Planar lattice subsets with minimal vertex boundary

A subset of vertices of a graph is minimal if, within all subsets of the same size, its vertex boundary is minimal. We give a complete, geometric characterization of minimal sets for the planar integer lattice X. Our characterization elucidates the structure of all minimal sets, and we are able to use it to obtain several applications. We characterize uniquely minimal sets of X: those which are congruent to any other minimal set of the same size. We also classify all efficient sets of X: those that have maximal size amongst all such sets with a fixed vertex boundary. We define and investigate the graph G of minimal sets whose vertices are congruence classes of minimal sets of X and whose edges connect vertices which can be represented by minimal sets that differ by exactly one vertex. We prove that G has exactly one infinite component, has infinitely many isolated vertices and has bounded components of arbitrarily large size. Finally, we show that all minimal sets, except one, are connected.


Introduction
The classical isoperimetric problem can be stated as follows: amongst all closed curves in the plane with fixed length, characterize those that enclose the maximal area. The solution to this isoperimetric problem is the circle. By a simple scaling argument, this problem is easily seen to be equivalent to the following dual problem: Problem 1. Amongst all closed curves in the plane that enclose a fixed area, characterize those that have minimal length.
The isoperimetric problem dates back to antiquity, as documented in Virgil's account of the founding of Carthage in the Aeneid. However, the first steps towards a solution of Problem 1 were given relatively recently by Steiner in the 19 th century. In this article, we give a solution to the discrete graph-theoretic analogue of Problem 1.
Discrete isoperimetric problems have been studied extensively in graph theory, and there are many applications in areas such as network design and the theory of error correcting codes [Har04,HLW06]. Given a graph X with vertex set V (X), the vertex boundary of A ⊂ V (X) is defined by The isoperimetric problem for the integer lattice in the plane. In this article we study the graph X = Z 2 1 with vertex set X 0 = Z 2 and edges connecting all pairs of vertices 1 -distance one apart. A nested sequence of minimal sets for X is given by Wang-Wang [WW77].
Our approach differs from the usual one of finding a sequence of minimal sets, in that we give a geometric characterization of every minimal set. While circles are the natural geometric solution to Problem 1, there can be many different congruence classes of minimal sets in X of a given size and our result exactly describes these solutions. This approach lets us prove many applications that allow us to better understand the collection of all minimal sets.
Before describing our results, we first establish some notation. We consider subsets of X 0 up to the following natural equivalence relation: we say two subsets A, B ⊂ X 0 are congruent if there is a graph automorphism φ of X such that φ(A) = B. It is clear that if A and B are congruent, then A is minimal if and only if B is minimal.
Given natural numbers α, β ∈ N, we define B(α, β) to be the set of vertices (x, y) ∈ X 0 that satisfy 0 ≤ y − x ≤ α and 0 ≤ y + x ≤ β. Similarly, given even integers α, β ∈ N, we definê B(α, β) to be the set of vertices (x, y) ∈ X 0 that satisfy 0 ≤ y − x ≤ α and −1 ≤ y + x ≤ β − 1. A box is a non-empty subset of X 0 that is congruent to either B(α, β) orB(α, β) for some α, β. Since boxes are parametrized by numbers α, β ∈ N, it is easier to determine whether a box is minimal than it is to determine whether an arbitrary set of vertices is minimal. Therefore, our broad strategy in solving Problem 2 is to compare an arbitrary set to its enclosing box. We show that the enclosing box can be obtained by "saturating" a set, i.e. by adding vertices that do not increase the boundary. In the course of our proof, we classify precisely which boxes are minimal, see Remark 5.14.
X 0 \ A1 is not a union of cones N = 1, E = 1 A1 is not minimal X 0 \ A4 is a union of cones N = 10, E = 10 A4 is minimal X 0 \ A2 is a union of cones A cone is a subset of X 0 congruent to the set {(x, y) | y − x ≥ 0, y + x ≥ 0}. As a precursor to our main result, we give a necessary condition for minimality: if a set A is minimal, then its complement X 0 \ A is a union of cones and furthermore, its enclosing box enc(A) is also minimal. Although this is far from a complete classification of minimal sets -which we give in Theorem A -it demonstrates the important role of the enclosing box in determining minimality.
If A 1 is the set shown in Figure 1.1, its complement X 0 \ A 1 is not a union of cones. This follows as any cone containing the white vertex in enc(A 1 ) \ A 1 must also contain a vertex of A 1 , and so X 0 \ A 1 is not a union of cones. Thus A 1 is not minimal by the preceding necessary condition. Similarly, A 2 can be seen not to be minimal since its enclosing box enc(A 2 ) is not minimal. However, to say whether or not A 3 and A 4 are minimal is a slightly more delicate matter, since both A 3 and A 4 satisfy the preceding necessary condition for minimality. To see why A 3 is not minimal and A 4 is, we use a numerical invariant of a box called its excess.
Denoted Ex(B), the excess of a box B measures how much larger a box is than the smallest minimal set A with |∂A| = |∂B|; see Definition 4.1. In particular, the excess of a box is nonnegative if and only if the box is minimal. An explicit formula for the excess of a box is given in Theorem 5.13. If a box has width r − k and length r + k, then its excess is approximately r−k 2 2 . Thus, boxes that are sufficiently close to being squares have positive excess, whilst boxes that are sufficiently long and narrow have negative excess.
Our main theorem, stated below, gives two related characterizations of minimal sets in terms of their enclosing boxes.
Going back to the examples in Figure 1.1, a straightforward application of Theorem 5.13the formula for the excess of a box -tells us that A 3 does not satisfy N ≤ E, but A 4 does. Since X 0 \ A 3 and X 0 \ A 4 are both unions of cones, Theorem A can be used to deduce that A 3 is not minimal and A 4 is minimal.
Applications. A natural question to consider is whether minimal sets of a fixed size are unique up to congruence. More formally, A ⊂ X 0 is uniquely minimal if A is minimal and any minimal set containing the same number of vertices as A is congruent to A. We completely classify uniquely minimal sets: Theorem B (Theorem 8.8). A subset of X 0 is uniquely minimal if and only if it is congruent to either B(2n, 2n) or B(n, n + 1) for some n ∈ N.
