Baxter Permutations, Snow Leopard Permutations, and Domino Tilings of Aztec Diamonds

Ben Caﬀrey, Gregory Michel, Kailee Rubin, & Jonathan Ver Steegh∗

March 3, 2014

Abstract In [2], Elkies, Kuperberg, Larsen, and Propp introduce a bijection between domino tilings of Aztec diamonds and pairs of alternating sign matrices whose sizes differ by one. In this paper we first study those smaller permutations (viewed as matrices) which are paired with doubly alternating Baxter permutations. We call these permutations snow leopard permutations, and we use a recursive decomposition to show they are counted by the Catalan numbers. We also describe a specific set of transpositions which generates these permutations. Lastly, we investigate the general case, where we conjecture that the number of larger permutations paired with a given smaller permutation is a product of Fibonacci numbers.

1 Introduction and Background

An alternating sign matrix (ASM) is an n × n matrix with entries in {0, 1, −1} whose nonzero entries in each row and column alternate in sign and sum to 1. An Aztec diamond of order n is a two dimensional array of unit squares with 2i squares in rows i ≤ n and 2(2n − i + 1) squares in rows n < i ≤ 2n, in which the squares are centered on each row. Figure1 shows the Aztec diamond of order 3.

Figure 1: The Aztec diamond of order 3.

Aztec diamonds can be tiled using 2×1 domino rectangles, which is to say they can be completely covered by disjoint dominoes whose union is the entire diamond. We abbreviate a tiling of an Aztec

∗Current address: c/o Eric Egge, Department of Mathematics, Carleton College, Northﬁeld, MN 55057; Email: [email protected]

1 diamond as a TOAD. In [2], Elkies, Kuperberg, Larsen, and Propp describe how to construct, for each TOAD of order n, a pair of ASMs of size n × n and (n + 1) × (n + 1). We can describe this construction most easily with an example. In Figure2, the vertices that compose the tiled Aztec diamond fall naturally into two matrices: the red vertices form an (n + 1) × (n + 1) matrix while the blue vertices form an n × n matrix. The tops of these matrices align with the northeast edge of the diamond. To construct an ASM on the red vertices, label each vertex of degree 4 with a 1, each vertex with degree 3 with a 0, and each vertex with degree 2 with a −1. The n × n matrix on the blue vertices can be constructed in the same way except the degree 4 and degree 2 rules are reversed. Given a TOAD T , we write LASM(T ) to denote the larger of the ASMs and SASM(T ) to denote the smaller of the two ASMs. Following [2], we say an (n + 1) × (n + 1) ASM A and an n × n ASM B are compatible whenever there exists a proper tiling T such that A = LASM(T ) and B = SASM(T ).

Figure 2: A domino tiling of the Aztec diamond of order 3

According to the rules deﬁned above, the TOAD in Figure2 has 0 0 0 1 0 0 1 1 0 0 0 LASM(T ) =   ; SASM(T ) = 1 0 0 . 0 0 1 0     0 1 0 0 1 0 0 In one-line notation, these matrices refer to the permutations 4132 and 312, respectively. In [2], Elkies, Kuperberg, Larsen, and Propp show that an order n+1 ASM with k entries equal to −1 is compatible with 2k ASMs of size n. As a corollary, each permutation matrix, which is an ASM with no −1 entries, is compatible with exactly one smaller ASM. Canary [1] gives an algorithm to construct the unique smaller ASM compatible with any given permutation. To implement this algorithm, suppose π is a permutation of length n + 1. Label the red vertices in a diagram for an Aztec diamond of the appropriate size with the entries of the matrix for π. For each blue vertex, if the two red vertices immediately to the northwest, and all red vertices northwest of those, are labeled with 0, then label the blue vertex 0. Now repeat this process in each of the other three directions (northeast, southeast, and southwest). Canary shows that each row and column of blue vertices will now contain an odd number of unlabeled vertices; there is a unique way to label these vertices with 1s and −1s to create an ASM.

2 Canary proves that this smaller ASM will also be a permutation matrix if and only if the original permutation matrix is Baxter. A Baxter permutation is a permutation that avoids 2 − 41 − 3 and avoids 3 − 14 − 2, which means that, in one-line notation, it has no subsequence ai, aj, aj+1, ak such that aj+1 < ai < ak < aj (for 3 − 14 − 2) or aj < ak < ai < aj+1 (for 2 − 41 − 3). For example, 174962835 is not Baxter because the subsequence 4−62−5 is an instance of 2−41−3. In contrast, 879164325 is Baxter because it contains no instances of 2 − 41 − 3 or 3 − 14 − 2. An interesting class of Baxter permutations are the doubly alternating Baxter permutations. We call a permutation π = a1a2 ··· an alternating whenever ai < ai+1 if i is odd and ai > ai+1 if i is even. This is to say that π starts with an ascent and the ascents and descents alternate. A doubly alternating permutation is an alternating permutation whose inverse is also alternating. We call permutations that are both doubly alternating and Baxter doubly alternating Baxter permutations th (DABPs). Guibert and Linusson show in [3] that Cn, the n Catalan number, counts both the DABPs of length 2n and the DABPs of length 2n + 1. The Catalan numbers are ubiquitous in 1 2n combinatorics and have a number of deﬁnitions. The simplest of these is the formula Cn = n+1 n , but they can also be thought of as the number of Catalan paths, which are paths from (0, 0) to (n, n) along an n × n square grid using north and east steps and not crossing below the main diagonal. Pn The Catalan numbers also satisfy the recursive deﬁnition C0 = 1 and Cn+1 = j=1 Cj−1Cn−j for n ≥ 0. In this paper, we introduce snow leopard permutations (SLPs), which are the permutations whose matrices are compatible with doubly alternating Baxter permutation matrices. In Section 2, we characterize these permutations recursively, and by doing so, we show that the compatibility relation in this case is one-to-one, which implies that the snow leopard permutations of length 2n are also counted by Cn, as are the snow leopard permutations of length 2n + 1. Then, we give a bijection between snow leopard permutations and Catalan paths in Section 3, and in Section 4 we describe a construction of snow leopard permutations based on transpositions. Lastly, in Section 5 we show that the identity permutation of length n is compatible with Fn permutations of length th n + 1, where Fn denotes the n Fibonacci number. We conclude by conjecturing that the number of length n + 1 permutations with which a length n permutation is compatible is a product of Fibonacci numbers.

2 Snow Leopard Permutations: a Recursive Characterization

In order to construct snow leopard permutations recursively, it is important to ﬁrst understand how to construct DABPs and TOADs recursively. To understand this process, we ﬁrst describe how to decompose DABPs and TOADs into smaller DABPs and TOADs, respectively.

2.1 Definitions and Notation Before we begin, we introduce some key definitions and specific notation for working with permutations in one-line notation.

C Deﬁnition 2.1. The complement of a permutation π = a1 ··· an is π = n − a1 + 1 n − a2 + 1 ··· n − an + 1. Lemma 2.2. The complement of a Baxter permutation is Baxter.

3 Proof. Assume π is a Baxter permutation. Since the patterns 2 − 41 − 3 and 3 − 14 − 2 are complements of each other, πC will avoid 2 − 41 − 3 since π avoids 3 − 14 − 2 and vice versa.

Deﬁnition 2.3. For any two sequences of positive integers, π1 = x1 ··· xi and π2 = y1 ··· yj, we write π1 ⊕ π2 to denote the concatenated sequence x1 ··· xiy1 ··· yj.

Definition 2.4. Given a sequence of positive integers π = x1 ··· xi, and two indices m and n with 1 ≤ m ≤ n ≤ i, we define πm→n to be the subsequence of consecutive entries given by xmxm+1 ··· xn. Definition 2.5. Given a sequence of positive integers π and a constant b, we define AUGC(π, b) to be the sequence of positive integers created by taking the complement of π and then increasing the value of each entry by b.

Example 2.6. Let π1 = 14325 and π2 = 768. We have π1 ⊕ π2 = 14325768, π12→4 = 432, and AUGC(π1, 5) = 10 7 8 9 6.

2.2 Doubly Alternating Baxter Permutation Decomposition In this subsection, we describe a decomposition algorithm that subdivides a single DABP into two smaller DABPs. In order to build to this algorithm, we must ﬁrst prove several facts about the structure of DABPs. We begin a Lemma of Ouchterlony from [4].

