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enter image description here

Those are the warmups. Here's the real challenge.

enter image description here

And yes, it is uniquely solvable.

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    $\begingroup$ Are we guaranteed that no puzzle comes in it's solved state? I.e puzzle 1 is not already solved. $\endgroup$ – TheGreatEscaper Mar 6 '17 at 11:25
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    $\begingroup$ Additionally, are we guaranteed that levels 1-4 have unique solutions? $\endgroup$ – TheGreatEscaper Mar 6 '17 at 11:33
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That 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:

enter image description here

And here's the 5th:

enter image description here


Working out (partial)

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:
enter image description here
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:
enter image description here
The red outline spot is easy to rule out:
enter image description here
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):
enter image description here 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.

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  • $\begingroup$ Proving uniqueness does take a long time :) And on a side note, a Halmiltonian path is different from a Euler walk. What I'm trying to say is that even though level 4 has lots of squares next to an odd number of squares, it may still be possible to have a path that goes through every square. For example take a 2xN grid. $\endgroup$ – Wen1now Mar 7 '17 at 7:50
  • $\begingroup$ What program did you use to draw this? $\endgroup$ – Wen1now Mar 7 '17 at 7:50
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    $\begingroup$ @Wen1now Ah oops, that was careless of me - indeed I was trying Hamiltonian path. As for drawing, Google Drawing actually. $\endgroup$ – Sp3000 Mar 7 '17 at 8:14

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