Maximal origami flip graphs of flat-foldable vertices: properties and algorithms

Flat origami studies straight line, planar graphs $C=(V,E)$ drawn on a region $R\subset\mathbb{R}^2$ that can act as crease patterns to map, or fold, $R$ into $\mathbb{R}^2$ in a way that is continuous and a piecewise isometry exactly on the faces of $C$. Associated with such crease pattern graphs are valid mountain-valley (MV) assignments $\mu:E\to\{-1,1\}$, indicating which creases can be mountains (convex) or valleys (concave) to allow $R$ to physically fold flat without self-intersecting. In this paper, we initiate the first study of how valid MV assignments of single-vertex crease patterns are related to one another via face-flips, a concept that emerged from applications of origami in engineering and physics, where flipping a face $F$ means switching the MV parity of all creases of $C$ that border $F$. Specifically, we study the origami flip graph ${\rm{OFG}}(C)$, whose vertices are all valid MV assignments of $C$ and edges connect assignments that differ by only one face flip. We prove that, for the single-vertex crease pattern $A_{2n}$ whose $2n$ sector angles around the vertex are all equal, ${\rm{OFG}}(A_{2n})$ contains as subgraphs all other origami flip graphs of degree-$2n$ flat origami vertex crease patterns. We also prove that ${\rm{OFG}}(A_{2n})$ is connected and has diameter $n$ by providing two $O(n^2)$ algorithms to traverse between vertices in the graph, and we enumerate the vertices, edges, and degree sequence of ${\rm{OFG}}(A_{2n})$. We conclude with open questions on the surprising complexity found in origami flip graphs of this type.


Introduction
When folding a piece of paper into a flat object, the creases that are made will be straight lines. This describes flat origami [7], which we formally model with a pair (C, P ), called the crease pattern, where P is a closed region of the plane (our paper), and the set of creases C = (V (C), E(C)) is a plane graph on P with straight line segments for the edges. (When the exact shape of the paper P is not important, we will refer to the crease pattern merely as C.) If there exists a mapping f : P → R 2 that is continuous, non-differentiable along the edges of C, and an isometry on each face of C, then we say that (C, P ) is locally flat-foldable. Also, folded creases come in two types when viewing a fixed side of the paper: mountain creases, which fold away in a convex manner, and valley creases, which are concave. We model this with a function µ : E(C) → {−1, 1}, called a mountain-valley (MV) assignment for the crease pattern C, where 1 (respectively −1) represents a mountain (respectively valley) crease. A MV assignment µ is called valid if µ can be used to fold C into a flat object without the paper intersecting itself.
Capturing mathematically how paper self-intersection works, and how it can be avoided, to achieve global flat-foldability is difficult. In fact, determining if a crease pattern (C, P ) is globally flat-foldable is NP-hard [1,4], even if a specific MV assignment is already given. In the special case where the crease pattern has only one vertex in the interior of P , called a single-vertex crease pattern or a flat vertex fold, determining if a MV assignment is valid is not straight-forward [7,13] but can be determined in linear time [5]. Indeed, there are many open questions that remain about enumerating valid MV assignments [8] and understanding their structure [12], even for very simple crease patterns.
Flat-foldability and valid MV assignments have been of interest to scientists in the study of origami mechanics and their application in constructing metamaterials [14], even in the case of single-vertex crease patterns [11]. A concept that has emerged from such applications is that of a face flip, where a valid MV assignment µ is altered by switching only the mountains and valleys that surround a chosen face F , denoting the new MV assignment (which may or may not be valid) by µ F . Face flips were first introduced in the literature by VanderWerf [16] and have been utilized in applications ranging from tuning metamaterials [14] to analyzing the statistical mechanics of origami tilings [3].
In this paper, we explore the relationships between valid MV assignments of a given crease pattern C using a tool called the origami flip graph, denoted OFG(C), which is a graph whose vertices are all valid MV assignments of C and where two vertices µ and ν are connected by an edge if and only if there exists a face F of C such that flipping F changes µ to ν (and vice-versa, i.e., ν = µ F ). We may view face flips and the paths they generate on OFG(C) as a reconfiguration setting for some problems in computational origami.
