Doubt verification regarding tiling.

Question

Question: Is it possible to cover a 10 × 10 board with the (3+1) type L-tetrominoes without them overlapping?

Sol: Color the columns white and black alternatingly. There are 50 white squares and 50 black squares. We can see that, regardless of how we place a piece, it always covers three squares of one color and one of the other. Let us call the piece black if it covers three black squares and white if it covers three white squares. Then the number of black pieces is equal to the number of white pieces. This tells us that the total number of pieces must be even. This would mean that the number of squares should be divisible by 8. Since there are 100 squares, there is no possible cover.

Now my doubt is that why did he have to color the entire columns (or maybe exchangeable with "row") to be W or B. I mean he could have colored alternatively I mean square by square Kind of both sided alternation (like a chessboard). In that case that piece would be having 2 black and 2 white squares. That means then it'd have equal no. of squares, so his argument won't be applicable here. Am I right?

Answer 1

Step through the solution in detail:

Sol: Color the columns white and black alternatingly. There are 50 white squares and 50 black squares.

Obvious enough.

We can see that, regardless of how we place a piece, it always covers three squares of one color and one of the other.

Either the long leg or the short leg of the tetromino has to be vertical, lying in a single column.

• If the long leg is vertical, then the three squares in it will all have the same color, while the remaining square of the short leg will be the other color.
• If the short leg is vertical, then the two squares in it will be the same color. On the long leg, the middle square falls in the neighboring stripe and will have the opposite color, while the final square falls in the 2nd stripe over and will have the same color as the short leg.

So as mentioned, this coloration in vertical stripes leads to every tetromino having 3 squares of one color and 1 square of the other color. This allows him to categorize the tetrominos by their majority color:

Let us call the piece black if it covers three black squares and white if it covers three white squares.

This would not be possible if the board were colored in a checkered pattern. Then every tetromino would have two squares of each color, so neither is a majority. There are other ways you could categorize the tetrominos, such as by the color of their corner square, but that doesn't work with the next part of the argument.

Then the number of black pieces is equal to the number of white pieces. This tells us that the total number of pieces must be even.

This is the crux of the argument. If $W$ is the number of white tetrominos and $B$ is the number of black tetrominos, then the total number of white squares is $3W + B$, while the total number of black squares is $W + 3B$. Since we know the board has equal numbers of squares in each color, $3W + B = W + 3B$, which leads to $W = B$. This argument doesn't work if all the tetrominos have the same number of black and white squares.

The total number of tetrominos is $W + B = 2W$, which is even.

This would mean that the number of squares should be divisible by 8. Since there are 100 squares, there is no possible cover.

Alternatively, one could note that there must be $\frac{100}4 = 25$ tetrominos, which is not even. In either case, this is a contradiction, meaning no such cover exists.