The Covering Radius and a Discrete Surface Area for Non-Hollow Simplices

We explore upper bounds on the covering radius of non-hollow lattice polytopes. In particular, we conjecture a general upper bound of d/2 in dimension d, achieved by the “standard terminal simplices” and direct sums of them. We prove this conjecture up to dimension three and show it to be equivalent to the conjecture of González-Merino and Schymura (Discrete Comput. Geom. 58(3), 663–685 (2017)) that the d-th covering minimum of the standard terminal n-simplex equals d/2, for every \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$n\ge d$$\end{document}n≥d. We also show that these two conjectures would follow from a discrete analog for lattice simplices of Hadwiger’s formula bounding the covering radius of a convex body in terms of the ratio of surface area versus volume. To this end, we introduce a new notion of discrete surface area of non-hollow simplices. We prove our discrete analog in dimension two and give strong evidence for its validity in arbitrary dimension.

since it is the unique non-hollow lattice d-polytope of minimum volume (see [4,Thm. 1.2]). Here 1 d = (1, . . . , 1) denotes the all-one vector in dimension d, and e i denotes the ith coordinate unit vector. 1 The covering radius of S (1 d+1 ) was computed in [10,Prop. 4.9]: Since the covering radius is additive with respect to direct sums (see Sect. 2.1), direct sums of simplices of the form S(1 l ) or lattice translates thereof also have covering radius equal to d/2. We conjecture that this procedure gives all the non-hollow lattice polytopes of maximum covering radius in a given dimension: In dimension three, translations and/or direct sums of the S(1 l )s produce nine pairwise non-equivalent non-hollow lattice 3-polytopes of covering radius 3/2, that we describe in Lemma 3.8.
One motivation for Conjecture A is as follows. The d-th covering minimum of a convex body K ⊆ R n with respect to a lattice ⊆ R n is defined as μ d (K , ) := max π μ(π(K ), π( )), where π runs over all linear projections π : R n → R d such that π( ) is a lattice. Covering minima were introduced by Kannan and Lovász [17] and interpolate between μ n (K ) = μ(K ) and μ 1 (K ), the reciprocal of the lattice width of K .
Since S(1 n+1 ) projects to S(1 d+1 ) for every d < n, we use (1) and get The converse inequality was conjectured in [10]: Conjecture B ( [10,Rem. 4.10]) For every n ∈ N and d ≤ n, In Sect. 3 we prove: Theorem 1.2 (equivalence of Conjectures A and B, Sect. 3.1) For each d ∈ N, the following are equivalent: (i) μ(P) ≤ /2 for every non-hollow lattice -polytope P and for every ≤ d.
(ii) Conjecture B holds for every ≤ d. That is, μ (S(1 n+1 )) = /2, for every , n ∈ N with ≤ d ≤ n. In Sect. 5 we derive the following closed formula for μ(S(ω)). Therein and in the rest of the paper we denote by Vol K the normalized volume of a convex body K with respect to a lattice , which equals the Euclidean volume vol K of K normalized so that a unimodular simplex of has volume one.
where π i : R d → R d−1 is the linear orthogonal projection along the line through the origin and the i-th vertex of S(ω).
Note that Vol π i (Z d ) (π i (S)) = Vol π i (Z d ) (π i (F i )), with F i being the facet of S opposite to the vertex v i . In this sense, the sum of these numbers is indeed a version of the "surface area" of S, except that the volume of each facet is computed with respect to the lattice projected from the opposite vertex. Motivated by this definition and Theorem 1.4 we propose the following conjecture, which is the main object of study in this paper: Conjecture C Let S be a d-simplex with the origin in its interior and with rational vertex directions. Then We formulate this conjecture only for simplices S rather than for arbitrary polytopes that contain the origin and have rational vertex directions, because without further study it is not clear how the discrete surface area Surf Z d S can be extended in a meaningful way. For example, we could project along the vertex directions as in the simplex case, but then the correspondence with the opposite facet is lost.
In Sect. 4 we give additional motivation for Conjecture C. We show that it implies Conjecture A (Corollary 4.3), that it holds in dimension two (Theorem 4.9), and that in arbitrary dimension it holds up to a factor of two (Proposition 4.4).
Covering criteria such as the one in Conjecture C are rare in the literature, but very useful as they reduce the question of covering to computing less complex geometric functionals such as volume or (variants of the) surface area (cf. [11,Sect. 31]). A classical inequality of this type is the following result of Hadwiger. We regard Conjecture C as a discrete analog thereof. Theorem 1.6 ( [13]) For every convex body K in R d , where vol K and surf K are the Euclidean volume and surface area of K .
Observe that the statement of Conjecture C is more intrinsic than Hadwiger's inequality. This is because the Euclidean surface area is not invariant under unimodular transformations, so that the bound in Theorem 1.6 depends on the particular representative of K in its unimodular class. Moreover, the inequality only holds for the standard lattice Z d and cannot easily be transfered to other lattices (cf. [24] for partial results for arbitrary lattices). In contrast, our proposed relation in Conjecture C is unimodularly invariant and there is no loss of generality in restricting to the standard lattice as we do (see Lemma 4.2 for details on these claims). Moreover, our proposed inequality in Conjecture C is tight for the large class of simplices S(ω). In Sect. 4.4, we complement our investigations on Conjecture C by extending it to the case where the origin lies in the boundary of the simplex S, rather than in the interior.
Another way to extend Conjecture A is to ask for the maximal covering radius among lattice polytopes with at least k ≥ 1 interior lattice points. The natural conjecture is:

Conjecture D Let k, d ∈ N be nonnegative integers. Then for every lattice d-polytope P with k interior lattice points we have
Equality holds for k = 1 if and only if P is obtained by direct sums and/or translations of simplices of the form S(1 l ), and for k ≥ 2, if and only if P is obtained by direct sums and/or translations of the segment [0, k + 1] and simplices S(1 l ).
In Sect. 6 we prove this conjecture in dimension two (see Theorem 6.3). Observe that no analog of Conjecture D makes sense for other covering minima. Indeed, the maximum d-th covering minimum μ d among non-hollow lattice n-polytopes with k interior lattice points does not depend on k or n, for d < n: It equals the maximum covering radius among non-hollow lattice d-polytopes, since every non-hollow lattice d-polytope can be obtained as the projection of a (d + 1)-polytope with arbitrarily many interior lattice points. In fact, assuming Conjecture A this maximum is given by for all n > d and k ∈ N.
Summing up, the relationship between our conjectures is as follows: A summary of our results is that all these conjectures hold in dimension two, that Conjecture A holds in dimension three, and that Conjecture C holds for the simplices of the form S(ω).

