Proof of a refinement of Blum’s conjecture on hexagonal dungeons

# Proof of a refinement of Blum’s conjecture on hexagonal dungeons

Tri Lai111This research was supported in part by the Institute for Mathematics and its Applications with funds provided by the National Science Foundation.
Institute for Mathematics and its Applications
University of Minnesota
Minneapolis, MN 55455
tmlai@ima.umn.edu
Mathematics Subject Classifications: 05A15, 05C70
###### Abstract

Matt Blum conjectured that the number of tilings of a hexagonal dungeon with side-lengths (for ) equals . Ciucu and the author of the present paper proved the conjecture by using Kuo’s graphical condensation method. In this paper, we investigate a 3-parameter refinement of the conjecture and its application to enumeration of tilings of several new types of the hexagonal dungeons.

Keywords: perfect matching, tiling, Aztec dungeon, hexagonal dungeon, graphical condensation.

## 1 Introduction

A lattice partitions the plane into fundamental regions. A (lattice) region considered in this paper is a finite connected union of fundamental regions. We call the union of any two fundamental regions sharing an edge a tile. We would like to know how many different ways to cover a region by tiles so that there are no gaps or overlaps; and such coverings are called tilings. We use the notation for the number of tilings of a region , and for the set of all tilings of .

Consider the lattice obtained from the triangular lattice by drawing in all attitudes of each unit triangle. The resulting lattice is usually called -lattice, since it is the lattice corresponding to the affine Coxeter group . On the -lattice, Blum investigated a variation of Aztec dungeon (see [1]) called hexagonal dungeon. In particular, we draw a hexagonal contour of side-lengths222The unit here is the side-length of the unit triangles. (in cyclic order, start by the west side) as the light bold contour in Figure 1.1. We draw next a jagged boundary running along the hexagonal contour (see the dark bold closed path in Figure 1.1), and denote by the region restricted by the boundary. Blum found a striking pattern of the numbers of tilings of the hexagonal dungeons, which led him to his well-known conjecture that the hexagonal dungeon has tilings, when . Fourteen years latter, Ciucu and the author proved the conjecture in [2] by using Kuo’s graphical condensation method [4]. However, the proof did not explain the (surprising) appearance of the numbers 13 and 14 in the Blum’s formula. In this paper, we consider generating function of the tilings of the hexagonal dungeons and give an explanation for the appearance of the numbers and .

The tiles in a hexagonal dungeon have three possible shapes: an obtuse triangle, an equilateral triangle, and a kite (see Figure 1.2). We consider the following generating functions

 F(x,y,z)=∑T∈M(HDa,2a,b)xmynzl, (1.1)

where , , are respectively the numbers of obtuse triangle tiles, equilateral triangle tiles, and kite tiles in the tiling of . We call the tiling generating function of the hexagonal dungeon. Our goal is to prove the following refinement of Blum’s conjecture.

###### Theorem 1.1 (Weighted Hexagonal Dungeon Theorem).

Assume and are two positive integers so that . Then the tiling generating function of the hexagonal dungeon is given by

 F(x,y,z)= γ(a)x3ab−2a2+3⌊a22⌋y6ab−a2+3a−b−2⌊a22⌋z9ab+3a2−7⌊a22⌋−2b−12a ×(x6+3x4y2+3x2y4+y6+2x3z3+2xy2z3+z6)2a2 ×((x4+2x2y2+y4+x3z+xy2z−y2z2+xz3+z4)(x2+y2−xz+z2))⌊a22⌋, (1.2)

where (i.e. is if is even, and if is odd).

This paper is organized as follows. In Section 2, we recall the definition of two important regions and , which were first introduced in [2]. In addition, we state a result on their tiling generating functions (see Theorem 2.1), which is the key to prove Theorem 1.1. Next, we prove Theorem 2.1 in Section 3, and use this to prove Theorem 1.1 in Section 4. Section 5 uses Theorem 1.1 to enumerate of tilings of several new types of hexagonal dungeons. Finally, Section 6 is devoted for an open problem on hexagonal dungeons with defects.

