Show HN: Lights Out: my 2D Rubik's Cube-like Game

There are two significant ways in which this differs from a Rubik's cube.

1. It's abelian. Moves can be done in any order, to the same result.

2. there's a simple algo to solve this. Working from top to bottom, left to right: click the first white cell, then click the cell below it. (Then there's a simple endgame at the end once the bottom row is reached.)

Hi everyone, I'm the OP and wanted to share a few comments that might come as spoilers. So read with caution!

1. The default game has a simple winning strategy: the app begins with a 5x5 board under the "Same Row & Column" variant. Some here have already figured out how to get all reds: [Memorize all the white squares on the board at some fixed moment, and click those cells in any order.] Some of the info below helps see why.

2. The game is always winnable: the app is set up to secretly begin with all reds and then perform many random clicks to mess it up; it's always reversible.

3. The order of clicks doesn't matter: the click actions commute.

4. Every click is self-inverse: clicking a cell twice under any variant leads to the same board as before.

5. A winning strategy need only list out which cells to click once: Because of the properties of commutativity and self-inverse, any winning strategy could be freely shuffled in order and have any duplicate clicks cancel out, leading to a strategy of cells meant to be clicked just once---the rest ignored.

6. Suppose "n" is odd and play the "Same Row & Column" variant. Then, the n-by-n board is solvable by the strategy of "click all the cells that were white." But when "n" is even, then this strategy fails on the n-by-n board. However, there indeed is another strategy that solves those systematically.

7. Very little is known about the other variants, nor about other sizes and dimensions of boards. And it is not true in general that any random pattern of white-and-red can be turned into all-reds. Try to find a good heuristic for separating out winnable vs. unwinnable boards!

8. I did forget to mention other versions of Lights Out besides the handheld game. Other physical games and video games used the 5x5 "Adjacent" variant, too.

Bonus Questions:

1. Can you think of an interesting clicking-rule variant that would not be commutative?

2. Can you come up with the winning strategy for winning n-by-n boards when n is even under the "Same Row & Column" variant?

3. Can you figure out any winning strategy for the other variants? I haven't found any good way to proceed without just memorizing the solutions.

Thanks for the comments!

Among the 2^n configurations, how many are solvable?

Here’s my solution: clicking four boxes that form the corners of a rectangle will flip them leaving the rest of the board unchanged. Using this move you can find sets of rectangle corners with more white than red and just click them. This will converge to a solution. If you can find a symmetric board where all rectangle corners have equal red and white then this method would fail. I haven’t found one yet.

EDIT: I found some positions where this technique cannot be directly applied.