Preliminaries
This section develops some tools that will be essential for our analyses. We first describe how the covering radius behaves with respect to projections, and more importantly, that it is an additive functional on direct sums of convex bodies and lattices. Afterwards we introduce and study the concept of tight covering which facilitates our characterizations of equality, for example the one in Theorem 1.3.

Projection and Direct Sum
Lemma 2.1 Let K ⊆ R d be a convex body containing the origin, and let π : R d → R l be a rational linear projection, so that π(Z d ) is a lattice. Let Q = K ∩ π −1 (0) and let L = π −1 (0) be the linear subspace spanned by Q. Then we have A particularly interesting case of the above result is when K decomposes as a direct sum. Let R d = V ⊕ W be a decomposition into complementary linear subspaces with dim V = and dim W = d − . The direct sum of two convex bodies K ⊆ V , L ⊆ W both containing the origin is defined as The direct sum of two lattices ⊆ V , ⊆ W is defined as With these definitions we can now formulate Corollary 2.2 Let R d = V ⊕ W be a decomposition as above, let K ⊆ V , L ⊆ W be convex bodies containing the origin, and let ⊆ V , ⊆ W be lattices. Then For the other inequality, let x ∈ V be a point not covered by cK + for some c < μ (K , ) and let y ∈ W be a point not covered bycL + for somec < μ d− (L, ). We claim that x + y ∈ V ⊕ W = R d is not covered by (c +c)(K ⊕ L) + ⊕ , and thus c +c ≤ μ d (K ⊕ L, ⊕ ). Since c andc were taken arbitrarily, this implies Assume, to the contrary, that x + y ∈ (c +c)(K ⊕ L) + ⊕ , that is, x + y = (c +c)(λ p +(1−λ)q)+w + z, for some λ ∈ [0, 1], p ∈ K , q ∈ L, w ∈ , and z ∈ . Since the sums are direct, we get x = (c+c)λ p+w and y = (c+c)(1−λ)q +z, which by assumption implies (c +c)λ > c and (c +c)(1 − λ) >c. These two inequalities cannot hold at the same time, and we arrive at a contradiction.

Tight Covering
Let P be a d-polytope, let F be a facet of P, and let p be a point that is last covered by P. We say that p needs F if p ∈ relint (μ · F) + .

Lemma 2.5
Let K ⊆ R d be a convex body of covering radius μ with respect to a lattice . Then, the following properties are equivalent: (i) K is tight for .
(ii) K is a polytope and for every facet F of K and for every last covered point p, p needs F. (iii) K is a polytope and every facet of every hollow translate of μ · K is non-hollow. (iv) Every hollow translate of μ · K is a maximal hollow convex body with respect to inclusion.
Proof The equivalence of (iii) and (iv) is the characterization of maximal hollow convex bodies by Lovász [20]. For the equivalence of (i) and (iv) observe that, by definition, μ is the largest constant such that (a) μ · K has a hollow lattice translate and (b) the inequality μ(K , ) < μ(K , ) in the definition of tightness is nothing but maximality of all such hollow translates. We now show the equivalence of (i) and (ii). Suppose there is a facet F of K that is not needed by some last covered point p. Let K = conv (K ∪ {x}), where x / ∈ K is a point beyond F, meaning that x violates the inequality that defines F, but satisfies all other facet-inducing inequalities of K . Then because p is still a last covered point of K (for the same dilate μ).
Conversely, if K is not tight let K be a convex body strictly containing K and that has the same covering radius. Let F be a facet of K with relint F ⊆ int K . Let p be a point that is last covered by K . Since the covering radii are equal and K K , p must also be last covered by K . Since we chose F so that relint F is in the interior of K , p does not need F.

Lemma 2.7 Every simplex is tight for every lattice.
Proof We use Lemma 2.5. Let be a simplex of covering radius μ with respect to a lattice , and let p be a point last covered by . That is, p / ∈ int(μ ) + . Let F 0 , F 1 , . . . , F d be the facets of , with interior facet normals v 0 , . . . , v d .
Every neighborhood of p is covered by μ + , and p can only lie in lattice translates of the boundary of μ . Suppose, in order to get a contradiction, that a certain facet F i is not needed by p. This implies that for every μ + z (z ∈ ) containing p there is a facet is the translation to p of the j-th facet-defining half-space of . This implies that we have a neighborhood of p covered by the d affine half-spaces with p in the boundary corresponding to the indices j = i. This is impossible since the corresponding d normals are linearly independent. Lemma 2.8 Let K 1 and K 2 be convex bodies containing the origin and let 1 and Proof Let μ = μ(K , ) and μ = μ(K , ). Then, μ K + ⊆ μ K + , so μ ≤ μ . Remark 2.10 (i) An example where equality holds in Lemma 2.9 is the following: Let K = [−1, 1] d and let be an arbitrary refinement of The inequality in Lemma 2.9 may not be strict, even for simplices. An example is the simplex (I ⊕ I ) ⊕ I of Lemma 3.8 below. It has the same covering radius as S(1 4 ) (equal to 3/2), yet it is isomorphic to S(1 4 ) when regarded with respect to the sublattice of index two generated by its vertices and its interior lattice point. This can easily be derived from its depiction in the bottom-center of Fig. 2, or from its coordinates in Table 1 (in these coordinates the sublattice is {(x, y, z) ∈ Z 3 : x ∈ 2Z}).

Equivalence of Conjectures A and B
As an auxiliary result we first reduce Conjecture A to lattice simplices.

Lemma 3.1 Every non-hollow lattice polytope contains a non-hollow lattice simplex of possibly smaller dimension.
Proof Let P be a non-hollow lattice polytope and p ∈ int(P) ∩ Z d . Applying Carathéodory's Theorem to an expression of p as a convex combination of the vertex set of P, we obtain an affinely independent subset of vertices that still has p as a convex combination. The vertices involved in that convex combination form a non-hollow simplex contained in P. Proof One direction is trivially true. We prove the other one by induction on d. Let P ⊆ R d be a non-hollow lattice polytope. In view of Lemma 3.1, we find andimensional non-hollow lattice simplex S ⊆ P. If = d, then we simply have μ(P) ≤ μ(S). So, let us assume that < d and assume that Conjecture A is proven for any dimension < d. Assume also that S contains the origin in its interior and write L S for the linear hull of S. We now apply Lemma 2.1 to the projection π onto L ⊥ S . Observe that S ⊆ P ∩ π −1 (0) = P ∩ L S , and that S is non-hollow with respect to Z d ∩ L S and π(P) is non-hollow with respect to the lattice π(Z d ). We get that Proof of Theorem 1.2 Suppose first that for ≤ d every lattice -polytope P has μ(P) ≤ /2. Since S(1 n+1 ) projects to S(1 +1 ), we have by (1), For the converse inequality, let π : R n → R be an integer projection along which the value of μ (S(1 n+1 )) is attained. Then, π(S(1 n+1 )) is non-hollow with respect to the lattice π(Z n ), and thus For the reverse implication (ii) ⇒ (i), suppose Conjecture B holds in every dimension ≤ d. Let P be a lattice -polytope with at least one interior lattice point, which without loss of generality we assume to be the origin 0. By Corollary 3.2 we can assume P to be a simplex, and we let v 0 , . . . , v be its vertices. Let (b 0 , . . . , b ) ∈ N +1 be a tuple of the barycentric coordinates of 0 in P; that is, assume that where N = i=0 b i ≥ + 1. Consider the (N − 1)-dimensional simplex S(1 N ), and the affine projection π : R N −1 → R that sends exactly b i vertices of S(1 N ) to v i , i = 0, . . . , . Expression (5) implies that π sends the origin to the origin, which in turn implies π to be an integer projection. In particular,

