Question

Suppose I’ve a square of size (2n+1)x(2n+1) for some n i.e. a square with odd side length.
Starting from the centermost cell I’m interested in counting the no. of ways of reaching any edge cell(as shown in the following figure).
Only non-overlapping paths are allowed i.e. we can not revisit a cell if it has been visited already.

Following Figure shows a square with side 9(n=4) and two possible paths of length 5.

I think all paths will be of length range: [n to (2n-1)^2+1 ]
Counting no. of Paths of Length:
1 – 0
2 – 0
3 – 0
4 – 4
5 – 32
6 – …?
But as the Path length increases I can’t seem to unwrap all the possibilities. I know the symmetry comes into play here but is there any structured way to count all the paths?

Thanks,

Answer 1

To find paths that start at the center of the square board and finish at the borders, you can use a tweaked DFS (Depth First Search) where you will store already visited tiles so you don’t step on them again.

There is indeed a lot of symmetry in the board. You can divide the number of searched paths by 4 just by noticing that:

• Starting from the center, you can go Up, Down, Left, Right
• All the paths that start going Up generate all other paths starting Down, Left, Right by rotation

Once you do that, you can go further by noticing that:

• Once you’ve gone U, you can go U, L or R
• All paths generated when you go L are the same ones generated when you go R by mirror symmetry.

You can repeat this several times until you reach the border of the square. The full amount of paths starting by going U will be:

• U-U-U-U (one path)
• 2 times paths starting with U-U-U-L
• 2 times paths starting with U-U-L
• 2 times paths starting with U-L

Once you count those, you have the full amount of paths by multiplying by four.