Tiling the UFW Symbol

Transcription

1 Tiling the UFW Symbol Jodi McWhirter George Santellano Joel Gallegos August 18, Overview In this project we will explore the possible tilings (if any) of the United Farm Workers (UFW) symbol. We will explore the tilings of the UFW symbol as a sub-shape of the Aztec diamond. One of our methods is to study the tilings of the Aztec diamond using the Lindström s-gessel Viennot method. While this method counts the tilings of the Aztec diamond, we hope to apply it to count the tilings for of our sub-shape. An additional method we hope to use is Kuperberg, Propp, and Wilson s tiling counting software to confirm in each example the amount of tilings that are possible. Alongside these two methods we will hand count the amount of tilings of our shape. After going through several examples, we later create a table to further explore where these numbers (the numbers are the number of tilings for each order of our shape) come from and see if there is any relationship or sequence that match with. It turns out there is a subsequence of our finding that matches with the squares of a sequence of numbers, and in our results we give an existence proof for these tilings and begin an attempt to try prove our conjecture. Before any of this, we begin with background of studying the tilings of the Aztec diamond and laying the ground work of definitions and terminology used through this project. We begin by defining how our shape will look inside the Aztec diamond for any order. Take for example this Aztec diamond of order 3. 1

2 In this example our UFW symbol is outlined in red where the wings will always be the middle row of the Aztec diamond. Notice we only went into the shape from the left and right by two squares while leaving a hole in our shape of two squares to create the shoulders of our bird. This will be the case for any order n shape, the n referring to the order of the Aztec diamond and also to the order of the UFW symbol inside it. 1 We define the cut of the shape to be the shape to be the number of squares to the side of the hole, the length of the wings. In the above example, the order 3 figure, the cut is 2. In general, the cut is the order 1. 2 History Before we begin working with our UFW shape, we believe that it is appropriate to have some idea of what this symbol means and where it comes from. The acronym UFW stands for United Farm Workers of America, and the symbol that this movement used was the Aztec eagle, which is the shape we wish to tile. The eagle was chosen because it represented dignity, and Cesar Chavez, a Chicano migrant farm laborer and one of the leaders and the face of this movement, felt that the movement needed a symbol; in particular, he felt that they needed a powerful symbol to give them some pride. This movement was begun by those with a desire to empower other migrant farm workers, increase their wages, and improve their working conditions. Although Cesar Chavez is heavily associated with this movement because he was more or less the face of the movement, it was actually organized early on by Dolores Huerta, a Chicano activist, and Larry Itliong, a Filipino migrant farm worker. An important part of the story is much less known is that this movement was not just for the Mexicans and Chicanos but also for the Filipino migrant farmworkers, who actually played a very big role in this movement. It began in Fresno, CA, around 1962 and spread across the whole United States and even across the Atlantic Ocean to England in the form of a boycott of grapes during the Delano grape strike, which lasted 5 years. 3 Background We begin with focusing on defining the terminology that will be used when discussing the number of tilings of the Aztec diamond, proof of Lindström s-gessel Viennot method, and the bijection 1 If there is still confusion on how we create our bird inside the Aztec diamond, please jump to the examples for clarification.

3 between large Schröder paths and the tilings of the Aztec diamond. We will call a cell a 1 1 unit squares that make up the Aztec diamond. A tile is the union of any two cells sharing an edge and a tiling of a region R (in the case our Aztec diamond) is a covering of the region by tiles such that there are no gaps or overlaps. For our non-intersecting graphs we will denote our on the left hand side of the Aztec diamond as a 1,..., a n and on the right hand side as b 1,..., b n, and our goal will be to find all the paths from a i to b i such that the path from a i to b i does not intersect the path from a j to b j for every i j. For clarity, we define the following terms. Definition 1. A source is a point on the southwest of the Aztec Diamond where we begin our path through our tiling to our sink point at the same height on the southeast side of the Aztec diamond. We can have set of source and sink points like S = {s 1,..., s n } denoting our sources and T = {t 1,..., t 2 } denoting our sink points. Here is an example to further explain the previous definition so that there is no ambiguity on what we are talking about. a 1 b 1 a 2 b 2 a 3 b 3 Notice this a picture of the bottom half of an order 3 Aztec diamond and our sources are S = {a 1, a 2, a 3 } and our sinks are T = {t 1, t 2, t 3 }. This now leads us to our next terminology. Our goal next is to count all the paths from S to T such that they do not intersect. Before that, we define what are paths and routings. Definition 2. Let S = {s 1,..., s n } be your sources and T = {t 1,..., t n } be your sinks. Thus a routing, denoted R, is a set of paths P 1,..., P n from S to T such that no two paths share a vertex. Here, we look at two different examples to build intuition of the previous definition: (0,0) (6,0) As we can see, this routing has paths that do not intersect intersect (or in other words, they do not share a vertex). Each path in this example is a Dyck path 2 where here our n = 3 because our intervals for Dyck paths always go from (0, 0) to (2n, 0). But we can also have paths like our next example that do not intersect as well. 2 A Dyck paths by definition is the same the thing as Schröder paths but restricted to the moves (1,1) and (1,-1)

