This document discusses using systems of linear equations to solve tiling problems. It provides an example of setting up variables to represent the number of tiles covering different cells in a 3x2 grid. This results in a system of equations that can be solved to determine if the grid can be tiled. The document also introduces the concept of "homology tilings" which allow for negative numbers of tiles, representing anti-tiles, and larger integers representing multiple copies of tiles. Solving equivalent homology tiling problems is presented as an easier approach than directly solving tiling problems.

1. SYSTEMS OF LINEAR EQUATIONS AND TILING
PROBLEMS
DAVE AUCKLY
As a warm up remember how to solve a system of linear equations. You
may manipulate any one equation by doing the same thing to each side of the
equals side, or you can add one equation to a different one. This will generate
a new system of equations. to make sure that the systems are equivalent,
you need to make sure that you can go backwards and get the original set of
equations from the new set. Here is an example fo ryou to solve:
5x + 2y = 22
3x − y = 11 .
A standard short-hand notation is to write out the equations without writng
the variables or equal signs. This is called the augmented sysem of the equa-
tion. For example the above system could be written as follows:
5 2 22
3 -1 11
Now you can just solve the system of equations on the back of this page. If
you don’t wish to do so now, don’t worry, we’ll do so later.
Time to change gears. Consider some tiling problems.
(1) Is is possible to tile an 8×8 square grid with 1×3 and 3×1 rectangles?
(2) If you remove one square from the 8 × 8 grid will it be possible to tile
it? If so which squares can you remove to tile it?
Since we talked about solving systems of equations, it might be nice to have
an application of systems of linear equations. Consider a tiling problem. Can
you tile a 3 × 2 grid using 2 × 1 and 1 × 2 rectangles? To answer this put the
lower left corner of the 3 × 2 grid at the origin in the coordinate plane with
sides parallel to the axes. Label each cell with the coordinates of the point in
the lower left of the cell. Now make the following variables:
1

2. 2 DAVE AUCKLY
n1 the number of tiles covering cells (0, 0) and (0, 1)
n2 the number of tiles covering cells (1, 0) and (1, 1)
n3 the number of tiles covering cells (2, 0) and (2, 1)
n4 the number of tiles covering cells (0, 0) and (1, 0)
n5 the number of tiles covering cells (1, 0) and (2, 0)
n6 the number of tiles covering cells (0, 1) and (1, 1)
n7 the number of tiles covering cells (1, 1) and (2, 1)
We then get one equation for each cell be requiring the number of tiles covering
each cell to be one. The resulting equations is the first system below. It can
be simplifie to equivalent systems as follows:
1 0 0 1 0 0 0 1
0 1 0 0 1 1 0 1
0 0 1 0 1 0 0 1
1 0 0 0 0 1 0 1
0 1 0 0 0 1 1 1
0 0 1 0 0 0 1 1
1 0 0 1 0 0 0 1
0 1 0 0 1 1 0 1
0 0 1 0 1 0 0 1
0 0 0 -1 0 1 0 0
0 0 0 0 -1 0 1 0
0 0 0 0 -1 0 1 0
1 0 0 0 0 1 0 1
0 1 0 0 0 1 1 1
0 0 1 0 0 0 1 1
0 0 0 1 0 -1 0 0
0 0 0 0 1 0 -1 0
0 0 0 0 0 0 0 0
The final equivalent system of equations reads: n1 = 1 − n6, n2 = 1 − n6 − n7,
n3 = 1 − n7, n4 = n6, and n5 = n7. Picking any value for n6 and any value for

3. EQUATIONS AND TILES 3
n7 gives a solution to the equations. Here are some sample solutions listed as
(n1, n2, n3, n4, n5, n6, n7):
(1, 1, 1, 0, 0, 0), (0, 0, 1, 1, 0, 1, 0), and (0, −1, 0, 1, 1, 1, 1).
OK, this last solution is rather odd. How can we have a negative number of
tiles? To have a real solution to the tiling problem all variables must take the
value 0 or the value 1 for we either do not have the tile, or we do. Adding the 0
or 1 constraint turns the problem into a subset of something known as Integer
Programming. Without this restriction we still have something interesting.
We can interpret −1 to represent the placement of an anti-tile. We could even
makes sense of larger integer values. A 3 would represent three copies of the
tile. To get back down to just one tile covering each square in the regioin, one
would then add come anti-tiles covering the same squares. We can define an
algebraic or homology tiling to be a stack of tiles and anti-tiles over the region
so each square of the region is covered by a total of 1 tile.
Any tiling is a homology tilling. Thus if it is impossible to find a homology
tiling for a region, it will be impossible to find a tiling of the region. It is much
easier to find equivalent homology tiling problems than it is to find equivalent
tiling problems.
(3) Can you tile the result of removing square (3, 4) from the 5 × 5 grid
using 1 × 2 and 2 × 1 rectangles?
(4) Translate the previous question into a system of equations. Does it
look familiar?
(5) Find simple, equivalent tile homology problems to it. Can you see the
solution to the big system of equations from problem (4)? What is it?
(6) Can you show that each grid region in the plane is equivalent to some
number of copies of the (0, 0)-square?
(7) Label each square in the plane with the number representing the equiv-
alent number of (0, 0)-squares.
(8) Use these ideas to solve our first tiling problem.
There is a different way to keep track of these ideas. Label the (a, b)-square
with xa
yb
. Given any region, add up the labels of the squares in it to obtain
a Laurent polynomial.
(9) What is the polynomial associated to any domino region in the grid?
(10) What can you say about the polynomial representing any region that
can be covered by dominoes?
(11) What is the polynomial associated to a 1 × 3 rectangle? What about
a 3 × 1 rectangle?
(12) Pick a small tile set, and use this idea to see what restrictions you can
find on grid regions that can be tiled using translates of tiles from your
set.
E-mail address: dav@math.ksu.edu