Square Tetris: covering a $4 \times 4$ square with (any) tetrominoes - Why no solutions with exactly $1$ $T$-piece?

Question

(Given a certain application of this problem, I'm surprised I couldn't find any discussion about it specifically. I'm probably just searching for the wrong terms. In any case, this one's been bothering me for a while.)

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Setup

(Unlike most "tetrominoes covering a bounded grid" problems I've seen so far,) let $M$ be the entire set of one-sided tetrominos: $$\{I,O,T,J,L,S,Z\}.$$

A "square" is a covering of a $4 \times 4$ grid with tiles in $M$, without overlapping, overhangs, or leaving any holes.

Let $n_K$ be the number of $K$-pieces in a covering, where $K \in M$. Since $4$ tetrominoes are needed to cover the grid, ($\because (4 \times 4) / 4 = 4$) it follows that for any square, $$\sum_{K \in M}n_K = n_I + n_O + n_T + n_J + n_L + n_S + n_Z = 4,$$ but this does not imply the reverse.

Here's an example of a square, where $n_I,n_L,n_J,n_Z = 1$, and $n_K = 0$ for all other $K \in M$: $$ \begin{array}{|c|c|} \hline I&I&I&I \\\hline L&J&J&J \\\hline L&Z&Z&J \\\hline L&L&Z&Z \\\hline \end{array} $$

Problem

For the case of $n_T = 1$ and $\sum_{K\in M, K\neq T}n_K = 3$, does a square exist?

Put informally, if you are trying to make a $4 \times 4$ square in Tetris with whole tetriminoes, if you place a $T$-piece in it, can you successfully complete the square with other pieces or are you forced to wait for another $T$-piece?

From my absent-minded pondering, it appears the answer is no, no such square exists. Why not?

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(My original observation was that a square must have an even number, including $0$, of $T$-pieces, but I then realized that the 3-alike case applies to any tetromino, so that's better suited for its own question.)

Answer 1

If you color the board like a checker board, every piece except the T covers 2 red and 2 black. The T piece covers 3 of one color. A 4x4 checker board has an equal number of red and black.

Answer 2

I want to expand on DanielV's answer, since I found an interesting follow up. That reasoning generalizes! Not only does it prove that there is no $4 \times 4$ square that only has $1$ $T$-piece, it proves that for any rectangular grid that can be covered with tetrominoes, there are no solutions with an odd number of $T$-pieces.

Put formally: Let $t \in \mathbb{N}$. For any covering of a rectangular grid with tiles in $M$, $$n_T = 2t.$$

Proof

Show grid can be covered with tetrominoes $\implies$ grid area is even

A tetromino, by definition, consists of $4$ cells. This means that any arbitrary grid that can be covered by them must have an area of $4m$, where $m \in \mathbb{N}$ is the number of pieces needed. $4m = 2(2m)$, $\therefore$ such area must be even. $(1)$

(DanielV) show(s) $T_b, T_r$ are odd, $T_b \neq T_r$, and $K_b = K_r = 2$ for all other $K \in M$

As DanielV shows, when you colour the grid like a checker-board (or a chessboard, but white doesn't show up well here), all the tetrominoes cover $2$ cells of one colour and $2$ of the other, both even numbers: $$ \begin{array}{|c|c|} \hline I & \color{red}I & I & \color{red}I \\\hline \end{array} \begin{array}{|c|c|} \hline O & \color{red}O \\\hline \color{red}O & O \\\hline \end{array} \begin{array}{|c|c|} \hline J \\\hline \color{red}J & J & \color{red}J \\\hline \end{array} \begin{array}{|c|c|} \hline & & L \\\hline \color{red}L & L & \color{red}L \\\hline \end{array} \begin{array}{|c|c|} \hline & \color{red}S & S \\\hline \color{red}S & S \\\hline \end{array} \begin{array}{|c|c|} \hline Z & \color{red}Z \\\hline & Z & \color{red}Z \\\hline \end{array} $$

...except the $T$-piece, the sole exception, which has $3$ and $1$ respectively, both odd numbers, and unequal: $$ \begin{array}{|c|c|} \hline T & \color{red}T & T \\\hline & T & \\\hline \end{array} $$

(Another way to put this is that the $T$-piece is the only tetromino that can't be split into $2$ dominoes.)

Let $K_b$ and $K_r$ be number of black and red cells, respectively, covered by the $K$-piece $\in M$.

Show $n_T = 2t$ for any rectangular grid of even area

A checker-board colouring of a grid with an even area gives $$c_b = c_r, \tag{2}$$ where $c_b, c_r$ are the number of black and red cells, respectively.

Observe that, when $n_T = 2t$, each pair of $T$-pieces can be arranged so one piece's $T_b = 3$ and the other's $T_b = 1$, so that both $$ \begin{align} T_{1b} + T_{2b} &= 4 \\ T_{1r} + T_{2r} &= 4, \end{align} $$ which satisfies $(2)$ above. Adding any other pieces in $M$ to this situation also satisfies $(2)$, since $K_b = K_r = 2$.

Suppose for contradiction there does exist a covering where $n_T = 2t + 1$. Since the $2t$ part satisfies $(2)$, and adding any pieces except $T$ also satisfies it, we can remove it without loss of generality to get $n_T = 1$. Also without loss of generality, let $T_b = 3$ and $T_r = 1$. We can then show: $$ \begin{align} c_b =& \overbrace{3}^{T_b} + (m - 1)\overbrace{2}^{K_b, K \neq T} \\ {\color{red}\lt}& \underbrace{1}_{T_r} + (m - 1)\underbrace{2}_{K_r, K \neq T} = c_r \end{align} \\ \\ c_b \,{\color{red}\lt}\, c_r$$ which contradicts $(2)$.

$\therefore n_T = 2t$ for any rectangular grid of even area. $(3)$

Putting it all together

• From $(1)$, for any arbitrary grid that can be covered by tetrominoes, its area is even.
• From $(3)$, for any rectangular grid of even area, $n_T = 2t$.

$\therefore$ for any rectangular grid that can be covered by tetrominoes, $n_T = 2t$ as required. $\blacksquare$