# 4-dimensional Knight's Tour

Imagine a 4-dimensional chess board, but with only 4 squares a side. So, $$4\times4\times4\times4=256$$ sub-hypercubes.

Your mission - should you choose to accept - is to:

Find a Knight's Tour

Four dimensional spaces are actually quite easy to visualize:

The orange squares represent the possible destinations of a knight that starts from the red square - it moves $$1$$ in a dimension and $$2$$ in another.

If the coordinates of a knight are $$(x,y,z,t)$$, then we can get to $$(x\pm1,y\pm2,z,t)$$ or $$(x,y\pm1,z,t\pm2)$$, or any other combination, as long as we remain on the board.

The coordinates of the knight shows are $$(1,1,1,1)$$, with depth representing the $$z$$-axis, and the $$t$$-axis being represented by the multiple grid show left to right.

Therefore the first few coordinates of the orange squares, read left-right, top-bottom, are:

$$(3,1,0,1), (1,1,0,3), (1,3,0,1), (3,0,1,1), (1,0,1,3), (3,1,1,0), \dots$$

## 1 Answer

This puzzle can be broken down relatively easily.

Start with

A 3-dimensional 4x4x2. This can be done as follows:

This is not as daunting as it looks. There are essentially just two patterns of four and we switch back and forth between them. The patterns are corner-middle-corner-middle, and side-side-side-side.

Note that we're starting from a corner and ending adjacent to that corner. And, of course, that we can do that in reverse.

So now we can put it together to finish the whole thing:

Do the red ones first, then the orange and then repeat the whole process for each column (hyper-column?)

Here I only show the first and last position of each 2x4x4 section. So we start at A and go to B, covering all the red squares. Then we jump to the next 2x4x4 section (B->C). Then we do C to D, covering all the orange squares. Similar logic takes us to E and the whole thing repeats.

So the overall loop is A---B-C---D-E---F-G---H-A'---B'-C'---D'-E'---F'-G'---H', where the "-" is a direct jump, and the "---" is the move that covers the corresponding 2x4x4.

• And $H^{'}\to A^{''}$ in the 5-th dimension, and so on... Well done! – JMP Aug 8 at 12:54