Origami flip graphs were introduced in [2], but only in the context of origami tessellations (crease patterns that form a tiling of the plane). In the present work, we focus on flat-foldable crease patterns that have a single vertex in the paper's interior, called flat vertex folds, with the additional requirement that the sector angles between the creases are all equal. We denote such a crease pattern by A 2n where 2n is the degree of the vertex. In Section 2, we provide background on flat origami and show that OFG(A 2n ) serves as a maximal "superset" graph for flat vertex folds-if C is any other flat vertex fold of degree 2n, then OFG(C) is a subgraph of OFG(A 2n ). In Sections 3 and 4, we prove that OFG(A 2n ) is connected using two different algorithms for finding paths in this graph, one of which further proves that the diameter of OFG(A 2n ) is n. In Section 5 we describe an algorithm for computing the size of OFG(A 2n ), generating a sequence that was not originally in the Online Encyclopedia of Integer Sequences, and find a formula for this as well as for the degree sequence of OFG(A 2n ). We conclude with open questions and a discussion of future work.
2 Background and maximality of OFG(A 2n ) Let (A 2n , P ) denote the crease pattern that contains only one vertex v in the interior of P , where v has degree 2n and the angles between consecutive creases around v are all equal (to π/n). We normally let P be a disc with v at the center. Let e 1 , . . . , e 2n denote the creases in A 2n and α i denote the face between e i and e i+1 (with the indices taken cyclically, so α 2n is between e 2n and e 1 ). A basic result from flat origami theory is Maekawa's Theorem, which states that, if v is a vertex in a flat-foldable crease pattern with valid MV assignment µ, then the difference between the number of mountain and valley creases at v under µ must be two, often denoted by M −V = ±2 [7]. However, in the case of the crease pattern A 2n Maekawa's Theorem is stronger: Theorem 1 is proved in [2,9], but a summary of the sufficient direction is: Find a pair of consecutive creases e i , e i+1 in A 2n with µ(e i ) ̸ = µ(e i+1 ) and fold them (making a "crimp") to turn the paper into a cone, on which we now have the crease pattern A 2(n−1) and a MV assignment that still has µ(e) = ±2. We have two options when forming this cone; it could be convex or concave, which will be preserved as we repeat this process until there are only two creases left,  Figure 1, although in the latter case only half of the vertices (those whose MV assignment satisfies µ(e) = −2) are shown. The vertices in these graphs are labeled with their corresponding valid MV assignment, where bold creases are mountains and non-bold means valley, a convention we will use throughout this paper. In [10], it is proved that OFG(C) is bipartite whenever C is a flat-foldable, single-vertex crease pattern, although we will not be making particular use of that here.
Theorem 1 tells us that any MV assignment of A 2n that satisfies M − V = ±2 will be valid. Therefore, there are 2 2n n−1 vertices in OFG(A 2n ). We will now show that the origami flip graph of A 2n has maximal size over all origami flip graphs of flat vertex folds of degree 2n, and further that such origami flip graphs are all subgraphs of OFG(A 2n ). The idea is that when all the sector angles of a flat vertex fold are equal, the only requirement for a MV assignment to be valid is that it satisfies Maekawa's Theorem. If, on the other hand, the sector angles are not all equal, then other restrictions will apply. For example, if a flat-foldable, single-vertex crease pattern C has consecutive sector angles α i−1 , α i , α i+1 where α i is strictly smaller than both α i−1 and α i+1 , then the creases e i and e i+1 bordering α i must have different MV parity, so µ(e i ) ̸ = µ(e i+1 ) must hold in any valid MV assignment µ of C. (This is known as the Big-Little-Big Lemma; see [7].) This implies that the faces α i−1 and α i+1 can never be individually flipped under a valid MV assignment µ, since doing so would make µ(e i ) = µ(e i+1 ). Other restrictions on when faces in a single-vertex crease pattern can be flipped are detailed in [10], but since A 2n does not have such restrictions, its origami flip graph will have the most edges possible. Examples of this when 2n = 4 are shown in Figure 2. We formalize and slightly expand this in the following Theorem.