As well as understanding individual minimal sets, we also want to understand the structure of the collection of all minimal sets. To do this, we initiate the study of the graph G of minimal sets. Vertices of G are congruence classes of minimal sets. Two vertices v and w in G are joined by an edge if there exist representative minimal sets A ∈ v and B ∈ w whose symmetric difference has size one. The graph G has a natural grading corresponding to the sizes of representative minimal sets. The induced subgraphs of G containing all congruence classes of minimal sets of size at most 10 and 41 are shown in Figures 1.2 and 1.3 respectively.
We exhibit the following features of G: Theorem C. Let G be the graph of minimal sets. Then: (1) G contains a single infinite component (Corollary 8.9); (2) G contains finite components of arbitrarily large height (Theorem 8.11), where the height of a component is the maximal length of a nested sequence of minimal sets in it; (3) G contains infinitely many isolated vertices (Theorem 8.11).
We note that the infinite sequence of nested minimal sets constructed by Wang-Wang is contained in the unique infinite component of G.
At the beginning of the introduction we mentioned two equivalent formulations of the isoperimetric problem in the Euclidean plane: maximizing the area enclosed by a curve of fixed length, The induced subgraph of G containing congruence classes of minimal sets of size at most 10. All boxes are labelled using Notation 5.5.
(a) The induced subgraph of G of congruence classes of minimal sets of size up to 41. The graded structure of the graph is shown where vertices representing sets of larger sizes appear above and are shaded with a darker color than those representing smaller sizes.  or minimising the length of a curve enclosing a fixed area. We thus consider the following discrete isoperimetric problem dual to Problem 2: Problem 3. Amongst all subsets of X 0 with a fixed vertex boundary, characterize those with maximal size.
We say a subset of X 0 is an efficient set if it is a solution to Problem 3. It turns out that Problems 2 and 3 are not equivalent for the integer lattice. Every efficient set is minimal (see Lemma 8.1), but it is not the case that every minimal set is efficient. We give an explicit solution to Problem 3.
Theorem D (Lemma 8.3). A subset of X 0 is efficient if and only if it is congruent to either B(n, n), B(n, n + 1) or B(2n, 2n + 2) for some n ∈ N.
While writing this article, we learned that Theorem D was essentially already known. Vainsencher-Bruckstein characterize sets that are both efficient and minimal, which they call Pareto optimal sets [VB08]. However, by Wang-Wang's result and an easy argument, it follows that efficient sets are always minimal, giving the above theorem. We note that our proof is independent of that of Vainsencher-Bruckstein.
The life expectancy of a minimal set A ⊂ X 0 is defined to be the maximum over all n such that there exists nested sequence (A i ) n i=0 of minimal sets with |A i | = |A| + i and A 0 = A. The life expectancy takes values in N ∪ {∞}. A minimal set is said to be mortal if it has finite life expectancy and immortal otherwise. A minimal set is said to be dead if its life expectancy is zero. We completely characterize mortal and dead sets: Theorem E. Let A ⊂ X 0 be a minimal set. Then: (1) A is dead if and only if it is a box that is not efficient (Theorem 8.5).
(2) A is mortal if and only if its enclosing box is dead (Proposition 8.6).
We note that by Theorem D we get an explicit characterization of mortal and dead sets in terms of box parametrizations.
A set A ⊂ X 0 is connected if its induced subgraph is connected. It can be seen in Figure 1.2 that the box B(0, 2) is minimal but is not connected. We show that, up to congruence, this is the only minimal set that is not connected.
Theorem F (Theorem 7.1). A minimal set in X is connected if and only if it is not congruent to B(0, 2).
Other Related Works. A complete solution to Problem 2 is known for very few graphs. Much of the literature has focused on exhibiting a sequence of minimal sets, i.e. a sequence (A n ) where each A n is a minimal set consisting of exactly n vertices. Finding such a sequence is NP-hard for a general graph (see [Har04]) and such sequences can typically be described only in special cases for graphs with an abundance of symmetry.
Harper exhibited a nested sequence of minimal sets for the d-dimensional hypercube Q d , where Q d is a graph on the vertex set {0, 1} d with an edge between a pair of binary strings that differ in a single coordinate [Har66]. This result was extended to (P q ) d , the d-fold product of paths on q vertices [Chv75, Mog83,BL91a] and to (K q ) d , the d-fold product of complete graph on q vertices [Har99]. In constrast with Q d , there is no nested sequence of minimal sets for (K q ) d .
The isoperimetric problem has also been studied on infinite graphs, including integer lattices. Let Z d 1 (respectively Z d ∞ ) be the graph on the vertex set Z d where two vertices are joined by an edge if their 1 -distance (respectively ∞ -distance) is 1. As already mentioned, Wang-Wang [WW77] exhibit a nested sequence of minimal sets in Z d 1 . Sieben gives a formula for the size of the vertex boundary of a minimal set of size n in this graph under the assumption that such sets are connected [Sie08]. This is then used to analyze strategies for what are called polyomino achievement games. Radcliffe-Veomett obtain a nested sequence of minimal sets in Z d ∞ [VR12]. The edge boundary of a subset of a graph is defined to be the set of edges that are incident to both a vertex of this subset and to a vertex outside this subset. The edge isoperimetric problem has also been well-studied for the various graphs mentioned above, namely, by [Har64,Lin64,Ber67,Har76] for Q d , by [Lin64] for (K q ) d and [BL91b] for (P q ) d and Z d 1 . Recently, [BE18] studied the edge isoperimetric problem for Z d ∞ and Z d with respect to any Cayley graph. Many of these preceding results use a technique called compression or normalization that replaces a vertex set A ⊂ V (X) with a set c(A) ⊂ V (X) such that |A| = |c(A)| and |∂A| ≥ |∂(c(A))|; see [Har66]. Whilst this technique is well-suited to finding a sequence of minimal sets, it does not generally allow one to give a structural characterization of all minimal sets.
Outline. In Section 2 we review the sequence of minimal sets constructed by Wang-Wang [WW77]. In Section 3 we introduce the notion of a saturated set. In Sections 4 and 5 we define the excess of a set and give an explicit formula for the excess of a box (Theorem 5.13). In Section 6 we show in Proposition 6.1 that all saturated minimal sets are boxes. Combined with our formula for the excess of box, we prove the first part of Theorem A, thus characterizing all minimal sets in terms of their enclosing boxes. In Section 7 we show that up to congruence, there is a unique disconnected minimal set, and we prove the second part of Theorem A. In Section 8 we study the graph G and classify which sets are efficient, uniquely minimal, dead and mortal.
