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 re: solution by brianjn)

the second line is continuing after the first. what i meant was after the four rolls the X would be facing up and 'adjacent' to the square that has 1 written on it, thus 'X''1' position after four rolls, then 1 down, 1 right, 1 up. for the X to be 'on' position 1.

and i am starting from the initial point everytime. thats why 7,9 moves etc. for subsequent points. staring from '1' one could reach '3' in 4 moves and '5' in 4 moves rather than 7 and 9 moves respectively.

Sorry, if i still am unclear, english is notmy first language.

Posted by Vishal Gupta on 2009-11-20 14:43:47