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Rolling a die across a tray

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A standard die has opposite faces summing to $7$ . It sits on the north-west corner cell of a tray with $3$ rows and $4$ columns, with $1$ on top, $2$ on the north-facing side and $3$ on the east-facing side. One move rolls the die over one of its bottom edges onto the neighbouring cell to the north, south, east or west, staying inside the tray. What is the minimum number of moves needed for the die to stand on the south-east corner cell with $2$ on top?

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Hint

Count the moves needed just to reach the corner, and check whether a shortest route can possibly show the face you want on top.

Solution

The corner is $2$ rows down and $3$ columns across, so at least $5$ rolls are needed, and since each roll changes the cell by one step the number of rolls is odd.
A route of exactly $5$ rolls consists of $2$ rolls south and $3$ rolls east in some order; trying all $10$ orders gives the top faces $1$ , $3$ , $4$ , $5$ and $6$ and never $2$ .
So $5$ rolls are not enough, and the next possible length is $7$ .
Rolling south, east, north, east, south, east, south gives the top faces $2$ , $4$ , $1$ , $5$ , $4$ , $6$ , $2$ after the successive rolls, so it ends on the corner with $2$ on top.
The minimum number of moves is $7$ .

Answer: $7$

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