4 (0,0) (6,0) This is an example of routing using a different type of paths where now our movement is not just restricted to just NE and SE; specifically, this is a routing of (large) Schröder paths where we can move E, or horizontally. We now define some of our new terms. Definition 3. A (large) Schröder path is a path from (0,0) to (2n,0) with steps (1,1) (North-East), (1,-1) (South-East), and (2,0) (East), while never going below the x-axis. It is similar to a Dyck path, the main difference being that a Schröder path allows the horizontal step in addition to the diagonal steps. Now that we know what (large) Schröder paths are, the most natural thing would be to count how many paths there are from any source to any sink within our graph. These are what we call (large) Schröder number, and we will count this based on the graph we have for our Aztec diamond (which we will show later). Definition 4. A (large) Schröder number, S n, counts the number of Schröder paths from (0,0) to (2n,0). The reason we are interested in these numbers is because eventually we want to use these numbers to find determinants and count the amount of tilings of the Aztec diamond (and eventually our sub-shape). First, consider this tiling of an Aztec diamond of order 3 along with its (large) Schröder path. a 1 b 1 a 2 b 2 a 3 b 3 As we can see, this is a (large) Schröder path through our tiling, due to the fact that we can move horizontally. But there are certain rules that we had that might not be clear just from the picture. The following rules were: Each path will start on the southwest border of the of the Aztec diamond. Each path starts and ends at the same height.

5 If you are at the top of a vertical tile, then move SE across the tile If you are at the bottom of a vertical tile, then move NE across the tile. If you are in the middle of the tile, then you move straight across the tile. Following these rules and counting all the possible ways to travel through the non-intersecting graph of our Aztec diamond will help us define our entries of our path matrix. Given any Aztec diamond of order n, we have a corresponding non-intersecting graph that describes all the possible paths through our Aztec diamond given the set of rules above. We define this term now. Definition 5. A non-intersecting graph is a graph G that shows all the possible paths from our set of sources to our set of sinks (i.e. from a i to b j ) given any Aztec diamond of order n. If we use the same Aztec diamond of order 3 from above, then our corresponding non-intersecting graph will be a 1 b 1 a 2 b 2 a 3 b 3 If we let S = {a 1, a 2, a 3 } be our sources and T = {b 1, b 2, b 3 } be our sinks with our corresponding matrix with entries m ij = # of (large) Schröder paths from a i to b j (note: here our whole Aztec diamond is above the x-axis, so we do not have to worry about going under it), we have the matrix AD 3, which will be our Path matrix for the Aztec diamond of order 3. If we compute the determinant of AD 3, we have det(ad 3 ) = 64 = 2 3(3+1)/2. The second equality comes from the fact that, given an Aztec diamond of order n, the number of tilings of that shape is 2 n(n+1)/2 (in our case, n = 3). We will come back to the proof of this later; for now, we stick with defining a path matrix. Note: in directed graphs, a path can have a weight, which is denoted as wt(p ) for each path P from s i to t j. For our problem, wt(p ) = 1 for all paths. Definition 6. Let S = {s 1,..., s n } be our sources and T = {t 1,..., t n } be our sinks. A path matrix, which we will denote as P, is a matrix whose entry p ij = wt(p ) P is a path from s i to t j In our case, wt(p ) = 1 for all P, so we are counting all the paths from s i to t j and letting p ij be that number. We now move onto Lindström s-gessel Viennot lemma, which says that we can count the amount of tilings of the an Aztec diamond of order n by computing the determinant of the path matrix of our Aztec diamond.

