Hmm how did you do this problem

IIRC this was around the time i first learned probabilistic method, and this was one of the problems i first used it on.

How would you code the right hand algorithm?

The only thing you know about the maze is that you can reach any square from any other square, so loops are okay.

This was the solution I presented to the judges; I was given this problem for AMSP Team Contest.

AoD, I don't think that exactly works. Imagine a scenario where the last move should be "up" but instead you hit a wall. Then, when you move down to begin the inverse sequence, your entire path is shifted down by one and you're screwed.

Yeah, unfortunately, simply taking the inverse of a sequence does not guarantee that you will return to the point you started at. I spent an hour trying to follow this idea, but I couldn't find anything useful.

Finite number of mazes, just solve at new starting point for each new maze.

^ waitwhat

You don't have to invert anything or use any symbols whatsoever.

List the mazes and pawn positions (that is, one copy of every possible maze with every possible pawn position (below, we'll use "maze" to mean "a maze with a starting pawn square chosen")). There are finitely many. Take the first maze and write down a (finite) solution that makes the pawn reach every square (obviously possible). Now if B selected that maze, your solution solves it. Also you can add as many future moves as you like after this solution without "unsolving" the maze.

Now, take the second maze and use your partial solution on it, moving the pawn around in whatever random path that results, possibly getting blocked lots of times. Whatever happens, the maze is still solvable. From the resulting position, make moves so that the pawn reaches every square (if that has not yet happened), and add all moves made, if any, to your partial solution.

Take the third maze and add more moves to solve it. Do this for all the mazes, in some order. The final solution solves any maze.

Awesome, we had this problem at Ideamath, except it was God vs. the Devil.

yeah i realized my mistake last night but i was too tired to edit
but i came up with math_explorer's solution afterwards (independently too; I'm so proud of myself)

yeah that's my solution

ah the freeness of arbitrarily long time solving a finite problem