A bit of background information here: The puzzles in this pack were computer-generated by a different program to the one which generated the rest of the puzzles in the app. The person who wrote this new program was a certain A. Ball, hence the name "The Ball Rooms". These puzzles are claimed to be even more difficult than the rest of the app; in fact, Mr. Ball himself was completely stumped on the first puzzle! I have therefore written this post to demonstrate checkerboard parity in action, show Mr. Ball that that puzzle is indeed solvable with pure logic, and also show you that checkerboard parity is to Alcazar what the Sledgehammer is to Sudoku.
Let us begin the puzzle. I am running this on BlueStacks Emulator, but that has nothing to do with the puzzle. If you don't have an Android device or an emulator, you can use this weblink instead, to follow along with these deductions.
Near the top-righthand corner, there is a small 2x2 region on which you can apply checkerboard parity. Some people would call this "wicking", "weak points", or "loop forcing", but it's really just checkerboard parity in disguise!
Next, consider the 5x4 region in the bottom left. There are an equal number of white and black cells, and one white exit is already used. Since there is only one black exit left, that exit must be used.
Checkerboard parity is a lot more useful than you might think, considering that most solving strategies are checkerboard parity in disguise. In fact, all of the steps in this puzzle come directly from checkerboard parity, so if you're creating an Alcazar solver, that's all you need to implement**!
I hope you leave this blog a little wiser on this subject than before (and able to complete the whole of "The Ball Rooms"). Especially you, Mr. Ball.
*"No closed loops" is also checkerboard parity in disguise. Consider the region formed by the squares used by the path in consideration. There is an equal number of each colour of cells (as the ends of the path are adjacent), so there is one white exit and one black exit. Therefore, each end of the path must extend out of the region, not into it, and so one can draw a wall between the two ends.
**There are a few advanced strategies, but for the most part, they have not yet been discovered. I believe I have discovered one which is definitely not equivalent to checkerboard parity (which I shall not reveal at this time), and one which may or may not be equivalent (I haven't thoroughly checked it yet).