## 2 The regions Da,b,c and Ea,b,c and their tilings generating functions.

First, we recall briefly the definition of the two regions and , which were first introduced in [2].

Let be six non-negative integers. Starting from a vertex of some unit triangle on -lattice, we travel along lattice lines units southwest, units southeast, units north, units northeast, and units northwest. We adjust so that the ending point is on the same vertical line as the starting point. Finally, we close the contour by go units north or south.

The contour is illustrated in Figure 2.1, for the case ; and in Figure 2.2, for the case . If some side of the contour has length , we assume that the side shrinks to a point. Denote by the resulting contour.

Based on the contour , we define two lattice regions and determined by the dark jagged boundaries as in Figure 2.1, for the case , and by Figure 2.2, for the case .

We already showed in [2] that the closure of the contour , the choice of , and the existence of tilings in the regions and require , , and . Moreover, it has been shown in [2] that the numbers of tilings of and is given by powers of and (see Theorem 3.1).

Next, we consider the tiling generating functions of and as follows. Define

 Φ(a,b,c):=Φ(a,b,c)(x,y):=∑T∈M(Da,b,c)xmyn, (2.1)

where and are respectively the numbers of obtuse triangle tiles and equilateral triangle tiles in tiling of . Similarly, we set

 Ψ(a,b,c):=Ψ(a,b,c)(x,y):=∑T′∈M(Ea,b,c)xkyl, (2.2)

where and are the numbers of obtuse triangle tiles and equilateral triangle tiles in tiling of .

Define three new functions as follows:

 g(a,b,c):=(b−a)(b−c)+⌊(a−c)23⌋, (2.3)
 q(a,b,c):=⌊(a−b+c)24⌋, (2.4)

and

 τ(a,b,c):={4a2+8b2+4c2−10ab−10bc+6ac−2b+cif 2a+c−2b>0;4a2+8b2+4c2−10ab−10bc+6ac+6a−8b+4cotherwise. (2.5)

Denote by and the two polynomials appearing in Theorem 1.1, i.e.

 P(x,y,z):=x6+3x4y2+3x2y4+y6+2x3z3+2xy2z3+z6 (2.6)

and

 Q(x,y,z):=(x4+2x2y2+y4+x3z+xy2z−y2z2+xz3+z4)(x2+y2−xz+z2). (2.7)

In addition, we define two simple functions and as

 h(a,b,c)=⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩(x4+2x2y2+y4+x)x/y if 3b+a−c≡5(mod6);(x3+xy2+1)x/y if 3b+a−c≡1(mod6);(x2+y2) if 3b+a−c≡4(mod6);y if 3b+a−c≡3(mod6);1 otherwise, (2.8)

and

 p(a,b,c)=⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩(x4+2x2y2+y4+x)x/y if 3b+a−c≡1(mod6);(x3+xy2+1)x/y if 3b+a−c≡5(mod6);(x2+y2) if 3b+a−c≡2(mod6);y if 3b+a−c≡3(mod6);1 otherwise. (2.9)

The generating functions and are given by the theorem stated below.

###### Theorem 2.1.

Assume that , and are three non-negative integers satisfying , and . Then

 Φ(a,b,c)=h(a,b,c)x2g(a,b,c)+3q(a,b,c)yτ(a,b,c)P(x,y,1)g(a,b,c)Q(x,y,1)q(a,b,c) (2.10)

and

 Ψ(a,b,c)=p(a,b,c)x2g(a,b,c)+3q(a,b,c)yτ(a,b,c)P(x,y,1)g(a,b,c)Q(x,y,1)q(a,b,c). (2.11)

Theorem 2.1 is the key in proving our main Theorem 1.1 in Section 4.

We conclude this section by noticing that Ciucu proved implicitly the formulas of and (under the form of tiling generating functions of the Aztec dungeons in Theorem 3.1 and Proposition 3.6 of [1]). In other word, our Theorem 2.1 generalizes Ciucu’s Theorem 3.1 and Proposition 3.6 in [1].

## 3 Proof of Theorem 2.1

Before proving Theorem 2.1, we need several definitions and terminology as follows.