Lemma 2.7. Suppose π is a DABP of length n. Then the following hold.

1. If n is odd then π(1) = 1.

2. If n is even and π(1) = 1 then π(n) = n.

3. If n is even, π(1) > 1, and k is the smallest integer for which π(k) < π(1), then i < k ≤ j implies π(i) < π(j).

This lemma leads to a number of interesting results, some of which are restatements of certain ideas of [4].

Proposition 2.8. If π is a doubly alternating Baxter permutation of odd length, then AUGC(π2→n, −1) is a doubly alternating permutation of even length.

Proof. Assume that π is a DABP of odd length. Then π(1) = 1 by Lemma 2.7a, so if we take 0 0 π = π2→n and decrease each entry by 1, we get a permutation. Since π is a consecutive sub- permutation of π which was Baxter, π0 will still avoid 3 − 12 − 4 and 2 − 41 − 3. Also, π0C will be Baxter by Lemma 2.2. Moreover, since the indices are offset by one from the original π, the new permutation π0 starts with a descent, and then has alternating descents and ascents. Since the complement operation turns ascents into descents and vice versa, π0C will be alternating. Moreover, since π(1) = 1, we know π−1(1) = 1 as well. This means that π0−1 also starts with a descent and then has alternating ascents and descents. The sequence of ascents and descents in the inverse of the complement will come from reversing the order of the initial sequence and swapping ascents and descents. (This can be most easily verified by looking at the geometry of the permutation matrices, in which the complement operation corresponds to a vertical reflection and the inverse operation corresponds to a reflection over the line y = −x). Since π0 has even length and starts with a descent,

4 after which the ascents and descents alternate, it will also end with a descent. This means that reversing the order and swapping ascents and descent will cause (π0C )−1 to start with an ascent while still alternating ascents and descents, making it an alternating permutation. Therefore, π0C 0C is a DABP. Since π =AUGC(π2→n, −1) by construction, we know that AUGC(π2→n, −1) is a DABP.

This result tells us that odd length DABP are equivalent to complements of DABPs that are one size smaller. As a result, we can understand DABPs of odd length by studying DABPs of even length. Lemma 2.9. If π is a DABP of even length, and we deﬁne the pair (x, y) so that y = π(x) and x is the smallest value such that π(x) < π(1), then π(i) < π(1) if and only if i ≥ x. That is, if we plot the permutation π as ordered pairs (x, π(x)) on a coordinate grid, then the plot will not cross the horizontal line at y = π(1) more than once. Proof. This is immediate from Lemma 2.7.

Lemma 2.10. If π is a DABP of even length n with π(1) = α, then π1 = π1→n−α+1 is a DABP augmented by π(1) − 1 with π1(1) = 1 and π2 = πn−α+2→n is a DABP. Proof. According to Lemma 2.9, all of the terms in π that are greater than or equal to α come before the terms that are less than α. Thus, the first n−α+1 terms of π must be some permutation of α, α + 1, ··· , n. So, if we subtract α − 1 from each term, we have a permutation π1 of length n − α + 1. Since we know the first value of this was originally α, the resulting permutation has π(1) = 1. Moreover, since π avoids 2 − 41 − 3 and 3 − 14 − 2 and π1 is a subsequence of π, we see that π1 also avoids these patterns, so it is Baxter. In addition, since π1 is a subsequence of π that starts with an odd index, it follows the same ascent/descent patterns as π, so it is alternating. The −1 −1 same holds for π1 , as it is a subsequence of π starting at an odd index (namely, α). Thus, π1 is a DABP. Similarly, all terms less than α in π are in the remaining α − 1 terms of π, so the final α − 1 terms of π are also a permutation of π. As with π1, this permutation π2 is a DABP. It avoids the −1 relevant patterns (since π does) so it must be Baxter. Moreover, π2 and π2 alternate as they are identical to the end of π and the beginning of π−1 respectively, both of which alternate.

Theorem 2.11 (DABP Decomposition). A permutation π is a DABP of even length if and only if, given π(1) = α and π(n − α − 1) = n, we have that π2→n−α is the complement of a DABP augmented by α and πn−α+2→n is a DABP.

Proof. Assume that π is a DABP with π(1) = α. By Lemma 2.10, we know that π1 = π1→n−α+1 is a DABP augmented by π(1) − 1 with π(1) = 1. Thus, π1(1) = n by Lemma 2.7. Therefore, π2→n−α is an arbitrary Baxter permutation (with each value increased by one). As in the proof of Proposition 2.8, by removing the ﬁrst entry in a DABP, we lose the doubly alternating property, but we can again restore this, while maintaining the Baxter property, by taking the complement. Thus, π2→n−α is the complement of a DABP augmented by a total of α. In addition, πn−α+2→n is a DABP as a result of Lemma 2.10. Now, suppose we have two even length DABPs π1 and π2 and we let α = |π2| + 1. We know C that π1 is Baxter. The complement operation changes ascents to descents and vice-versa. If C C we increase each value of π1 by α, π1 will still avoid the necessary patterns to be Baxter and its

5 ascent/descent pattern will not change. We have constructed AUGC(π1, α), which is a permutation ∗ of α + 1, ··· , |π1| + |π2| + 1. Now note that π1 = α⊕AUGC(π1, α) ⊕ |π1| + |π2| + 2 will avoid the ∗ necessary patterns to be Baxter but it will also begin and end with an ascent. Now π = π1 ⊕ π2 ∗ will have the correct ascent/descent pattern because both π1 and π2 starts and ends with an ascent and we are adding a descent in the middle by concatenating them. The same holds for π−1. Also, ∗ since both π1 and π2 were Baxter, π will be as well. Thus, π is a DABP of length |π1| + |π2| + 2, which is even.

As a note, this result gives an alternate proof of Guibert and Linusson’s result from [3].

Theorem 2.12. The doubly alternating Baxter permutations of length 2n are counted by Cn.

Proof. Recall that the Catalan numbers satisfy the recursive definition C0 = 1 and Cn+1 = Pn j=1 Cj−1Cn−j for n ≥ 0. We know there is trivially one DABP of length 0 (the empty permutation, which is trivially a DABP). Furthermore, from this decomposition result, we get a DABP of length 2n by merging two specific DABPs of length 2i and 2n − 2i. Since any pair of such DABPs corresponds to a DABP as in Theorem 2.11, the number of doubly alternating Baxter permutations satisfies the same recursive relationship as the Catalan numbers.

2.3 Aztec Diamond Decomposition Suppose D is a TOAD whose LASM is the matrix for a doubly alternating Baxter permutation. Then because LASM(D) contains no −1s, D is completely determined by its DABP. Thus, D inherits an analogous decomposition from its DABP.

Deﬁnition 2.13. We call a tiling of an Aztec diamond D a doubly alternating Aztec diamond (DAAD) if LASM(D) is the matrix for a doubly alternating Baxter permutation.

Deﬁnition 2.14. A block is a pair of two dominoes adjacent along a long edge, forming a 2-by-2 box.

The DAAD shown in Figure3 contains 7 blocks.

Lemma 2.15. The blocks of a DAAD are in bijection with the 1s in its SASM.

Proof. Let D be a DAAD of order n that contains a block B. By Canary’s algorithm, this point may correspond to a 1 in SASM(D) or a −1 in LASM(D). However, because LASM(D) is a permutation, it cannot contain a −1. Thus, a block must correspond to a 1 in SASM(D). Conversely, a 1 in SASM(D) must label a vertex of degree 2, which creates a block in D.

As a note, Lemma 2.15 implies that exactly one block will lie on each diagonal of a DAAD since each row and column of its SASM contain exactly one 1. Now we are ready to introduce a composition algorithm. First, we must deﬁne a process HS which takes a DAAD of order n and returns a DAAD of order n + 2.

Algorithm 2.16 (HS). Let D1 denote a DAAD of order n ≥ 0.

