Function get_potential_left_corner_tiles 

fn get_potential_left_corner_tiles(
    sorted_tiles: impl Iterator<Item = [u32; 2]>,
) -> (Vec<[u32; 2]>, Vec<[u32; 2]>)

This function filters sorted_tiles into two lists, one containing all tiles that could be the top left corner of the largest rectangle (assuming the largest rectangle has a top left corner), and the second containing all tiles that could be the top right corner.

It assumes sorted_tiles is sorted in ascending “y” values, or, to get the top right and bottom right corners, that sorted_tiles is sorted in descending “y” order.

It works (for the top left corners, for illustration) by only returning tiles (from the set of all tiles, “T”) within the region:

R = { (x, y) ∈ ℝ² : ∀ (tx, ty) ∈ T, tx ≤ x ⇒ ty ≥ y }

Tiles outside of this region cannot possibly be a corner of the largest rectangle. Assume, for proof by contradiction, that the top left corner of the largest rectangle is in the complement of the set “R”:

R’ = { (x, y) ∈ ℝ² : ¬ (∀ (tx, ty) ∈ T, tx ≤ x ⇒ ty ≥ y) } = { (x, y) ∈ ℝ² : ∃ (tx, ty) ∈ T, tx ≤ x ∧ ty < y }

That is, for the corner (x, y), there exists another tile (tx, ty) that is to the left and above the corner tile, which means the tile isn’t the corner of the largest possible rectangle, completing the proof by contradiction.

The top_tiles and bottom_tiles are the corner points of this region R, built up by scanning through tiles in either left to right or right to left order.