Acknowledgements. The authors are thankful for helpful discussions with Joseph Briggs, who introduced us to Harper's Theorem. We also thank Nir Lazarovich for helpful comments and suggestions.
RG was supported by Israel Science Foundation Grant 1026/15 and EPSRC grant EP/R042187/1. IL was supported by the Israel Science Foundation and in part by a Technion fellowship. AM was supported by the Israel Science Foundation Grant No. 1562/19. ES was supported by the Azrieli Foundation, was supported in part at the Technion by a Zuckerman Fellowship, and was partially supported by NSF RTG grant #1849190.

Wang-Wang sets
We recall the nested sequence of minimal sets in X constructed by Wang-Wang.
Throughout this article, we fix the graph X with vertex set X 0 = Z 2 , where two vertices (x, y), (x , y ) ∈ X 0 are joined by an edge if and only if |x − x | + |y − y | = 1.
Wang-Wang gave a nested sequence, W W 1 ⊂ W W 2 ⊂ . . . , of minimal sets in X such that |W W n | = n for all n ≥ 1 [WW77]. Throughout this article, a Wang-Wang set is a subset A ⊂ X 0 that is congruent to W W n for some n. In the upcoming sections, we utilize them to prove our characterization of minimal sets in X.
We use Figure 2.1 to define the vertices x r(n)+1 , . . . , x r(n+1) . We first set x r(n)+1 to be the specific vertex adjacent to W W r(n) shown as vertex a in Figure 2.1. The vertices x r(n)+1 , . . . , x r(n)+n are those along the oriented line segment − → aA ordered by the given orientation. The next set of vertices are those along the segment − → bB, then those on − → cC, and finally those on − → dD (where each such sequence is again ordered by the given orientation).

Saturated sets
We define the notion of a saturated set, a subset of X 0 with the property that if any additional vertex is added to this set, then its boundary must increase.

Excess
We introduce excess, a number associated to a subset A ⊂ X 0 that we will use in later sections to characterize minimal sets and to study the structure of the graph of minimal sets.
The following lemma shows that the excess of A is well-defined.

Lemma 4.2. For any finite
Proof. We first claim that if n ≥ 6, then there exists a minimal set B with |∂B| = n. Indeed, as noted in the proof of Lemma 2.1, for every m ≥ 2 either If |A| = 1, then A is minimal. Otherwise, |A| ≥ 2 = |W W 2 |, so Lemma 2.1 ensures that |∂A| ≥ |W W 2 | = 6. By the preceding claim, there exists a minimal set B with |∂A| = |∂B|.
The next two lemmas follow almost immediately from the definition of excess.

Boxes
In this section, we define boxes. These are sets that are bounded by lines of slope 1 and −1. We prove some key facts regarding these sets and give an explicit formula for their excess. As a consequence, we determine which boxes are minimal sets.
Proposition 5.6 demonstrates that the following definition of a box is equivalent to the one given in the introduction.
Convention 5.2. There is an ambiguity when giving coordinates for boxes, which only arises for degenerate boxes of the form B(a, b, c, d) where either a = b or c = d. For example, B(0, 0, 0, 3) and B(0, 0, 0, 2) are the same box. To remedy this issue and ensure boxes can be uniquely parametrized, given a box B(a, b, c, d) we implicitly assume that a and c are maximal and that b and d are minimal out of all possible choices.
Given a subset A ⊂ X 0 , we define N (A) := A ∂A. We next show that boxes are saturated. In the next section, we prove a converse to this statement for minimal sets (see Proposition 6.1).
Lemma 5.3. Every box is saturated.
x and y do not satisfy one of the four defining equations of B. Without loss of generality, we assume x + y > d. It follows that (x + 1, y) and (x, y + 1) are adjacent to v and not contained in N (B).  When Convention 5.2 is followed, it is evident that the modulus of a box is well-defined and is invariant under congruence. We intuitively expect that a box of modulus {α, β} is minimal when |α − β| is small. We precisely quantify how small |α − β| must be in Theorem 5.13 and Remark 5.14.
We now show that a box of modulus {α, β} is congruent to a "standard box" B(α, β) or B(α, β) as defined below.
Proof. Suppose first that B contains no corners. Then the intersection of the line y − x = a with the line y + x = c does not have integer coordinates, so a and c have opposite parity. Similarly, we deduce that a and d have opposite parity and that b and c have opposite parity. Thus b − a and d − c must both be even. Let u be the vertex of B which lies on the the line y = x + a and has minimal y-value of all such possible choices. We can apply a translation which sends u to the origin (0, 0). The resulting box is then On the other hand, suppose that B contains a corner v. Then there exists an automorphism of X sending v to the origin that maps B to the box B(b − a, d − c).
The next two lemmas calculate the size of a box and its boundary.
We thus assume that α ≥ 1. If α = β = 1, then B is congruent to B(1, 1) and the formula |∂B| = α + β + 4 = 6 clearly holds. We thus also assume that β ≥ 2 and proceed by induction on β. We assume that the lemma is true for all boxes of modulus {α , β }, where max(α , β ) < β. Let L and L + be the lines with equations y = −x + d and y = −x + d + 1 respectively. Let Moreover, if α and β are both even, then Proof. Let p : X 0 → Z be the projection map given by (x, y) → y − x. Let B be a box and let I be the interval p(B). Thus |B| = i∈I |p −1 (i) ∩ B|. We break the proof into cases depending on the type of box B and the parity of α and β.
We first analyze the case where B = B(α, β). Suppose β is even. It follows that |p −1 (i)| = β 2 +1 for even i ∈ I and |p −1 (i)| = β 2 for odd i ∈ I. Thus, if α is even, then If α is odd, we have: The last equality follows since αβ+α+β+1 2 is equal to |B| and hence it is an integer. On the other hand, suppose that β is odd. In this case, it follows that |p −1 (i)| = β+1 2 for all i ∈ I. Thus, Finally, let B =B(α, β) where both α and β are even. It follows that |p −1 (i)| = β 2 for even i ∈ I and |p −1 (i)| = β 2 + 1 for odd i ∈ I. Thus, Remark 5.9. If α, β ∈ N are both even, then by Lemma 5.7 and Lemma 5.