Conjecture A in Dimensions 2 and 3
We here prove Conjecture A in dimensions two and three, including the case of equality.   We now show that every other non-hollow lattice polygon contains a (unimodularly equivalent) copy of one of these three, which implies Conjecture A. For this let us consider the following auxiliary family of lattice triangles with k interior lattice points: For each k ∈ N, and α ∈ {0, 1}, let

Conjecture A in dimension two
Observe that

Indeed, this implies
by Corollary 2.2 and Theorem 1.4, respectively. We see that their covering radius equals 1 for k = 1, and is strictly smaller for k ≥ 2.
The following statement might be known in the literature on lattice polygons. In absence of a clear reference we give a detailed proof for the sake of a complete presentation. Proof Without loss of generality, assume the origin is in the interior of P. Consider the complete fan whose rays go through all non-zero lattice points in P. We call this the lattice fan associated to P, and it is a complete unimodular fan. Since a 2-dimensional fan is uniquely determined by its rays, we denote by F {v 1 , . . . , v m } the fan with rays through v 1 , . . . , v m ∈ R 2 . In particular, the lattice fan of P is denoted by F {P ∩ Z 2 }.
By the classification of complete unimodular fans, see [9, Thm. V.6.6], F {P ∩ Z 2 } can be obtained (modulo unimodular equivalence) by successively refining the lattice fan of either S(1 3 ) or for some l ∈ Z ≥0 . Observe that F 0 is the lattice fan of I ⊕ I , F 1 refines the lattice fan of S(1 3 ) ∼ = M 1 (1) and, for every l ≥ 2, we have that F l is unimodularly equivalent to the lattice fan of independently of which interior lattice point of M k (α) we consider the rays of its lattice fan emanating from.
This, together with the fact that M 1 (α) ⊆ M k (α) for every k ≥ 1, implies that P contains one of M 1 (0), M 1 (1), or I ⊕ I . Proof By Lemma 3.5, unless P is one of S(1 3 ), I ⊕ I , or I ⊕ I , it strictly contains one of them. If the latter happens then its covering radius is strictly smaller than 1, since the three of them are tight by Lemmas 2.7 and 2.8.

Conjecture A in dimension three
For the three-dimensional case we introduce the following concept: Definition 3.7 A minimal d-polytope is a non-hollow lattice d-polytope not properly containing any other non-hollow lattice d-polytope.
In this language, our results in dimension 2 can be restated as: There are exactly three minimal 2-polytopes, they have covering radius 1, and every other non-hollow lattice 2-polytope has strictly smaller covering radius.
In dimension three things are a bit more complicated. To start with, instead of three direct sums of (perhaps translated) simplices of the form S(1 i ) there are nine, that we now describe.
The last five polytopes are illustrated in Fig. 2, which is borrowed from [5, p. 123]. Observe that the last three can equally be written as where (I ⊕ I ) denotes I ⊕ I translated to have the origin as a vertex and (I ⊕ I ) is I ⊕ I translated to have the origin at an endpoint of its edge of length two.
Proof That all the described direct sums are non-hollow follows from the following more general fact: The direct sum of two or more non-hollow lattice polytopes containing the origin is non-hollow if (and only if) all but at most one of the summands has the origin in its interior. Indeed, if the summand exists then its interior point(s) are interior in the sum; if it does not then the origin is an interior point in the sum. With this in mind, we only need to check that the nine described polytopes are pairwise unimodularly non-equivalent, which is left to the reader.
A second difference with dimension two is that these nine non-hollow lattice 3polytopes are no longer the only minimal ones. Minimal non-hollow 3-polytopes have been classified and there are 26 with a single interior lattice point (see [18,Thm. 3.1] and Tables 2 & 4 therein) plus the infinite family described in Theorem 3.10.
To prove Conjecture A in dimension three we show that, on the one hand, the covering radii of the 26 with a single interior lattice point can be explicitly computed and/or bounded, giving the following result, the proof of which we postpone to Appendix C. On the other hand, all the (infinitely many) minimal non-hollow 3-polytopes with more than one interior lattice point have covering radius strictly smaller than 3/2, as we now prove. For any k ∈ N and α, β ∈ {0, 1}, we define M k (α, β) to be the following lattice tetrahedron: Proof Let P be a minimal lattice 3-polytope with more than one interior lattice point, and let L be a line containing two of them. Without loss of generality we assume that L = {(0, 0, z) : z ∈ R} and L ∩ P is the segment between (0, 0, z 1 ) and (0, 0, z 2 ), with z 1 ∈ [0, 1) and z 2 ∈ (r , r + 1] for some r ∈ {2, . . . , k}, so that L contains r interior lattice points of P.

Claim 1
The minimal faces of P containing respectively (0, 0, z 1 ) and (0, 0, z 2 ) are non-coplanar edges. Let F 1 and F 2 be those faces. If one of them, say F 1 , had dimension two, then conv (F 1 ∪ {(0, 0, r )}) would be a non-hollow lattice polytope strictly contained in P. If one of them, say F 1 , had dimension zero then necessarily to be a non-hollow lattice polytope strictly contained in P. Thus, F 1 and F 2 are both edges of P. They cannot be coplanar, since otherwise there would be vertices p and q of P, one on either side of the hyperplane aff (F 1 ∪ F 2 ), and the polytope conv(F 1 ∪ {(0, 0, r ), p, q}) would be non-hollow and strictly contained in P. Hence, conv (F 1 ∪ F 2 ) is a nonhollow lattice tetrahedron and, by minimality, P = conv (F 1 ∪ F 2 ). We denote as v i and w i the vertices of F i , for i = 1, 2.
Claim 2 All the lattice points in the tetrahedron P other than the four vertices are on the line L. Let H i be the plane containing the line L and the edge F i , for i = 1, 2. and H ± 2 , all lattice points of P except its four vertices must lie on L. In particular, we have r = k.