6 Figure 3: The DAAD corresponding to the DABP 37564812 and its compatible SLP 3654721.

Case 1: If n = 0, then return

Case 2: If n > 0, then to construct HS(D1):

1. Reﬂect D1 over the line y = −x.

2. Draw D2, an empty Aztec diamond of order n + 2.

3. Place D1 at the center of D2.

4. Fill the remainder of D2 with vertical dominoes.

Note that HS(D1) will always contain vertical blocks at its N and S edges, which we call a hat and shoe, respectively. The name HS stems from this terminology. In Figure 4d we outline the hat in blue and the shoe in red. Now we are ready to deﬁne a merge function which, given two DAADs of order n1 and n2, respectively, returns a DAAD of order n1 + n2 + k, for 1 ≤ k ≤ 3. Note that when composing one empty diamond with one non-empty diamond, we assume that D1 is empty.

Algorithm 2.17. Given DAADs D1 and D2 of order n1 and n2, respectively, where n1 and n2 are either odd or zero, to ﬁnd MERGEDAAD(D1,D2) we do the following.

7 (a) D1 (b) Step 1

Figure 4: An illustration of the HS process. HS takes the DAAD corresponding to DABP 3412 and SLP 321 to the DAAD corresponding to the DABP 153426 and SLP 14325.

Case 1: If n1 = 0, then return HS(D2).

Case 2: If n1 6= 0 and n2 6= 0, then

1. Draw D3, an empty Aztec diamond of order n1 + n2 + 3.

2. Place D2 inside D3 so their west edges align.

3. Place HS(D1) inside D3 so their east edges align. 4. Fill the remainder of the diamond with horizontal dominoes.

Note that the ﬁnal step of Case 2 is the only possible way to tile the remainder of D3. D2 and HS(D1) contain n2 and n1 + 2 blocks, respectively. As Figures 5d and 5e indicate, the placement of D2 and HS(D1) forces a horizontal block at the east edge of D2, which we call a connector. At

8 this point D3 contains all of its n1 + n2 + 3 blocks. Recall that each block corresponds to a 1 in SASM(D3). Since SASM(D3) is the matrix for a permutation of length n1 + n2 + 3 and the tiling so far has already placed n1 + n2 + 3 1s in SASM(D3), the remainder of the matrix must be all 0s. By Canary’s algorithm this forces us to tile the rest of the diamond with horizontal dominoes.

(b) D2

(a) D1

(d) Steps 2 and 3 (e) Step 4

Figure 5: An illustration of the composition of DAADs. We outline the connector in red.

Now, suppose we want to invert the process.

Algorithm 2.18. Let D be a DAAD of order n ≥ 1. To ﬁnd the unique DAADs D1 and D2 such that MERGEDAAD(D1,D2) = D :

1. Locate the block that lies along the northeast edge of D and call this B1. Recall that a single block lies in each diagonal of a DAAD, so B1 is unique.

9 2. If B1 is not at the east corner of D:

(a) Locate the southernmost block that lies on the same vertical as B1 and call this B2. Then B1 and B2 correspond to the hat and shoe created by the HS process.

(b) Draw the Aztec diamond D1 such that the north edge of D1 borders the south edge of B1 and the south edge of D1 borders the north edge of B2. This process eﬀectively inverts HS(D1) when D1 is not the empty diamond.

3. If B1 is at the east corner of D, then D1 is empty. This process inverts HS(D1) when D1 is the empty diamond.

4. If any blocks inside D lie outside of and to the left of D1 then call the rightmost of these blocks C, the connector. Draw the diamond D2 such that the east edge of D2 borders the west edge of C and the west edge of D2 borders the west edge of D.

5. If no blocks lie outside of D1, then D2 is the empty diamond.

Theorem 2.19 (DAAD Decomposition). An Aztec diamond D is a DAAD of order n if and only if there exist two DAADs, D1 and D2, such that MERGEDAAD(D1,D2) = D.

Proof. We need to show that both MERGEDAAD and its inverse are well-deﬁned maps. First, we show that given two DAADs, D1 and D2, of order n1 and n2, respectively, D = MERGEDAAD(D1,D2) is also a DAAD.

Case 1: Suppose that n1 = n2 = 0. Then MERGEDAAD(D1,D2) = HS(D1) is the Aztec diamond corresponding to the DABP 12, so D is a DAAD.

Case 2: Suppose that n1 6= 0 and n2 = 0. Then D = HS(D1), where D1 is not the empty diamond. Using Canary’s algorithm, we can overlay D1 with its corresponding DABP matrix LASM(D1). Let π denote the permutation encoded by LASM(D1). To create HS(D1), we first reflect D1 over the line y = −x. At the permutation level, this corresponds to taking the complement of π. Secondly, we place the reflected D1 at the center of D3, an Aztec diamond of order n2 + 2 and fill the remainder of D3 with vertical dominoes. At the matrix level, this process corresponds to placing LASM(D1) at the center of LASM(D3), adding a 1 to the beginning of the first row and end of the last row, and filling the remainder with 0s. At the permutation level, we now have 0 C 0 π = 1 ⊕ π ⊕ n2 + 2. Since π is a DABP, it follows from the decomposition of DABPs that π must also be a DABP and thus HS(D1) is a DAAD of order n2 + 2.

Case 3: Suppose that n1 6= 0 and n2 6= 0. Then to merge D1 and D2 we first create an empty diamond D3 of order n1 + n2 + 3, in which we place HS(D1) to the right and D2 to the left. Let πi denote the permutation encoded by LASM(Di) and let D4 = HS(D1). At the matrix level this placement corresponds to placing LASM(D2) at the bottom left corner of LASM(D3) and LASM(D4) at the top right corner of LASM(D3). By shifting LASM(D3) to the right, we augment π3 by n2. When we fill the remainder of D4 with horizontal tiles, we fill the remainder of C LASM(D3) with 0s. From case 2, we know that π4 = 1 ⊕ π1 ⊕ n1 + 2 so at the permutation level, we now have π3 = AUGC(π1, n1) ⊕ π4. Because π1 and π4 are both DABPs, π3 is also a DABP so D is a DAAD.

10 2.4 Snow Leopard Permutation Decomposition The decomposition algorithm described above can also be understood as a decomposition algorithm for the snow leopard permutation that underlies the DAAD. In order to understand this decomposition algorithm’s eﬀect on these permutations, it is more helpful to consider the corresponding MERGE operation. Given two snow leopard permutations π1 and π2 with lengths n1 and n2 respectively, we can use an algorithm analogous to the MERGEDAAD operation to construct a snow leopard permutation with length n1 + n2 + k, 1 ≤ k ≤ 3 in the following way.

Algorithm 2.20. Given two snow leopard permutations π1 and π2 with lengths n1 and n2 respectively, where n1, n2 are either odd or zero, we deﬁne MERGESL(π1, π2) as follows.

1. If n1 = n2 = 0, then MERGESL(π1, π1) = 1.

2. If n1 = 0 and n2 6= 0, then MERGESL(π1, π1) = n2 + 2 ⊕ n2 + 1 ⊕ π2.

3. If n2 = 0 and n1 6= 0, then MERGESL(π1, π1) = 1 ⊕ AUGC(π1, 1) ⊕ n1 + 2.

4. If n1 6= 0 and n2 6= 0, then MERGESL(π1, π1) = n2 + 2 ⊕ AUGC(π1, n2 + 2) ⊕ n1 + n2 + 3 ⊕ n2 + 1 ⊕ π2.

Proposition 2.21. π is a snow leopard permutation of odd length if and only if there exist snow leopard permutations π1, π2 of odd length (or length 0) so that π = MERGESL(π1, π2). Proof. Since the 1 entries in the permutation matrix that describe a given snow leopard permutation are in bijection with the blocks in the corresponding DAAD, the merging and decomposition process of the snow leopard permutations should exactly mirror that of the MERGEDAAD operation from the previous subsection. The reﬂection over the y = −x axis that happens in the DAAD merge is represented by taking the complement of π1, and the values we add that surround π1 in the merged snow leopard permutation represent the two blocks added in the HS algorithm. The next entry added corresponds to the connector block, and then we add in the π2 to complete the permutation just as we added in the second DAAD to complete the tiling. Therefore, these merge operations are equivalent, so the merging on the snow leopard level falls directly from that in the previous section.

As with the MERGEDAAD operation, we deﬁne an algorithm to invert the process. In other words, given a snow leopard permutation π, we can identify the two unique smaller permutations that can be merged to form π.

