On the board below, someone has placed a die the same size as one square on the upper-left corner, with an X on its only marked face, on the top.

You proceed to roll the die, rotating it 90° about an edge that's on the board so as to get to the next square, with a new face of the cube now exposed. Continue doing this until the die is on the square marked 1, but with the X again on the top. What's the shortest way of doing this?

What's the shortest way if the goal is to get to the square marked 2, also with the X on top again? Likewise for the goals marked 3, 4, 5 and 6.

bonus:

What if, instead of an X, a Y was on the top face, in the correct orientation, and the goal was to get to the same numbered squares, in the correct orientation again? Is it possible? How?

X					1
					2
					3
					4
					5
					6

(In reply to solution by Vishal Gupta)

I'm unsure how ou interpreted this, but firstly, rolling the die over 4 times has the "X" facing upward and one place short of the intended cell.

Secondly it does not appear that to reach each destination you always began at the "X" which was the problem intention.

Wonder what such a journey would be had Charlie required you to visit each cell in order with the stipulation that "X" would be facing up, and also what that journey might have been had each leg to be a unique distance.

Posted by brianjn on 2009-11-19 19:49:05