Theorem 2 Let C be a flat-foldable, single-vertex crease pattern of degree 2n that is not A 2n . Then OFG(C) is isomorphic to at least 2n distinct, labeled subgraphs of OFG(A 2n ).
Proof: Suppose we have an arbitrary flat vertex fold C with degree 2n, creases c 1 , . . . , c 2n , and angles β i . Let ν be a valid MV assignment of C. Then, ν also represents a valid MV assignment for A 2n . Specifically, if e 1 , . . . , e 2n are the creases in A 2n and we define µ by µ(e i ) = ν(c i ), then µ will be a valid MV assignment on A 2n by Theorem 1 (since ν must satisfy Maekawa's Theorem).
Thus, we have a mapping f between all MV assignments ν of C and some MV assignments and ν(c) = ν βi (c) for all other creases c of C. The same relationship holds true between f (ν) and f (ν βi ). That is, flipping the corresponding face α i (between e i and e i+1 in A 2n ) in f (ν) will result in f (ν βi ). This can be written as Furthermore, our labeling of the creases e i in A 2n was arbitrary, and by the rotational symmetry of A 2n we had 2n different ways we could have done this, resulting in at least 2n distinctly-labeled copies of OFG(C) (since C ̸ = A 2n ) that may be found in OFG(A 2n ). □ If µ is a valid MV assignment for a crease pattern C, then we say that a face F of C is flippable under µ if µ F is also a valid MV assignment for C. In what follows, we will make extensive use of the following Lemma.
Proof: By Theorem 1, µ α k will be an invalid MV assignment if and only if µ α k (e i ) ̸ = ±2. This will only happen if µ(e k ) = µ(e k+1 ) (i.e., the creases that border α k have the same MV assignment under µ) and this value, µ(e k ), is different from the majority of the creases in µ. For example, if µ(e i ) = 2 and µ(e k ) = µ(e k+1 ) = −1, then µ α k (e i ) = 6, meaning that µ α k violates Theorem 1 and thus is invalid. All other possibilities for µ(e k ) and µ(e k+1 ) preserve the MV summation invariant and thus allow α k to be flippable under µ. □ We will utilize the following definition in Section 4: given two MV assignments µ and ν of A 2n , let S(µ, ν) denote the set of creases e 1 , . . . , e 2n with µ(e i ) ̸ = ν(e i ). This set is useful because it provides us with a quantity that is face-flip invariant.

Lemma 2
The parity of |S(µ, ν)| (the size of S(µ, ν)) is invariant under face flips. That is, if µ and ν are valid MV assignment of A 2n , then the |S(µ, ν)| will have the same even/odd parity as |S(µ αi , ν)| for any face α i of A 2n .
Proof: Suppose we flip a face α i of A 2n under µ. Then we are changing the MV assignments of two creases. This will change the size of S(µ, ν) by either 0 (if exactly one of e i and e i+1 is different between µ and ν) or 2 (if e i and e i+1 are both the same or both different between µ and ν). Therefore, the parity of |S(µ, ν)| is invariant under face flips. □

Connectivity of OFG(A 2n )
In this section, we present an algorithm for face-flipping between any two valid MV assignments µ and ν of A 2n . This will prove that OFG(A 2n ) is connected. In contrast, if C is an arbitrary flat vertex fold, then OFG(C) is not always connected. We invite the reader to verify that the degree-6 flat vertex fold with sector angles (45 • , 15 • , 60 • , 85 • , 75 • , 80 • ) has two disconnected 4-cycles for its origami flip graph. (Determining the connectivity of OFG(C) for general flat vertex folds C is quite convoluted and beyond the scope of this paper; see [10] for details.) Note that we will provide a different algorithm in Section 4 that also proves the connectedness of OFG(A 2n ), but the one given here, which we will call the FEA-Shwoop algorithm, is useful for other reasons, such as in an elegant proof of Lemma 3 below and as a warm-up for what follows. In the algorithm, we start with crease e 1 . If µ(e 1 ) = ν(e 1 ), then we move on to crease e 2 . If µ(e 1 ) ̸ = ν(e 1 ), then we would like to flip the face α 1 , since µ α1 (e 1 ) = ν(e 1 ), and then continue the algorithm on crease e 2 comparing µ α1 with ν.
