Those are the warmups. Here's the real challenge.
And yes, it is uniquely solvable.
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Sign up to join this communityThat was a nice puzzle. It took a while to get on track, but when it clicked everything got a lot easier.
To allow others to try, my solution will get progressively more hintier, starting with two things I tried that didn't work:
Forming a path through every cell - this quickly becomes impossible on level 4. If you colour the cells like a checkerboard, there's 11 of one parity and 9 of the other, a difference of 2, but since you alternate parity with each move the difference should be at most 1 for a solution to exist.
Splitting the areas into 2 congruent parts. I tried ignoring the fact that level 1 is only one square and noted that all other levels had an even number of squares. There's a unique way to split level 3 into two disjoint congruent areas (both areas need to be disconnected for this to work), but it was impossible to do the same for level 4.
Here's the third thing I tried:
Failing the above, I noticed that the number of white squares in level N is divisible by N. Specifically, the levels have 1, 2, 6, 20 and 60 squares respectively. I thus tried to split level N into N congruent areas, which also failed.
That was close, and led to the key to the puzzle:
Looking up 1, 2, 6, 20, 60 on OEIS reveals that this is sequence is the total number of squares covered if you take all N-ominoes for a given N. For example, there are 5 unique tetrominoes (counting reflections/rotations as the same), and 5*4 = 20. Ditto for 12 pentominoes giving 12*5 = 60. The obvious thing to do now is split up the areas into one of each polyomino.
Here are the solutions to the first 4:
And here's the 5th:
I'm finding it very hard to succinctly show working without it turning into a case bash, so here's a partial working out. I'll try to stick to the following colour order when filling:
Let's try starting with the plus piece, since that's rotationally symmetric. There's only three places it can go since all other positions fence off chunks which aren't a multiple of 5 squares in size:
The red outline spot is easy to rule out:
By looking at places where the 5-long piece can go, here's roughly why it can't be the orange spot (click for full size):
So the plus piece must be on the left, which also locks in an L-shaped piece further left of it.
I'd continue, but this took a very long time to draw and there's a lot more case bashing left, so I think I'll just say you get the gist of it by now :) (I'd show uniqueness but this took much longer than I thought). I might continue if I get time some day.