Edit: limiting it to square flips was a great idea. There are just enough moves to make the answer non-obvious (after lvl 15), but not so many possible moves that you get overwhelmed.
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Edit 2: I just remembered I made a similar "game"[1], where you select columns to XOR with other columns and try to reach the target pattern. Use the scroll wheel and shift+wheel to change the pattern and size.
That was actually part of a real research project in optimizing circuits for computing binary finite fields, where the "game" was a sandbox to try different algorithms. The best algorithm was actually found by someone playing in this sandbox and coming up with an efficient strategy.
Although interestingly some other people in the comments here say they liked how the progression goes.
And that's an interesting little game you made.
And I spent quite some time creating an algorithm and solver to find the par for Unflip. I'm planning to release a blog post about it soon
Took ages to find an algorithm that produces mostly interesting puzzles in the goldilocks difficulty zone.
Also, the game does give you a hint about how to bruteforce the level if you only have one square left. Good job.
https://archive.org/details/Quadromania_1987_CP_Verlag_de-en...
It took me a while to get past the instruction screen (I needed to press space, then "E" for english, then press escape) Now I can't seem to figure out the controls for the actual game.
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Edit: and the right answer was so obvious, too
edit: found a way to solve it in 6, but not in 4
I was thinking of making weekly levels be very big (like 12x12) where I don't know what the par is myself, and have it be a challenge to see who in the community can figure out a way to get the lowest number of moves.
And daily levels would probably be more similar to the regular levels.
I'll be particularly curious to hear the pattern(s) in how many solutions there are, and/or the probability of a random board being solvable.
The haptic feedback on mobile is really on point in this implementation.
Regarding the number of solvable boards. It is actually possible to calculate the exact number. The solvable boards correspond to the image of the matrix of moves. It will be a vector (sub)space over Z/2Z, so it will always be a power of 2, and its size will be determined by the rank of the matrix. For example for 5x5, there are 2^22 valid boards.
And regarding the number of solutions to a given board, (assuming you ignore the order of moves, because it doesn't matter) it will always be the exact same number, and in the case of 5x5 it will be 2^8. 8 is the nullity of the matrix of moves.
Note that 22+8=30, which is 44+33+22+11, which is the number of possible moves.
Well done!