This article is from the Puzzles FAQ, by Chris Cole chris@questrel.questrel.com and Matthew Daly mwdaly@pobox.com with numerous contributions by others.

Count the ways to tile an MxN rectangle with 1x2 dominos.

geometry/tiling/count.1x2.s

The number of ways to tile an MxN rectangle with 1x2 dominos is

2^(M*N/2) times the product of

{ cos^2(m*pi/(M+1)) + cos^2(n*pi/(N+1)) } ^ (1/4)

over all m,n in the range 0<m<M+1, 0<n<N+1.

[Exercises:

0) Why does this work for M*N odd?

1) When M<3 the count can be determined directly;

check that it agrees with the above formula.

2) Prove directly this formula gives an integer for all M,N, and

further show that if M=N it is a perfect square when 4|N and

twice a square otherwise.

]

Where does this come from? For starters note that, with the usual checker-

board coloring, each domino must cover one light and one dark square. Assume

that M*N is even (but as it happens our formula will work also when both

M,N are odd --- see exercise 0 above). Form a square matrix of size

M*N/2 whose rows and columns are indexed by the light and dark squares,

and whose (j,k) entry is 1 if the j-th light and k-th dark square are

adjacent and zero otherwise. There are now three key ideas:

First, the number of tilings is the number of ways to match each light

square with an adjacent dark square; thus it is the _permanent_ of our

matrix (recall that the permanent of a rxr matrix is a sum of the same

r! terms that occur in its determinant, except without the usual +1/-1

sign factors).

Second, that by modifying this matrix slightly we can convert the

permanent to a determinant; this is nice because determinants are generally

much easier to evaluate than permanents. One way to do this is to replace

all the 1's that correspond to vertical adjacency to i's, and multiply the

whole thing by a suitable power of i (which will disappear when we raise

it to a fourth power).

[Exercise 3: check that this transformation actually works as advertised!]

Third, that we can diagonalize the resulting matrix A --- or, more

conveniently, the square matrix of A' order M*N whose order-(M*N/2)

blocks are 0,A;A-transpose,0 , whence det(A') = +-(det(A))^2. Then

the rows and columns of A' are indexed by squares of either hue on our

generalized checkerboard, and its entries are 1 for horizontally adjacent

squares, i for vertically adjacent ones, and 0 for nonadjacent (including

coincident) squares. This A' can be diagonalized by using the trigonometric

basis of vectors v_ab (a,b as in the formula above) whose coordinate at

the (m,n)-th square is sin(a*m*pi/(M+1)) * sin(b*n*pi/(N+1)).

Exercise 4: verify that these are in fact orthogonal eigenvectors of A',

determine their eigenvalues, and complete the proof of the above formula.

(None of this is new, but it does not seem to be well-known: indeed

each of the above steps seems to have been discovered independently

several times, and I'm not sure whom to credit with the first discovery

of this particular application of the method. For different approaches

to exactly solvable problems involving the enumeration of domino tilings,

see the two papers of G.Kuperberg, Larsen, Propp and myself on

"Alternating-Sign Matrices and Domino Tilings" in the first volume of

the _Journal of Algebraic Combinatorics_.)

--Noam D. Elkies (elkies@zariski.harvard.edu)

Dept. of Mathematics, Harvard University

geometry/tiling/rational.sides.p

A rectangular region R is divided into rectangular areas. Show that if

each of the rectangles in the region has at least one side with

rational length then the same can be said of R.

geometry/tiling/rational.sides.s

"Fourteen proofs of a result about tiling a rectangle" (Stan Wagon)

_The American Mathematical Monthly_, Aug-Sep 1987, Vol 94 #7. There

was also a fifteenth proof published a few issues later, attributed to

a (University of Kentucky?) student.

geometry/tiling/rectangles.with.squares.p

Given two sorts of squares, (axa) and (bxb), what rectangles can be tiled?

geometry/tiling/rectangles.with.squares.s

A rectangle can be tiled with (axa) and (bxb) squares, iff

(i) gcd(a,b)=1 , and any of the following hold:

either: both sides of the rectangle are multiples of a;

or: both sides of the rectangle are multiples of b;

or: one side is a multiple of (ab), and the other is any length EXCEPT

one of a finite number of "bad" lengths: those numbers which are

NOT positive integer combinations of a & b. { By Sylvester's theorem

there are (a-1)(b-1)/2 of these, the largest being (a-1)(b-1)-1. }

(ii) gcd(a,b) = d .