6 Before we give the lemma, we need a couple more definitions. A directed graph is a graph for which every edge of the graph has a direction from one vertex and to the other vertex. A directed cycle is a directed graph starting and ending at the same point. Lemma 7. Lindström s-gessel Viennot lemma goes as follows: let G be a directed graph with no directed cycles, s 1,.., s n = S be the sources, and t 1,..., t n = T be the sinks. Let P be a path matrix for G. Then det(p ) = the number of routings from S to T This is only true if and only if wt(p ) = 1 (which we are assuming for the purposes of our project). Sketch of proof: First, take a path matrix P of some graph directed graph G with no directed cycles. Thus det(p )is summing over ALL path systems R = {P 1,.., P n }; by path systems in this situation, we mean all possible paths from S to T going from any s i to any t j : hence routings. For each path system R, we must make sure it is a routing. Thus for any R, lexicographically we find the first P i and P j that intersect, and we call their first intersection at a vertex v. Now, we replace the sub-path of P i from s i to v with the sub-path of P j from s j to v. Now we have new paths P i and P j. We check for other intersections and perform the same process. After this, we have a new path system R from S to T. We replace R with R and note that R will not be in our previous list of paths. Thus if we do this for every path R that is not routing, then we will have our desired result. Note: this is a more specific version of the Lindström s-gessel Viennot lemma where each weight of an edge is 1. It is possible to put weights on your graph for different purposes, but we need only to count the edges, so we do not need weights on our edges. We have now seen that the number of tilings is the same as counting all directed paths from S to T across our graph G made by our Aztec diamond. But these paths are exactly the large Schröder paths; hence there is a bijection between the large Schröder paths and the the tilings of the Aztec diamonds. We will explore this more in the next theorem. Theorem 8. There exists a bijection between the set of domino tilings of the Aztec diamond of order n, and the set, Π n, of n-tuples (π 1,..., π n ) of large Schröder paths satisfying: i) Each π i goes from ( 2i + 1,0) to (2i 1,0) for 1 i n. ii) No two paths π i and π j can intersect. Sketch of proof: Given a tiling T n of AD n, an Aztec diamond of order n, we can associate an n-tuple (τ 1,..., τ n ) of non-intersecting paths as follows. We label the rows of AD n as 1, 2,..., 2n, indexing from the southwest border up. For each i, 1 i n, we start a path τ i at the center of the left edge of the ith row. The path will move to the right edge of the ith row by crossing successive tiles that it encounters following a line through the center of the tile. Each n-tuple of paths corresponds to a unique domino tiling. Call this set of paths associated to a unique domino tiling Λ n. NTS: There exists a bijection between Λ n, the set of non-intersecting paths (τ 1,..., τ n ) associated with a unique domino tiling, and Π n, the set of n-tuples (π 1,..., π n ) of large Schröder paths. For a tiling T n of AD n with associated path (τ 1,.., τ n ), define ψ : λ n Π n such that ψ(t n ) = (π 1,..., π n ). We construct each π i from τ i with i 1 up steps at the beginning of τ i and i 1 up steps at the end. This raises our paths above the x-axis and each now satisfies conditions i) and ii) such that our shifted (τ 1,.., τ n ) Π n, and we call them (π 1,..., π n ). This is clearly bijective and completes this sketch of Eu and Fu s A simple proof of the Aztec diamond theorem.

7 Now that we have seen the bijection of domino tilings of the Aztec diamond and large Schröder paths, we want to use this information about Aztec diamonds and apply it to our sub-shape. Before we continue, we conclude with actually seeing why the number of tilings of the Aztec diamond of order n is 2 n(n+1)/2. This is our best attempt at sketching the proof in our own words. Definition 9. In the proof Elkies, Kuperberg, Larsen and Propp s we have that AD(n), the number of tilings of the Aztec diamond with order n, is 2 n(n+1)/2. Sketch of proof: We first need background on height functions, alternating sign matrices and skewed summations of alternating sign matrices. To define a Height Function, H T (v), we need a conceptual understanding. Imagine that we extend a tiling T n for a given Aztec diamond by tiling the complement of T n, the rest of the plane, with a horizontal tiling. We ll call this extended tiling T +. Let G be the graph with vertices {(a, b) Z 2 : a + b n + 1} with an edge between (a,b) and (a, b ) when a a + b b = 1. We color the lattice squares of Z 2 in a checkerboard coloring so that the line {(x, y) : x + y = n + 1} that bounds the upper-right border of the Aztec diamond passes through only white squares. This is the standard/even coloring. We orient every edge of G so that it has a black square on its left and a white square on its right. This is called the standard orientation of G with arrows circulating clockwise around white squares and counterclockwise around black squares. Let u v denote an edge uv whose standard orientation is from u to v. We ll call v = (a,b) a boundary vertex of G if a + b = n or n+1. The boundary cycle is a closed zigzag path. If we travel across the 6 edges of the boundary of a domino, our path goes up three edges and down three edges. Every vertex of our graph G lies on the boundary of one domino of T +. Thus, if we label the leftmost vertex ( n 1, 0) of G, then we can uniquely assign integer valued heights, H T (v), to all the vertices of G with the constraint that if an edge uv (with u v) belongs to the boundary of some tile in T +, then H T (v) = H T (u) + 1. These height functions have two properties: i) H(v) takes on the successive values 0, 1, 2,..., 2n + 1, 2n + 2, 2n + 1,..., 0,..., 2n + 2,..., 0 as v travels along the boundary cycle of G; ii) if u v, then H(v) is either H(u) + 1 or H(u) 3. An Alternating Sign Matrix (ASM) is a square matrix of entries 0, -1 and 1 such that the nonzero entries in any row or column alternate between 1 and -1 and each row and column sum is 1. Here is an example: If A is an n x ASM with entries a ij, 1 i, j n, we may define the skewed summation of A, A, with entries a ij = i + j 2( i j a i j ), an (n + 1) (n + 1) matrix. i =1 j =1 There exists the relation a ij = 1 2 (a i,j 1 + a i,j 1 a i 1,j 1 a i,j ) so we can recover A from A*. We now have enough artillery to sketch the proof.