A perfect matching of a graph is a collection of disjoint edges covering all vertices of . The dual graph of is the graph whose vertices are fundamental regions in and whose edges connect precisely two fundamental regions sharing an edge. The tilings of a region can be identified with the perfect matchings of it dual graph. In the view of this, we use the notation for the number of perfect matchings of .

In the weighted case, defines the sum of weights of perfect matchings of , where the weight of a perfect matching is the product of weights of all constituent edges. Define similarly the weighted sum of tilings of a weighted region . Each edge of the dual graph of has the same weight as that of its corresponding tile in .

Consider the following recurrences (R1)–(R5), where the notations and have been used to indicate some polynomials in .

 ★(a,b,c)★(a−3,b−3,c−2)=xy2★(a−2,b−1,c)★(a−1,b−2,c−2)+x2y4★(a−1,b−1,c−1)★(a−2,b−2,c−1). (R1)
 ★(a,b,c)★(a−2,b−2,c)=x2y4★(a−1,b−1,c)2+★(a,b,c+1)★(a−2,b−2,c−1). (R2)
 ★(a,b,0)★(a−2,b−2,0)=x2y4★(a−1,b−1,0)2+★(a,b,1)★(3b−2a,2b−a,1). (R3)
 ★(a,b,c)★(a−2,b−3,c−2)=xy2★(a−1,b−1,c)★(a−1,b−2,c−2)+xy2★(a−2,b−2,c−1)★(a,b−1,c−1). (R4)
 ⎧⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪⎩★(a,b,c)★(a−2,b−3,c−2)=xy2◊(c,b−1,a−1)★(a−1,b−2,c−2)+xy2★(a−2,b−2,c−1)★(a,b−1,c−1);\par◊(a,b,c)◊(a−2,b−3,c−2)=xy2★(c,b−1,a−1)◊(a−1,b−2,c−2)+xy2◊(a−2,b−2,c−1)◊(a,b−1,c−1). (R5)

We notice that the recurrence (R1) implies the recurrence in Lemmas 4.1 and 5.1 in [2] by specializing . Similarly, (R2) and (R3) are weighted versions of the recurrence in Lemmas 4.2(a) and 5.2(a) and the recurrence in Lemmas 4.2(b) and 5.2(b) in [2], respectively. Finally, the specializations of (R4) and (R5) give the recurrences in Lemmas 4.3 and 5.3 of [2].

Next, we show that and satisfy the above recurrences (R1)–(R5) (with certain constraints) by using the following Kuo’s Condensation Theorem.

###### Theorem 3.1 (Kuo’s Condensation Theorem [4]).

Assume that is a planar bipartite graph, and that and are its vertex classes with . Let be four vertices appearing in a cyclic order on a face of . Assume in addition that and . Then

 M(G)M(G−{u,v,w,t})=M(G−{u,v})M(G−{w,t})+M(G−{u,t})M(G−{w,v}). (3.1)
###### Lemma 3.2.

Let , and be non-negative integers so that , , , and . Assume in addition that . Then and both satisfy the recurrence (R1), i.e. we have

 Φ(a,b,c)Φ(a−3,b−3,c−2)=xy2Φ(a−2,b−1,c)Φ(a−1,b−2,c−2)+x2y4Φ(a−1,b−1,c−1)Φ(a−2,b−2,c−1) (3.2)

and

 Ψ(a,b,c)Ψ(a−3,b−3,c−2)=xy2Ψ(a−2,b−1,c)Ψ(a−1,b−2,c−2)+x2y4Ψ(a−1,b−1,c−1)Ψ(a−2,b−2,c−1). (3.3)
###### Proof.

We assume that each obtuse triangle tile of and is weighted by , each equilateral triangle tile is weighted by , and kite tiles have weight 1. To specify the weight assignment, we use the notations and for the corresponding weighted versions of and . Thus, the functions and are exactly and , respectively.

Apply Kuo’s Condensation Theorem 3.1 to the dual graph of the weighted region with the four vertices chosen as in Figure 3.1(a). More precisely, the black triangles on the west and south corners of region correspond respectively to the vertices and ; the shaded triangles on the west and north corners correspond to the vertices and .