The following lemma allows us to take a nested sequence of subsets of a box, all with the same size boundary. This lemma will be used in Sections 7 and 8.3 as well as here.
Lemma 5.10. Let B be a box of modulus {α, β}, and let L be an extremal line of B. Set n = |L ∩ B| − 1. Then the following are true: Proof. Without loss of generality, we may assume that B = B(a, b, c, d) and that L is the line with equation y = −x + d.
We now suppose that β := d − c ≥ α := b − a and that α ≤ 1, and we prove (2) for this remaining case. As before, let L be the line with equation y = −x + d. Since α ≤ 1, L intersects B in a single vertex v. It follows that B := B \ v is a box and is of strictly smaller modulus. Thus Lemma 5.7 implies that |∂B | < |∂B|. It follows from Lemma 2.1 that any minimal set of size at most |B| − 1 has boundary of size at most |∂B| − 1. This implies Ex(B) ≤ 0 as required. If Q is any other extremal line of B, then |Q ∩ B| ≥ |L ∩ B| = 1 and we also get that Ex(B) ≤ |Q ∩ B| − 1. Thus, (2) follows.
Finally, to see (3), suppose L is a standard line that intersects B, and let L be the extremal line of B that is parallel to L . It follows that |L ∩ B| ≤ |L ∩ B| + 1. By what we have shown, we get that Ex(B) ≤ |L ∩ B| − 1 as required.
Remark 5.11. By considering the box B = B(2, 2), which has excess one, we see that the bounds for Ex(B) given in the previous lemma are sharp.
To calculate the excess of an arbitrary box, we first compute the excess of a box with modulus {α, β}, where |α − β| ≤ 1.
Proof. We first remark that if A ⊂ X 0 is minimal, then |∂A| = |∂W W |A| |. Thus for any minimal Let Y be one of B(2n, 2n), B(2n + 2, 2n + 3) , B(2n + 1, 2n + 1) or B(2n + 1, 2n + 2). Then Y is congruent to a Wang-Wang set W W m for some m. We are now ready to calculate the excess of any box.
Setting r = 2m + where ∈ {0, 1} and m ∈ Z, we get the equation: It follows from Lemma 5.7 that |∂B(r, r)| = |∂B(r − k, r + k)|. Lemma 4.4 then implies When r is even (and so = 0), Ex B(r, r) = r 2 by Lemma 5.12. Substituting this into the above equation yields When r is odd (and so = 1), Ex B(r, r) = r−1 2 by Lemma 5.12. Thus, Case B: α and β have different parity. We can write r = s + 1 2 and k = t + 1 2 , for some s, t ∈ Z. Note that r = s, so we need to show The last equality follows since t 2 + t is even. Lemma 5.8 now yields |B (s, s + 1)| = s 2 + 3s + 3 2 where the last equality follows again because t 2 + t is even.
Remark 5.14. By combining Lemma 4.3 and Theorem 5.13, we have a complete characterization of which boxes are minimal.

Characterizing minimal sets
In this section we prove Theorem 6.4, giving our first characterization of minimal sets in the graph X = Z 2 1 . We also prove Proposition 6.7, which characterizes boxes as precisely the sets that are both saturated and ∞ -connected.
We first explain how to deduce Theorem 6.4 from the following proposition, whose proof occupies the remainder of this section.
Proposition 6.1. If A ⊂ X is minimal and saturated, then it is a box.
Definition 6.2. Given a finite set A ⊂ X 0 , the enclosing box, denoted enc(A), is the smallest box that contains A.
The enclosing box of a set is well-defined, as the intersection of boxes is itself a box. The enclosing box of a minimal set is the unique smallest saturated set containing it: Lemma 6.3. Let A ⊂ X 0 be a minimal set, and let A = A 0 ⊂ A 1 ⊂ A 2 ⊂ . . . be a (possibly finite) maximal sequence of nested minimal sets such that |A i+1 | = |A i | + 1 for each i. Then A N = enc(A) for some N ≥ 0 and |∂A| = |∂(enc(A))|.
Proof. Let N ≥ 0 be the largest integer such that |∂A N | = |∂A|. Such an integer exists as there are minimal sets with arbitrarily large boundaries (see the proof of Lemma 4.2 for instance). To prove the lemma, it suffices to show that A N = enc(A).
We first claim that A N must be saturated. For a contradiction, suppose otherwise. It follows there exists a set A N ⊃ A N such that |∂A N | = |∂A N | and |A N | = |A N | + 1. Furthermore, by Lemma 2.1, A N is minimal. By the maximality of our nested sequence, it contains a set A N +1 . As A N +1 is minimal and |A N +1 | = |A N |, it follows that |∂A N +1 | = |∂A N |, contradicting our choice of N .
As A N is saturated and minimal, it is a box by Proposition 6.1. If enc(A) = A N , then enc(A) must be a proper subset of A N . However, in this case, we then have that |∂(enc(A))| < |∂A N | by Lemma 5.7, contradicting our choice of N . Thus A N = enc(A), and the lemma follows.
We are ready to prove our first characterization of minimal sets, using Proposition 6.1. We now begin our proof of Proposition 6.1. We first establish some terminology regarding two metrics on X 0 : the 1 -metric and the ∞ -metric. We say that two vertices in X 0 are adjacent if their distance is exactly 1 in the 1 -metric. An 1 -path is a sequence (u i ) n i=0 of elements of X 0 such that u i−1 and u i are adjacent for every 0 < i ≤ n. We say A ⊂ X 0 is connected if any x, y ∈ A are the endpoints of an 1 -path contained in A.
Two vertices u, v ∈ X 0 are said to be ∞ -adjacent if d ∞ (u, v) = 1. An ∞ -path is a sequence (u i ) n i=0 of elements of X 0 such that u i−1 and u i are ∞ -adjacent for every 0 < i ≤ n. A subset A ⊂ X 0 is ∞ -connected if any pair of vertices in A are the endpoints of an ∞ -path contained in A. An ∞ -component of A is a maximal ∞ -connected subset of A.
A standard line is a set of the form {(x, y) | y − x = w} or {(x, y) | y + x = w} for some w ∈ Z. The next two lemmas will be needed to prove Proposition 6.7, our characterization of boxes.
Lemma 6.5. Suppose A ⊂ X 0 is saturated and C ⊆ A is an ∞ -component of A that is not contained in a standard line. Then C contains a pair of adjacent vertices.