8 Our motivation is to show that there is a bijection between the domino tilings of the Aztec diamond of order n and the pairs (A, B) such that A A n, B A n+1. Sketch of Elkies et al. proof: Given a tiling T n of an order-n Aztec diamond, we construct A that records H T (v) for v odd, and B that records H T (v) for v even. The parity of a vertex v is the parity of x + y + n + 1. Let a ij = H T ( n+i+j, i+j) for 0 i, j n and b ij = H T ( n 1+i+j, i+j) for 0 i, j n+1. We use this to construct A, B. It follows that if we fix A, the number of (n + 1) (n + 1) ASMs B such that the pair (A, B) results in a height function, satisfying the conditions listed in the background, is 2 N +(A) where N + (A) is the number of +1s in A. That means AD(n) = 2 N +(A). A A n Similarly for B, where N (B) is the number of -1s in B, AD(n) = 2 N (B). If we take B A n+1 AD(n) = 2 N (B) and replace n by n 1 and B by A, we get 1) AD(n 1) = 2 N (A). B A n+1 A A n It is known N + (A) = N (A) + n, A A n, so we get 2) AD(n) = 2 n 2 N (A). Combining 1) A A n and 2), we derive AD(n) = 2 n AD(n 1) and thus AD(n) = 2 n(n+1)/2 recursively. We are now at the point where we have explored the Aztec diamond enough, gained enough background to now apply these methods (theorems and definitions) to our UFW sub-shape to see what works and does not work. 3.1 Example 1: Order 2 We begin by trying to tile the shape first; then, we will use the non-intersecting paths methods in order to confirm our findings. 3 We will then use the Sage code to find out how many possible tilings there actually are to determine whether or not the amount we found using the first two methods is correct. First, we begin tiling by hand. We can see that the two wings must have vertical tilings, which means that what remains is the inside square. There are two ways to tile that figure (two verticals or two horizontals), hence there are only 2 tilings of our figure of order 2. Now, we move onto its non-intersecting graph: a 1 b 1 a 2 b 2 3 Note: the non-intersecting path methods has been proven for the original shape (the Aztec diamond), but we have not proven it for our shape. We are simply testing out the method to see if it does work for our specific sub-shape.

9 We count all the paths from a 1, a 2 to the vertices b 1, b 2, beginning with the paths from a 1 to b 1. Starting from a 1, there is one way to start: straight across to the middle vertex. From the middle vertex, there is one path to b 1, thus there is one unique path going from a 1 to b 1. We will denote this number m 1,1 = 1; we proceed by counting the paths from a 1 to b 2. There is only 1 path from a 1 to b 2 making m 1,2 = 1. Starting from a 2, there is only one path from a 2 to b 1 as well, hence m 2,1 = 1. If we are going to b 2 instead, then there are two ways to start from a 2, and those [ are ] 1 1 the only 2 paths, giving us that m 2,2 = 2. Putting these together, we get the matrix M 2 = 1 2 with det(m 2 ) = 1. This indicates that perhaps the non-intersecting path method does not work for the order 2 shape, but there is a possibility still that it could work for our Aztec diamond sub-shape of order greater than 2. Our next step is to check these answers using our code in Sage to determine the correct number of tilings. vax_string = "X X\nXXXX\n XX " printvax_string printmatchings ( vax_string ) X X XXXX XX 2 With this confirmation from the Sage code, we can say now that we are confident that there are 2 tilings of our order 2 sub-shape. 3.2 Example 2: Aztec Diamond of order 3 We proceed with this example in the same way as we did in Example 2. Thus, we begin with tilings and note that the wings, which create two 2 2 squares, have a fixed tiling of either two verticals or two horizontals. If the wings are not tiled in this way, then there cannot be a tiling of our sub-shape. This forces the rest of the shape to be fixed horizontally; thus there are 4 tilings of our sub-shape. We now continue with our investigation of the order 3 UFW symbol using the non-intersecting subgraph of our sub-shape in this example:

10 a 1 b 1 a 2 b 2 a 3 b 3 We have to count all the possible paths from the vertices a 1, a 2, a 3 to vertices b 1, b 2, b 3 to create a matrix M 3 whose m i,j entry denotes all the paths from vertex a i to vertex b j. We begin with the paths from a 1 to b 1. Starting from a 1, there are 2 ways to start, diagonally up (and down again) or to the horizontally to the right, both of which end up at the second vertex, the vertex that is directly horizontally to the right of a 1. WLOG, choose the horizontal path. From this second vertex, there are two paths to the third vertex, the vertex horizontally to the left of b 1 and a necessary stop to b 1 - horizontally to the right or diagonally down (and up again) - and there are two paths from this third vertex to b 1, diagonally up (and down) or horizontally to the right. Thus there are 4 paths from the second vertex to b 1 and 2 4 = 8 paths starting from a 1 and ending at b 1. We will denote this number m 1,1 = 8. We will count all the paths in a similar manner. Thus, our matrix looks like M 3 = and det(m 3 ) = 8(10 4) 6(12 4) + 2(12 10) = = 4, which matches what we found out when we tiled it ourselves. We now use the Sage code to confirm our findings. vax_string = "XX XX\nXXXXXX\n XXXX \n XX " printvax_string printmatchings ( vax_string ) XX XX XXXXXX XXXX XX 4 Although our methods of using non-intersecting graphs and finding the determinants of its matrix did not work for the first example, it seems to have worked here. This could be the starting point for which the non-intersecting graphs methods will work for the rest of our problem.