Consider the region corresponding to the graph . The region has some tiles, which are forced to be in any tilings. By removing these edges, we get region (see the region restricted by the bold contour in Figure 3.1(b)) and obtain

 M(G−{u,v})=W⋅M(Da−2,b−1,c(x,y))=W⋅Φ(a−2,b−1,c),

where is the product of weights of all forced tiles. Thus, by collecting the weights of the forced tiles, we get

 M(G−{u,v})=xa+f−3y2a+2f−7Φ(a−2,b−1,c), (3.4)

where as usual.

Similarly, we get four more identities, which are respectively illustrated in Figures 3.1(c)–(f):

 M(G−{v,w})=xc+f−2y2c+2f−4Φ(a−1,b−1,c−1), (3.5)
 M(G−{w,t})=xb+c−3y2b+2c−7Φ(a−1,b−2,c−2), (3.6)
 M(G−{t,u})=xa+b−3y2a+2b−8Φ(a−2,b−2,c−1), (3.7)

and

 M(G−{u,v,w,t})=xa+b+c+f−7y2a+2b+2c+2f−16Φ(a−3,b−3,c−2). (3.8)

Substituting the above five equalities (3.4)–(3.8) into the identity (3.1) in Theorem 3.1, we obtain (3.2).

Similarly, we get (3.3) by applying Theorem 3.1 to the dual graph of the weighted region with the four vertices chosen as in Figure 3.2. ∎

One readily sees that Lemma 3.2 implies Lemma 4.1 in [2] by specializing .

Applying also the Kuo’s Theorem 3.1 to the dual graphs of and with the four vertices chosen as in Figure 3.3, we get the following weighted version of Lemma 4.2 in [2].

###### Lemma 3.3.

Let , and be non-negative integers satisfying , , , and .

a. If , then and both satisfy the recurrence (R2).

b. If , then and both satisfy also the recurrence (R3).

A similar generalization of Lemma 4.3 in [2] is obtained by applying Kuo condensation to the dual graphs of and with the four vertices selected as in Figure 3.4.

###### Lemma 3.4.

Assume that are three non-negative integers satisfying , , , , and . Assume in addition that .

a. If , then and both satisfy (R4).

b. If , then the pair of functions satisfies the double-recurrence (R5).

We now denote by and the expressions on the right hand sides of the equalities (2.10) and (2.11) in Theorem 2.1, respectively. It is routine to verify that the two functions and satisfy the same recurrences (R1)–(R5) for any (If one wants a detailed verification, we recommend the proofs of Lemmas 5.1–5.3 in [2]). This yields an inductive proof of Theorem 2.1 on the perimeter of the contour .

The base cases are the situations when at least one of the following hold:

1. Perimeter of the contour if at most 14.

2. .

3. .

Similar to the proof of Theorem 3.1 in [2]. These base cases can be verified by the help of the computer package vaxmacs written by David Wilson333This software is available at http://dbwilson.com/vaxmacs/, together with Maple by Maplesoft.

For induction step, we assume that the theorem is true for any regions and with perimeter greater than or equal to (it is easy to check that the perimeter is always even). We consider several cases ans show that the number of tilings of the regions and the expressions on the right hand sides of (2.10) and (2.11) satisfy the same recurrence: (R1), (R2), or (R5). However, all arguments here are essentially the same as that in the induction step of the proof of Theorem 3.1 in [2] (with the recurrences are replaced by their corresponding weighted versions). Therefore, we can finish our proof of Theorem 2.1 here.

## 4 Proof of Theorem 1.1

Before presenting the proof of Theorem 1.1, we quote the following lemma.

###### Lemma 4.1 (Graph-Splitting Lemma; Lemma 3.6(a) in [5]).

Let be a bipartite graph, and let and be the two vertex classes.

Assume that an induced subgraph of satisfies following two conditions:

1. There are no edges of connecting a vertex in and a vertex in .

2. .

Then

 M(G)=M(H)M(G−H). (4.1)
###### Proof of Theorem 1.1.