Proof
The projection of M k (α, β) along the z direction is I ⊕ I and the fiber over the origin is the segment Moreover, the last inequality is met with equality only in the case k = 2, α = β = 1. But for M 2 (1, 1) we can consider the projection (x, y, z) → x, whose image is I and whose fiber is Thus, by Lemma 2.1 and Theorem 1.4, we have In fact, we can be more explicit:

Remark 3.12
The covering radius of M k (α, β) admits a closed expression: The first formula directly follows from Lemma 2.1. The second one also does, using We are now ready to prove Conjecture A in dimension three: Theorem 3.13 Let P be a non-hollow lattice 3-polytope. Then with equality if and only if P is unimodularly equivalent to one of the nine polytopes in Lemma 3.8.
Proof Let P be a non-hollow lattice 3-polytope, and let T be a minimal one contained in it. If T is not one of the nine in Lemma 3.8 then T , and hence P, has covering radius strictly smaller than 3/2 by either Corollary 3.11 or Theorem 3.9. If T is one of the nine and P = T then since these nine are tight by Lemmas 2.7 and 2.8.

Conjecture C
We here focus on Conjecture C. We show that it implies Conjecture A, we prove it up to a factor of two in arbitrary dimension, and we prove it in dimension two. Finally, in Sect. 4.4, we investigate how the proposed bound changes if we allow the origin to be contained in the boundary of the given simplex. As a preparation, let us first reinterpret Conjecture C in terms of (reciprocals of) certain lengths. To this end, let S = conv {v 0 , . . . , v d } be a d-simplex with the origin in its interior, and assume that it has rational vertex directions, that is, the line through the origin and the vertex v i has rational direction, for every 0 ≤ i ≤ d.
As in Conjecture C, let π i be the linear projection to dimension d − 1 vanishing at v i . Finally, let i be the lattice length of S ∩ π −1 i (0). Put differently, let u i be the point where the ray from v i through 0 hits the opposite facet of S and let i be the ratio between the length of [u i , v i ] and the length of the primitive lattice vector in the same direction. In formula: In particular, Conjecture C is equivalent to the inequality Proof By construction, we have Prop. 1.2.9]). Hence, We now also detail the claim in the introduction, that the discrete surface area defined in Definition 1.5 is invariant under unimodular transformations.

Lemma 4.2 Let S be a d-simplex with the origin in its interior and with rational vertex directions. Let A be an invertible linear transformation. Then
Proof As before we write S = conv {v 0 , . . . , v d } and we let π i be the projection vanishing at v i , for 0 ≤ i ≤ d. Clearly, AS = conv {Av 0 , . . . , Av d } and the corresponding projectionπ i vanishing at Av i can be written asπ i = π i A −1 . Therefore, we get as claimed.

Conjecture C Implies Conjecture A
Proof In view of Corollary 3.2, it suffices to consider lattice simplices. Therefore, let S = conv {v 0 , . . . , v d } be a lattice d-simplex containing the origin in its interior. Furthermore, let ω i be the lattice length of the segment [0, v i ]. Then, 1 − ω i / i is the i-th barycentric coordinate of the origin with respect to the vertices of S, so that On the other hand, for a lattice simplex we have ω i ≥ 1. Thus, assuming Conjecture C holds for S, we have

Conjecture C Holds up to a Factor of Two
In the formulation of Lemma 4.1, Conjecture C is easily proven inductively up to a factor of two.
with the lattice lengths i defined as above.
Proof As above, let u i be the intersection of the line Note, that u i lies in the relative interior of F. Also, let π i be the linear projection vanishing at v i . By the assumptions on S, the projection π i is rational and thus π i (S) is a (d − 1)-dimensional simplex having the origin in its interior and with rational vertex directions with respect to π i (Z d ).
Using Lemma 2.1 and the induction hypothesis for π i (S), we get where the j are the corresponding lattice-lengths in π i (S). Thus, to prove the proposition we only need to show that j ≥ j , for all j = i. In fact, since the one-dimensional Here, the last inequality comes from the fact that Remark 4.5 Corollary 4.11 in the next section proves Conjecture C in the plane. So we can base the inductive proof above on the stronger assumption that μ(S ) ≤ c d−1 for all d-simplices S with the origin in its interior and with rational vertex directions. This is a good bound in R 3 since c 3 = 5/8.

Conjecture C in Dimension Two
In this section we prove Conjecture C in dimension two. Our first remarks are valid in arbitrary dimension. Throughout, let S = conv {v 0 , . . . , v d } be a simplex with the origin in its interior and with rational vertex directions.
Denoting the Euclidean length of a vector x ∈ R d by x , and writing

Remark 4.6
The fact that the p i 's are primitive imposes some condition on the vector α ∈ N d+1 . Namely, for each i ∈ {0, . . . , d}, we have Indeed, let be the lattice generated by { p 0 , p 1 , . . . , p d }, and let i be the sublattice generated by { p j : j = i}. Then, the primitive vector of i in the direction of p i is which is an integer multiple of p i if, and only if, gcd(α j : j = i) = 1.
As in the previous sections, for each i let i be the lattice length of S ∩ Rv i . The following lemma says that the vectors α and β = (β 0 , β 1 , . . . , β d ) contain all the information about S needed to compute the right-hand side in (6).

Lemma 4.7 The lattice length of S ∩ Rv i equals
Proof To slightly simplify notation, we do the computations for i = 0. For this, let us use the vectors p 1 , . . . , p d as the basis for a linear coordinate system in R d . In these coordinates, p 0 becomes On the other hand, the equation of the facet of S opposite to v 0 is so that this facet intersects the line spanned by p 0 at the point Thus, the segment S ∩ Rv 0 has endpoints (α 0 /β 0 ) p 0 and (8), which implies the statement. Specializing to dimension two Our proof of Conjecture C in two dimensions is based on applying Lemma 2.1 to the projection π : R 2 → R along the direction of v i , for some fixed i ∈ {0, 1, 2}. Then, with the notation above, (i) α 0 , α 1 , and α 2 are pairwise coprime, by Remark 4.6; (ii) the lattice length of S ∩ π −1 (0) is i ; (iii) the lattice length of π(S) equals Here we use that the projection of the segment [0, v j ] = (α j /β j )[0, p j ] has length α k α j /β j , since gcd(α j , α k ) = 1 implies that π( p j /α k ) is a primitive lattice point in the projection.
We now prove Conjecture C for d = 2 which also gives another proof of Conjecture A in the plane. If (α 0 , α 1 , α 2 ) = (1, 1, 1), then consider the lattice generated by p 0 , p 1 , p 2 . Let A be the linear transformation sending e i to p i , for i = 1, 2. Then, = AZ 2 and S = AS(ω) for a suitable ω ∈ R 3 >0 . Moreover, since the p i s are primitive, the lattice lengths i are the same for every pair (S, Z 2 ), (S, ), and (S(ω), Z 2 ). Observing that ⊆ Z 2 is a sublattice, we may therefore apply Theorem 1.4 and get So, we assume that (α 0 , α 1 , α 2 ) = (1, 1, 1) and thus we can apply Theorem 4.9, which provides us with an index i ∈ {0, 1, 2} such that the inequality (9) holds. As we saw above, this implies the desired bound.

