Size of the map relative to the bot is (n - 3) * (m - 3) To find cell the bot can move to by comparing every 3x3 to a 3x3 of '.' create a graph while comparing the 3x3 cells, must keep track of directions Every step has the same cost so bfs gives shortest path to destination Start is always top-left Destination is always bottom-right The graph is non-directional and every node has a maximum of 4 edges, one for every cardinal direction

Read input file and generate graph (every 3x3 is a node, every node has a type EMPTY or WALL, two EMPTY nodes next to each other have an edge that connects them). Bfs from source (always [0][0]), if [n-3][m-3] has distance < 0 the destination can't be reached. Populate a char array with the opposite of the direction of travel during path traversal from dest to source (recursive like print_path from bfs.c) Print distance of destination and path from source to dest.

Expand wall by moving one block and checking 3x3 with with current block as center, if there's a wall put wall in current block Do Bfs on map with expanded walls. Check distance of destination if >= 0, it can be reached. Go back by finding previous node (the one that has distance == current distance - 1).

[https://virtuale.unibo.it/mod/assign/view.php?id=800757]