To specify the weight assignment on tiles of the hexagonal dungeon , we use the notation for its weighted version with obtuse triangle tiles, equilateral triangle tiles, and kite tiles are weighted by , respectively. It means that is now . We now divide the weight of each tile in the region by . We get the new weighted region and obtain

 M(HDa,2a,b(x,y,z))=26(3ab+2a2)−3(3a+b)M(HDa,2a,b(xz,yz,1)), (4.2)

where is the total number of tiles in the region . Thus, the theorem can be reduced to the case when .

We now prove the theorem in the case . Apply the two zigzag cuts to the dual graph of the weighted region as in Figure 4.1 to divide it into three disjoint subgraphs , , and (in order from left to right). These subgraphs satisfy the conditions in Graph-splitting Lemma 4.1, so we obtained

 M(HDa,2a,b(x,y,1))=M(G1)M(G2)M(G3). (4.3)

The graph and are both isomorphic to the dual graph of the region , where each obtuse triangle tile is weighted by , and each equilateral tile is weighted by (see the regions with bold boundaries in Figure 4.2). Thus, . Moreover, the graph corresponds to the middle region in Figure 4.2, which has only one tiling (all tiles are forced with the pattern as in Figure 4.2). By counting the number of each type of tiles in the unique tiling of the middle region in Figure 4.2, one obtains that the weight of the perfect matching in is . By (4.3), we get

 M(HDa,2a,b(x,y,1))=x3a(b−2a)y3a(b−2a+1)+(3a−1)(b−2a)Φ(2a,3a,2a)2. (4.4)

By definition, we obtain , , and if is even, and if is odd (we are assuming ). Thus, we get

 M(HDa,2a,b(x,y,1))=γ|z=1(a)x3ab−2a2+3⌊a22⌋y6ab−a2+3a−b−2⌊a22⌋P(x,y,1)2a2Q(x,y,1)⌊a22⌋, (4.5)

where is the restriction of at . This implies the theorem for , and (4.2) in turn deduces the theorem in the general case. ∎

## 5 New hexagonal dungeons

In this section, we investigate several new hexagonal dungeons.

The -lattice can be partitioned into rhombi with two diagonals drawn in (see the top picture in Figure 5.1). Apply the replacing rule (a) in Figure 5.1 to all the rhombi, and denote by the resulting lattice. We have a variant of the hexagonal dungeon as in Figure 5.2 (see the region restricted by the dark bold jagged contour). Denote by the new hexagonal dungeon.

###### Corollary 5.1.

Assume that and are two positive integers, so that . Then

 M(HD(1)a,2a,b)=γ(1)(a)29a2−6a−3⌊a22⌋33a2+2⌊a22⌋+6ab+3a−b+γ172a2, (5.1)

where is 1 if is even, and if is odd.

Next, we quote a well-known subgraph replacement trick called urban renewal, which was first discovered by Kuperberg.

###### Lemma 5.2 (Urban renewal).

Let be a weighted graph. Assume that has a subgraph as one of the graphs on the left column in Figure 5.3, where only white vertices can have neighbors outside , and where all edges have weight . Let be the weighted graph obtained from by replacing by its corresponding graph on the right as in Figure 5.3, where all dotted edges have weight . Then we always have .

We have a small observation as follows. If there are several parallel edges connecting the same vertices and in a graph , then the number of perfect matchings of doesn’t change if the we can replace these parallel edges by a new single edge connecting and with the weight equal to the sum of weights of the original edges (see Figure 5.4).

###### Proof of Corollary 5.1.

Consider the dual graph of (see the upper picture in Figure 5.5). Apply suitable replacements in Lemma 5.2 around all shaded rectangles in . These replacements create several pairs of parallel edges (consisting of an edge of weight and an edge of weight ) in the resulting graph (see the lower picture in Figure 5.5). Next, we replace each pair of parallel edges by a new single edge of weight . This way the dual graph of is transformed into the dual graph of the weighted hexagonal dungeon . Since there are shaded rectangles in , we have

 M(HD(1)a,2a,b)=M(G)=2CM(HDa,2a,b(1/2,3/2,1)). (5.2)

Then the corollary follows from Theorem 1.1. ∎

Next, we consider the application of the replacement rule (b) in Figure 5.1 to all rhombi of -lattice, and denote by the resulting lattice. On the -lattice, we have a new version of the hexagonal dungeons illustrated by the region restricted by the dark bold contour in Figure 5.6.

