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.

Firstly, you can fill in all the "trivial" cells. That is, any cell with exactly two walls surrounding it.

Now apply parity on the whole puzzle. Because you can only use two exits, one of which is already known to be black, and there are an equal number of white and black cells, you can remove all the other black exits.

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.

Also, any other white exits in that region cannot be used.

Now we're back to trivial cells. Note, of course, that this could be considered as parity on a 1x1 region. So almost all Alcazar strategies are equivalent to checkerboard parity*!

Now, consider the 2x2 region just below the right centre. It has two white exits in use, and can only use two black exits. So it must use those, and can't use any other white exits.

More trivialties.

Note this irregularly-shaped region. It has an equal number of white and black cells, and only has one available black exit, so it must be used. Also, in the bottom right, you can add a white exit, by parity.

Do this next one yourself.

This should also be pretty obvious.

And now trivial deductions solve the puzzle.

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.

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*"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).

I have a strategy I have not heard named yet, it involves sub-rooms that can be entered and exited more than once. If there are multiple ways of connecting the same doors in a sub-room, none of them can be correct, as that would mean the puzzle could have more than one solution. The correct connections of the sub-doors will be the one that can only be done in exactly one way.

ReplyDeleteThis is a type of uniqueness logic, which I do not consider to be valid logic. For one thing, when testing a puzzle, you cannot first assume that there is one solution, as that's what you're trying to prove in the first place.

Delete"Also, any other white exits in that region cannot be used."

ReplyDeleteStupid question: I'm curious how you come to this conclusion. When I first read this, I couldn't figure out how to rule out the possibility that this region was entered and exited twice. Eventually I realized for that to be true, it would require one path of even length and one of odd length, which wouldn't fit into a region with an even square count. Is that how you get there, or something else?

This conclusion is proven in the previous post. The gist is that the number of black exits and white exits used in a region with equal amounts of black cells and white cells must be the same.

Delete