However, if e 1 and e 2 have the same MV assignment under µ, then α 1 might not be flippable under µ if it falls under Lemma 1; such a µ and α 1 are shown in Figure 3. Since α 1 is not flippable, we move to α 2 and check to see if it is flippable under µ. If so, then we flip it, and doing so will make α 1 flippable (since it will no longer satisfy Lemma 1). Then we have µ α2,α1 (e 1 ) = ν(e 1 ), and we may proceed with crease e 2 comparing µ α2,α1 and ν. If α 2 is not flippable, then we try to flip the next face, α 3 . Eventually we will find some face α i that can be flipped (otherwise µ would be all mountain or all valley creases and violate Maekawa's Theorem) and then we can flip the sequence of faces α i , α i−1 , α i−2 , . . . , α 1 . We call this sequence of flipping faces in order to make µ αi,...,α1 (e 1 ) = ν(e 1 ) a shwoop, and an example of such a shwoop is shown in Figure 3.
Thus, our algorithm is to start by comparing µ(e 1 ) and ν(e 1 ), flipping α 1 or performing a shwoop to make them agree on e 1 if needed, and then moving on to e 2 , and so on. We call this algorithm FEA-Shwoop(A 2n , µ, ν), and pseudocode for it is shown in Algorithm 1. (FEA stands for Flipping Equal Angles.)

Theorem 3
The FEA-Shwoop(A 2n , µ, ν) algorithm inputs two valid MV assignments for A 2n and outputs a sequence of faces that, when flipped in order, will provide a sequence of valid MV assignments that start with µ and end with ν.
Proof: As previously described, the algorithm uses single face-flips and shwoops to generate a sequence of valid MV assignments of A 2n that, starting with µ, make the MV parity of creases e 1 , e 2 , e 3 , . . ., in order, agree with that of ν. We need to prove that (1) finding faces to perform a shwoop is always possible and (2) that when the algorithm terminates after i = 2n − 1, the resulting MV assignment will be ν.
Suppose we are at stage i = k in the algorithm where we have valid MV assignments µ F and ν for A 2n where F is the sequence of faces we've already flipped, µ F (e i ) = ν(e i ) for i = 1, . . . , k − 1, and µ F (e k ) ̸ = ν(e k ).
Then, if α k is flippable under µ F , we flip it so that µ F ∪{α k } (e k ) = ν(e k ) and move on to i = k + 1.
If we cannot flip α k under µ F , then that means, for example, that µ F is majority-mountain and e k and e k+1 are both valleys under µ F . So we look to see if we can flip face α k+1 under µ F . If that's not possible, then we look at face α k+2 , and continue in search of a flippable face α k+j under µ F for some j.
Suppose we get all the way to α 2n−1 without finding a flippable face under µ F . That means that µ F = ν on creases e 1 , . . . , e k−1 and, assuming µ F is majority-mountain, that µ F = −1 (valley creases) on e k , . . . , e 2n (since face α 2n−1 borders the creases e 2n−1 and e 2n ). Since µ F is a valid MV assignment, this means that ν must also be all valleys on e k , . . . , e 2n , for if it were anything else, then ν would have fewer valley creases than µ F and thus violate Maekawa's Theorem. This contradicts our assumption that µ F and ν disagreed on crease e k , and so our supposition is false.