###### Corollary 5.3.

Assume that and are two positive integers, so that . Then

 M(HD(2)a,2a,b)=γ(1)(a)29a2−6a−⌊a22⌋39ab−3a2+⌊a22⌋+3a−b52a2134a2+⌊a22⌋109⌊a22⌋, (5.3)

where is defined as in Corollary 5.1.

###### Proof.

Similar to the previous corollary, the dual graph of can be transformed into the dual graph of by applying urban renewal around all shaded squares ( is defined as in Corollary 5.1), and replacing each pair of parallel edges by a single edge with weight (see Figure 5.7). We have

 M(HD(2)a,2a,b)=2CM(HDa,2a,b(3/2,3/2,1)), (5.4)

and the corollary follows again from Theorem 1.1. ∎

Finally, application of the replacement (c) in Figure 5.1 gives us a new lattice as in Figure 5.8. On the new lattice, we get the variant of region as the region restricted by dark bold contour in Figure 5.8. The tilings of the above new regions are enumerated by powers of , , , and as follows.

###### Corollary 5.4.

Assume that and are two positive integers, so that . Then

 M(HD(3)a,2a,b)=γ(3)(a)29a2−6a−2⌊a22⌋33ab−2a2+5⌊a22⌋5⌊a22⌋1932a2, (5.5)

where is 1 if is even, and if is odd.

###### Proof.

This corollary can be obtained similarly to Corollaries 5.1 and 5.3. We apply urban renewal at all shaded rectangles, where is defined as in Corollary 5.1 (see Figure 5.9). Then we replace each pair of parallel edges by a single edge of weight . This way, we transform the dual graph of into the dual graph of . Again, Theorem 1.1 deduces (5.5). ∎

## 6 An open question on hexagonal dungeons with defects.

We conclude this paper by considering a hexagonal dungeon where some fundamental regions have been removed as follows. Similar to the original hexagonal dungeon, we consider a similar region that is restricted by a jagged boundary running along the hexagonal contour of side-lengths (in cyclic order, starting from the west side). Next, we remove triangles from the region from each of the west and east sides, and denote by the resulting region. Figure 5.10 shows the region ; the black triangles indicate the fundamental regions removed.

We consider the tiling generating function of as

 K(x,y,z):=∑T∈M(HHa,b)xmynzl, (6.1)

where , , are respectively the numbers of obtuse triangle tiles, equilateral tiles, and kite tiles in the tiling as usual. It seems that is given by a simple product similar to in Theorem 1.1.

###### Conjecture 6.1.

Assume that and are two positive integers, so that . Then the generating function always has form

 xXyYzZP(x,y,z)L,

where depend only on and .

Since several triangles have been removed along the boundary of the region, our method in the paper seems does not work for .

Finally, we notice that ones can create new versions of the region on the lattices , and as we did for the hexagon dungeons in the previous section.

## References

• [1] M. Ciucu, Perfect matchings and perfect powers, J. Algebraic Combin. 17 (2003), 335–375.
• [2] M. Ciucu and T. Lai, Proof of Blum’s Conjecture on Hexagonal Dungeons, J. Combin. Theory Ser. A 125 (2014), 273–305.
• [3] N. Elkies, G. Kuperberg, M. Larsen, and J. Propp, Alternating-sign matrices and domino tilings (Part I), J. Algebraic Combin. 1 (1992), 111–132.
• [4] E. H. Kuo, Applications of Graphical Condensation for Enumerating Matchings and Tilings, Theor. Comput. Sci. 319 (2004), 29–57.
• [5] T. Lai, Enumeration of hybrid domino-lozenge tilings, J. Combin. Theory Ser. A 122, 2014, 53–81.
• [6] J. Propp, Enumeration of matchings: Problems and progress, New Perspectives in Geometric Combinatorics, Cambridge Univ. Press, 1999, 255–291.
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