{'fraction_non_alphanumeric': 0.08348161328588374, 'fraction_numerical': 0.02372479240806643, 'mean_word_length': 3.3181818181818183, 'pattern_counts': {'":': 0, '<': 4, '<?xml version=': 0, '>': 1, 'https://': 0, 'lorem ipsum': 0, 'www.': 0, 'xml': 0}, 'pii_count': 0, 'substrings_counts': 0, 'word_list_counts': {'cursed_substrings.json': 6, 'profanity_word_list.json': 0, 'sexual_word_list.json': 0, 'zh_pornsignals.json': 0}}
Minimal polygons with fixed lattice width

We classify the unimodular equivalence classes of inclusion-minimal polygons with a certain fixed lattice width. As a corollary, we find a sharp upper bound on the number of lattice points of these minimal polygons.

Our main result is a complete classification of minimal polygons up to unimodular equivalence, see Theorem 2.4. As a corollary, we provide a sharp upper bound on the number of lattice points of these minimal polygons. First, we show in Lemma 2.3 that each minimal polygon ∆ satisfies ls (∆) = lw(∆), where = conv{(0, 0), (1, 0), (1, 1), (0, 1)}. The latter can also be proven using results on lattice width directions of interior lattice polygons (see [4,Lemma 5.3]), but we choose to keep the paper self-contained and have provided a different proof. Moreover, we use the technical Lemma 2.2 in the proofs of both Lemma 2.3 and Theorem 2.4.
In the joint paper [4] with Castryck and Demeyer, we study the Betti table of the toric surface Tor(∆) ⊂ P (∆∩Z 2 )−1 for lattice polygons ∆. In particular, we present a lower bound for the length of the linear strand of this Betti table in terms of lw(∆), which we conjecture to be sharp. For showing this conjecture for polygons of a fixed lattice width, it essentially suffices to prove it for the minimal polygons (see [4,Corollary 5.2]). Hence, Theorem 2.4 allows us to check the conjecture using a computer algebra system. Remark 1.2. Of course, the question of classifying minimal polytopes can also be asked in higher dimensions. For instance, it can be shown that each threedimensional minimal polytope ∆ ⊂ R 3 with lw(∆) = 1 is equivalent to a tetrahedron of the form conv{(0, 0, 0), (1, 0, 0), (0, 1, 0), (1, y, z)} with 1 ≤ y ≤ z and gcd(y, z) = 1. These include the Reeve tetrahedrons (where y = 1). For comparison, there is only one minimal polygon with lattice width one up to equivalence, namely the standard simplex conv{(0, 0), (1, 0), (0, 1)}. In all dimensions k ≥ 2, among the minimal polytopes we will find back the so-called empty lattice simplices ∆ ⊂ R k , i.e. convex hulls of k + 1 lattice points without interior lattice points. If k ≥ 4, not all empty lattice simplices have lattice width 1. For more information, see [1,6,7].

The classification of minimal polygons
Throughout this section, we will use the notations from Section 1. The following result appears already in [2, Remark following Lemma 5.2], but can be proven in a shorter way.
Proof. If v and w do not form a Z-basis of Z 2 , we take a primitive vector u ∈ conv{(0, 0), v, w} such that v and u form a Z-basis. Let Q, Q be lattice points of ∆ such that Q , u − Q, u = lw u (∆). Write u = λv + µw with 0 < λ, µ and λ + µ ≤ 1. Now After applying a unimodular transformation, we may assume that u = (0, 1) and v = (1, 0), and that ∆ fits into d , hence ls (∆) = d.
By replacing v by −v, we may assume that we are in the first case. Moreover, we may choose v such that the difference min Q∈∆ P v, Q − v, P is minimal but greater than 1, and such that lw v (∆ P ) < d.
We apply a unimodular transformation so that P = (0, 0) and v = (0, 1). Let y m (resp. y M ) be the smallest (resp. greatest) y-coordinate occurring in ∆ P . Note that y m = min Q∈∆ P v, Q and y M = max Q∈∆ P v, Q , hence y m > 1 and y M − y m < d.
From now on, suppose that V is not equal to ∅ or Denote by S the set of sides that ∆ and H have in common. Then S can not contain two adjacent sides S 1 , S 2 : otherwise for the vertex P = S 1 ∩ S 2 , each side of H would have a non-empty intersection with ∆ P , contradicting the fact that there is a v ∈ {(0, 1), (1, 0), (1, −1)} with lw w (∆ P ) < d.
Assume that S ≥ 2 and take S 1 = [Q 1 , Q 2 ] ∈ S. Its adjacent sides of H contain no points of ∆ except from Q 1 and Q 2 . This implies that S = {S 1 , S 2 } where S 1 , S 2 are opposite edges of H , and that ∆ is the convex hull of these two edges. Hence ∆ is equivalent to the quadrangle conv{( , 0), (d, d − ), ( , d), (0, d − )} ⊂ H , which is of type (T 2).
If S consists of a single side S, we may assume that S = [Q 1 , Q 2 ] is the bottom edge of H . Let P 1 (resp. P 2 ) be the vertex of ∆ on the upper left diagonal side (resp. the right vertical edge) of H . If P 1 is also on the top edge of H (i.e. P 1 = ( , d)), then ∆ has only four vertices, namely Q 1 , Q 2 , P 1 , P 2 . Applying the transformation (x, y) → (x, −x + y + ), we end up with a quadrangle of type (T 2). By a similar reasoning, if P 2 is on the top edge of H (i.e. P 2 = (d, d)), we end up with type (T 2). If neither P 1 nor P 2 are on the top edge of H , then there is a fifth vertex P 3 on that top edge, and we are in case (T 3).
The only remaining case is when S = ∅, hence each edge of H contains only one point of ∆. If H and ∆ have no common vertex, then ∆ is of type (T 5). If they share one vertex, we can reduce to type (T 4) using a transformation if necessary.
Note that two common vertices of H and ∆ can never be connected by an edge of H as that edge would be in S, so there are at most three common vertices. If there are three shared vertices, then ∆ is a triangle of type (T 1), again using a transformation if necessary. So assume H and ∆ share two vertices. Together these two points occupy four edges of H and each of the other two edges of H (call them A and B) contains exactly one vertex of ∆. Take two pairs of opposite sides of H (so four sides in total) that together contain A and B, then they contain all vertices of ∆: since any common vertex of H and ∆ lies on two sides of H , they cannot lie both on the sides we didn't choose, as they are parallel. We can find a unimodular transformation mapping these sides into the four sides of d , hence ∆ is of type (T 2).
Proof. Note that there exist minimal polygons attaining the bound (see Figure 2): the simplex conv{(0, 0), (d, 0), (0, d)} is of type (T 1) and has (d + 1)(d + 2)/2 lattice points, and the quadrangle conv{(1, 0), (d, 1), (d − 1, d), (0, d − 1)} is of type (T 2) and has (d − 1) 2 + 4 lattice points. This also holds for triangles of type (T 1) with x and y non-zero. If ∆ is of type (T 3), we obtain the same upper bound (d − 1) 2 + 4 after applying a unimodular transformation that maps the bottom edge of ∆ to the left upper diagonal edge of H . We are left with triangles of type (T 1) with either x or y zero. Assume that y = 0 (the case x = 0 is similar). Then ∆ has the edge [(0, 0), (d, 0)] in common with d and its other vertex is (x, d). For each k ∈ {0, . . . , d}, the intersection of ∆ with the horizontal line on height k is a line segment of length d − k, hence it contains at most d − k + 1 lattice points. So in total, ∆ has at most d k=0 (d − k + 1) = (d + 1)(d + 2)/2 lattice points.