Algorithm 2.22 (Snow Leopard Decomposition). Given a snow leopard permutation π with length n, we can obtain the two component snow leopard permutations π1 and π2 for which π = MERGESL(π1, π2) as follows.

1. If π(1) = n, then π(2) = n − 1. In this case, π1 is the empty permutation and π2 is equal to π3→n.

2. If π(1) = 1. Now π2 is the empty permutation and π1 = AUGC(π2→n−1, −1).

−1 3. If 1 < π(1) < n, then let x = π (n). Now π1 = AUGC(π2→x−1, −π(1)) and π2 = πx+2→n.

11 There are some immediate results from the decomposition and merge algorithm described above.

Corollary 2.23. The identity permutation In = 123 . . . n and the reverse identity permutation Rn = n n − 1 ... 1 are both snow leopard permutations for all odd n. Proof. We proceed by induction. For n = 1, the permutation of length 1 is compatible with 12, which is a DABP. Thus, 1 = I1 = R1 is snow leopard. For n > 1, assume that In and Rn are snow leopard. Then Rn+2 = MERGESL(∅,Rn) is a snow leopard permutation and In+2 = MERGESL(Rn, ∅) is a snow leopard permutation where ∅ is the empty permutation. Corollary 2.24. If π is a snow leopard permutation of odd length, then π preserves parity, which is to say π(i) is odd if and only if i is odd.

Proof. We proceed by strong induction on n, the length of π. If n = 1, then π = 1, which preserves parity. For n > 1, assume all snow leopard permutations of length k ≤ n preserve parity. Let π be a snow leopard permutation of length n + 2. We can write π = MERGESL(π1, π2) where π1 and π2 are snow leopard permutations. By the inductive hypothesis, π1 and π2 preserve parity. We consider 3 cases.

Case 1: If π1 = ∅, then π = n + 2 ⊕ n + 1 ⊕ π2. For i = 1 and i = 2, π preserves parity since n + 2 is odd and n + 1 is even. Since the indices of π2 are shifted by two, the remaining permutation also preserves parity.

Case 2: If π2 = ∅, then π = 1 ⊕ AUGC(π1, 1) ⊕ n + 2. Both π(1) and π(n + 2) are odd. Moreover, for 1 < i < n + 2,

C π(i) = AUGC(π1, 1)(i − 1) = π1 (i − 1) + 1 = n + 3 − π(i − 1) + 1.

If i is odd, then π(i) = n + 4 − π(i − 1) is an odd integer plus an even integer minus an even integer, so π(i) is odd. On the other hand, if i is even, then π(i) = n + 4 − π(i − 1) is an odd integer plus an even integer minus an odd integer, so π(i) is even. Thus, π preserves parity.

Case 3: If π1 and π2 have non-zero odd lengths n1 and n2 with n1 + n2 + 1 = n, then π = n2 + 2 ⊕ AUGC(π1, n2 + 2) ⊕ n1 + n2 + 3 ⊕ n2 + 1 ⊕ π2. For i = 1, π(i) = n2 + 2, which is odd. For 1 < i ≤ n1 + 1, an identical argument as that made in Case 2 shows that π is parity preserving for these i. For i = n1 + 2, which is odd, π(i) = n1 + n2 + 3 = n + 2, which is odd. For i = n1 + 3, which is even, π(i) = n2 + 1, which is even. Finally, for i > n1 + 3, π(i) = π2(n1 + 3 − i). Since n1 + 3 − i preserves the parity of i, so does applying π2. Thus π preserves the parity of these values of i. Therefore π is parity preserving for all i.

Theorem 2.25. The number of snow leopard permutations of length 2n − 1 is Cn. Proof. As in the proof of Theorem 2.12, recall that the Catalan numbers satisfy the recursive def- Pn inition C0 = 1 and Cn+1 = j=1 Cj−1Cn−j for n ≥ 0. Trivially, we know that the permutation 1 is a snow leopard permutation as it is compatible with 12, which is a DABP. This tells us that the number of snow leopard permutations of length 1 is C1. For completion, we will say that the number of snow leopard permutations of length −1 is also 1, which is C0.

From the snow leopard permutation decompositon, we know that we can merge two non-empty

12 permutations whose sizes add up to 2n − 4 to create a snow leopard permutation of length 2n − 1. This means that a snow leopard permutation of length 2n − 1 comes from an SLP of length 2j − 1 and an SLP of length 2n − 4 − 2j + 1. If we inductively assume that the former are counted by Cj and the latter are counted by Cn−j−1, we get that the number of snow leopard permutations of Pn size 2n − 1 is given by j=0 CjCn−j−1, which is identical to the recursive deﬁnition of the Catlan numbers.

3 Snow Leopard Permutations: a Bijection with Catalan Paths

Thus far, we have showed that the snow leopard permutations are counted by the Catalan numbers through a recursive decomposition. However, we lack a more direct proof. In this section we conﬁrm this result by introducing a non-recursive bijection from snow leopard permutations of length 2n+1 and 2n + 2 to CP2n+2, the set of Catalan paths of length 2n + 2. Algorithm 3.1 (Ascent/Descent Map). Given a snow leopard permutation π of length 2n + 1, we can construct a corresponding Catalan path p of length 2n + 2 by a function we denote κ. We use th pj to denote the j step of p.

Case 1: Suppose π has length 2n + 1 (n ≥ 0). We look at consecutive entries π(i) and π(i + 1) to deﬁne κ : SL2n+1 → CP2n+2 by  π(i) < π(i + 1) and i is odd   N or   π(i) > π(i + 1) and i is even  κ(π)i = for 0 ≤ i ≤ 2n + 1   π(i) < π(i + 1) and i is even  E or   π(i) > π(i + 1) and i is odd

By convention we treat the empty entries π(0) and π(2n + 2) as 2n + 2 and 0, respectively. In particular, for every odd length SL permutation we have κ0 = N and κ2n+1 = E.

Example 3.2. Suppose π = 1432567. Then κ(π)0 = N trivially. Continuing, κ(π)1 = N because π(1) = 1 < π(2) = 4 and 1 is odd. Similarly, κ(π)2 = N because π(2) = 4 > π(3) = 3 and 2 is even. Continuing in this manner, κ takes us to the path in Figure6.

Case 2: Now suppose π has length 2n + 2 (n ≥ 0). Then π = 1 ⊕ AUGC(π0, 1), where π0 is a snow leopard permutation of length 2n + 1. We deﬁne κ(π) = κ(π0).

Example 3.3. Let π = 18567432. Then π = 1 ⊕ AUGC(1432567, 1). Now κ(π) = κ(1432567), which again yields the path in Figure6.

Although it is not immediately obvious that the Ascent/Descent algorithm maps every snow leopard permutation to a Catalan path, we argue by induction that this algorithm is well-deﬁned.

13 4

1 2 3 4

Figure 6: The path returned when κ is applied to the snow leopard permutations 1432567 and 18567432.

Theorem 3.4. Suppose π is a snow leopard permutation of length 2n − 1. Then κ(π) = p is a Catalan path of length 2n.

Proof. If n = 0 then π = 1. Treating π(0) as 2, we get that κ(π)0 = N and κ(π)1 = E, so p is a Catalan path of length 2. We continue by strong induction on n. Assuming κ(π) is a Catalan path of length 2n for all n ≤ i, we will show that κ(π) is a Catalan path for n = i + 1. Let π be a snow leopard permutation of length 2(i + 1) − 1 = 2i + 1. We can decompose π = MERGESL(π1, π2), where π1 and π2 are snow leopard permutations of length n1 and n2, respectively, and n1 and n2 are both odd. This gives us three cases.

