Mathematical solution to https://cdn.cs50.net/2020/x/events/puzzles/puzzles.pdf -> Putting it all together
First, we encode tiles into binary in low-byte order, assuming an 8-bit row-size.
For example:
J
UN
H
would be encoded as 00000001 00000011 00000010 => 66306
To solve the puzzle then, we would expect to see all (8) rows and all (8) columns filled.
This would mean 64 bits would be filled, which can be represented as 2^64.
We can move a tile around the puzzle by "shifting" it in the binary space.
If we label the position of each tile a-r, then we can represent the "shift" action of tile a as 2^a.
All solutions to this puzzle satisfy this equation:
2^64 =
66306 * 2^a +
66305 * 2^b +
769 * 2^c +
771 * 2^d +
65795 * 2^e +
1796 * 2^f +
1794 * 2^g +
1539 * 2^h +
515 * 2^i +
66305 * 2^j +
770 * 2^k +
65793 * 2^l +
770 * 2^m +
65793 * 2^n +
131587 * 2^o +
196865 * 2^p +
259 * 2^q +
770 * 2^r
All solutions to this puzzle are also integers, and must fit within the board, thus we can constrain a-r as such:
a-z ∈ ℤ {0, 63}
We will then still have some solutions which are not possible, as a piece may be positioned on a row-boundary.
To solve this, we will encode just the width of each tile in similar fashion to the tile encoding.
For example:
J
UN
H
would be encoded as 00000011 => 3
We can then add the constraints (TODO - these are wrong - fix)
0 = (2^8 - 1) AND (3 x 2^a % 2^8) - (3 x 2^a) +
(2^8 - 1) AND (3 x 2^b % 2^8) - (3 x 2^b) +
(2^8 - 1) AND (3 x 2^c % 2^8) - (3 x 2^c) +
(2^8 - 1) AND (3 x 2^d % 2^8) - (3 x 2^d) +
(2^8 - 1) AND (7 x 2^e / 2^8) - (7 x 2^e) +
(2^8 - 1) AND (7 x 2^f / 2^8) - (7 x 2^f) +
(2^8 - 1) AND (7 x 2^g / 2^8) - (7 x 2^g) +
(2^8 - 1) AND (3 x 2^h / 2^8) - (3 x 2^h) +
(2^8 - 1) AND (3 x 2^i / 2^8) - (3 x 2^i) +
(2^8 - 1) AND (3 x 2^j / 2^8) - (3 x 2^j) +
(2^8 - 1) AND (3 x 2^k / 2^8) - (3 x 2^k) +
(2^8 - 1) AND (1 x 2^l / 2^8) - (1 x 2^l) +
(2^8 - 1) AND (3 x 2^m / 2^8) - (3 x 2^m) +
(2^8 - 1) AND (1 x 2^n / 2^8) - (1 x 2^n) +
(2^8 - 1) AND (3 x 2^o / 2^8) - (3 x 2^o) +
(2^8 - 1) AND (3 x 2^p / 2^8) - (3 x 2^p) +
(2^8 - 1) AND (3 x 2^q / 2^8) - (3 x 2^q) +
(2^8 - 1) AND (3 x 2^r / 2^8) - (3 x 2^r)