11 3.3 Example 3: Order 4 We proceed with our last example in the same manner as the previous two examples. This figure will be a lot more complicated to tile and will have to be done in cases. We start out by trying to count all of them by hand. In our initial attempt, we came up with 42 tilings; however, running the Sage code gave us a different answer, so we start over, this time with a number in mind. We proceed by looking at sections of the figure and seeing how many ways we can tile each section given a tiling of the rest of the graph. Using this method and keeping an eye out for double-counting, we can get most of them, and then by looking at a certain section of the figure and seeing how many times that tiling appears, we can find the missing tilings. Pictured below are some of the subsections we can consider with the rest of the tiling fixed. Note that counting all of these counts some tilings multiple times, so we have to watch out for the double-counting.

12 In particular, when we started over, we first got 46, then 48, and then 50. At 50, we knew we had only 4 left, so we looked at the possible tilings of the right wing along with the tile directly below it, and we got 4 tilings of this region; we then looked through the tilings we already had of the whole shape, we realized that two of these regional tilings had two fewer tilings than the other two, and since neither of the missing tilings were symmetric, we had found our last 4 tilings. This process led us to believe that brutally counting the tiles, while sufficient for smaller tilings, is not a feasible way to count the tilings of anything bigger than order 3, unless you know ahead of time how many tilings there are, and even that is difficult. a 1 b 1 a 2 b 2 a 3 b 3 a 4 b 4 Now that we have brutally counted as many tilings as we can by hand, we look at the nonintersecting graph of our shape and count all the possible paths from vertices a 1, a 2, a 3, a 4 to vertices b 1, b 2, b 3, b 4. We will count the paths in the same manner as done in Example 2, and we get the matrix

13 M 4 = It follows det(m 4 ) = 4. We can see already that this method breaks again, and it is possible that counting all the non-intersecting paths and finding the determinant of the corresponding matrix only works for the original shape (Aztec diamond) and not our sub-shape. We continue with our Sage code to confirm the amount of tilings for our sub-shape. vax_string = "XXX XXX\nXXXXXXXX\n XXXXXX \n XXXX \n XX " printvax_string printmatchings ( vax_string ) XXX XXX XXXXXXXX XXXXXX XXXX XX Example 4: Order 5 In order to manually count the tilings of the order 5 shape, we look first at the 2 4 rectangles on the wings. The rest of the shape can be tiled all horizontally, and each of the wings has 5 tilings, since 5 is the 5th (or, the (4+1)st) Fibonacci number. This gives 25 tilings for the whole shape. Since, using the computer code, we find that the shape has 36 tilings, we know that we should be able to find 11 more. We first try to find which parts can be tiled with vertical tiles and discover that the majority of the rest of the shape cannot. We then consider the following division of the UFW symbol:

14 We observe that making the bottom four tiles of the shaded region vertical fixes the rest of that shaded area, and that gives us the 11 missing tilings. This also led us to a conjecture, explained more fully in the Results section, that the wings plus the the section below them (shaded part) are the sub-shapes that can be tiled with vertical tiles and the remaining (white) section can only be tiled horizontally. a 1 b 1 a 2 b 2 a 3 b 3 a 4 b 4 a 5 b 5 With our graph of our UFW shape with the usual set of sources and sinks, S = {a 1,..., a 5 } and T = {b 1,..., b 5 }, we again count all the Schröder paths from a i to b j ; at this point, we decide that counting these manually is also difficult and look into either finding or writing a program to do this, just as there is one to count the tilings of a figure. The matrix we get from counting the Schröder paths in the order 5 shape is M 5 = Det(M 5 ) = 36. This is the number of tilings we found for our order 5 shape, both in counting manually and using the program; it appears that this method of counting is correct for shapes of odd order.

15 vax_string = "XXXX XXXX\nXXXXXXXXXX\n XXXXXXXX \n XXXXXX \n XXXX \n XX " printvax_string printmatchings ( vax_string ) XXXX XXXX XXXXXXXXXX XXXXXXXX XXXXXX XXXX XX 36 4 Results It seems that using Lindström s-gessel Viennot method of counting the non-intersecting paths, used for the original shape (Aztec diamond), works well for sub-shapes of odd orders, but it does not work for our sub-shape of even orders. Our conjecture for that is that for an even-ordered figure, the length of the cut is odd, so the tops of the wings are comprised of 2 (an odd number) rectangles. When one tiles these and overlays the graph of paths, one will find that one will always end up getting stuck at the gap in the figure. We have not found an accurate way to count the even ordered tilings without the Sage tiling program, so we will change our focus to the odd ordered subshapes. Thus we will further explore the shapes of odd order, but we will start with an existence proof. We wish to prove that for any n 2, there will always be a tiling for our sub-shape of order n. Theorem 10. For n 2, there exists a tiling of the UFW symbol of order n. To prove this theorem, we will use the result that every Aztec diamond has an all-horizontal domino tiling. This is true since every row of the diamond is symmetric and thus has an even length. Proof. Let n be the order of the UFW symbol. There are two cases to consider: n is odd, and n is even. We, at the start of this paper, defined the cut of the UFW symbol to be the length of the wings, which turns out to be the order 1. For the purpose of this proof, we will enumerate the rows of the UFW symbol starting at the top, such that the top row is row 1 and the bottom row is row n + 1. Case 1: n is odd. Then the cut is even. Shown for reference is a symbol of order 3.

