Exploring Polyominoes

Exploring Polyominoes
This section is about tiles made out of squares called
polyominoes. You have probably met them before; dominoes are
a type of polyomino that is made from 2 x 1 squares like this:
The prefix poly- means many, so polyominoes are any shape made
out of any number of squares. Here is a tetromino, which is made
from 4 squares (from the Greek work tetra meaning four). You
might recognize it from the game Tetris.
Note how all the squares are joined edge to edge, so for
example this would not be a polyomino:
Here are a number of different puzzles and problems involving polyominoes.
Multilated chessboards
A domino is made from 2 squares. There is only one type of domino (1x2, shown
Can you place dominoes so that completely cover a normal 8x8 chessboard? How
about other sizes of chessboard such as 7x7, or 6x6?
Here is a 5x5 chessboard with a square removed – can
you cover this board with dominoes?
Can you always cover a 5x5 board with dominoes if you
remove any single square?
Here is a 6x6 board with two squares removed. It
seems to be impossible to cover this board with
dominoes, but can you explain why?
Is it always not possible to cover a 6x6 board if you
remove any two squares? Can you explain what is
Straight polyominoes
We saw in the previous section that a normal chessboard can be covered with
dominoes. Which p x q boards can be covered with dominoes?
Investigate the same question for other straight 1 x n polyominoes on r x s boards.
Can you find a rule?
Fault lines
This investigation is about placing dominoes on a grid so that there are no fault lines.
You could think of this like building a wall out of bricks; you do not want fault lines as
the wall would not be strong.
What is a fault line? It is when
you can draw a line through the
grid without cutting through a
domino like this:
For which grids is it possible to place dominoes without creating fault lines?
A tromino is a polyomino made from 3 squares. There are only 2 of them:
Which boards can be covered with both types of
tromino? How about combinations of the two?
Can this mutilated chessboard be covered with either
type of tromino?
Investigate the same question if other squares are removed. Can any squares be
removed to allow both trominoes to be placed?
There are different 5 tetrominoes – can you find them all?
Now can you cover a 5 x 4 grid with one of each?
If not, can you explain why?
It is easy to cover the 8 x 8 chessboard with just the straight 1 x 4 tetromino. Is it
possible to cover an 8 x 8 board with each of the other tetrominoes?
Investigate for other sized boards.
You may have guessed that a pentomino is one made from 5 squares.
Can you find all the different pentominoes?
Can you cover a 5 x 4 rectangle with 4 different pentominoes? How about a 5 x 5
square with 5 different pentominoes? How about a 5 x 6 rectangle with 6 different
ones? Can you cover a 6 x 10 rectangle with all different pentominoes? How about a
3 x 20 rectangle?
Here are four of the pentominoes put in pairs so that the outline of each shape is
exactly the same (congruent).
Can you use the remaining 8 pentominoes to make two more congruent pairs?
Challenge 1
Suppose you want to make it impossible to put a given pentomino on an 8 x 8 board
by placing a minimal number of monominoes.
For example, to make it impossible to put the Tpentomino on the board, you could do it by placing
16 monominoes on the squares shown in red on the
In fact, this arrangement also prevents placement of
three other pentominoes – which ones?
Now, can you prevent placement of the T-pentomino
with a different arrangement of fewer monominoes?
Can you find the minimum arrangements for all the other pentominoes?
It turns out that we can prevent placement of all pentominoes with just three
different arrangements of 16 monominoes. The arrangement above is one of them;
can you find the other two? [Hint: they both have reflection symmetry]
A good method for finding these arrangements might be to start with a smaller
board and work up. Can you extend this investigation to find the minimum number
of monominoes required to prevent placement of other polyominoes.
Challenge 2
How many different ways are there to put a single monomino
on an 8 x 8 board? Here by different we mean that the board
will not look the same if we rotate or reflect it.
Again, it might be useful to start with a smaller board and work
up from there. Can you see a pattern in the number of different
Can you extend this to placing two monominoes?