Case 1: If n1 6= 0 and n2 6= 0, then π = n2 + 2 ⊕ AUGC(π1, n2 + 2) ⊕ 2i + 1 ⊕ n2 + 1 ⊕ π2. By convention, κ(π)0 = N. Every entry in AUGC(π1, n2 + 2) is larger than n2 + 2, so π(1) < π(2), C which gives us that κ(π)1 = N. Now π2→n1+1 = AUGC(π1, n2 + 2). However, augmenting π1 does C C not affect the ascents and descents of π1 , so κ(AUGC(π1, n2 + 2)) = κ(π1 ). Since π1 is a snow leopard permutation with length n1 < 2i + 1, κ(π1) = p1 is a Catalan path of length n1 + 1 by our inductive hypothesis. Taking the complement of π1 turns all ascents into descents and all descents into ascents, which consequently replaces all N steps in p1 with E steps and vice versa. However, C C π1 starts at index 2 of π instead of index 1. Thus, when insering π1 into π we increase each index C of π1 by 1, which changes the parity of each index. Effectively, we again replace each N step in C 0 0 C κ(π1 ) with an E step and vice versa, creating a Catalan path p1. Note that p1 6= p1. Because π1 does not fall at the beginning or end of the decomposition of π, κ no longer applies the two trivial 0 steps to π1. Thus, π1 = N, π1,E. Continuing along, 2i+1 is the largest entry of π so π(n1 +1) < π(n1 +2) = 2i+1, where n1 +1 is even. This gives us κ(π)n1+1 = E. Finally, πn1+3→2i+1 = π2. Since n2 < 2i+1, we know that κ(π2) = p2 is a Catalan path by the inductive hypothesis. When inserting π2 into π, we must ignore the first

N step of κ(π2) since π2 does not fall at the beginning of π. Thus, κ(π)n1+3, κ(π)n1+4, . . . , κ(π)2i+1 = 0 0 0 0 p2 where p2 = N, p2. Overall we have that κ(π) = N, N, p1, E, E, N, p2 = N, p1, E, p2, where p1 and p2 are Catalan paths of length n1 + 1 and n2 + 1, respectively. Thus, κ(π) is a Catalan path of length n1 + n2 + 4 = 2i + 2.

Case 2: If n1 = 0 and n2 6= 0. Then π = n2 + 2 ⊕ n2 + 1 ⊕ π2. By convention, κ(π)0 = N. Since π(1) = n2 + 1 < π(2) = n2 + 2 and n2 + 2 is odd, κ(π)1 = E. Next, since every entry of π2 is smaller than n2 + 1, π(2) = n2 + 1 < π(3) which means that κ(π)2 = N. Finally, κ(π2) = p2

14 where p2 is a Catalan path of length n2 + 1. Inserting π2 into π, we must ignore the ﬁrst step 0 of p2 because π2 does not fall at the beginning of π. Thus, κ(π)3, κ(π)4, . . . , κ(π)n2+2 = p2 where 0 0 p2 = N, p2. Overall, we have that κ(π) = N, E, N, p2 = N, E, p2, where p2 is a Catalan path of length n2 + 1. Thus, κ(π) is a Catalan path of length n2 + 3 = 2i + 2.

Case 3: If n1 6= 0 and n2 = 0, then π = 1 ⊕ AUGC(π1, 1) ⊕ 2i + 1. By convention, κ(π)0 = N. Since 1 is the smallest entry of π, π(1) = 1 < π(2) so κ(π)1 = N. Now π2→2i+1 = AUGC(π1, 1). 0 0 Following the same logic from Case 1, κ(π)2, κ(π)3, . . . κ(π)1+n1 = p1, where κ(π1) = p1 = N, p1,E. Next, 2i + 1 is the largest entry of π, so π(2i) < π(2i + 1) = 2i + 1, giving us that κ(π)2i = E. 0 Finally, κ(π)2i+1 = E by convention. Overall, we have that κ(π) = N, N, p1,E,E = N, p1,E, where p1 is a Catalan path of length n1 + 1. Thus, κ(π) is a Catalan path of length n1 + 3 = 2i + 2. Now that we have a method to construct a Catalan path from a snow leopard permutation, we need a method to construct a snow leopard permutation from a Catalan path. In particular, we show how to invert the Ascent/Descent map.

Algorithm 3.5 (First-Touch Algorithm). This algorithm takes us from Catalan paths of length 2n to snow leopard permutations of length 2n + 1. We will abbreviate the ”First-Touch” function to FT. Here, we assume that Catalan paths are drawn on a ﬁnite lattice, such that each path begins at (0,0) and includes N and E steps of length 1. We write di to denote the diagonal line y = x + i. Note that no path can cross below the main diagonal, d0. Let p denote a Catalan path of length 2n. Case 1: If n = 0, then FT (p) = ∅ Case 2: If n = 1, then FT (p) = 1 Case 3: If n > 1, then to construct FT (p):

1. Label the vertices down d0 with consecutive odd integers 1, 3,..., 2n−1. Suppose k is the label where the path ﬁrst touches d0.

2. Define p1 as the subpath of p before the first touch that lies above d1. Define p2 as the subpath of p after the first touch.

3. If k > 1, then FT (p) = k ⊕ AUGC(FT (p1), k) ⊕ 2n − 1 ⊕ k − 1 ⊕ FT (p2). If k = 1, then p2 is empty so FT (p) = 1 ⊕ AUGC(FT (p1), 1) ⊕ 2n − 1.

15 4 4 1 1

3 3 3 3

2 2 5 5

1 1 7

1 2 3 4 1 2 3 4

(a) A Catalan path p of length 8. (b) Since p ﬁrst touches d0 at 1, FT (p) = 1 ⊕ AUGC(FT (p1), 1) ⊕ 7, where p1 is the red subpath above.

3 1

1 2 3 4

(c) The subpath p1 ﬁrst touches d1 at 3, so FT (p1) = 3 ⊕ AUGC(FT (p2), 3) ⊕ 5 ⊕ 2 ⊕ FT (p3), where p2 is the subpath in red and p3 is the subpath in green. FT (p2) = FT (p3) = 1, so FT (p1) = 34521 and FT (p) = 1432567.

Figure 7: The First-Touch algorithm takes the path in Figure 6 and returns the snow leopard permutation 1432567.

Theorem 3.6. κ and FT are inverses of each other. Proof. Let p be a Catalan path of length 2n. If n = 1, then p = N,E is composed of exactly one north step and one east step. Thus κ(FT (p)) = κ(1) = N,E = p. Thus, for a trivial path, κ and FT are inverse maps. We proceed by strong induction on n. That is, assuming that κ(FT (p)) = p for all paths of length n ≤ i, we will show that κ(FT (p)) = p for all paths of length n = i + 1. Let p be a path with length 2i + 2. We can decompose p = N, p1, E, p2 where p1 is a Catalan path that doesn’t touch the diagonal, p2 is a Catalan path, and the divide between p1 and p2 happens exactly when the path ﬁrst touches the diagonal (it is possible for either p1 or p2 to be the empty Catalan path). If we apply FT to p, we will be in Case 3, since we are assuming p has length greater than 2. This gives us two cases.

Case 1: If p2 = ∅, then k = 1 (where k is deﬁned in Case 3 of FT ). Thus, FT (p) = k ⊕ AUGC(FT (p1), k) ⊕ 2n − 1. We can consider the string of ascents and descents that correspond to FT (p). Since every value of AUGC(FT (p1), k) is greater than k, the permutation starts with an ascent. Likewise, since it is a permutation of length 2n − 1 that ends with 2n − 1, it must also end with an ascent. Also, the string of ascents and descents that come from the AUGC(FT (p1), k) component of FT (p) are exactly the opposite of those that come from FT (p), since taking the

16 complement reverses ascents and descents. Now consider κ(FT (p)) = κ(k ⊕ AUGC(FT (p1), k) ⊕ 2n − 1). This Catalan path starts with a north step, since FT (p) starts with an ascent. Then we apply κ to the AUGC(FT (p1), k) component of FT (p), however we are off by one index (since we already considered one ascent). Being off by one index will turn even indices into odd indices and odd indices into even indices. Therefore, the steps that are added to our path are exactly the reverse of those that would have been added had we taken only κ(AUGC(FT (p1), k), which are the reverse of those that come from κ(FT (p1)). The two reversals undo each other and we add the steps that correspond to κ(FT (p1)), which is p1 by our inductive hypothesis. The final step is an ascent in an even index, which corresponds to an E step, so all together, κ(FT (p)) = N, p1,E = p.

