 21:53
  • $\begingroup$ I believe that retaining the requirement that "a move may not undo the previous move" can also be done in ten steps (not 16 as you suggest). As in F, F, F, F, R, B, B, B, B, L. $\endgroup$
    – Penguino
    Jun 15 at 23:16
  • $\begingroup$ @Penguino nice. I'll edit that in. $\endgroup$
    – cap
    Jun 16 at 2:47
  • $\begingroup$ @Penguino's solution is what I was hinting at: F, L, L, L, L, B, R, R, R, R is essentially the same. $\endgroup$ Jun 16 at 9:02
  • $\begingroup$ @Daniel Mathias but your version has a better tune... $\endgroup$
    – Penguino
    Jun 18 at 3:17

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