Notes on polyominoes
Mutilated chessboards
Of course, we can only cover chessboards with dominoes if there is an even number
of squares.
It will be possible to cover this 5x5 board after removing the corner (black) square. In
fact, removing any black square will mean it is possible to cover the board. This is
because we are now left with 12 black and 12 white squares, and each domino
covers one of each. By this same logic, it will not be possible to cover this
chessboard after removing any white square!
A similar argument works for the multilated 6x6 board. We have removed 2 black
squares, leaving an unequal number of black and white squares. As each domino
covers exactly one of each, we will not be able to cover this board. However, if we
remove one of each colour, it will now be possible.
Straight polyominoes
Clearly we can’t cover an r x s board with dominoes if both r and s are odd. However,
if one of them is even, then it is possible.
Investigating further, we find a similar thing happens with all 1 x n straight
polyominoes. They will cover any r x s board if (a) n divides r x s and (b) n divides
either r or s.
Fault lines
The smallest board that can be covered in dominoes with no fault lines is 5 x 6.
However, this does not mean that all larger boards can be covered. For example, it is
not possible to cover a 6 x 6 board.
There is quite a nice impossibility proof for this: There are 10 possible fault lines
across the board, each of which divides the board into two parts containing an even
number of squares. If a domino crosses a fault line, then one of its squares will be in
each part of the board, creating an odd number of squares remaining to be filled in
each part. For this reason, at least two dominoes must cross any fault line. But this
means we need 2 x 10 = 20 dominoes over fault lines. Unfortunately, we only have
18 dominoes available!
This suggests that a 6 x 8 board is possible; there are 5 + 7 = 12 fault lines, requiring
24 dominoes, and we have exactly 24 available… so is it possible?
It is possible to cover the mutilated board with the L-shaped tromino, but not with
the straight tromino.
Consider a 3-colouring of a board like this:
There are 22 red squares, and 21 blue and 21 white
ones. If we remove any non-red square then we are
definitely in trouble, but we haven’t here so we’re in
with a chance.
However, if you try it out you soon begin to realize it
isn’t possible. Why not?
Well, a corner square is not a ‘true’ red in the sense
that the corner square could have been a white or blue
(if we rotate the board round 90 degrees). So only by
removing ‘true’ red squares (those that remain red
after a 90 degree rotation) will it be possible; these are
the black ones shown on the right.
There are 5 tetrominoes:
It is not possible to cover the 5x4 grid with one of each. This can be seen if we
consider the parity of each tetromino. Here this means how many of each colour
they cover when placed on a chessboard. As you can see from the colouring above,
they all have even parity (cover an even number of each colour) apart from the Tshaped tetromino, which has odd parity. For this reason, we will not be able to cover
the 5x4 grid as it even parity.
It is possible to cover the 8x8 board with all the tetrominoes apart from the skew
one (it’s fairly obvious why if you try it – you can’t even cover an edge with them!).
There are 12 different pentominoes arranged in a 6 x 10 rectangle. All the other
arrangements stated in the text are possible and are left as an exercise to the reader
(note that the 3x20 one is really difficult!)
Regarding the congruent pairs: the second one is given below. The third pair, made
from the four remaining pentominoes, is left to the reader [Hint: there is a hole in
the middle of the shape].
Challenge 1
You can prevent placement of any pentomino with only 16 monominoes, as can be
seen from these three arrangements (which pentomino does each arrangement
However, we can use fewer monominoes to prevent
placement of individual pentominoes. For example,
here is a (rotationally symmetric) arrangement of 10
monominoes that prevents placement of the Xpentomino.
What is the minimum needed for prevention of each
I will not consider the other polyominoes in depth here, although I found it
interesting that we need 32 monominoes to avoid placement of the domino and the
L-tromino! Similarly, we need 21 to avoid placement of straight trominoes and the Ltetromino. Is this a coincidence or does this have to be the case?
Challenge 2
There are 10 different positions for the monomino:
It turns out that the number of different monomino
placements is the nth triangle number for a 2n x 2n
(or 2n-1 x 2n-1) board.