16 Consider row 1 and rows 2 - (n+1) separately. Rows 2 - (n+1) (white) form half of the Aztec diamond of order n; since every Aztec diamond has an all-horizontal tiling, this section can be tiled. Row 1 (colored) is in two 1 (n 1) rectangles, and since n-1 is even, each of these sections can be tiled. Thus the UFW symbol can be tiled for order n odd. Case 2: n is even. Then the cut is odd. Shown for reference is a symbol of order 4. Consider rows 1-2 and rows 3 - (n+1) separately. Rows 3 - (n+1) (white) form half of the Aztec diamond of order n 1, which has an all-horizontal tiling. Consider rows 1-2 (colored) as the sum of three parts: the 1 2 rectangle beneath the hole and the two 2 (n 1) rectangles that make up the wings. Clearly, the 1 2 rectangle can be tiled with exactly one domino. Each of the 2 (n 1) has F n+1 tilings, where F n+1 is the (n + 1)st Fibonacci number. Thus the UFW symbol can be tiled for order n even. Theorem 11. The number of tilings of the UFW symbol of odd order n correspond to the square of the n 1 2 term of the large Schröder numbers. We used Sage to calculate the first 25 large Schröder numbers using an algorithm from the On-line Encyclopedia of Integer Sequences and then squared each of them; note that S 0 = 1 and S 1 = 2: Input: # Generalized algorithm of L. Seidel def A006318_list(n) : D = [0]*(n+1); D[1] = 1 b = True; h = 1; R = [] for i in range(2*n) : if b : for k in range(h, 0, -1) : D[k] += D[k-1] h += 1; else : for k in range(1, h, 1) : D[k] += D[k-1] R.append(D[h-1]); b = not b return R A006318_list(25) # Peter Luschny, Jun

17 Output: # the large Schroeder numbers [1, 2, 6, 22, 90, 394, 1806, 8558, 41586, , , , , , , , , , , , , , , , ] Input: [i^2 for i in A006318_list(25)] Output: # the large Schroeder numbers squared [1, 4, 36, 484, 8100, , , , , , , , , , , , ,

18 , , , , , , , ] The table below lists our findings. The order refers to the order of the UFW symbol, and the Tilings column lists the number of tilings each shape has that we found using the Sage code. We found the prime factorization for each and noticed that the number of tilings for each odd-order shape is a square. We added a Square Root column for the odd orders; these are the large Schröder numbers. Other than all being even, we have not been able to find a pattern for the even orders. Order Tilings Prime Factorization Square Root Our next goal is to figure out why the large Schröder numbers appear in our count of the tilings of odd-ordered figures; in particular, why do we get the large Schröder numbers squared? We are still working on figuring out why it is the large Schröder numbers, but we have a lead on why they are squared. Consider the UFW symbol in three pieces: the middle section, as an Aztec diamond of order n+1 2 missing one of the longest middle rows, and the two wings, what remains, which is like a UFW symbol of order n 1 2, but without the missing part on top. Shown below are two examples, the first, the order 3 shape; and the second, the order 5 shape.

19 Our next claim is that the number of tilings of each colored region of the figures above and for every odd-ordered figure is the large Schröder number we said in Theorem 11 corresponds to this figure, that is, for a UFW symbol of order n, the new region has the n 1 2 th large Schröder number of tilings. Using the tiling software, we have confirmed this for up to order 9. We also have started to try applying the Linström s-gessel Viennot method of counting paths, which is so far working. The order 3 figure is trivial (1 1 matrix, M 3sub = [ 2 ], and det(m 3sub ) = 2 = S 1 ). For the order 5 figure, our sub-sub-shape, the wing, has the graph: a 1 b 1 a 2 This produces the matrix M 5sub = [ ] 5 2, which has determinant 6 = S a 3 b 1 a 4 b 2 a 5 b 3