Case 2: If p2 is non-empty, then FT (p) = k⊕AUGC(FT (p1), k)⊕2n−1, k−1⊕FT (p2). The exact logic of Case 1 gives us that κ(FT (p)) starts with N, p1,E,.... This is followed by a N step, since the 2n−1 to k−1 gap is a descent at an odd index. This is then followed by an E step, since k−1 to the start of FT (p2) will be a gap that corresponds to a descent in an even index. After that, we are exactly in an odd index, so we add the result of κ(FT (p2)), excluding its trivial ﬁrst N step because FT (p2) does not fall at the beginnging of FT (π). By our inductive hypothesis, κ(FT (p2)) = p2, so combining this with the north step we just added, we get κ(FT (p)) = N, p1, E, p2 = p.

4 Snow Leopard Permutations: a Characterization with Transpo- sitions

It is well known that every permutation is a product of transpositions and even a product of adjacent transpositions. In this section, we show that all snow leopard permutations of odd length can be constructed using a particular set of transpositions. We begin by deﬁning these transpositions.

Deﬁnition 4.1 (Snow Leopard Transpositions). For any substring xi, ..., xj of a permutation π, deﬁne the following transpositions.

1. If xi and xj are odd and xi−1, xi, ..., xj, xj+1 or xi−1, xj, ..., xi, xj+1 is a decreasing sequence of consecutive integers, then we write τ1(π) to denote the permutation obtained from π by interchanging xi and xj.

2. If xi and xj are even and xi−1, xi, ..., xj, xj+1 or xi−1, xj, ..., xi, xj+1 is an increasing sequence of consecutive integers, then we write τ2(π) to denote the permutation obtained from π by interchanging xi and xj.

Note that if either xi or xj occurs at π(1) or π(n) we then waive any requirement for the behavior of π(0) or π(n + 1) respectively (possibly both).

For example, given the permutation π = 983654721 we can see the substring 36547 is one transposition of even integers away from being a consecutive sequence of increasing integers, so τ2(π) = 983456721. Similarly, given π = 567894321 we see the substring 4321 is a decreasing sequence of consecutive integers (here we waive any requirement for the behavior of the (n + 1)st entry, so τ1(π) = 567894123. Proposition 4.2. A permutation of odd length is a snow leopard permutation if and only if it can be obtained from Rn using a sequence of the transpositions described in Deﬁnition 4.1.

17 To prove this result we describe how each transposition of a snow leopard permutation corresponds with a transformation of the associated DABP. In particular, we prove these transpositions maintain the property of being snow leopard by showing that the accompanying DABP transfor- mations preserve both the doubly alternating and Baxter properties. Then we show inductively that every snow leopard permutation is a product of transpositions applied to π = Rn. But ﬁrst we must understand a basic property of the relationship between a DABP and its compatible snow leopard permutation.

Figure 8: Graphs showing every snow leopard permutation of length 3, 5, and 7 as a product of the transpositions described in Deﬁnition 4.1.

Lemma 4.3. Let π1 be a snow leopard permutation and let π2 be its corresponding DABP. Now let m, m − 1, ..., n + 1, n be a decreasing substring of consecutive integers in π1 with π1(k) = m and m, n both odd. Then π2 must have the following substring starting at π2(k): m, m + 1, m − 2, m − 1, ..., n + 2, n + 3, n, n + 1.

Similarly, if π1 contains an increasing substring of consecutive integers, n, n + 1, ..., m − 1, m, with π1(k) = n and m, n both even, then π2 must have the substring n + 1, n, n + 3, n + 2, ..., m − 1, m − 2, m + 1, m beginning at position k − 1.

18 For example, if π1 = 7654321, then the compatible DABP is π2 = 78563412. Similarly, if a snow leopard permutation contains the substring 45678 beginning at position k, then the compatible DABP must contain the substring 547698 beginning at position k. Note that because snow leopard permutations map even integers to even integers and odd integers to odd integers, the parity of m and k are the same.

Figure 9: Small examples of the DABP patterns corresponding to increasing (left) and decreasing (right) strings of consecutive integers in our snow leopard permutation.

Proof. We can appeal to the one-to-one correspondence between snow leopard permutations and DABPs. Suppose π2 is a DABP containing the substring

S = m, m + 1, m − 2, m − 1, . . . , n + 2, n + 3, n, n + 1, where m = π2(k) and k is odd. The ascents and descents of this sequence alternate, with the ascents starting at odd indices, so it is possible to construct the remainder of π2 such that π2 is alternating. It is tedious, though straight-forward to show that we can construct π2 such that π2 is also doubly alternating and Baxter. Let π1 denote the snow leopard permutation compatible with π2. Using Canary’s algorithm we can then show that S forces π1 to contain the substring m, m − 1, . . . , n + 1, n, where π1(k) = m. We leave this as an exercise for the reader. Snow leopard permutations preserve parity, so k being odd implies that m and n are also odd. Once we have constructed π2, the one-to-one correspondence between DABPs and snow leopard permutations ensures that π2 is unique. The second part of the lemma follows from a similar argument. Now we have the tools to show that if any of the transpositions described above are applied to a snow leopard permutation, the result is another snow leopard permutation.

Lemma 4.4. Let π1 be a snow leopard permutation with π1(k) = m decreasing consecutively to π1(l) = n with m and n both odd. Further let, π1(k − 1) = m + 1 and π1(l + 1) = n − 1. Now let π2 be the compatible DABP with π2(k) = m, π2(k + 1) = m + 1, π2(l) = n, and π2(l + 1) = n + 1. We can transform π2 by setting π2(k) = n, π2(k + 1) = m, π2(l) = n + 1, and π2(l + 1) = m + 1.

19 This resulting matrix is still doubly alternating and Baxter and the resulting compatible matrix is τ1(π1).

Proof. Notice that because k is odd, there must exist a descent from π2(k−1) to π2(k). Because n < m this descent is maintained in the transformed matrix. π2(k) = n then ascends to π2(k + 1) = m, then descends to π2(k +2) = m−2. At the other end of our substring, we see that π2(l −1) = n+3 falls to π2(l) = n + 1 which then rises to π2(l + 1) = m + 1. Finally, because m + 1 > n + 1, we know the descent to π2(l + 2) is maintained. Therefore our transformed matrix is alternating. −1 A look at the inverse of the transformed matrix reveals that, because k < l, π2 (n−1) descends −1 −1 −1 to π2 (n) = k. Then π2 (n) = k ascends to π2 (n + 1) = l, followed by a descent to the unaltered −1 −1 π2 (n + 2) = l − 2. At the other end of the permutation we observe π2 (m − 1) = k + 3 descending −1 −1 to π2 (m) = k + 1, which in turn ascends to π2 (m + 1) = l + 1. Because l + 1 > k + 1 we know −1 −1 our descent from π2 (m + 1) to π2 (m + 2) is maintained. Thus the inverse of our transformed matrix is alternating and we can conclude that our transformed matrix is doubly alternating. To show that the transformed permutation is Baxter, we must show there exist no occurences of the pattern 2-41-3 or 3-14-2. Note that for the transformation to have created one of these patterns, our “41” or “14” must involve one of the altered entries. If one of the forbidden patterns appears in our transformed permutation and the “41” or “14” exist entirely to either side of our altered entries, the forbidden pattern must have appeared in our original permutation. Within our altered entries only four adjacent pairs vary by more than two. Therefore we need only to check four cases for the possibility of a forbidden pattern. Assume to the contrary that our transformation has created a forbidden pattern. Case 1: The “41” of the forbidden pattern occurs at π(k − 1) and π(k). Because all entries n to m + 1 appear to the right of π(k), we see that the “2” and “3” of our pattern must be greater than m + 1. But if that were the case, the “2-41-3” pattern would have already existed when π(k) = m, which is a contradiction. Case 2: The “14” of the forbidden pattern occurs at π(k) and π(k + 1). We see all entries between π(k) and π(k + 1) are to the right of π(k + 1), so we cannot have created a forbidden pattern here. Case 3: The “14” of the forbidden pattern occurs at π(l) and π(l + 1). We see all entries between π(l) and π(l + 1) are to the left of π(k + 1), so we cannot have created a forbidden pattern here. Case 4: The “41” of the forbidden pattern occurs at π(l + 1) and π(l + 2). Because all entries n to m + 1 appear to the left of π(l + 1), we see that the “3” and “2” of our pattern must be less than n. But if that were the case, the “2-41-3” pattern would have already existed when π(l + 1) = n, which is a contradiction. Therefore, in all cases, the transformation of π2 is a DABP compatible with τ1(π1). As such, τ1(π1) is a snow leopard permutation, which shows directly that the set of snow leopard permutations is closed under τ1.

