I'm actually writing my (high school) research paper on this. Here are my conventions:
- We are trying to turn the lights off.
- Buttons and lights refer to the same physical object. However, we refer to the thing as a buttons in the context of pressing it, and as a light in the context of a toggle in state (We press the button to toggle the light).
- Two lights are neighbors if pressing one's button will toggle the other (In your variation, two lights are neighbors if they lie in the same row or column). By convention, we let a light be a neighbor of itself.
- Denote an easy-run as the action of pressing all on buttons at once (I believe Bruce Torrence came up with this term).
- We let an initial configuration of lights be easy if you can win with one easy-run.
- We let an initial configuration of lights be easy-iterable if you can win with several easy-runs.
I prove that in the two-state alien tiles variation,
- Boards with dimension $2n \times 2n$ are always solvable.
- Boards with dimension $2n \times 2n$ are always easy-iterable (In at most two easy runs) for any initial configuration.
- Boards with dimension $2n+1 \times 2n+1$ are easy for any solvable initial configuration.
It's really easy to prove that boards of even dimension are always solvable. It suffices to show that you can toggle any individual light. To do this, just press all buttons in that light's row and column. Some basic parity analysis will show that this always works (keep in mind that the dimensions are even).
Unfortunately, this is not the same for boards of odd dimensions (Ignore the $1 \times 1$ exception). For those who are not familiar with Lights Out analysis, here is how we can make this argument: If there are $N$ buttons, then there are $2^N$ possible initial configurations of lights (legal or not). There are also $2^N$ possible solutions (ways to press the buttons. Just like how a light can be on or off in an initial configuration, a button is either pressed or not pressed in a solution).
If we claim that all initial boards must be solvable, then for each of the $2^N$ possible initial boards, there is a solution. But there's also $2^N$ solutions, so there is a bijection between initial boards and solution.
Now take an empty board of odd dimensions. A solution to this board is to press nothing. If we find another solution, then there is no bijection between initial boards and solutions, so this would prove that not all initial configurations are solvable.
There are actually many other solutions. Just pick two rows, two columns, or a row and a column minus their intersection, and press all buttons in there. It's easy to show that this will still result in an empty board.
But that's not very interesting. I think the most interesting result here is that you can win with just easy-runs. For the odd-dimension case, you only need one easy-run, and you win. So, winning is a pretty mindless process (though, note that this solution is not necessarily optimal). For the even-dimension case, you need at most two-easy runs, which is not as mindless, but still really easy to execute. So unless you are trying to be optimal, this observation absolutely kills the difficulty of this variation. Let's prove that the killing actually works.
The odd-dimensions case is really easy. I used induction. Here is a brief sketch: In the inductive step, we can traverse through all legal configurations of lights by pressing any button. Basically, given an initial configuration that we know is easy, we just prove that after we press a button, the configuration is still easy. We can use the obvious all-off board as our base case. The meat of the proof is not hard: Just use the fact that an initial configuration is easy if each light that is on has an odd number of neighbors that are also on, and each light that is off has an even number of neighbors that are on. I leave the details to the reader.
The even-dimensions case is tougher. To reword the desired: We want to prove that after an easy-run, we get an easy game. I leave the bulk of this proof to the reader. Here is a rough sketch:
- Prove this lemma: Consider some initial configuration of lights, and let $X$ be the set of lights toggled when all lights that are on are pressed. Then $X$ is the unique set of buttons to press that solve the board.
- Prove this lemma: All solution buttons have an odd number of neighbors that are on, and the rest of the buttons have an even number of neighbors that are on. (A solution button is a button that needs to be pressed in the unique solution).
- Case bash on the four types of lights arising from the properties of on/off and solution button/not a solution button.
In case you are wondering whether or not I actually did these proofs, here is an excerpt from my not-final-draft paper: https://www.dropbox.com/s/e312cri57z4egap/deleted_Lights_Out.pdf?dl=0
(I am a high school student that is bad at writing, pls dun judge)
So yeah, this Lights Out variation is pretty dead. Dead in a really cool way, though.
(To make the Alien Tiles thing, check both wrap-around boxes, and be a bit careful with the move-shape: If the board is $4 \times 4$, don't make the move shape the biggest cross possible, because some buttons might be pressed twice per move)