Analogs to Conjecture C with the Origin in the Boundary
As we said in the introduction, the question analogous to Conjecture A for general lattice polytopes has an easy answer: the maximum covering radius among all ddimensional lattice polytopes equals d and is attained by, and only by, unimodular simplices. This phenomenon generalizes to analogs of Theorem 1.4 and Conjecture C, which admit easy proofs. The generalization concerns the simplices S(ω), except we now allow one of the entries of ω (typically the first one) to be zero so that the origin becomes a vertex: Then

is the linear projection that forgets the i-th coordinate.
Proof S(0, ω) can be redescribed as In this form it is clear that μ (S(0, ω)) equals the unique μ ∈ [0, ∞) such that 1 d lies in the boundary of μ · S(0, ω), which equals i 1/ω i , as stated.
with equality if and only if S is unimodularly equivalent (by a transformation fixing the origin) to S(0, ω) for some ω ∈ R d >0 .
Then the point 1 is the only point in the unit cube [0, 1] d that is last covered by Z d + μ · S(0, ω). Since 1 lies in the interior of p + μ · S(0, ω), the covering radius of S(0, ω) is strictly smaller with respect to than it is with respect to Z d .
Our next results say that Proposition 4.12 and Corollary 4.13 are not only analogs (without the factor of two) of Theorem 1.4 and Conjecture C, but also a limit of them when we make one of the vertices tend to zero. We consider this as additional evidence for Conjecture C. Formally: with equality if and only if the primitive lattice vectors parallel to v 0 , . . . , v d add up to zero.
Observe that the condition for equality includes, but is more general than, the case when S is of the form S(ω).
Proof For each i, let u i be the primitive lattice vector parallel to v i , and let U = {u 0 , . . . , u d }. We choose j to be an index minimizing the (absolute value of the) determinant of U \ {u i } among all i. Observe that S is of the form S(ω) if and only if all those determinants are equal to 1.
To simplify notation, in the rest of the proof we assume j = 0 and we drop the superindex from the notation S ( j) t . Since the volume functional is continuous, we have and, for each i = 1, . . . , d, Thus, the only thing to prove is that The volume on the left-hand side does not depend on t because the vertex of S t that depends on t is projected out by π 0 . Moreover, this volume equals . Hence, the inequality follows from Both sides of (11) with equality if and only if the primitive lattice vectors parallel to v 0 , . . . , v d add up to zero.
Proof This follows from Theorem 4.14 since where the last inequality is Corollary 4.13. (10) is not true for all choices of j. Without any assumption on j the proof of Theorem 4.14 carries through up to the point where we say that (10) would follow from (11), but the latter inequality is not true in general. For a specific example, let S = conv {(0, −1), (1, 1), (−1, 1)} and consider j = 0. Then for i = 1, 2,

Remark 4.16 Equation
This gives We finally look at the intermediate case where 0 is in the boundary of S = conv {v 0 , . . . , v d } but not a vertex. We can generalize Conjecture C to

Proposition 4.17 Conjecture E ⇔ Conjecture C.
Proof The implication Conjecture E ⇒ Conjecture C is obvious, since the latter is the case I = ∅ of the former. For the other implication, for each i = 0, . . . , d, let This holds because π I is an affine bijection from S I to π I (S), so that π I (S) can be considered to be the same as S I except regarded with respect to a (perhaps) finer lattice.
Proof of Theorem 1. 4 The simplex S(ω) can be triangulated into the d + 1 simplices the half-open parallelotope spanned by the primitive edge directions of S i incident to the origin. Let i ∈ [d] 0 be fixed. Then, for any x ∈ R d there is a lattice point v i ∈ Z d such that x ∈ v i + λS i and the dilation factor λ ≥ 0 is the smallest possible. Let L i (x) be the set of all such lattice points v i . For a fixed v ∈ Z d , we define to be the region of points that are associated to v in this way. Explicitly these regions are translates of theP i , more precisely we claim that , and let λ ≥ 0 be smallest possible with x ∈ v + λS i . By the definition of S i , we can write x − v = j∈[d] 0 \{i} α j e j , for some α j ≥ 0. If there were an index j such that α j ≥ 1, then x ∈ v + e j + λS i and the intersection of this simplex and v + λS i would be a smaller homothetic copy of S i containing x. Thus, λ would not be minimal and this contradiction implies that The smallest common refinement A d of the arrangements A 0 d , . . . , A d d is known as the alcoved arrangement (see [3,Chap. 7] for a detailed description). The fulldimensional cells of A d , also called its chambers, are lattice translations of the simplices C π = conv 0, e π(1) , e π(1) + e π (2) , . . . , e π(1) + . . . + e π(d) , where π is a permutation of {1, . . . , d}. Each chamber of A d is the intersection of regions R i (v). More precisely, Therefore, the chambers C π are exactly those regions of points in R d that, for each i ∈ [d] 0 , are associated to the same lattice point, that After these preparations, we are ready to compute the covering radius of S(ω). Note that, since [0, 1] d is a fundamental cell of Z d , we only need to find the smallest dilation factor μ so that the lattice translates of μS(ω) cover the unit cube. Moreover, we may focus on what happens within one chamber C π , and by symmetry we assume that π = Id. Among all points in C Id = conv {0, e 1 , e 1 + e 2 , . . . , e 1 + . . . + e d }, we are looking for a point y which is last covered by dilations of S i + e [i] , for some i ∈ [d] 0 , and the factor of dilation needed. Here, we write e [i] = e 1 + . . . + e i . If we let i : R d → R be the linear functional which takes value 1 on the facet F i of S(ω) that is opposite to ω i e i , this is equivalent to The key observation is that y is the point where all the values | i (y − e [i] )|, 0 ≤ i ≤ d, are equal. This is because i (x − e [i] ) is nonnegative for x ∈ C Id and because there is a positive linear dependence among the functionals i , so there cannot be a point y where they all achieve a larger value than at a point where they all achieve the same value. Therefore, y satisfies the conditions The explicit expression of the functionals i is Thus we need to solve the following system of equations: This system is solved by y = (y 1 , . . . , y d ) with The value that the functionals take at y is by what we said above the covering radius of S(ω), and it is given by Observe that Theorem 1.4 says that (4) in Conjecture C is an equality for simplices of the form S(ω). Other simplices may also produce an equality, as the triangle T = S(1 2 ) ⊕ S (1 2 ) shows:

The Covering Product Conjecture
The following conjecture was proposed in [10], which was the initial motivation to compute the covering minima of the simplex S(1 d+1 ).