Thus, we will find a face α k+j that is flippable under µ F where k + j is no more than 2n − 1. We then flip α k+j and perform a shwoop to be able to make a new MV assignment µ F ∪{α k+j ,α k+j−1 ,...,α k } that will agree with ν on crease e k .
We now examine how the algorithm terminates. The last face that could be flipped in this algorithm is α 2n−1 . Let µ x be the last MV assignment produced up to this point (so, after step i = 2n − 2 in the algorithm). For step i = 2n − 1, suppose that µ x (e 2n−1 ) = ν(e 2n−1 ). This means µ x and ν agree on all the creases e 1 , . . . , e 2n−1 , which implies that they must also agree on e 2n , for otherwise one of µ x or ν would not satisfy Maekawa's Theorem despite both being valid. Thus µ x = ν and the algorithm completes successfully.
Similarly, if µ x (e 2n−1 ) ̸ = ν(e 2n−1 ) then we must also have that µ x (e 2n ) ̸ = ν(e 2n ). Then flipping face α 2n−1 will make µ x,α2n−1 = ν, and this face-flip must be possible because ν is a valid MV assignment for A 2n . Thus the algorithm completes successfully in this case as well. The FEA-Shwoop(A 2n , µ, ν) algorithm uses a nested loop, each of which are O(n), and therefore the running time of the whole algorithm is O(n 2 ). Note that this algorithm does not necessarily produce the shortest path in OFG(A 2n ) between µ and ν, since the choice of the first crease e 1 to consider might not be the most efficient, for instance.

Diameter of OFG(A 2n )
There is a different algorithm that we could use to flip between any two valid MV assignments µ and ν of A 2n , one that also proves that the diameter of OFG(A 2n ) is n. We call this algorithm FEA-Halves(A 2n , µ, ν).
Recall from Section 2 that, if µ and ν are two valid MV assignments of A 2n , then S(µ, ν) is the set of creases e i with µ(e i ) ̸ = ν(e i ).
Lemma 3 If µ and ν are two valid MV assignments of A 2n , then |S(µ, ν)| is even.
Case 2: ν is majority-mountain. Then, Equation (3) will have +2 on its right-hand-side, and subtracting from Equation (1) gives This means that we have at least two valley creases in S(µ, ν) under µ. Let e im ∈ S(µ, ν) be a valley crease under µ, and let α j be the face in B(µ, ν) that borders e im . We claim that α j is a flippable face under µ: If the other crease bordering α j is also a valley under µ, then since µ is majority-valley, µ αj will be majority-mountain and still satisfy Maekawa's Theorem. If the other crease bordering α j is a mountain under µ, then µ αj is still majority-valley and satisfies Maekawa because µ did. In both cases we have that µ αj is a valid MV assignment. Then B(µ αj , ν) will have one fewer face than B(µ, ν), the missing face being α j , and the Lemma is proved. Find α ∈ L such that L is flippable under η Append α to S Replace η with η α and L with L \ {α} Output S Therefore, the FEA-Halves algorithm (see Algorithm 2) will input two valid MV assignments, µ and ν for A 2n and compute the set of faces B(µ, ν) = k j=1 B(e i2j−1 , e i2j ) as well as the complement set of faces (in A 2n ) B(µ, ν) = B(e i 2k , e i1 ) ∪ k−1 j=1 B(e i2j , e i2j+1 ). Since these form a disjoint union of all the faces in A 2n , one of B(µ, ν) and B(µ, ν) will have size ≤ n. Pick that set, say it's B(µ, ν), and apply Lemma 4 repeatedly to generate a sequence of at most n face flips that will transform µ into ν. This proves most of the following theorem.

Theorem 4 The flip graph OFG(A 2n ) is connected and has diameter n.