Then merely apply (i) to the problem with a,b replaced

by a/d, b/d and the rectangle lengths also divided by d.

i.e. all cells must appear in (dxd) subsquares.

------

PROOF

It is clear that (ii) follows from (i), and that simple constructions give

the "if" part of (i). For the "only if" part, we prove that...

(S) If one side of the rectangle is not divisible by a, and the other is

not divisible by b, then the tiling is impossible.

The results in (i) follow immediately from (S).

To prove (S): ( Chakraborty-Hoey style ).

~~~~~~~~~~~~~~~~

Let the width of the rectangle be a NON-(a-multiple). Then the number of

bxb squares starting (i.e. top edge) at row 1 must be a NON-a-multiple.

Thus the number of bxb starting at row 2 must BE an a-multiple. Similarly

for the number starting at rows 3,4,...,b . Then the number starting at

row (b+1) must be a NON-a-multiple again.

Similarly the number starting at rows (2b+1), (3b+1), (4b+1),... must all be

non-a-multiples. So if the number of rows is NOT a multiple of b, (call it

bx+r), then row (bx+1) must have a NON-a-multiple of bxb squares starting

there, i.e. at least one, and there is no room left to squeeze it in. [QED]

----

A Rickard-style proof of (S) is ..BBB....BBWWW...WBBB....BBWWW...W(..etc) ~~~~~~~ also possible, by ..BBB....BBWWW...WBBB....BBWWW...W coloring the rectangle in ..BBB....BBWWW...WBBB....BBWWW...W vertical strips as shown here: <- a ->< b-a ><- a ->< b-a >

Every square tile covers an a-multiple of black squares. But if the width

is a NON-b-multiple, and the number of rows is a NON-a-multiple, then there

are a NON-a-multiple of black squares in total. [QED]

(Note: the coloring must have 1 column of blacks on the right, and any

==== spare columns of whites on the left.)

===================

Bill Taylor. wft@math.canterbury.ac.nz

>A Rickard-style proof of (S) is ..BBB....BBWWW...WBBB....BBWWW...W(..etc) > ~~~~~~~ also possible, by ..BBB....BBWWW...WBBB....BBWWW...W >coloring the rectangle in ..BBB....BBWWW...WBBB....BBWWW...W >vertical strips as shown here: <- a ->< b-a ><- a ->< b-a > >

>Every square tile covers an a-multiple of black squares. But if the width

>is a NON-b-multiple, and the number of rows is a NON-a-multiple, then there

>are a NON-a-multiple of black squares in total. [QED]

>

>(Note: the coloring must have 1 column of blacks on the right, and any

> ==== spare columns of whites on the left.)

This statement of how to position the colouring isn't good enough, I'm

afraid. Take a=4, b=7 and consider e.g. a 19x10 rectangle. Coloured your

way, you get:

BWWWBBBBWWWBBBBWWWB BWWWBBBBWWWBBBBWWWB ::::::::::::::::::: BWWWBBBBWWWBBBBWWWB BWWWBBBBWWWBBBBWWWB

The result has 10*10=100 black squares in it, which *is* a multiple of a=4,

despite the fact that 19 is not a multiple of 7 and 10 is not a multiple of

4.

Of course, there is an alternative offset for the pattern that does give you

the result you want:

WWBBBBWWWBBBBWWWBBB WWBBBBWWWBBBBWWWBBB ::::::::::::::::::: WWBBBBWWWBBBBWWWBBB WWBBBBWWWBBBBWWWBBB

To show this happens in general: because the width of the rectangle is a

non-multiple of b, it is possible to position it on the pattern so that the

leftmost column in the rectangle is white and the column just right of the

right edge of the rectangle is black. Suppose N columns are black with this

positioning. Then the rectangle contains N*H black cells, where H is the

height of the rectangle.

If we then shift the rectangle right by one, the number of black columns

increases by 1 and it contains (N+1)*H black cells. The difference between

these two numbers of black cells is H, which is not a multiple of a.

Therefore N*H and (N+1)*H cannot both be multiples of a, and so one of these

two positionings of the pattern will suit your purposes.

David Seal

dseal@armltd.co.uk

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