If you haven't read the complete solution of the first puzzle, you should do so now

At first sight, it seems that since every path can be traversed in two directions, forward and backward, the number of paths needs simply to be divided by 2.

At second sight, things are not that easy, as it often happens...

There are those paths that have already been counted just once, not twice,
because their first part, from border to center, is exactly equal to their second part, from center to border, except for the direction.

Let us call them "symmetric paths" and count them separately.

The number P we are looking for is the sum of the number of symmetric paths P_{1} and the number of non-symmetric paths P_{2}:

P = P_{1} + P_{2}

The number of symmetric paths is simply the number of paths from border to center, since for any one of them there is only one way to return to the border from the center:

P_{1} = C

For any one of the C half-paths that lead from the border to the center, there are C − 1 ways to return to the border in a non-symmetrical way.

But since we don't want to count all these paths twice, once forward and once backward, we have to divide by two:

P_{2} = C (C − 1) / 2

Now it's only a matter of putting the pieces together. It's just boring school mathematics. The result is:

P = C (C + 1) / 2

We already know that

C = 4 (Q − 1) = 4 (2^{S} − 1) = 4 (2^{(L − 1) / 2} − 1)

We should stop here, unless we want just one big final formula, that is maybe reassuring, but less practical for a real calculation.

Here is the big final formula:

P = 2^{(L + 2)} − 7 × 2^{(L + 1) / 2} + 6

If all of this was too easy for you, you can go ahead and generalize the solutions of the 1st and 2nd puzzle to D dimensions.

We have just seen D = 2. For D = 3 the letters lie in a octahedron. For D = 4, well, ...