Can I fix a tiled floor with only one wrong tile left?

Question

If I have a $n\times m$ rectangular floor completely tessellated with $2\times 2 $ and $1 \times 4$ tiles and it now happens that I accidentally break one of those (no matter which one)- Can I then fix my misfortune if I only have one tile of the other kind left? (supposed that it is no problem to rearrange the remaining tiles) Any help would be appreciated.

Edit: Sorry for my bad wording.

Answer 1

Draw a grid of $1 \times 1$ squares on the floor and number them as follows:

\begin{bmatrix}1&i&-1&-i&1&i&-1&-i& \cdots \\ i&-1&-i&1&i&-1&-i&1& \cdots \\ -1&-i&1&i&-1&-i&1&i& \cdots \\ -i&1&i&-1&-i&1&i&-1& \cdots \\ 1&i&-1&-i&1&i&-1&-i& \cdots \\ i&-1&-i&1&i&-1&-i&1& \cdots \\ -1&-i&1&i&-1&-i&1&i& \cdots \\ -i&1&i&-1&-i&1&i&-1& \cdots \\ \vdots & \vdots & \vdots& \vdots& \vdots& \vdots& \vdots& \vdots & \ddots\end{bmatrix}

Define a tile's "tile-sum" as the sum of the numbers it covers.

Each $1 \times 4$ tile has a tile-sum of $0$ no matter how it is placed.

Each $2 \times 2$ tile can have a tile-sum of $2$, $2i$, $-2$, or $-2i$ depending on where it is placed.

The sum of the tile-sums of all the tiles is $(1+i+\cdots+i^{m-1})(1+i+\cdots+i^{n-1}) = \dfrac{(1-i^m)(1-i^n)}{(1-i)^2} = \dfrac{i}{2}(1-i^m)(1-i^n)$

If $m$ or $n$ is a multiple of $4$, then the sum of tile-sums is $0$. Thus, we need $a$ tiles with tile-sum $2$, $a$ tiles with tile-sum $-2$, $b$ tiles with tile-sum $2i$, and $b$ tiles with tile-sum $-2i$. Thus, the total number of $2 \times 2$ tiles is $2(a+b)$ which is even.

If $m$ and $n$ are both $2 \pmod 4$, then the sum of tile-sums is $2i$. Thus, we need $a$ tiles with tile-sum $2$, $a$ tiles with tile-sum $-2$, $b+1$ tiles with tile-sum $2i$, and $b$ tiles with tile-sum $-2i$. Thus, the total number of $2 \times 2$ tiles is $2(a+b)+1$ which is odd.

If $m$ and $n$ are both not multiples of $4$ and only one is a multiple of $2$, then you can't tile the floor.

Therefore, given an $m \times n$ floor, the pairity of the number of $2 \times 2$ tiles needed is constant. Do you see the problem?

Answer 2

So you break $1\times 4$ and then take the broken piece out. Take $3$ more of the $1\times 4$ pieces out and then you should have enough room for a $4\times 4$ depending on the arrangement. If you break the $4\times 4$ you will never have enough pieces because all you have left is a $1\times 4$. So your mistake is there.

Answer 3

Not in general. Say you had a $2 \times 2$ floor tessellated by a single $2 \times 2$ tile. If you broke that tile you'd be SOL, since your $1 \times 4$ tile would be too long. Similar situations could occur for a $1 \times n$ floor.