Proof. As C is ∞ -connected and not contained in a standard line, it contains an ∞ -path γ = (u 0 , . . . u n ) such that u 0 and u n do not lie on the same standard line. We assume no u i is adjacent to u i+1 , otherwise we are done. Thus there exists an i such that u i−1 , u i and u i+1 do not lie in the same standard line. By applying an automorphism of X, we may assume u i−1 = (1, −1), u i = (0, 0) and u i+1 = (1, 1). As A ⊇ C is saturated, Lemma 3.2 ensures that (1, 0) is contained in A. As C is an ∞ -component of A, (1, 0) ∈ C. We are done as (1, 0) is adjacent to u i . Lemma 6.6. If A ⊂ X 0 is a finite saturated set, then every ∞ -component of A is a box.
Proof. Let C be an ∞ -component of A. If C is contained in a standard line, then we are done. Otherwise, by Lemma 6.5, C contains two adjacent vertices. In particular, C contains a box that is not contained in a standard line. Thus, up to congruence, C contains a box B = B(a, b, c, d) that is not contained in a standard line and is maximal, i.e. is not contained in any other box contained in C. We will show that B = C.
Assume for a contradiction that B is a proper subset of C. As C is ∞ -connected, there exists an ∞ -path from B to C \ B which is contained in C. Thus, there are vertices w = (x, y) ∈ B and v = (x , y ) ∈ C \ B which are ∞ -adjacent. By symmetry, we may assume that x + y > d.
We claim that we may assume that x + y = d. For suppose that x + y = d, then as w is ∞ adjacent to v, we have x + y = d − 1 and v = (x + 1, y + 1). At least one of (x, y + 1) or (x + 1, y) is contained in B, since B is not contained in a standard line. By replacing w with such a vertex, we may assume x + y = d. Figure 6.1 We now show that the box B := B(a, b, c, d+1) is contained in C, contradicting the maximality of B. We first show that C contains a vertex of B \ B. If v ∈ B , then we are done. Otherwise, either x + y = d + 2, or y − x < a, or y − x > b. If y − x < a and x + y = d + 2 -as is shown in Figure 6.1 -then v = (x + 1, y) and y − x = a. As B is not contained in a standard line, (x − 1, y + 1) ∈ A. Thus by Lemma 3.2, z := (x, y + 1) ∈ A. Since C is an ∞ -component of A, we have that z ∈ C ∩ (B \ B). If y − x > b and x + y = d + 2, then a similar argument demonstrates that B \ B contains a vertex of C. Finally, if x + y = d + 2, then v = (x + 1, y + 1). As B is not contained in a standard line, either (x + 1, y − 1) ∈ B or (x − 1, y + 1) ∈ B. By Lemma 3.2, (x + 1, y) ∈ A in the first case and (x, y + 1) ∈ A in the second. In either case, using the fact that C is an ∞ -component of A, we deduce there exists a vertex in C ∩ (B \ B).  B(0, 8, 0, 4). The three overlapping Swiss crosses can be used to deduce, via Lemma 3.2, that if C contains a single vertex of B \ B, then B ⊂ C.
The following proposition characterizes ∞ -connected, saturated subsets of X 0 as boxes.
Proposition 6.7. A finite subset A ⊂ X 0 is a box if and only if it is ∞ -connected and saturated.
Proof. If A is a box, then it is finite, ∞ -connected and saturated by Lemma 5.3. For the other direction, Lemma 6.6 implies that an ∞ -connected saturated set is a box.
The next two lemmas, together with Proposition 6.7, show that a finite minimal saturated subset of X 0 is a box (Proposition 6.1).
Lemma 6.8. Let C 1 and C 2 be distinct ∞ -components of a finite, saturated subset A ⊂ X 0 . By Lemma 6.6, C 1 and C 2 are boxes. Let (x, y) ∈ ∂C 1 ∩ ∂C 2 . Up to applying an automorphism of X, we may assume that (x − 1, y) ∈ C 1 . Then (x − 1, y) is a corner of C 1 and (x + 1, y) is a corner of C 2 . Moreover, if C is an ∞ -component of A with (x, y) ∈ ∂C, then C is equal to either C 1 or C 2 .
Proof. Note that (x, y) / ∈ A and (x, y ± 1) / ∈ A, for otherwise C 1 and C 2 would be joined by an ∞ -path in A. Thus it must be the case that (x + 1, y) ∈ C 2 . In particular, the only vertices of A adjacent to (x, y) are (x − 1, y) and (x + 1, y). It follows that the only ∞ -components of A which contain (x, y) in their boundary are precisely C 1 and C 2 . Moreover, (x − 1, y ± 1) / ∈ A since A cannot contain configuration (d) in Figure 3.1 by Lemma 3.2. We deduce that (x − 1, y) and (x + 1, y) are corners of C 1 and C 2 respectively.
For the next lemma, we let c(A) denote the number of ∞ -components of a subset A ⊂ X 0 . Lemma 6.9. Let A be a finite saturated subset of X 0 . Then there exists a set A ⊂ X 0 such that |A | = |A| and |∂A | ≤ |∂A| − (c(A) − 1). If in addition A is minimal, then it is ∞ -connected.
Proof. We first recall that every ∞ -component of A is a box by Lemma 6.9. We prove the lemma by induction on c(A) = n. The base case n = 1 trivially follows by setting A = A. When n = 2, then A is obtained from A by translating one of the two ∞ -components to reduce the boundary of the set by 1 (this is possible as these components are boxes). Now suppose c(A) > 2 and that for all finite saturated sets S, with c(S) < c(A) there exists a set S ⊂ X such that |S | = |S| and |∂S | ≤ |∂S| − (c(S) − 1).
