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This is a variant on The Tiled Labyrinth.

The rules are the same, except that the goal has changed.

You will probably want to use this script which was created for the initial puzzle but applies equally well to this version.

Summarizing the rules for those too lazy to read the original problem:

  • You have a board consisting of a set of 5x5 dice, each of which has paths on the different sides.
    enter image description here
  • The starting position is
    enter image description here
  • You may rotate all the dice in a given column up or down
  • You may rotate all the dice in a given row left or right
  • You may not rotate the dice in the row or column corresponding to your current location
  • After any rotation you may move to any other location that is connected to your current location.

The new goal: Create a single path that connects all four corners of the board. A sample solution might look like this:

enter image description here

A few notes:

  • Any solution that creates a path between all four corners will be accepted, It does not have to be the same as my sample solution above.
  • I will be looking for the solution that involves the fewest rotations (I will not count the number of moves)
  • Here is a 28-rotation sequence that results in the end position above, but your goal is to do better. It also includes a hint about how I generated the sequence

(B+), (C+), (D+), (D+), (2-), (D+), (2+), (D-), (5-), (C+), (5+), (C-), (5+), (D-), (5-), (D+), (B+), (5+), (B-), (5-), (B-), (5+), (B+), (5-), (5+), (E+), (5-), (E-), [A1 -> E5]
The idea behind the subsequence "(5+), (E+), (5-), (E-)" is to rotate the die at E5 without changing the rest of the board

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    $\begingroup$ To make this puzzle more stand alone, you'd need to show the pattern of the dice and the starting position. $\endgroup$ – Dr Xorile May 23 at 16:07
  • $\begingroup$ Edited the problem to make it more stand alone $\endgroup$ – Eugene Styer May 23 at 21:26
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To get an upper bound, here's a solution using 19 13 moves that doesn't require the movement of ball at all (which isn't a great optimization since ball movements are free!).

(B+), (C+), (D+), (2+), (5+), (D+), (E+), (C+), (C+), (5-), (C+), (C+), (5+)

I used Kevin's script (http://jsfiddle.net/vwe9kw7j/2/)

Final position:

Screenshot of Kevin's script with 13-move solution

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    $\begingroup$ Two notes: 1. The "Undo" line gives away your solution, 2. Keep working, I have a solution with 16 moves $\endgroup$ – Eugene Styer May 23 at 21:36
  • $\begingroup$ @EugeneStyer, noted! I found a 13 move solution (and put it under spoiler tags!) $\endgroup$ – Dr Xorile May 27 at 3:53
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Update: A computer search has come up with a 8-move solution. (Actually multiple 8-move solutions but I'll only give this one):

(5+), (4+), (2-), (3+), (C+), (D-), (D-), (B+)

The final position for this solution is:
enter image description here

Original Post:
I'll add this as an answer since I found a different 13-move solution:

(B+), (C+), (D+), (D+), (2-), (3+), (4+), [A1 -> B1], (2+), (4-), (A+), (2-), (4+), (A-)

With a final position of:
enter image description here

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