Lemma 4.5. Let π1 be a snow leopard permutation with π1(k) = m increasing consecutively to π1(l) = n with m and n both even. Further let, π1(k − 1) = m − 1 and π1(l + 1) = n + 1. Let π2 be the compatible DABP with π2(k) = m + 1, π2(k + 1) = m, π2(l) = n + 1, and π2(l + 1) = n. We can transform π2 by setting π2(k) = n + 1, π2(k + 1) = m + 1, π2(l − 1) = n, and π2(l) = m. This resulting matrix is still doubly alternating and Baxter and the resulting compatible matrix is τ2(π1).

20 Figure 10: An example of the transposition of a snow leopard permutation and corresponding transformation of the compatible DABP described in Lemma 4.4.

Proof. By switching all instances of “increase” and “decrease” we can turn our proof of Lemma 4.4 into a proof of Lemma 4.5.

We have proven that the set of snow leopard permutations of odd length is closed under τ1 and τ2. The only remaining task is to show that each snow leopard permutation of odd length can be constructed from Rn by applying τ1 and τ2. We can do this inductively using the recursive decomposition described in section 2 and shown in Figure 11. Figure8 shows the base cases for n = 3, n = 5, n = 7.

Figure 11: The non-trivial recursive structure of a snow leopard permutation.

Proof. Assume we can construct every odd snow leopard permutation up to length n. Using the snow leopard permutation decomposition, which is pictured in Figure 11. To generate any

21 snow leopard permutation of length n, ﬁrst we transpose the hat (h) and shoe (k) into place using τ1. Then, using our inductive hypothesis, we can immediately construct the snow leopard permutation comprising of the entries 1, 2, . . . , k−2. Generating the component of the snow leopard permutation that came from the complement of a smaller snow leopard permutation and spans k + 1, k + 2, . . . , h − 1 does not immediatly follow from the inductive hypothesis. But notice that taking the complement of a snow leopard permutation changes all increases to decreases and vice versa. Additionally, notice that because k and h are odd, SLC has parity reversed. By repeated application of τ1, we can transform the all-decreasing substring h − 1, h − 2, . . . , k + 1 into the all-increasing substring k + 1, k + 2, . . . , h − 1. Now note that we can construct SLC in same way we inductively assumed we could generate SL, but with the opposite transposition at every step. Again, this is due to the fact that all increases and decreases have reversed and the parity has C reversed, so if we apply τ1 where we would have applied τ2 and vice-versa, we can constuct SL . In the other cases of the snow leopard permutation decomposition, which arise when one of the two permutations that make up the larger snow leopard permutation is empty, we can use a similar argument to contruct the permutation with τ1 and τ2. Therefore, we have constructed the entirity of this snow leopard permutation of odd length by placing the hat and shoe using τ1 and by placing the decomposed portions of the snow leopard permutation using sequences of τ1 and τ2 that were initially used to construct them.

5 Compatible Permutation Pairs and Products of Fibonacci Num- bers

So far, we have considered doubly alternating Baxter permutations, which have the property that for each DABP of length n + 1, there is a unique compatible permutation of length n. Conversely, we know that for each snow leopard permutation of length n there is a unique compatible DABP of length n + 1. However, if we consider, for a given permutation of length n, all compatible permutations of length n + 1, regardless of whether they are doubly alternating, we get diﬀerent results. Here are a few examples of permutations of length n listed with all of their compatible permutations of length n + 1.

1. 123 is compatible with 1234, 1324, 2134, 2143, 1243. 2. 213 is compatible with 2314, 3214, 3124. 3. 1234 is compatible with 13254, 13245, 12354, 12345, 21354, 21345, 12435, 21435. 4. 1342 is compatible with 14352, 13452, 13542. 5. 3421 is compatible with 43521, 35421, 34521, 43512, 35412, 34512.

We have veriﬁed the following conjecture for permutations up to length n = 11. Conjecture 5.1. Suppose σ is a permutation of length n for which there exists at least one compatible permutation of length n + 1. Then the number of permutations of length n + 1 compatible with σ is a product of Fibonacci numbers. In this section, we prove some results aimed at eventually proving this conjecture. The overall strategy is as follows.

22 1. We show that transposing adjacent consecutive entries in any larger permutation maintains the same smaller compatible permutation.

2. We show that these swaps give us all of the permutations compatible with a length n identity permutation In = 123 . . . n, or a length n reverse identity permutation Rn = n n − 1 ... 1, and that the number of these permutations is Fn+1. 3. We conjecture that any permutation of length n + 1 compatible with a permutation of length n can be changed, using transpositions of adjacent consecutive entries, into a permutation whose TOAD consists of disjoint strips of blocks where each strip resembles either the identity permutation or its reverse.

4. We conjecture that entries cannot be swapped between strips of blocks while maintaining the same smaller permutation. Given this property, each strip must represent a Fibonacci number of possible larger compatible permutations. So, if we have m total strips of blocks in the TOAD of the permutation described in step 3, then the total number of permutations of length n + 1 compatible with the given length n permutation is a product of m Fibonacci numbers.

We begin with step 1. Theorem 5.2. Suppose π is a Baxter permutation of length n + 1 compatible with a permutation σ of length n. Then the permutation obtained by transposing any two consecutive adjacent entries in π is also compatible with σ. Proof. We want to show that if we have two adjacent consecutive entries in π that they create a block in its TOAD with a 1 in the center that, when rotated 90 degrees, has no eﬀect on the center 1 (which represents an entry in σ) or any surrounding tiles. Say we have adjacent consecutive entries. If there is a 0 in the smaller matrix between our adjacent 1’s in the larger matrix, then we know from [1] that this forces an adjacent −1 entry, which is not allowed in the permutation case. So we have one possibility. There is a 1 in the smaller matrix between our adjacent 1’s in the larger matrix. In this case, there are only two possible tilings.

(a) Increasing sequence (b) Decreasing sequence

Clearly, these are blocks that, when rotated 90 degrees, have no eﬀect on any outside entries and that maintain the same entry of the smaller matrix. Additionally, (a) is the rotated version of (b) and vice versa.

23 Now, we move on to step 2. For this step, we must ﬁrst introduce a lemma.

Lemma 5.3. A 1 entry in the larger of a compatible pair of permutation matrices must be adjacent to a 1 entry in the smaller matrix. Equivalently, a 1 entry in the larger matrix cannot be adjacent to only 0 entries.

Proof. Up to a reﬂection, there is only one tiling that allows a 1 in the larger matrix to border only 0s in the smaller, shown here.

Figure 13: A lonely 1 in the larger matrix.

But we know from [1] that this “windmilled” construction violates our Baxter condition, and is therefore not allowed.

With this lemma, we can prove the following theorem.

Theorem 5.4. The identity permutation In = 123 . . . n and the reverse identity permutation Rn = n n − 1 ... 1 are both compatible with Fn+1 permutations of length n + 1. Proof. We describe a bijection using the following algorithms. Algorithm 5.5 (P-T). This algorithm takes us from permutations of length n + 1 compatible with the identity permutation of length n to tilings of a strip of size n+1 with monominoes and dominoes. Given a permutation, read from left to right looking at adjacent pairs of indices (ie. 12, 23, 34, ...). For each pair, we add dominoes or monominoes to our strip going from left to right (for the reverse of the identity, go from right to left instead). There are four cases.

1. The pair is increasing and consecutive. Add 2 monominoes to the strip, and skip the next pair.

2. The pair is decreasing and consecutive. Add a domino to the strip, and skip the next pair.

3. The pair is increasing and non-consecutive. Add a monomino to the strip, and go to the next pair (no skip).

4. The pair is decreasing and non-consecutive. This cannot happen while maintaining the identity in the smaller permutation. Lemma 5.3 tells us that any 1 entry in our larger matrix must border a 1 entry in the smaller matrix. Given coordinates (i, i) for a 1 entry in the smaller identity matrix, the only available spots for a 1 entry in the larger matrix are (i, i), (i, i + 1), (i + 1, i), or (i + 1, i + 1). If the pair is decreasing and non-consecutive, this means there is

24 a 1 entry in the larger matrix whose horizontal and vertical coordinates diﬀer by more than one, which means the entry violates Lemma 5.3. This is illustrated in Figure 14.