Given a box B = B(a, b, c, d) we say that the line y = −x + d is the NE extremal line of B and that the line y = −x + c is a SW extremal line of B. Consider the smallest box containing A and let L be its NE extremal line. Let C be an ∞ -component of A which contains a vertex of L. Since A is saturated, so is C and hence by Proposition 6.7, the set C is a box. LetC = A \ C. By Lemma 6.8, any vertex of ∂C ∩ ∂C is adjacent to a corner of C that does not lie on L. Thus, |∂C ∩ ∂C| ≤ 2, and we get that: |∂C| + |∂C| ≤ |∂A| + 2. We will now show that we can replaceC by a different set so that the union of this set with C satisfies the claim. If |C| = 1, then c(A) = 2 and we reduce to a base case. If |C| = 2, then set W := {(0, 0), (−1, 1)}. Otherwise set W = W W |C| . Now since A is saturated,C is also saturated and we have c(C) = c(A) − 1 = n − 1. By the induction hypothesis, there exists a set P such that |P | = |C| and |∂P | ≤ |∂C| − (n − 2). As W is minimal, |∂W | ≤ |∂P |. Thus, |∂W | ≤ |∂C| − (n − 2).
We now translate W and C appropriately in order to define A . Let Q be the line given by y +x = 0. Since W is a Wang-Wang set of size greater than two or is congruent to {(0, 0), (−1, 1)}, we can apply an automorphism so that W ⊂ {(x, y) | y + x ≤ 0} and |W ∩ Q| ≥ 2. Thus there exists some v ∈ ∂W that is adjacent to two vertices of W ∩ Q. Let u ∈ C be a vertex on the SW extremal line of C. By translating C, we can suppose that u and v coincide. Setting A := W ∪ C, we get (see Figure 6.3) |∂A | ≤ |∂C| + |∂W | − 3. Thus and This concludes the proof of the lemma.
Proof of Proposition 6.1. Suppose A ⊂ X 0 is minimal and saturated. By Lemma 6.9, A is ∞ -connected. Lemma 6.6 now implies that A is a box.

Minimal sets are connected
Recall that a subset A ⊂ X 0 is connected if the subgraph of X induced by A is connected. In this section we prove that (almost) all minimal sets are connected. Additionally, we use similar ideas to prove another characterization of minimal sets in terms of cones.
Theorem 7.1. A minimal set in X is connected if and only if it is not congruent to B(0, 2). Before proving the above theorem, we need to first define cones and prove a series of lemmas.
Definition 7.2. A cone is a subset of X 0 congruent to C 0 := {(x, y) | y − x ≥ 0, y + x ≥ 0}. An extremal ray of the cone C 0 is the intersection of C 0 with either the line y = x or y = −x. An extremal ray of a cone is the image of an extremal ray of C 0 under the given congruence. A cone at the vertex v ∈ X 0 is a cone whose two extremal rays intersect at v. If v ∈ X 0 , then the cone above v is a cone at v that is translation-equivalent to C 0 ; the cone to the right of v is a cone at v that is translation-equivalent to C 0 rotated clockwise 90 • . The cones below v and to the left of v are defined analogously.
Lemma 7.3. Let A ⊂ X 0 be a finite set such that |enc(A) \ A| < Ex(enc(A)) and X 0 \ A is a union of cones. Suppose v ∈ A and there is a cone C based at v such that C ∩ A = {v}. Without loss of generality, we can suppose v = (0, 0) and that C is the cone above v. Then (−1, −1), (0, −1), (1, −1) ∈ A.
w 0 ∈ A. If w 0 / ∈ A, then w 0 is contained in a cone C 0 disjoint from A, which we may assume is at w 0 . Since v / ∈ C 0 , C 0 must either be below, to the left or to the right of w 0 . In either of the latter two cases, C 0 ∪ C contains either the intersection of the extremal line y − x = b with B or the intersection of the extremal line y + x = d with B. As, |(C 0 ∪ C) ∩ B| ≤ |B \ A | ≤ Ex(B), this contradicts Lemma 5.10(2). Thus we may assume that C 0 is the cone below w 0 .
Note that v does not lie on the extremal line y − x = a nor the extremal line y + x = c, for otherwise C contains the intersection of an extremal line with B and we get a contradiction as in the previous paragraph. Let L be the line of slope −1 passing through v. Let φ : X 0 → X 0 be the translation (x, y) → (x − 1, y). As v does not lie on y − x = a or y + x = c, given a vertex u ∈ L ∩ B it follows that either u ∈ C ∩ B ∩ L or φ(u) ∈ C 0 ∩ B. It follows that As v does not lie on y − x = a or y + x = c, either L is an extremal line of B or (0, 1) ∈ B. If L is an extremal line, then this contradicts Lemma 5.10(2). If not, then (0, 1) ∈ C ∩ (B \ L), so |C ∩ L ∩ B| < |C ∩ B|. Thus the inequality in (7.1) is strict, contradicting Lemma 5.10(3). We deduce that w 0 ∈ A. w 1 ∈ A. Suppose w 1 / ∈ A. Then there is a cone C 1 based at w 1 that does not intersect A. Since v / ∈ C 1 , C 1 must be either below or to the right of w 1 . In either case, if L is the line through v and w 1 , then B ∩ L ⊆ C C 1 and so If L an extremal line of B, then (7.2) contradicts Lemma 5.10(2). We thus assume L is not an extremal line of B.
We claim that the inequality in (7.2) is strict. If we show this, then we get a contradiction by Lemma 5.10(3), and we can deduce that w 1 ∈ A as required. We first observe that as L is not an extremal line of B and as w 0 = (0, −1) ∈ B (by what we have already shown), we must have that (1, 0) ∈ B. There are now two cases depending on whether or not w 1 ∈ B. If w 1 / ∈ B, then the line y − x = −1 = a through (0, −1) and (1, 0) is an extremal line of B. Furthermore, as Ex(B) > 0, it follows from Lemma 5.10(2) that b − a ≥ 2. Thus, we conclude that (0, 1) ∈ (B ∩ C) \ L. We thus deduce that |C ∩ L ∩ B| < |C ∩ B| and that the inequality in (7.2) is strict, proving the claim when w 1 / ∈ B. If w 1 ∈ B \ A, then Ex(B) ≥ |B \ A| + 1 ≥ 2. Lemma 5.10(2) then implies that every extremal line of B contains at least three vertices, and so b − a, d − c ≥ 4. As v, w 1 , (0, −1), (1, 0) ∈ B, it follows that B \ L must intersect at least one of C or C 1 , and so either |C ∩ L ∩ B| < |C ∩ B| or |C 1 ∩ L ∩ B| < |C 1 ∩ B|. In either case, we deduce as before that the inequality in (7.2) is strict as required.
Lemma 7.4. Up to congruence, B(0, 2) is the only disconnected box that is a minimal set. In particular, if B is a box with Ex(B) > 0, then B is connected.