Proof: To see that the diameter of OFG(A 2n ) equals n, let µ be any valid MV assignment of A 2n and consider the complement MV assignment µ which is µ but with all the mountains and valleys reversed. To transform µ to µ, every crease needs to be flipped, and (since there are 2n creases and each face flip switches two creases) doing this this requires at least n face flips. Since every crease is flipped, this gives us a lower bound on the required number of face flips between nodes of OFG(A 2n ). The FEA-Halves algorithm guarantees at most n face flips to flip from µ to µ, so the diameter of OFG(A 2n ) is n. Examples that require n face flips can be readily found

Counting edges of OFG(A 2n )
We saw in Section 2 that OFG(A 2n ) has 2 2n n−1 vertices. Counting the edges in OFG(A 2n ) is not as straight-forward. We first perform this enumeration using the method shown in Algorithm 3, which we call Edge-Count(n). This takes each valid MV assignment µ of A 2n and uses Lemma 1 to compute the degree of µ in OFG(A 2n ): each vertex µ will have degree 2n unless there are nonflippable faces (bordered by "VV" if µ is majority-mountain or by "MM" if µ is majority-valley) which must then be subtracted from 2n. We then take the sum of the vertex degrees and divide by two to find the number of edges. When we first encountered it, this sequence Edge-Count(n) did not appear in the Online Encyclopedia of Integer Sequences. It has now been added by one of the authors and is sequence A352626 [15].
The running time of this algorithm is super-exponential in n, since it needs to calculate and check every valid MV assignment of A 2n . Fortunately, we can do better.

Theorem 5 The number of edges in OFG(
Note that the formula in Theorem 5 matches the output of Edge-Count(n). We prove this formula using a probabalistic approach.
Proof: In any uniformly chosen at random MV assignment of A 2n , some faces will be flippable and some will not be flippable. Define random variables G = the number of flippable faces in a MV assignment of A 2n (or "good" faces) and B = the number of unflippable faces (or "bad" faces). Also let 1 αi denote the indicator random variable for α i being a bad face. That is, B = 1 α1 + 1 α2 + · · · + 1 α2n . Then linearity of expectation gives us where V and E are the vertices and edges in OFG(A 2n ), respectively. Thus we have The Edge-Count(n) algorithm can be used to generate the degree sequence for OFG(A 2n ). Let f k (2n) denote the number of vertices of degree k in OFG(A 2n ), so that Edge-Count(n) = (1/2) 2n k=n−2 kf k (2n). The values for f k (2n) for 2 ≤ n ≤ 6 and the possible degrees k are shown in Table 1, and studying these led to the following formula.
Theorem 6 The number of vertices of degree k in OFG(A 2n ) is for n + 2 ≤ k ≤ 2n.  Table 1: Values for f k (2n) generated by running Edge-Count(n).
We provide a combinatorial proof of this result developed by Jonah Ostroff.
Proof: We will enumerate the number of valid MV assignments µ of A 2n that are majoritymountain with b non-flippable faces; such a vertex in OFG(A 2n ) will have degree k = 2n − b, and this enumeration will equal f k (2n)/2. In this situation we will have n + 1 mountains, n − 1 valleys, and by Lemma 1 there should be exactly b pairs of consecutive creases around A 2n that are "VV" under µ. That means there are exactly n − b − 1 valley creases that are not followed by a valley (say, going clockwise around the vertex). Therefore we are counting the number of ways to arrange mountains and valleys so that there are exactly n − b − 1 runs of consecutive valleys.
We can construct such MV assignments as follows: First we place the n + 1 mountains around a circle and mark one of them as the "start" point.
Then we place boxes in n − b − 1 of the n + 1 spaces between the mountains. This gives us a MV assignment with the required conditions, but we've only counted ones that "start" with a mountain crease. Call the set of these MV assignments A. We rotate each member of A around the A 2n crease pattern to get a bigger set of MV assignments, B, with 2n|A| elements. We claim that each MV assignment we are looking for (valid, majority-mountain with exactly b non-flippable faces) appears in B exactly (n + 1) times. To see this, let µ meet our required conditions and suppose µ has no rotational symmetry (meaning that each rotation of µ in A 2n is a MV assignment distinct from µ). Then a rotated version of µ will appear in A exactly (n + 1) times, since there are (n + 1) mountains in µ. These rotations of µ in A will result in exactly (n + 1) copies of µ appearing in B.
