I have a normal 6-sided die and I would like to find a model for how it transforms under rotatins (north, south, east, west), that can help me determine which side is up. Just to make sure we are using the same die, here's how mine is numbered:
To give a few examples, when the die is in that position (1 facing up, 2 facing at me), if I rotate it south it will end up with 5 facing up and 1 facing me. Similarly, a rotation to the west gives 3 facing up and 2 facing at me.
I tried representing the die and its rotations as a graph where each node is one of the states the die can be in and there are four transitions for every node (corresponding to each of the four rotations). The problem is that there is a total of $24$ states and $4 * 24 = 96$ transitions and writing down all of them is laborious and very error prone.
The reason there are $24$ states is because when 1 is facing up and 2 is facing at me is quite different from when 1 is facing up and 3 is facing me. So a state of the die is a 2-tuple of the form $(\text{number facing up},\ \text{number facing me})$. (There are $6$ numbers that can face up and for each of them there are are $4$ numbers that can face me, so $24$.)
I need some help in finding a nicer mathematical model for determining what face is up, given a starting position and a sequence of rotations.
See the expanded view of the cube.
I added an imaginary square (the half one) to the expanded view. Let's define two sets such that;
$$\left( { y }_{ 0 },{ y }_{ 1 },{ y }_{ 2 },{ y }_{ 3 } \right) =\left( 1,2,6,5 \right) \\ \left( { x }_{ 0 },{ x }_{ 1 },{ x }_{ 2 },{ x }_{ 3 } \right) =\left( 1,3,6,4 \right) $$
The values in the first and third indices in both series are equal whenever you rotate, not ${ x }_{ 0 }={ y }_{ 0 }$ and ${ x }_{ 2 }={ y }_{ 2}$, i.e everytime you rotate the values comes to that indices will be equal.
Let's n be number of south or north. If it is south say $n>0$, if it is north say $n<0$. Every four rotation in that directions the series $\left( { y }_{ 0 },{ y }_{ 1 },{ y }_{ 2 },{ y }_{ 3 } \right) =\left( 1,2,6,5 \right)$ will repeat. Before completing four rotations if you pass east-west rotations you have to work with $ \left( { x }_{ 0 },{ x }_{ 1 },{ x }_{ 2 },{ x }_{ 3 } \right) =\left( 1,3,6,4 \right) $ by changing the first and third indices. So I will just explain for south-north rotation. By the way the first indices in the series is what you see on the upper face of the cube.
For any $n$ rotations in south-north direction your series will be:
$$\\ \left( { y }_{ 3n\; { mod }\; 4 },{ y }_{ 3n+1\; { mod }\; 4 },{ y }_{ 3n+2\; { mod }\; 4 },{ y }_{ 3n+3\; { mod }\; 4 } \right) $$
So what you see on the upper face will be $ { y }_{ 3n\; { mod }\; 4 }$. If you past to east-west rotations don't forget, the first and third indices of series $y$ should pass series $x$.