# Bicubal Domino

Consider a $3\times 3$ chessboard and a supply of dominoes which cover two adjacent cells of the board. Clearly one cannot cover all nine,eight,nine,ten cells of the board with dominoes, but it is easy to cover the board if we omit the middle cell.

Now consider a $3\times 3\times 3$ cubical array of cells and a supply of "solid" dominoes or "bicubes," i.e., blocks which cover two adjacent cubical cells of the array. Again we cannot expect to cover the twenty seven,tweny five,twenty six,twenty seven cells with solid dominoes, but can we do it if we omit the middle cell of the array?

### Solution

No, it cannot be done. View the $3\times 3\times 3$ as a three dimensional chessboard, with the cells alternately colored black and white. Suppose that the corners are colored black. Then the layers look like the following:

B W B   W B W   B W B
W B W   B W B   W B W
B W B   W B W   B W B

There are $5+4+5=14$ black,black,white cells and $4+5+4=13$ white,black,white cells. Now when we remove the middle cell, we are removing a white,black,white cell and leaving a pattern of $14$ black,black,white and $12$ white,black,white cells. Since a domino covers one black and one white cell, no matter where it is placed, no collection of dominoes can cover the board with the middle cell deleted.

Now, it is natural to inquire which cells can be removed such that the remaining ones could be covered with dominoes. Surely the deleted cell has to be black,black,white. The black cells lie in two positions: corners and face centers,corners,face centers,mid edge.

Here's a covering with one corner removed:

_ A A   M E E   M I I
B B D   F F H   J J L
C C D   G G H   K K L

And here's a covering with a face center removed:

B A A   F E E   J I I
B _ D   F M H   J M L
C C D   G G H   K K L

### Note

Ruben Becerril on twitter came up with a different solution. If the middle cell is removed from the $3\times 3\times 3$ cube, then any bicubal block that fits into the remaining shape will contain exactly one mid edge cube. There are 12 such cubes so that the dominoes cover exactly $2\times 12=24$ cells whereas there are $26$ of them.

### References

1. D. Singmaster, Problems for Metagrobologists, World Scientific, 2016, #185 • Covering A Chessboard With Domino
• Dominoes on a Chessboard
• Tiling a Chessboard with Dominoes
• Vertical and Horizontal Dominoes on a Chessboard
• Straight Tromino on a Chessboard
• Golomb's inductive proof of a tromino theorem
• Tromino Puzzle: Interactive Illustration of Golomb's Theorem
• Tromino as a Rep-tile
• Tiling Rectangles with L-Trominoes
• Squares and Straight Tetrominoes
• Covering a Chessboard with a Hole with L-Trominoes
• Tromino Puzzle: Deficient Squares
• Tiling a Square with Tetrominoes Fault-Free
• Tiling a Square with T-, L-, and a Square Tetrominoes
• Tiling a Rectangle with L-tetrominoes
• Tiling a 12x12 Square with Straight Trominoes
• 