Equality is attained for the simplex S(1 d+1 ).
Conjecture F is known to hold for d = 2 [25]. We show it in arbitrary dimension for the simplices S(ω).
Proof Since every permutation of the vertices of S(1) is a unimodular transformation, and since the considered product functional is invariant under unimodular transformations, we can assume that ω 0 ≤ ω 1 ≤ . . . ≤ ω d . By Theorem 1.4, the covering radius of S(ω) is given by . . , d}, |I | = j, we project onto the j-dimensional coordinate plane indexed by I and obtain μ j (S(ω)) ≥ μ j (S(ω I )). In particular, choosing I = {1, . . . , j}, we have Next, in view of ω j ≥ ω j−1 ≥ . . . ≥ ω 0 , we get with strict inequality unless ω j = ω j−1 = . . . = ω 0 . Finally, computing the volumes of the pyramids over the d + 1 facets of S(ω) with apex at the origin, we obtain Vol Z d S(ω) = σ d (ω 0 , ω 1 , . . . , ω d ). Combining this with (13) and (14) yields The case i = 0 coincides with the triangle M k (0); the cases i > 0 produce kite-shaped quadrilaterals. Observe that the triangle M k (1) ∼ = S(k, 1, 1) is very similar to M k (0) but has smaller area. One could expect it to achieve a larger covering radius but it does not, as computed in Remark 3.4: Since Conjecture A holds in the plane (Corollary 3.6), to prove Conjecture D in dimension two it suffices to consider lattice polygons with at least two interior lattice points. More precisely, we show Theorem 6.3 Let P be a non-hollow lattice polygon with k ≥ 2 interior lattice points. Then μ(P) ≤ 1/2 + 1/(k + 1), with equality if and only if P is the direct sum of two lattice segments of lengths 2 and k + 1.

Example 6.2 In dimension three we can have
Remember that a lattice polytope P has (lattice) width ω ∈ N if there is an affine integer projection from P to the segment [0, ω] but not to [0, ω − 1]. Equivalently, the width is the reciprocal of the first covering minimum. Every non-hollow lattice polytope has width at least two. Our next two lemmas deal with the case of width exactly two.

Lemma 6.4 For a non-hollow lattice polygon P the following are equivalent:
(i) P has width equal to two. (ii) The interior lattice points of P are collinear.
Proof The fact that width two implies that all interior lattice points are collinear is straightforward to check. For the converse, without loss of generality assume that the k interior lattice points of P are (0, 1), . . . , (0, k), with k ≥ 2. We claim that P ⊂ [−1, 1] × R, which implies that P has width two with respect to the first coordinate. Suppose to the contrary that P has a lattice point (x, y) with |x| ≥ 2. Then the triangle with vertices (0, 1), (0, 2), and (x, y) is not unimodular, which implies that it contains at least one lattice point other than its vertices, by Pick's formula (cf. [3,Chap. 1.4]). That point is necessarily in int P and not on the line containing (0, 1) and (0, 2), a contradiction.

Lemma 6.5 Theorem 6.3 holds if P has width two.
Proof We keep the convention from the previous proof that the interior lattice points in P are given by (0, 1), . . . , (0, k), which implies that P ⊂ [−1, 1] × [0, k + 1]. Let S be the segment P ∩ ({0} × R), which contains all the interior lattice points. Observe that one endpoint of S is either (0, 0) or (0, 1/2) and the other is either (0, k + 1) or (0, k + 1/2). We distinguish three cases, depending on whether none, one, or both of them are lattice points: • If exactly one endpoint is a lattice point, then P contains a copy of M k (1), whose covering radius is strictly smaller than 1/2 + 1/(k + 1) (see Example 6.1). • If no endpoint is a lattice point, then S = {0} × [1/2, k + 1/2] and P is the convex hull of its two edges containing the endpoints of S. Without loss of generality we assume where a and b are nonnegative integers with a+b = 2k. There are two possibilities: If a = b = k, then P is a parallelogram of covering radius at most 1/2, because (1/2) P contains a fundamental domain of Z 2 . If a = b, then one of them, say a, is at least k + 1. In this case, P contains the triangle conv {(−1, 0), (−1, a), (1, 1)} whose covering radius is bounded by 1/2+1/a ≤ 1/2+1/(k +1). Since triangles are tight by Lemma 2.7, equality can only hold when P coincides with this triangle, implying b = 0. But in that case a = 2k and 1/2 + 1/a < 1/2 + 1/(k + 1), since k ≥ 2. • If both endpoints of S are lattice points, then they are given by (0, 0) and (0, k +1).
Applying Lemma 2.1 to the projection that forgets the second coordinate gives the upper bound: The fiber S has length k + 1 and the projection of P has length 2.
For the case of equality, observe that if P has lattice points u ∈ {−1} × R and v ∈ {1} × R such that the mid-point of uv is integral, then P contains (an affine image of) the direct sum of [−1, 1] and a segment of length k + 1. Since that direct sum is tight by Lemma 2.8, P either is given by this direct sum or it has strictly smaller covering radius. Thus, we can assume that P does not have such points u and v. This implies that P has a single lattice point on each side of S. Without loss of generality we can assume for an odd a ∈ [1, 2k + 1]. We claim that the proof of Lemma 2.1 implies that μ(P) is strictly smaller than λ := 1/2 + 1/(k + 1). Indeed, that proof is based on the fact that λP contains the following parallelogram Q, which is a fundamental domain for Z 2 : (−1/2, 1), (1/2, a/2), (1/2, 1 + a/2