On the other hand, suppose µ has rotational symmetry, say r j (µ) = µ for some j that divides 2n, where r(µ) is µ rotated by π/n in A 2n . Let 2n = qj. Then a rotated copy of µ will appear in A exactly (n + 1)/q times (that is, it would be (n + 1) times, one for each mountain in µ, but every qth one is a duplicate because of the rotational symmetry). Each of these rotated copies of µ are rotated a full 2n times in B, each giving us q copies of µ in B. That's a total of q(n+1)/q = (n+1) copies of µ in B.
Therefore, the number of valid MV assignments of A 2n that are majority-mountain and have exactly b non-flippable faces is 2n n + 1 To include the majority-valley cases, we multiply by two. Substituting b = 2n − k and simplifying gives the desired result. □

Conclusion
We have seen how the origami flip graph of A 2n has the largest size among the flip graphs of flat vertex folds of degree 2n, that it contains all such origami flip graphs as subgraphs, and that it is a connected graph with diameter n. Furthermore, the algorithms used to prove these facts could be useful in further studies of origami flip graphs. For example, the FEA-Shwoop algorithm has the interesting property that it provides a way to flip between any two valid MV assignments of A 2n without flipping the face α 2n . Since the labeling of the faces was arbitrary, this means that we can always avoid flipping a chosen face and still traverse the origami flip graph. This feature is used in the forthcoming paper [10] to help classify when OFG(C) will be connected for arbitrary flat vertex folds C. Indeed, [10] also explores when the FEA-Shwoop algorithm can be used in other situations besides the crease pattern A 2n . Despite A 2n being, in a sense, the most simple case of all degree-2n flat vertex folds, as it requires only Maekawa's Theorem to determine if a MV assignment will be valid, its origami flip graph nonetheless exhibits surprising complexity. Further details on the structure of OFG(A 2n ) remains unexplored. For instance, Theorem 2 does not tell the whole story about the number of copies of OFG(C) that can be found in OFG(A 2n ).
Open Problem 1 If C is a flat vertex fold of degree 2n, how do we determine the exact number of distinct labeled subgraphs of OFG(A 2n ) that are isomorphic to OFG(C)?
There are various computational origami reconfiguration problems other than those already considered in this paper that have yet to be explored. Perhaps the next most basic one to consider is the following: Open Problem 2 What is the computational complexity of finding a shortest path between two vertices µ and ν in OFG(A 2n )?
Readers familiar with closed meanders might suspect a connection between them and flat foldings of A 2n . A closed meander of order n is a closed curve in R 2 that crosses a given directed line 2n times [6]. The cross-section of any flat folding of A 2n can be viewed as a closed meander if we draw a directed line through (and perpendicular to) the cross-section. However, the number of homeomorphically-distinct closed meanders of order n, denoted M n , is not equal to the number of vertices in OFG(A 2n ), since different layer orders of the paper give us different meanders but not different MV assignments, and the latter is all we care about in origami flip graphs. Still, there could be a way to extend our work in this paper to include layer orderings.
Open Problem 3 Is there a way to create a flip graph for the space of closed meanders of order n where the number of vertices is M n ? Or, equivalently, an origami flip graph for A 2n that also considers different paper layer orders as distinct foldings?
Also, we have seen that determining the degree sequence of OFG(A 2n ) involves the different ways to separate the valleys (assuming we're majority-mountain) into runs of consecutive valleys. In other words, if we have n − 1 valleys we are considering the integer partitions of n − 1. The different integer partitions affect f k (2n) for different k, so their influence is lost in Theorem 6. However, perhaps another connection is possible.
Open Problem 4 Can we clarify the role that integer partitions of n − 1 play in OFG(A 2n )?
This is further evidence, also seen in [7], that the single-vertex case of flat origami continues to possess more combinatorial richness than one would originally expect.