20 The order 7 graph of the wing, pictured above, produces the matrix M 7sub = which has determinant 22 = S 3. We also confirmed the determinant for the order 9 sub-shape, getting 90 = S 4. It seems, then, that Lindström s-gessel Viennot method works for this sub-shape of our shape. The middle part has exactly one tiling, the all-horizontal tiling, since the squares on each side give only one option for being covered, a horizontal tiling, and this fixes every tile in the shape. The number of tilings of the entire symbol, we hope to say, is the product of the tilings of each of these sub-shapes, which would give us the n 1 2 th large Schröder number squared. 4.1 Future Direction We conjecture that Lindström s-gessel Viennot method works for our sub-shape of odd order and for their sub-shapes, but proof by example is not sufficient enough. A kind person (Jodi s husband, Richard), towards the end, wrote a Python program for us that counts the number of paths from any a i to each b j, so we can more quickly check the determinants of higher order shapes and subshapes. We also conjecture that the odd-ordered shapes can be thought of as three sub-shapes, two of which have the same number of tilings and the other, a single, all-horizontal tiling. While dividing these shapes into those separate tiling regions has worked so far, we do not rigorously show that this is the case. We therefore plan to continue in the future to explore these shapes more and hope to come up with something more rigorous to prove our conjectures. We invite our audience to further explore this problem along with us for the benefit of greater mathematical understanding and the delight of another conjecture, proven. To prove Theorem 11, we will first prove three lemmas. Lemma 12. For any tiling of our UFW shape of odd order n, there exist three sub-shapes of our UFW shape separated by a hard boundary such that no tiling of the UFW shape has a tile that crosses this boundary. Proof. Take any UFW shape of order n and now we consider our hard-boundary. We now define our hard-boundary to be the two lines that start on both sides of the hole of our shape (where the eagle s head is suppose to be) and lines to continue going diagonally down and outwards (down, over, down, over, etc.) until you reach the boundary of the UFW shape. This forms a body of a skinny Aztec diamond (as opposed to fattened ) and two wings. We now continue by focusing on our upper left wing shape and shading our most upper left cell, then continue by shading every other cell until you finish the whole upper left wing shape. It should create a checkerboard pattern. Consider the following examples:

21

22 Claim: No tiling of the UFW symbol can have a tile that crosses this hard-boundary. To prove this claim, we will examine the left boundary and the wing sub-shape it forms. Since the UFW logo is symmetric, the same argument will hold for the right boundary. First note that for an odd n, the cut the length of the top two rows of our wing sub-shape has an even length. Since every subsequent row down from the top row decreases in length by two, each row will have an even length. Thus, every row can be tiled by a horizontal tile which means that the entire wing shape has an all-horizontal tiling. Since each tile must contain both a shaded and white cell, and an all-horizontal tiling exits, there is a perfect matching between shaded and white cells in our wing; that is, there is the same number of shaded cells as there are white cells. Since there exists a perfect matching within our wing, we cannot take away any shaded cells without destroying the matching which would mean a tiling of it does not exist (Ardila, 2005) nor can we add any white cells without doing the same. Since any tile that crosses the hard boundary would do either of the above, we cannot have any tiling of our UFW shape that has such a tile. Lemma 13. The skinny Aztec diamond has only one tiling. (or can only be tiled all-horizontally) Proof. We shall prove this by induction. For reference, we will use the same shading as in the previous proof. Our base case, the order 1 skinny Aztec diamond, is a single horizontal tile, which clearly has exactly one, all-horizontal tiling. Thus our base case is proven. Suppose for some n, the skinny Aztec diamond of order n has exactly one tiling, and this tiling is the all-horizontal tiling.

23 Consider the order n + 1 shape. Notice that the far left cell of the skinny Aztec diamond body (a white cell) can only be tiled by a horizontal tile, which covers the shaded cell to its right. Tiling that tile forces the cells directly above and below the covered shaded cell to be covered by horizontal tiles. This pattern continues up until the top and bottom sets of cells, so now the whole left side two cells deep has been tiled all-horizontally. This leaves an order n skinny Aztec diamond, which has a single, all-horizontal tiling. Thus the skinny Aztec diamond of any order n has only one tiling, and that tiling is all-horizontal. Lemma 14. Given our UFW sub-shape of odd order n and after overlaying our hard-line, each wing will have S n 1 number of tilings. Here S n represents the nth Large Schröder Number. 2 Proof. Let n be an odd number; then n = 2k + 1 for some k > 0. S k is the kth Schröder Number. Consider the UFW symbol of order n; its wing, by our definition has order k = n 1 2.