)}.
But we can argue that, moreover, the vertices of Q are its only points not contained in the interior of λP, and that each of these vertices is in the interior of some lattice translation of λP because the vertical offset of the left and right edges of Q is not an integer. This implies λ is strictly larger than μ(P).
For the rest of the proof of Theorem 6.3 we can assume ω ≥ 3. Let m be the maximum number of collinear lattice points in our polygon P. Applying Lemma 2.1 to the projection along the line containing those m points gives Another useful fact is that along the direction that attains the width ω there are ω − 1 parallel lines intersecting the interior of P, each of them contains at most m lattice points, and with every lattice point of P lying on one of those lines. Thus These two bounds are enough to show that Proof By (15), the statement is trivial unless There are five integer solutions of this equation with ω ≥ 3 and m − 1 ≥ 3: We only need to look at the last four: • If (ω, m − 1) ∈ {(3, 5), (5, 3)} then (15) gives μ(P) ≤ 1/3 + 1/5 = 1/2 + 1/30. This is smaller than 1/2 + 1/(k + 1), because (16) gives, respectively, k ≤ 12 and k ≤ 16. (3,4), (4, 3)} then μ(P) ≤ 1/3 + 1/4 = 1/2 + 1/12. For (3,4) this is enough since (16) gives k ≤ 10. For (4, 3), however, (16) gives k ≤ 12, so we still need to consider the cases k = 11 or 12. For these we use the following argument: ω = 4 implies that, along the direction where ω is attained, we have three intermediate lattice lines intersecting P. Along these lines we have to place our k ≥ 11 points, and no more than four on each line (because m = 4). Thus, each line gets at least three points. This makes P contain a parallelogram Q with two parallel edges of lattice length two and of width two with respect to the direction of those edges. We have that Q is a fundamental domain of (2Z) 2 , which implies μ(P) ≤ μ(Q) ≤ 1/2.
Thus, the cases that remain are m ≤ 3 or (ω, m) = (3, 4). These can be proven with a case study that we only sketch here. The details can be found in the arXiv version of this article [7]. The case study goes as follows: • For the case (w, m) = (3,4), in [7,Lem. 6.7] we show that one of the following three things happen: k ≤ 5, in which case μ(P) ≤ 1/3 + 1/3 ≤ 1/2 + 1/(k + 1).
-P contains a fundamental domain Q of (2Z) 2 . As in the last paragraph of the previous proof, this implies μ(P) ≤ μ(Q) ≤ 1/2. -P has four collinear lattice points along one of the two intermediate lines in the direction attaining the width, and (at least) three of them are interior to P. In this case the intersection of P with that line has length at least 3 + 1/3 = 10/3, so (15) can be strengthened to This gives the statement if k ∈ {5, 6}. In the case k ≥ 7 we must have four collinear lattice points in one of the two intermediate lines, so that we can further improve (15) using 11/3 for the length. Then This is enough since (w, m) = (3, 4) implies k ≤ 8, by (16).
• The case m ≤ 2 is trivial: it implies that P does not have two lattice points in the same class modulo 2Z × 2Z, so it has at most four lattice points. The only non-hollow lattice polygon with at most four lattice points is S(1 3 ). • For the case m = 3, in [7, Lemma 6.8] we show that ω ≥ 3 and m = 3 imply that P cannot have three collinear interior lattice points. Since the interior lattice points cannot all be collinear (by Lemma 6.4), they must form either a unimodular triangle, a unit parallelogram, or S(1 3 ). Thus, P is contained in one of the three polygons of Fig. 3. From there, ad-hoc arguments show that always μ(P) ≤ 1/2 + 1/(k + 1).

Remark 6.7
Lattice polygons with m ≤ 3 contain at most nine lattice points in total, since they cannot have two points in the same residue class modulo (3Z) 2 . In particular, they have k ≤ 6. On the other hand, the polytopes with (ω, m) = (3, 4) have k ≤ 8 by (16). Thus, the cases not covered by Lemmas 6.5 and 6.6 have between three and eight interior lattice points. Castryck [6] enumerated all lattice polygons with k ≤ 30 up to unimodular equivalence, and showed that there are 120 + 211 + 403 + 714 + 1023+1830 of them with k equal to 3, 4, 5, 6, 7, and 8. Hence, the arguments sketched above can be replaced by a computer-aided computation of the covering radius of these 4301 polygons. (In fact, the covering radius needs only to be computed for those with m ∈ {3, 4}).

The constant M > 0 is chosen large enough such that every non-active inequality involving M is redundant.
Proof By the periodicity of the arrangement μP + Z d , we get that Hence, the covering radius equals the minimal μ ≥ 0 such that for all x ∈ [0, 1] d there exists an ∈ N P such that x ∈ μP + . This gives a mixed-integer program with infinitely many constraints. In order to turn it into a finite program, we may also interpret the covering radius as the supremum among μ ≥ 0 such that there exists an x ∈ [0, 1] d such that x / ∈ μP + N P . Modeling this non-containment condition can be done as follows: For a fixed ∈ N P , we have x / ∈ μP + if and only if there exists a defining inequality of P that is violated, that is, there exists an i ∈ {1, . . . , m} such that a i x > μb i +a i . Introducing the binary variable y i for each 1 ≤ i ≤ m and each ∈ N P , and using a large enough constant M > 0, this is modeled by the first two lines in the program, as the condition m i=1 y i ≥ 1 ensures that at least one inequality is violated for . We can replace the supremum by a maximum and the strict inequality a i x > μb i + a i by a non-strict one, since P is compact and the covering radius is in fact an attained maximum.
In order to make this formulation effective, we need to find a suitable finite subset is, hence, a finite abelian group of order V d−1 . The Cayley digraph G associated with the quotient group Z d / is defined as the directed graph with vertex set Z d / and edges (x + , x + e i + ), for x ∈ Z d and 1 ≤ i ≤ d. The following is a particular case of [21,Lems. 3 & 4] (cf. also [10,Thm. 4.11]):

