Number to the left/right on a dice? - math

Question

I was wondering if there is a formula/trick to calculate what number is to the right or to the left on a standard 6-sided die if you know which number is on top and which is facing you.
Need it to solve a problem, but I don't feel like listing all 24 possibilities in an if-statement...:)

Answer 1

There aren't 24 possibilities.
From Wikipedia:
The sum of the numbers on opposite
faces is seven.
So as you already know two numbers, there are 4 possibilities left.
I'm not sure what you meant with "on the top" and "facing you", but I think you meant two neighbour faces here, so there are only 2 possibilities left as you know there opposite faces numbers, too (and those 2 possiblities only differ by the two numbers being left/right or right/left).
So, for example, using a unfolded dice, you've got "1" on the top, "2" facing you:
X
X1X
2
X
You now know that the opposite faces will be "6" (bottom) and "5" (facing away from you):
5
X1X
2
6
So there are these both possibilites:
5 5
314 413
2 2
6 6
There is only one possibility left when you know if your die is "left-handed" or "right-handed" (again, Wikipedia):
This constraint leaves one more
abstract design choice: the faces
representing 1, 2 and 3 respectively
can be placed in either clockwise or
counterclockwise order about this
vertex. If the 1, 2 and 3 faces run
counterclockwise around their common
vertex, the die is called
"right-handed"; if they run clockwise
it is called "left-handed". Standard
modern Western dice are right-handed,
whereas Chinese dice are often
left-handed
So, for the example above, the left one is a "left-handed" die, the right one is a "right-handed" die.

Answer 2

Opposite sides of a die always add up to 7 (at least, this is the convention).
By process of elimination you can tell what the "invisible" pair will be:
1/6
2/5
3/4
So, for every pair above, if you can see a number from it, remove it. The remaining pair is the one you are looking for.
Since there is no way to determine the "handedness" of the die, it is impossible to tell which of the pair will be to the right and which to the left.

Answer 3

I have no idea why this works, but:
For each pair (top/facing) there is of course only two possibilities, as both the top value and the facing value each eliminate itself and it's opposite from being left and or right: for e.g., If 1 is on the top, and a 2 is facing you, then the left and right must be either a 3 or a 4...
So for each combination, if the sum of the values is odd, and less than 7, or even and greater than 7, the lower of the two possible values is on the left and the higher is on the right.
.. and vice versa
As I am in America, I guess this rule is for a "right-handed" die, reverse it for a left-handed die.

Answer 4

Programattically you could do this quite easily. It is probably a little overkill (because you could use an if statement), you could also create a class such as
public class Dice {
private DiceFace[] sides = new DiceFace[]{DiceFace.ONE, DiceFace.TWO, DiceFace.THREE, DiceFace.FOUR, DiceFace.FIVE, DiceFace.SIX};
class DiceFace {
// set the face number and connecting faces...the below figures are wrong, I don't have a set of dice to check agains
static final DiceFace ONE = new DiceFace(1, 2, 3, 4, 5);
private int face;
private int northFace;
private int southFace;
private int eastFace;
private int westFace;
public DiceFace(int face, int northFace, int southFace, int eastFace, int westFace) {
// set the values
}
}
}
You could then write a method to give you the possible values for your answer by checking the dice and the values to the left and right of it.