24 Let us examine the wing, and in particular, the wing upside-down. The paths from s to t are exactly the Schröder paths from (0,0) to (2k,0), so there are S k of these paths. Claim: there is a bijection between tilings of the wing of order k and the Schröder paths from (0,0) to (2k, 0). Consider a tiling of a wing of order k. We will construct a Schröder path, starting at s. A Schröder path has three moves: (2, 0), (1, 1), and (1, -1). The nodes shown on the figure above show the possible movements. Note that the position of these nodes is fixed and will have the same pattern on any order wing. Because of this, the only nodes that will be on the edge of the figure (so, the only places the Schröder path can begin or end) are at s and at t. We will associate the (2,0) movement with moving (left to right) across a horizontal tile, the (1,1) movement with starting at the lower left side of a vertical tile and moving to its upper right side, and the (1,-1) movement with starting at the upper left side of a vertical tile and moving to its lower right side. Starting at s, you will either be on the left side of a horizontal tile or the lower left side of a vertical tile, so the path will start with one of those two movements. Note that after every movement, you will land on another node, and that node, unless it is t, will either be on the left side of a horizontal tile, the lower left side of a vertical tile, or the upper left side of a vertical tile. Since those are exactly the options we have given ourselves for movement, the path can continue until it reaches t, which is our desired end location. Thus every tiling corresponds with exactly one Schröder path. Consider a Schröder path from (0,0) to (2k,0). By using the same movement association listed in the previous direction, we can overlay a tiling on the Schröder path. We must show that the area below and the area above the Schröder path can be tiled, and also that this tiling is unique. First, we will show that every vertical tile must be on the Schröder path. Suppose that there was a vertical tile not on the Schröder path. Every vertical tile has exactly two nodes on its border, whereas a horizontal tile can have either two nodes, one on each side, or one node in the middle of the tile. Thus if there is a vertical tile, a Schröder path can be drawn through it. By the previous direction, we have that this Schröder path can continue until it reaches t. However, this contradicts that this tile is not on the Schröder path, since there is only one Schröder path per tiling. Thus every vertical tile must be on the Schröder path, or, from another perspective, every tile that is not on the Schröder path must be horizontal, and not only horizontal, but the horizontal that has a node in its middle, since a horizontal tile with nodes on its sides will encounter the same contradiction as the vertical tile. Since all the vertical tiles must be on the Schröder path, and since the Schröder path starts and ends on the same level, there must be an even number of vertical tiles, since each vertical tile either brings the path up one level or down one level. Note also that every horizontal tile has a length of

25 2. Thus there will always be an even number of cells between each vertical tile that is on the same level, so each of these segments can be tiled by horizontal tiles. Thus the area under the path can be tiled, and there is only one tiling: all horizontally. Consider the region above the path. We d like to prove that that a tiling above the Schröder path is possible. WLOG, the only tiles that fit along the left boundary of our shape are vertical tiles with nodes on the lower left/upper right, or horizontal tiles with a node in the middle. We can disregard having vertical tiles in this region because we have shown previously that vertical tiles lie on the path. Placing a horizontal tile on the left boundary above the path forces a horizontal tiling of the left and right boundaries above our path. Now we must consider how the path can affect the tiling in the region above it and show that a tiling for the region exits. If the path is horizontal, the region above the path can be tiled horizontally since we showed earlier that our shape has a unique all horizontal tiling. Now we must consider if our path includes vertical tiles. As explained earlier, tiles on our path will connect at their nodes. When vertical tiles are present in a path, there are edges without nodes as shown in the example below. Only the horizontal tile with node in the middle can be adjacent to these edges since the other tiles have nodes on their borders.

26 We showed earlier that vertical tiles on a path come in pairs i.e. for every vertical tile with nodes on the bottom left/top right, there will be a vertical tile with nodes on the top left/bottom right at the same level. Thus, every path is of even length, any gaps between vertical tiles are of even length, and the rows above the path must be of even length. Thus, these spaces above our path can be tiled horizontally and, more specifically, must be tiled by the horizontal tile with node in the middle. Note that every Schröder path is different. Each Schröder path corresponds to exactly one tiling of this wing. Thus there is a bijection between tilings of our order k wing and Schröder paths from (0,0) to (2k, 0), so there are exactly 2k tilings of the wing. We now return to our original conjecture, Theorem 11, to finally prove it. Proof. By Theorem 10, there exists a tiling for every UFW sub-shape of odd order n. By Lemma 12, we have that there is a hard boundary that divides our UFW symbol into three sub-shapes such that no tiling of our UFW sub-shape can cross that boundary; in particular, these sub-shapes are two wings of order k = n 1 2 and a body of a skinny Aztec diamond. Since these areas are thus tiled independently, the number of tilings of the whole UFW symbol will be the product of the number of tilings of each of these sub-shapes. Let us first consider the body. By Lemma 13, the skinny Aztec diamond has a single, all-horizontal tiling. Since the body has only one possible tiling, the number of tilings of our UFW symbol depends completely on the number of tilings of the wings. Since the wings are exactly the same shape and size, the number of tilings of the UFW symbol will be the square of the number of tilings of one wing. According to Lemma 14, a wing of order k = 2n + 1 of an odd-order n UFW symbol has S k tilings, where S k = S n 1 2 is the kth large Schröder number. Thus the total number of tiling of our UFW sub-shape is ( ) 2. S n 1 2 S n 1 2 = S n References Code: The code we used to count the number of tilings was written by Greg Kuperberg, Jim Propp, and David Wilson and adapted for Sage use by Frederic Chapoton. We used the On-line Encyclopedia of Integer Sequences to find the large Schröder numbers: Federico Ardila, Algebraic and geometric methods in enumerative combinatorics, Electronic. Retrieved: March 1, Noam Elkies, and Greg Kuperberg, Michael Larsen, and James Propp, Alternating-sign matrices and domino tilings. I, J. Algebraic Combin. 1 (1992), no. 2, Date Retrieved: May 5, Sen-Peng Eu and Tung-Shan Fu, A simple proof of the aztec diamond theorem, Electronic. J. Combin. Retrieved: May 5, Ardila, F., and Stanley, R. P. (2005, January 25). Tilings. Retrieved April 08, 2017, from