Proof. Suppose B is a disconnected box that is a minimal set. Then it contains more than one vertex and is contained in a standard line. Thus B is congruent to B(0, 2n) for some n > 0. Note that Ex(B) = n(1−n) 2 by Theorem 5.13. Lemma 4.3 implies n = 1 and so B = B(0, 2). Furthermore, as Ex(B) = 0, the second claim follows.
The next lemma describes the geometry of a minimal set. We now state a characterization of minimal sets using cones.
Let For contradiction, suppose A is not minimal. Then A D, so pick v ∈ D \ A. As X 0 \ A is a union of cones, there exists a cone C containing v and disjoint from A. Without loss of generality, we may suppose C faces upwards (i.e. it lies above some vertex) and that v ∈ D ∩ C has maximal y-coordinate out of all vertices in D ∩ C . Let C be the cone above v. Since C ⊂ C and v has maximal y coordinate, C is a cone above v such that C ∩ D = {v}.
Since D is a minimal set, Lemma 7.5 ensures X 0 \ D is a union of cones. Moreover, since Thus A is minimal as required.

The graph of minimal sets
In this section, we study the graph of minimal sets, G (see the introduction for a definition).
8.1. Dead and mortal sets. A finite subset A ⊂ X 0 is efficient if for every B ⊂ X 0 , |∂B| = |∂A| implies |A| ≥ |B|, and we say that A is inefficient otherwise. Equivalently, A is efficient if |∂B| = |∂A| implies that Ex(A) ≥ Ex(B) by Lemma 4.4. The main result of this subsection is Theorem 8.5, which characterizes dead sets (defined in the introduction) as inefficient sets, which in turn are classified in terms of specific boxes (Lemma 8.3), and Theorem 8.6 characterizing mortal sets.
We first show that efficient sets are boxes and are minimal: We say A ⊂ X 0 is a Wang-Wang box if it is simultaneously a box and a Wang-Wang set.
Remark 8.2. A subset of X 0 is a Wang-Wang box if and only if it is congruent to either B(m, m) or B(m, m + 1) for some m ∈ Z (recall Notation 5.5).
We now characterize efficient sets. Proof. Let B be an efficient set. By Lemma 8.1, B is a box. Remark 5.9 implies that no box of the formB(α, β) is efficient. It thus follows from Proposition 5.6 that B is congruent to B(α, β) for some α, β ∈ Z. Without loss of generality, we may assume that β ≥ α, and we set r = α+β 2 and k = β−α 2 . Note that α = r − k and β = r + k. Suppose first that α and β have the same parity. Then r and k are both integers. By Lemma 5.7, |∂B(r − k, r + k)| = |∂B(r, r)|. Since B is efficient, Theorem 5.13 gives: Ex(B(r − k, r + k)) = r − k 2 2 ≥ Ex(B(r, r)) = r 2 .
By the above equation, either k = 0, or r is odd and k = 1. Thus B is either the Wang-Wang box B(α, α) or the box B(m − 1, m + 1) for m odd, respectively.
Before proving the next result, we first show that one can always add a vertex to a box such that the resulting set has boundary one larger than the box, as long as the box contains at least two vertices.
Proof. Let (x, y) be a vertex of B with y maximal. Suppose first that (x + 1, y) ∈ B. Since B is saturated by Lemma 5.3, it follows from Lemma 3.2 that (x − 1, y) / ∈ B. The claim now follows for this case by noting that |∂ B ∪ {(x, y + 1)} | = |∂B| + 1. A similar argument shows the claim when (x − 1, y) ∈ B.
We now give a characterization of dead sets.
Theorem 8.5. Let A ⊂ X 0 be a minimal set. The following are equivalent: (1) A is dead.
(2) A is an inefficient box.
(3) A is a box that is not congruent to either a Wang-Wang box or to a box of the form B(m − 1, m + 1) for odd m ∈ Z.
Proof. Suppose A is dead. We first show that A is saturated. Let v ∈ X 0 \ A. If |∂(A {v})| ≤ |∂A|, then Lemma 2.1 would imply that A {v} is minimal, contradicting the hypothesis that A is dead. Thus |∂(A {v})| > |∂A| for all v ∈ X 0 \ A, ensuring that A is saturated. Since A is saturated and minimal, Proposition 6.1 implies it is a box. Since a set with one vertex is not dead, |A| > 1. Hence Lemma 8.4 ensures that there exists a vertex v ∈ X 0 \ A such that |∂(A ∪ {v})| = |∂A| + 1. Since A is dead, A ∪ {v} is not minimal and so there exists a minimal set C such that |C| = |A| + 1 and |∂C| < |∂A| + 1. As A is minimal, Lemma 2.1 implies |∂C| = |∂A|. Thus A is an inefficient box. Now suppose A is an inefficient box. Since A is saturated by Lemma 5.3, |∂(A ∪ {v})| > |∂A| for every v ∈ X 0 \ A. As A is inefficient, there exists a set C such that |C| > |A| and |∂C| = |∂A|. Lemma 2.1 now implies that A ∪ {v} isn't minimal for any v ∈ X 0 \ A. Hence, A is dead.
The equivalence of (2) and (3)  Consequently, in order to prove the theorem, we need to show that a box B is mortal if and only if it is dead. Since dead sets are mortal, this reduces to demonstrating that a box which is not dead is immortal. By Theorem 8.5 we only need to show that Wang-Wang boxes and B(m − 1, m + 1), for odd m ∈ Z, are immortal sets. Wang-Wang boxes are immortal because they are contained in the infinite nested sequence of minimal sets (W W n ) ∞ n=1 . Let B = B(m − 1, m + 1) for some odd m ∈ Z. Let v = (0, m) and B := B {v}. Since v is not contained in B but is adjacent to (0, m − 1), (1, m) ∈ B, it follows that |∂B | = |∂B| + 1. By Lemma 5.7, Lemma 5.8 and as m is odd, |B| = |B(m, m)| and |∂B| = |∂B(m, m)|. Since B(m, m) is a Wang-Wang box and in particular, it is saturated, then any minimal set of size |B| + 1 must have boundary strictly greater than |∂B|. Thus, B is a minimal set. By Lemma 6.3, there exists a sequence of minimal sets B ⊂ B ⊂ · · · ⊂ enc(B ) such that the size of the symmetric difference between consecutive sets in this sequence is one. As enc(B ) = B(m, m + 1) is a Wang-Wang set (see Remark 8.2), it is immortal. Thus, B is immortal as well.