Figure 14: Since our smaller permutation is the identity permutation, there must be blocks arranged vertically in its TOAD as shown. Lemma 5.3 tells us that the only available places for 1s in the larger matrix are those labelled with green dots. Each red line represents the index of the entry in our larger compatible permutation. So, we know that the ﬁrst entry must be a 1 or a 2, and the last entry must be an n or an n+1. We also know that there are only three options for each other entry, and that there is no way to arrange these options such that our larger compatible permutation has a decreasing non-consecutive adjacent pair.

We now have a set of rules that brings us from each permutation of length n + 1 compatible with the identity permutation of length n to a unique tiling of a strip of size n+1 with monominoes and dominos. Next we describe the reverse direction. Algorithm 5.6 (Reverse of P-T). This algorithm takes us from tilings of a strip of size n + 1 with monominoes and dominoes to permutations of length n+1 compatible with the identity permutation of length n. Given a tiling, read from left to right looking at adjacent pairs of indices. For each pair, we add integers to our permutation going from left to right (for the reverse of the identity, go from right to left instead). There are four cases. 1. The pair consists of two monominoes. Add the indices corresponding to the monominoes in increasing order to the permutation, and skip the next pair. This creates the increasing consecutive pair of digits from case 1 of Algorithm 5.5. 2. The pair consists of a single domino. Add the indices in decreasing order to the permutation, and skip the next pair. This creates the decreasing consecutive pair of digits from case 2 of Algorithm 5.5. 3. The pair consists of a monomino followed by the ﬁrst half of a domino. Add the index of the monomino to the permutation, and go to the next pair (no skip). So, the next pair will

25 be a domino, and this creates the increasing non-consecutive pair of digits from case 3 of Algorithm 5.5.

4. The pair consists of the last half of a domino followed by a monomino. This is impossible given rules 1, 2, and 3.

Because of Lemma 5.3, and as illustrated in Figure 14, each entry in the larger compatible matrix must be one of those indices listed in the rules of Algorithm 5.6. Also, since these rules do not allow for repeated entries in the resulting permutation, they take us from each tiling of a strip of size n + 1 with monominoes and dominoes to a unique permutation. By construction, each such permutation is of length n + 1 and is compatible with the identity permutation of length n. So, the rules in these algorithms describe a bijection between permutations of length n + 1 compatible with the identity permutation of length n and monomino/domino tilings of a strip of size n + 1, which are known to be counted by Fibonacci numbers.

Here are a few examples of length 5 permutations compatible with 1234 and their corresponding monomino/domino tiling by the algorithms above.

(a) 12345 (b) 21345

This concludes parts 1 and 2 of the proof of Conjecture 5.1. Now, we outline what we know about the conjectures in parts 3 and 4. First, we introduce some deﬁnitions that will be helpful in understanding how In and Rn can be used to interpret less orderly permutations. Deﬁnition 5.7. A strip of blocks of size n in a TOAD is a set of n blocks arranged so that each block shares a border with another block in the strip, and one of the two following properties hold.

1. Every block has the same horizontal coordinate. In this case, we call the strip a vertical strip.

2. Every block has the same vertical coordinate. In this case, we call the strip a horizontal strip.

In Figures 16a and 16b are a couple of examples of vertical and horizontal strips.

Deﬁnition 5.8. We call a vertical strip of blocks in a TOAD an identity strip if it has the property that all 1s in the strip have the same horizontal coordinate.

Deﬁnition 5.9. We call a horizontal strip of blocks in a TOAD a reverse identity strip if it has the property that all 1s in the strip have the same vertical coordinate.

In Figures 17a and 17b are a couple of examples of identity and reverse identity strips. With these deﬁnitions, we introduce two conjectures that more formally describe parts 3 and 4 of our proposed proof of Conjecture 5.1.

26 (a) Vertical strip of size 4 (b) Horizontal strip of size 4

Figure 16: A vertical strip and a horizontal strip.

Conjecture 5.10. Given a permutation π of length n and a compatible permutation σ of length n + 1, the blocks in the TOAD for π and σ must be arranged in disjoint strips. Furthermore, we can convert each strip into an identity or reverse identity strip without changing π. Conjecture 5.11. Given a permutation π of length n and a compatible permutation σ of length n+1, any swap of 1 entries between disjoint strips in the TOAD for π and σ will result in a change in π (or, equivalently, a new arrangement of strips). We think that the proofs of these conjectures must rely on the limited number of ways that 1s in a larger compatible matrix may surround 1s in the smaller matrix. Here are a few questions to consider.

1. If we have a 1 in the smaller matrix, and therefore a block in our TOAD, that only borders 0 entries in the larger matrix, must the block be a part of a strip of size 2 or more? What can or must happen around this block?

2. If we have a 1 in the smaller matrix that borders a single 1 in the larger matrix, what does this mean for the adjacent rows and columns of the TOAD? What if it borders two 1s?

3. What can/must happen at the ends of a strip and how can/must disjoint strips aﬀect each other/ﬁt together in a TOAD?

4. Finally, can we use these results to prove that there exists no set of transpositions of 1 entries in a larger matrix that moves a 1 entry outside of the strip it lies on? In other words, can we prove that only the adjacent consecutive transpositions will maintain the same smaller matrix?

We have veriﬁed Conjecture 5.1 by computer data for n ≤ 11, where n is the length of the smaller permutation. Ideally, we would like our conjectures to give us an exact formula for the number of compatible larger permutations given any smaller permutation. However, in larger cases, it is harder to predict how many 1 entries of our larger matrix lie in each strip.

27 (a) Identity strip of size 4 (b) Reverse identity strip of size 4

Figure 17: An identity strip and a reverse identity strip.

For instance, given a strip of size n, we have seen cases where this strip has n + 1, n, or n − 1 larger matrix 1 entries. Here, we need to know how the relative location of each strip aﬀects this number of 1 entries. We have observed that those strips that do not border the edge of the Aztec diamond tend to be the ones with fewer total 1 entries, and that those that do border the edge of the diamond require more 1 entries to form legal tilings.

6 Questions and Open Problems

1. Can we classify the snow leopard permutations non-recursively? In this paper, we have given a recursive characterization of snow leopard permutations, and shown how to generate them with transpositions. What other classiﬁcations exist? Is it possible to look at any given permutation and decide whether or not it is a snow leopard permutation without our decomposition?

2. Is Conjecture 5.1 true? The details related to this question are shown at the end of Section 5.

3. What smaller permutations are compatible with larger alternating Baxter permutations? This paper deals with larger doubly alternating Baxter permutations. What if the larger permutation is alternating and Baxter, but not doubly alternating? Are the smaller permutations compatible with these counted by products of Catalan numbers?

7 Appendix: Useful Code

Some of the programs we wrote to generate data for this project can be found in a zip ﬁle at apps.carleton.edu/curricular/math/assets/AztecGenerator.zip, while others are in a Mathematica ﬁle at apps.carleton.edu/curricular/math/assets/SnowLeopardCode.nb.

28 8 Acknowledgements

First and foremost, we would like to thank our advisor, Eric Egge, for his guidance, encouragement, and ever-present positivity throughout this project. We would also like to thank Jim Propp for meeting with us personally to discuss the trajectory of our research. Lastly, we would like to thank all members of the Carleton Mathematics department for their support and wisdom.

References

[1] Canary, Hal. Aztec diamonds and Baxter permutations. The Electronic Journal of Combina- torics 17.1 (2010): R105.

[2] Elkies, Noam, Kuperberg, Greg, Larsen, Michael, and Propp, James. Alternating-sign matrices and domino tilings (Part I). Journal of Algebraic Combinatorics 1.2 (1992): 111-132.

[3] Guibert, Olivier and Linusson, Svante. Doubly alternating Baxter permutations are Catalan. Discrete Mathematics 217.1 (2000): 157-166.

[4] Ouchterlony, Eric. Pattern avoiding doubly alternating permutations. gar- sia.math.yorku.ca/fpsac06/papers/83.pdf