I'm try to design a game in which the board is made up of a 3×3 grid of square tiles. Each tile is a 5×5 grid of spaces. Each tile has 4 exit spaces each located on 1 of the middle 3 spaces along each edge of the tile. All exits must be connected along a path of orthogonally adjacent spaces. The paths can only be 1 space wide and must all lead to a single intersecting space in any of the middle 9 spaces of a tile. Spaces in this path that are not exits cannot be on an edge space. How many possible tiles can exist within these parameters?
It is not necessary that paths connect from tile to tile. I'm curious to know the math behind this in case I need to adjust tile sizes.
Best Answer
If I understand things correctly, each tile has three possible exit spots per edge, and there's a 3x3 grid of possible internal intersection spots. So that the naive answer would be $3^6=729$. However, I assume a tile is the same if you rotate it.
For counting problems with symmetry, we can use Burnside's Lemma. I assume the only symmetries are rotation by multiples of a quarter turn.
Not rotating fixes (doesn't change) all $3^6$ tiles. Rotating by a half-turn fixes those tiles with a center intersection point and opposite exits, so you can really only choose exits on two adjacent edges, and there are $3^2$ tiles in all. Rotating by a quarter-turn (which can be done in $2$ ways) fixes those tiles with a center intersection point, and the exits on three edges are rotations of the first: $3$ tiles (based on the exist choice for one edge) in all.
By Burnside's Lemma, the number of rotationally-distinct tiles should be the average of the numbers of fixed tiles: $$\frac{3^6+3^2+2*3}4=186$$
If the tiles are transparent and can be flipped over, then there would be fewer distinct tiles.