aa r X i v : . [ m a t h . M G ] F e b TILINGS WITH THE MINIMAL TILE PROPERTY
IWAN PRATON
Abstract.
A square tiling of the unit square is said to have theminimal tile property if the smallest tile can tile all the other tiles.We show that in such a tiling, the smallest tile cannot be too small.
Start with a unit square and fix a positive integer n . Tile the unitsquare with n smaller squares. (In other words, put n non-overlappingsmall squares—which do not have to be the same size—inside the unitsquare, with no space left over). This can be done for n = 4 and n ≥ n for which regular tilings exist. But they do, as shown in the figurebelow. k · · · ... k · · · ... Figure 1.
MTP tilings for n = 2 k and n = 2 k + 3,where k ≥ possibilities to check. One way to do this is to show that for a givenvalue of n , the smallest tile cannot be too small. Then there are onlyfinitely many possible values for the side lengths of all tiles, so we onlyneed to check a finite number of configurations. The theorem below isa result of this nature. Theorem 1.
In an MTP tiling with n squares, the side length of thesmallest tile is at least / n . The proof uses the electrical network theory of Brooks et al [2]. Itturns out to be convenient to adjust our scale so that the smallest tilehas side length 1. Then all tiles in an MTP tiling has integer sidelengths; the side length of the big tiled square is also an integer.We now summarize the electrical network approach to analyzingtilings. Details are in [2] or [5]. Suppose we have a rectangle of height h and width w tiled by squares whose side lengths are integers. We placethe rectangle and its tiles so that all sides are horizontal or vertical.First we convert the tiling into a graph G ( V, E ), where the vertex set V consists of all connected components of the horizontal boundariesof the tiling, and the edge set E is the set of all squares in the tiling.Since every tile has two horizontal boundaries, it is natural to say thattwo vertices is connected by an edge if the corresponding tile has thetwo corresponding horizontal components as boundaries. (Note that itis possible to have multiples edges connecting two vertices.)Let m denote the number of vertices in the graph G . We label thevertices of G with integers 1 to m . The vertex corresponding to the topedge of the rectangle is vertex 1; similarly, the vertex corresponding tothe bottom edge is vertex m .In the simple example below, the rectangle has height 2 and width4, tiled with 5 squares. The corresponding graph is shown on the right.1 5 Figure 2.
A rectangle tiled by squares and its corre-sponding network graph.We now think of G as an electric network. The edges are wiresthrough which electric currents can flow. We’ll assume that each edge ILINGS WITH THE MINIMAL TILE PROPERTY 3 has unit resistance. Let’s say that one unit of current flows into ver-tex 1, goes through the network, and leaves from vertex m (which isgrounded, i.e., vertex m has zero electric potential). Then it is possible,using Kirchoff’s and Ohm’s Laws, to calculate the amount of currentin each wire and the electric potential in each vertex. The calculationscan be conveniently summarized in matrix form.Define an m × m matrix L = ( a ij ), where a ii is the degree of vertex i , and for i = j , a ij = ( i and j are not connected; − r if there are r edges connecting i and j .It turns out that L is a singular matrix with rank m −
1. For ourexample, the matrix L is L = − − − − − − − − − − − . Now let p = ( p , . . . , p m ) T , where p i is the electric potential at vertex i . Note that p m = 0 since vertex m is grounded. Ohm’s Law indicatesthat the i th component of L p is the total amount of current flowing into(and out of) vertex i ; by Kirchoff’s Law (which is a current preservationlaw), this amount is 1 when i = 1; − i = m ; and 0 when i = 1 , m .Therefore L p = e − e m where e i is the i th standard basis vector. Thusto find p k we need to solve the matrix equation L v = e − e m ; we wantthe unique solution with v m = 0.We can get this solution by first solving the equation L ′ v ′ = e ,where L ′ is the matrix L without its last row and last column. Thesolution we want is v ′ with a 0 appended as the last entry. The matrix L ′ is invertible, so there is a unique solution.In our example, L ′ = (cid:18) − − − − − − (cid:19) , whose determinant is 32. Thesolution to L ′ v ′ = e is v ′ = (4 , , , T , so the vector of potentials is p = (4 , , , , T .We can also find the potentials a different way. Go back to the tilingof the rectangle. Place the bottom edge of the rectangle at y = 0and let q i denote the y -coordinate of the horizontal boundary thatcorresponds to the vertex i . Define the vector q as ( q , . . . , q m ) T . Wenote that q m = 0, q = h , the height of the rectangle, and each q i is aninteger. In our example, q = (4 , , , , IWAN PRATON
We define potentials and currents in G using q as follows; we’ll showthat these potentials and currents satisfy Ohm’s and Kirchoff’s Laws.We define the electric potential at vertex i to be q i . Previously weimposed 1 unit of current flowing through G from vertex 1 to vertex m , but now instead of 1 unit we will use w units, where w is the widthof the rectangle. The currents in the interior of G are defined as follow.If there is an edge connecting vertex i and vertex j , we say there is acurrent of q i − q j units flowing from i to j . These currents and potentialsclearly satisfy Ohm’s Law (since each edge has unit resistance). Nowsuppose vertex j corresponds to an interior horizontal boundary. Atile above and on this horizontal boundary contributes q i − q j , the sidelength of the tile, to the current coming into vertex j . The total amountfrom such tiles is the sum of the side lengths of these tiles, so it is equalto the length of the boundary. Similarly, the total contribution of thetiles on and below the horizontal boundary the negative of the lengthof this boundary. Therefore the total current coming into and out ofvertex j is zero, in accordance to Kirchoff’s Law. Of course, these lawsare also satisfied at vertex 1 and m . Therefore the potentials q i andthe currents q i − q j are exactly the values attached to the electricalnetwork, except that the external current from vertex 1 to vertex m isnow w units.We conclude that L q = w e − w e m , and so L ( q /w ) = e − e m . Since potentials are unique, we have p = q /w , providing an explicit solution to the problem of finding the electricpotentials of the network.This is sufficient to provide a proof of Theorem 1. Proof of Theorem 1.