8.2. Uniquely minimal sets. In this subsection, we characterize uniquely minimal sets in X. Recall from the introduction that the grading of a vertex of G is the size of one of its representatives, and uniquely minimal sets correspond exactly to vertices of G that are unique out of vertices of the same grading.
Lemma 8.7. Let W W n be a Wang-Wang set that is not a box. Then there exists a minimal set A such that |A| = n and A is not congruent to W W n .
Thus, we have that |B| − 1 ≤ |B | ≤ |B| in both the case that α = β and that β = α + 1. Note that the line y = x contains β+1 2 + 1 vertices of B . Since k ≤ α 2 < β+1 2 , by Lemma 5.10 there exists some set A such that |∂A| = |∂B| = |∂W W n |, |A| = |B| − k = |W W n |, and enc(A) = B . In particular, A must be minimal. Since B = enc(A) and B = enc(W W n ) are not congruent, A and W W n are not congruent.
Theorem 8.8. A subset of X 0 is uniquely minimal if and only if it is congruent to either B(2n, 2n) or B(n, n + 1) for some n ∈ N.
Proof. Suppose A is uniquely minimal. Since W W |A| is minimal and |W W |A| | = |A|, A must be congruent to W W |A| . By Lemma 8.7, A must be a box. We note that A cannot be congruent to B(r, r) for odd r ∈ N, since Lemma 5.7 and Lemma 5.8 imply that |B(r − 1, r + 1)| = |B(r, r)| and |∂B(r − 1, r + 1)| = |∂B(r, r)|. Thus A is a box of the form B(2n, 2n) or B(n, n + 1) for some n ∈ N by Remark 8.2.
For the converse, suppose B is congruent to B(2n, 2n) or B(n, n + 1) for some n ∈ N. In particular, B is congruent to a Wang-Wang set, so it is minimal. Suppose A is another minimal set such that |A| = |B|. It follows that |∂A| = |∂B|. By Lemma 8.3 B is efficient. Since |A| = |B|, we get that A is also efficient. Furthermore, Lemma 8.3 also implies that any efficient set of size |B| is actually congruent to B, hence A is congruent to B.
Corollary 8.9. The graph G contains exactly one infinite connected component.
Proof. Let C be an infinite component of G. As there are only finitely many sets (up to congruence) of any given size, there exists a number m 0 such that C contains a vertex of grading m for every m ≥ m 0 . By Theorem 8.8, B(2n, 2n) is uniquely minimal for every n. Thus, C contains B(2n, 2n) for every n sufficiently large, and so C is the unique infinite component of G.
8.3. Finite components. In this subsection, we show that G contains infinitely many isolated vertices and finite components with arbitrarily many vertices. To do so, we prove the following more general result that gives sufficient conditions for a vertex of G to be contained in a finite component, i.e. a component of G that is a finite subgraph. Moreover, the possible gradings of vertices in this component is exactly described. Recall from the introduction that the height of a component is the maximal length of a nested sequence of minimal sets in it.
(4) B is a dead set Let C be the component of G containing the vertex representing B. Then, for any vertex in C representing a set A, we have that |B| − d ≤ |A| ≤ |B| and that enc(A) is congruent to B. Furthermore, C contains a vertex representing a set of size k for every |B| − d ≤ k ≤ |B|. In particular, C is finite and has height exactly d + 1.
Proof. We first prove the final claim of the theorem. Let B = B 0 ⊃ B 1 ⊃ · · · ⊃ B n be the sets given by Lemma 5.10 where |∂B i | = |∂B| and n = |L ∩ B| − 1 ≥ d + 1 (where L is an extremal line of B). As Ex(B) = d, we have that B i is minimal for each i ≤ d. The claim follows. Now let C be the set of all vertices in C represented by a set C such that there exists a path C = C 0 , . . . , C n = B in G with n ≤ d and |C i+1 | = |C i | + 1 for all 0 ≤ i < n. As Ex(B) = d, for all C ∈ C we must have that |∂C| = |∂B| and, consequently by Lemma 6.3, we have that enc(C) = B. Thus, to prove the remaining claims of the theorem, it is enough to show that C = C. Additionally, as C is connected, it is enough to show that given any vertex v in G, represented by a set A, that is adjacent to a vertex in C , represented by a set C, then v ∈ C . Let A and C be such sets.
Suppose first that A ⊂ C. Then |A| = |C| − 1 ≥ |B| − d − 1 (by the definition of C ) and by (3) it follows that A contains a vertex in every standard line which has non-empty intersection with B. Thus enc(A) = B. However, by Theorem 6.4, we must have that |A| ≥ |B| − d. Consequently, v ∈ C . On the other hand, suppose that C ⊂ A. As enc(C) = B and as B is dead, we must also have that v ∈ C . Thus, C = C as claimed.
Theorem 8.11. The graph G has finite components of arbitrarily large height and it contains infinitely many isolated vertices.
Proof. Let α = 2l 3 + l 2 + l and β = 2l 3 + l 2 − l for some integer l ≥ 3. Note that α and β are always positive. Consider the box B = B(α, β). Then B is not congruent to the box B(m − 1, m + 1) for any odd integer m, and B is not congruent to a Wang-Wang box. Therefore by Theorem 8.5, B is a dead set.
By Proposition 5.13, we have that Ex(B) = l 3 . Furthermore, given any standard line L, B ∩ L is either empty or contains at least min( α 2 + 1, β 2 + 1) = l 3 + l 2 2 − l 2 + 1 vertices. In particular, as l ≥ 3, B ∩ L is either empty or contains at least l 3 + 2 = Ex(B) + 2 vertices. Thus by Proposition 8.10, the component of G containing B is finite and has height at least l 3 . As this is true for any l ≥ 3, the first claim follows. Now let α = k 2 + k and β = k 2 − k for some integer k ≥ 4. Then we claim that the box B = B(α, β) is an isolated vertex of G. By Theorem 5.13, Ex(B) = 0. As above, for any standard line L, B ∩ L is either empty or contains at least 2 vertices. Also B is a dead set by Theorem 8.5. Therefore by Proposition 8.10, B is an isolated vertex of G. Thus G contains infinitely many isolated vertices.