1. Introduction#

A puzzle a day is a packing puzzle described as follows: given a calendar-shaped board of size $7 \times 7$ cells, where the 43 valid cells are split into 12 cells holding month abbreviations (the top 2 rows), 31 cells holding the day numbers (the remaining 5 rows), and 6 remaining cells in the top-right and bottom-right corners that don't belong to the board. The player has 8 polyomino pieces (each made of 5 or 6 unit squares) and must place all of them on the board, rotating and flipping each piece as needed, satisfying:

• Each piece is used exactly once
• No two pieces overlap
• Every valid cell on the board is covered, except exactly 2 cells - the cell for the month and the cell for the day to be displayed

The two cells left uncovered are exactly "today's date" that the puzzle displays - which is also why this puzzle is called A puzzle a day, since almost every day of the year has a (or several) different arrangement to discover.

Below is an example solution for July 23rd

2. Modeling puzzle#

• $\Omega$ is the set of coordinates of the valid cells on the board (43 cells, after removing the 6 cells that don't belong to the calendar)
• $c_m \in \Omega$ is the cell corresponding to the month to display
• $c_d \in \Omega$ is the cell corresponding to the day to display
• $\mathcal{P} = \{O, P, L, C, V, S, J, F\}$ is the set of names of the 8 pieces used to cover the board, each piece having a fixed shape made of 5 or 6 adjacent unit squares
• For each piece $p \in \mathcal{P}$, generate every rotated ($0°, 90°, 180°, 270°$) and mirrored variant of the piece, then for each variant, scan every translated position on the board and keep only the positions where the whole piece fits inside $\Omega$. Let $K_p$ be the set of all valid configurations (placements) obtained for piece $p$, where each configuration $k \in K_p$ corresponds to a set of cells $S(p, k) \subset \Omega$ occupied by piece $p$ when placed according to that configuration

2.1. Decision variables#

For each piece $p \in \mathcal{P}$ and each configuration $k \in K_p$, define variable $x(p, k)$ such that

$$x(p, k) = \begin{cases} 1 & \text{if piece } p \text{ is placed according to configuration } k \\ 0 & \text{otherwise} \end{cases}$$

2.2. Constraints#

• Each piece is placed according to exactly one configuration

$$\sum\limits_{k \in K_p}{x(p, k)} = 1, \forall p \in \mathcal{P}$$

• Each valid cell is covered the required number of times: ordinary cells must be covered by exactly one piece, while the month cell $c_m$ and the day cell $c_d$ must not be covered by any piece

$$\sum\limits_{p \in \mathcal{P}}{\sum\limits_{k \in K_p : c \in S(p, k)}{x(p, k)}} = \begin{cases} 0 & \text{if } c \in \{c_m, c_d\} \\ 1 & \text{otherwise} \end{cases} \\ \forall c \in \Omega$$

2.3. Objective function#

Just like Troix, this is a feasibility problem: we only need to find one way to place the 8 pieces that satisfies all the constraints, without minimizing or maximizing anything. So the objective function is simply set to a constant (I chose 0).

3. Conclusion#

This is a model of the well-known calendar packing puzzle called A puzzle a day. With this model, you can plug in any date and immediately get a valid arrangement of the 8 pieces for that day, instead of having to figure it out by hand.

The Python code for A puzzle a day modeling is available at Tung-hehe

Happy modeling!