As above, we scale our tiling so that the smallesttile has side length 1. It suffices to prove that the side length of thebig tiled square is at most 2 n . In the notation above, we want to showthat h = w ≤ n .First we note that in this case, there is a tile of side length 1, so q , . . . , q m are integers with no common factors (other than 1). Let d = det L ′ . Then d p is a vector of integers that is a scalar multipleof q . Since the components of q are relatively prime, in fact d p is aninteger multiple of q . In particular, dp > q >
0, so dp ≥ q . Note that q = h and p = q /w = h/w , so dh/w ≥ h or w ≥ d .By the Matrix-Tree theorem [5], d is the number of spanning treesof the graph G . Since a spanning tree is a subset of the edges of G , wehave the crude estimate d ≤ n , where n is the number of edges of G , ILINGS WITH THE MINIMAL TILE PROPERTY 5 i.e., the number of squares in the tiling. Thus we have w ≤ n , whichis what we want. (cid:3) Note that the proof does not rely on the minimal tile property: whatwe need is that the components of q are relatively prime. It is straight-forward to verify that this can be realized for any tiling of the unitsquare. Thus a similar result holds for all tilings, not just MTP tilings.Of course, the bound here is quite crude and not useful for compu-tational purposes unless n is small. For specific values of n there arebetter bounds.Here is an example. First we introduce some notation. If T is atiling of the unit square with n tiles, denote by σ ( T ) the sum of theside lengths of the tiles in T . We want to find an optimal tiling T M such that σ ( T M ) ≥ σ ( T ) for all MTP tilings T . More generally, wewould like to find T M such that σ ( T M ) ≥ σ ( T ) for all tilings T , notjust MTP tilings.Suppose now n = k + 3 where k is a positive integer. The standard k × k grid, where one tile is further divided into 4 equal pieces, isconjectured to be optimal over all tilings [3]. It also happens to be anMTP tiling. The figure below shows an example with k = 3. Figure 3.
MTP tiling for n = k + 3 (here k = 3).The sum of the side lengths of this MTP tiling is k + 1 /k . Thus σ ( T M ) ≥ k + 1 /k . Theorem 2.
Suppose T M is an optimal tiling with n = k + 3 (where k ≥ ). Then the smallest tile in T M has side length at least / (3 k ) .Proof. We recall a result from [4]: if we have m nonoverlapping tileswith total area A , then the sum of their side lengths is at most √ mA .Now let x denote the length of the smallest tile in a tiling T . Thenits area is x and the other k + 2 tiles have area 1 − x . Thus σ ( T ) ≤ x + p (1 − x )( k + 2) < x + √ k + 2 < x + k + 1 /k − / (3 k ) , so if x < / (3 k ), then T is not optimal. (cid:3) IWAN PRATON
References [1] J. Alm, personal communication.[2] R.L. Brooks, C.A.B. Smith, A.H. Stone, and W.T. Tutte, The dissection ofrectangles into squares, Duke Math. J. vol. 7 (1940), 312–340.[3] P. Erd˝os and A. Soifer, Squares in a Square. Geombinatorics, vol. IV, issue 4(1995), 110–114.[4] W. Staton and B. Tyler, On the Erd˝os Square-Packing Conjecture. Geombina-torics, vol. XVII, issue 2 (2007), 88–94.[5] D. Wagner, Combinatorics of Electrical Networks, lecture notes available at
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