Lemma B.1 In these conditions, let δ(G) be the (directed) diameter of G. That is, δ(G)
is the maximum distance from 0 to any other node of G. Then Proof  (a 1 , . . . , a d ), since p has the same barycentric coordinates in T as p has in f (T ). Also, where is the ambient lattice of T . We only need to show that = f ( ) and for this it is enough to check that the normalized volume of f (T ) with respect to equals V . This normalized volume is the order of the quotient /V Z d , and this quotient is a cyclic group generated by p + V Z d . Thus, the normalized volume is the smallest k ∈ N such that k(a 1 , . . . , a d ) ∈ V Z d . We have k = V since V = a 0 + · · · + a d gives gcd (V , a 1 , . . . , a d ) = gcd (a 0 , a 1 , . . . , a d ) = 1.
Lemma B.2 implies that a lattice simplex is determined, modulo unimodular equivalence, by the volume vector (a 0 , . . . , a d ) of any lattice point p in it, as long as gcd (a 0 , . . . , a d ) = 1. Since /V Z d is then cyclic with generator p + V Z d , we call T the cyclic simplex generated by (a 0 , . . . , a d ). We denote it by T (a 0 , . . . , a d ) and denote by G (V ; a 1 , . . . , a d ) the digraph in the statement.
In what follows we are interested in cyclic tetrahedra T (a, b, c, d). When a = 1, Lemmas B.1 and B.2 are particularly easy to apply, since then G (V ; a, b, c) coincides with the Cayley digraph of (Z V ) 2 with respect to the generators (1, 0), (0, 1), and (−b, −c). That is, G (V ; a, b, c) has (Z V ) 2 as vertex set and from each vertex (i, j) we have the following three arcs: Figure 4 shows the computation of the covering radii of T (1, 1, 1, 2) (V = 5) and T (1, 1, 2, 3) (V = 7): a grid with V 2 cells represents the nodes of  1, 1, 1, 2)) = 7/5 and μ(T (1, 1, 2, 3)) = 9/7 G(5; 1, 1, 1) and G(7; 1, 1, 2) with the origin at the south-west corner. The grid has to be regarded as a torus, so that every cell has an east, west, north, and south neighbor. A step north or east increases the distance from the origin by one, unless the cell that we move to can be reached by a shorter path. When this happens, the corresponding arc of G(V ; a, b, c) is not used in any shortest path from the origin, and we highlight in bold the corresponding wall between cells. Using this idea, one can compute the distance from the origin to each cell in a breadth-first search manner. Observe that, by commutativity, we only need to consider paths that first use edges with step (−b, −c) and then east or north steps. Thus, in order to verify that the distances we have put are correct in the whole diagram, only the distances along the path with steps (−b, −c) starting at the origin need to be checked. The cells along that path have their distances also in bold, and they coincide with the cells with bold south and west walls. The path finishes when it arrives in a cell that can be more shortly reached from the origin by only east and north steps.  Figure 5 shows that δ(G(4k; 1, 2k − 1, 2k − 1)) = 4k − 1, from which Lemmas B.1 and B.2 give μ(M k (1, 1)) = (4k + 2)/4k. The 26 minimal non-hollow lattice 3-polytopes with a single interior lattice point were classified by Kasprzyk [18]. We list them in Tables 1 and 2, in the same order as they  appear in Kasprzyk's Tables 2 and 4. Table 1 contains the 16 that are tetrahedra and  vector (a, b, c, d), consisting of the normalized volumes of the pyramids from the origin over the facets. When the volume vector is primitive, our tetrahedron equals the cyclic tetrahedron T (a, b, c, d) of Lemma B.2. In some cases, an additional "description" of the example is given, which helps us later to bound its covering radius. Table 1 The 16 minimal non-hollow tetrahedra with exactly one interior  For example, via this description we can identify the nine polytopes from Lemma 3.8 that have covering radius 3/2. The exact covering radius, computed with the algorithm of Appendix A using the SCIP solver in exact solving mode [8], is also shown. All except those of Lemma 3.8 have μ < 3/2, which provides a computer proof of Theorem 3.9. In the rest of this section we include a computer-free proof.

C.1 The Sixteen Tetrahedra
For most of the tetrahedra in Table 1 we are going to bound the covering radius based solely on the volume vector of its interior point. We need the following auxiliary result about the covering radius of some (perhaps non-lattice) triangles.
If v 2 > 1, then v strictly contains the triangle w , for some w = (w 1 , 1) ∈ R 2 ≥1 . By Lemma 2.7 triangles are tight for every lattice, so that μ( v ) < μ( w ) and it thus suffices to consider v = (a, 1), for a ≥ 1. Let F 0 be the edge of v not containing v, and let F 1 and F 2 be the edges of v not containing e 1 and e 2 , respectively. Further, let = {(x, y) : x + y = 1} be the line containing F 0 . An elementary calculation gives , a a + 2 , ∩ (F 2 + e 2 ) = 1 a + 2 , a + 1 a + 2 , This already shows that the translates {0, 1} 2 + v cover the unit cube [0, 1] 2 , for every a ≥ 1, so that μ( v ) ≤ 1 as claimed.
In order to decide the equality case, observe that in the covering of [0, 1] 2 by these four translates, the point (2/(a + 2), a/(a + 2)) is covered last, and is not contained in the interior of any of the four triangles. However, the translate (2, 1) + v may contain this point in the interior. Noting that this happens if and only if 4 − a < 2, that is, a > 2.
Remark C.2 Every non-hollow lattice triangle is isomorphic to some v considered with respect to a superlattice of Z 2 . (Let (a, b, c) be the volume vector of an interior point, with a ≤ b ≤ c, and take v = (b/a, c/a)). With this, Proposition C.1 provides another proof of Conjecture A in the plane. This approach fails in higher dimensions since, for example, we have computed that the tetrahedron (3/2,1,1) has covering radius 14/9 > 3/2.
Let T be a lattice tetrahedron with the origin 0 in its interior and let (a, b, c, d) ∈ N 4 be its volume vector, written with a ≤ b ≤ c ≤ d. Let A, B, C, D be the vertices of T labeled in the natural way (so that a is the determinant of BC D, etc.). Lemma C. 3 With this notation, suppose that the triangle OC D is unimodular. If either of the conditions (i) or (ii) below holds, then μ(T ) < 3/2. ≤ c and (a, b, c) = (1, 1, 2). (ii) a + b ≥ 4, 3c ≥ a + b + d, and (a, b, c, d) = (2, 2, 2, 2).
Proof Since the triangle OC D is unimodular, there is no loss of generality in taking C = (1, 0, 0) and D = (0, 1, 0). Once this is done, A and B must have z coordinate equal to b and −a, in order for the determinants of BC D and AC D to be a and b, respectively. In order for the determinants of AB D and ABC to be c and d, the segment AB must intersect the plane z = 0 at the point (−c/(a + b), −d/(a + b)). That is, T ∩ {z = 0} is the triangle (c/(a+b),d/(a+b)) of Proposition C.1. Then, Lemma 2.1 applied to projecting along the z coordinate gives μ(T ) ≤ μ (c/(a+b),d/(a+b)) + 1 a + b .
The condition that OC D is unimodular follows again from gcd(a, b) = 1, except for the tetrahedron (2, 2, 3, 5). • The remaining tetrahedron (2, 2, 3, 5) is marked "Pyr 4 (S(1 3 ))" because it has a facet isomorphic to S(1 3 ) (the facet in the plane x + z = 2y + 1) and the opposite vertex is at distance four from that facet. Lemma 2.1 applied to the projection along the base of the pyramid gives  Table 2 The ten minimal non-hollow non-tetrahedra with exactly one interior lattice point, with their covering radii ⎛

The Ten Non-Tetrahedra
For the ten polytopes in Table 2 we use the following direct arguments: • Four of them are the non-tetrahedra in Lemma 3.8, of covering radius 3/2.
• There are another four that are affinely equivalent to the previous four, except considered with respect to a finer lattice. They are marked as Bipyr i ( · ), where i is the index of the superlattice, since they are also (skew) bipyramids over their intersection with the plane z = 0. This intersection is, in the four cases, one of the three non-hollow lattice polygons with μ = 1. Lemma 2.1 for the projection π onto the z-coordinate gives μ(P) ≤ μ(P ∩ {z = 0}) + μ(π(P)) ≤ 1 + 1 4 = 5 4 , since π(P) has length at least four in all four cases.
• The remaining one is marked I ⊕ Q 4 because it decomposes as