# The Fourth Dimension

A three-dimensional solution to this problem comprises *n* two-dimensional solutions arranged along the third dimension to form a cube. It therefore follows
that a four-dimensional solution must comprise *n* three-dimensional solutions arranged along the fourth dimension to form a Tesseract (the four-dimensional
equivalent of a cube). So how do we go about finding a 4D solution?

In November 2008, as soon as I had identified what I believed (and later proved) to be the first possible 11³ solution, I started looking at the relationship
between the 11² solutions (or "boards") that fitted together to build it. What I found was that while each of the boards is unique within the cube
(i.e. it isn't a rotation or mirror of a board in any other Layer), they are all essentially **variations** of the same 11² solution. For example,
consider the following board that occupies Layer 1 of the cube:

1, 3, 5, 7, 9, 11, 2, 4, 6, 8, 10

Expressed in our 2D solution notation, each number represents the Column in which a Queen sits, with the number's position in the sequence representing the Row. If we take the final number – representing a Queen in Column 10 of Row 11 – and move it to the beginning of the sequence, we find that we have produced another unique 11² solution. Specifically, we have produced the board that occupies Layer 6 of the cube:

10, 1, 3, 5, 7, 9, 11, 2, 4, 6, 8

From the point of view of the Queens themselves, what we have done is taken the Queen from Column 10 on Row 11, shifted all the other Queens up one row, and then replaced that Queen in Column 10 on Row 1. If we did this a further nine times we would end up producing all eleven boards that make up the first 11³ solution.

Extending this idea to the first possible four-dimensional solution, we simply need to take the first 11³ solution and rotate its layers in the same way to produce the other ten cubes that will fill our tesseract. For example, here is the first 11³ solution expressed in our 3D solution notation using 2D board numbers:

1, 976, 2358, 98, 1371, 2366, 326, 1781, 2593, 579, 2109

Taking the board out of Layer 11, shifting the other ten boards up one Layer, and then replacing that board in Layer 1, we get the following unique 11³ solution:

2109, 1, 976, 2358, 98, 1371, 2366, 326, 1781, 2593, 579

And just like with the 11² board above, if we did this a further nine times we would end up producing all eleven cubes necessary to build the first 4D solution.

### But how do we arrange eleven cubes to form a tesseract?

To build a 3D solution, the eleven 2D boards were arranged along the third dimension such that no Queen could attack another, even between boards. Similarly, to build a 4D solution we must arrange our eleven 3D cubes along the fourth dimension such that no Queen may attack another, even between cubes. Without a lot of rather scary mathematics it's almost impossible for us to perceive how those cubes can be arranged along the fourth dimension. Fortunately, however, the arrangement of the boards within our cube gives us the order in which we need to arrange the cubes within our tesseract.

In Layer 1 of our first 11³ solution, the first Queen was in Row 1, Column 1. In Layer 2, the first Queen was in Row 1, Column 5. This is the first cell that the Queen could occupy where she wasn't at risk of attack from any Queen in Layer 1. Therefore, as the first Queen in Cube 1 of our first 4D solution is in Layer 1, Row 1, Column 1, the first Queen in Cube 2 must need to be in Layer 1, Row 1, Column 5 to be safe from attack by any other Queen in Cube 1. Put simply, the arrangement of the cubes based on Layer 1 is identical to the arrangement of the boards in the first cube.

### So how many 4D solutions are there?

Just as we found that we could create eleven unique variations of the first 2D 11² solution, and eleven unique variations of the first 3D 11³ solution, so
we can create eleven unique variations of the first 4D solution by rearranging its cubes in exactly the same way (removing the last cube, shifting the others along,
and then replacing that cube as the first one). Interestingly, the same appears to be true of the solutions to the 11-based problem in **any higher number of
dimensions**. However, while there are always eleven **unique** solutions, as the number of available views (rotations, mirrors, etc) increases
with the number of dimensions there will no doubt be more and more **total** solutions found by an algorithm equivalent to the one we used for the 3D
puzzle.