Forbidden Patterns in Tropical Plane Curves

Tropical curves in $\mathbb{R}^2$ correspond to metric planar graphs but not all planar graphs arise in this way. We describe several new classes of graphs which cannot occur. For instance, this yields a full combinatorial characterization of the tropically planar graphs of genus at most five.


Introduction
A tropical plane curve is a metric graph, G, embedded in R 2 , which is dual to a regular subdivision ∆ of some lattice polygon P . In the smooth case ∆ is a unimodular triangulation of P , and the planar graph G is trivalent of genus g, where g agrees with the number of interior lattice points of P . Such graphs have been called tropically planar by Coles et al. [5]. Here we are concerned with the question: Which graphs G occur in this way?
For any fixed g ≥ 1, Castryck and Voight gave a procedure for finding finitely many lattice polygons with exactly g interior lattice points whose triangulations suffice to produce all such graphs [4]. This was employed in [2] and [5] to computationally determine the tropically planar graphs of genus g ≤ 7. Despite its success that approach is severely limited by the combinatorial explosion of the number of planar graphs and the number of triangulations of the relevant polygons. While it is doubtful that this can be pushed much further, here we seek to find obstructions for a given trivalent graph to be realizable as a tropical plane curve. It was known previously that graphs with a sprawling node [3,Proposition 4.1], crowded graphs [12,Lemma 3.5] and TIE-fighter graphs [5,Theorem 3.4] cannot occur; cf. Figure 1. Here we add further forbidden patterns to this list. As an immediate consequence, for the first time, we provide a complete combinatorial characterization of the tropically planar graphs of genus g ≤ 5, in contrast to the proof in [2] which rests on substantial computer support.
More importantly, we identify a particularly interesting "boundary case": if it is realizable, a graph with a heavy cycle forces a number of geometric restrictions for the polygon P and the triangulation ∆. As an example, this leads us to studying a natural family of triangulations, which we call anti-honeycombs. In this way we get a glimpse of tropically planar graphs of genus g = 8 and beyond.
While this work is motivated by the desire to understand tropical curves and their moduli, here we concentrate on the combinatorial aspects, which means we do not take edge lengths of metric graphs into account. This comes with the advantage that our results also hold for triangulations which are not regular. Acknowledgment. We are grateful to Dominic Bunnett and Marta Panizzut for helpful comments on an early draft of this article and for requiring a minor correction.

Lattice Polygons and Tropical Plane Curves
Let V ⊂ R d be a finite set of points. A polytopal subdivision of V is a polytopal complex which covers the convex hull conv V and uses (a subset of) the given point set V as its vertices. If that subdivision is induced by a height function on V , it is called regular. A subdivision which consists of simplices is a triangulation. A comprehensive reference on the subject is the monograph [6] by De Loera, Rambau and Santos. Here we will be concerned with the following case. Let P be a (convex) lattice polygon, i.e., P is the convex hull of finitely many points in Z 2 , and V = P ∩ Z 2 is the set of lattice points in P . A triangulation ∆ of V is unimodular if each triangle in ∆ has normalized area one, i.e., Euclidean area 1 2 . This is the case if and only if ∆ uses all points in V . We write ∂P for the boundary of P and int P for its interior. The number g(P ) = #(int P ) of interior lattice points is the genus of P ; cf. Figure 2 for an example.
Let ∆ be a not necessarily unimodular triangulation of V . The dual graph Γ = Γ(∆) is the abstract graph whose nodes are the triangles of ∆; they form an edge in Γ if two triangles share an edge in ∆. The dual graph is necessarily connected and planar, and each node has degree at most three.
Next we describe a procedure to obtain a certain graph minor of Γ. First, if there is a node of degree one, we delete it together with the unique incident edge. We repeat this step until no nodes of degree one are left. The remaining edges are nonredundant. Second, if there is a node of degree two, we delete the node and join its neighbors by an edge. Again we repeat until there are no more nodes of degree two. The resulting graph is the skeleton, which we denote G = G(∆). By construction the skeleton is a trivalent planar graph, and it does not depend on the ordering in which the edge deletions and contractions are performed. In this way each edge of G arises as an edge path in Γ. This yields a surjective map, which we denote η, from the nonredundant edges of Γ onto the edges of G. Note that this contraction map η is undefined for any edge which is redundant. This elementary description of the skeleton is algorithmic in nature and serves our purposes.
A split is a subdivision of V with exactly two maximal cells; it is necessarily regular [9,Lemma 3.5]. If U and W are the two maximal cells of a split, then the intersection U ∩ W is a common edge of the two convex polygons U and W . It spans the corresponding split line. Two splits of V are weakly compatible if there is a triangulation which refines them both. Moreover, two splits are (strongly) compatible if their split lines do not meet in the interior of P . Compatibility implies weak compatibility. The split lines of two weakly compatible splits which are not strongly compatible must meet in a point in V , i.e., an interior lattice point of P . An edge e of the connected graph G is a cut edge if deleting e creates two connected components. Otherwise e lies in some cycle of G.
The following three technical results are extracted from the proof of [5,Theorem 3.4], where cut edges are called "bridges". Lemma 1. Let e be a cut edge of G. Then η −1 (e) contains a cut edge of Γ, which is dual to an edge, s, of ∆. Moreover, the vertices of s lie on the boundary ∂P , and s spans a split line of V .
Note that η −1 (e) in Lemma 1 may contain several cut edges of Γ, and so the edge s of the triangulation ∆ may not be unique.

Lemma 2.
Let e be an edge between v 1 and v 2 in a cycle, C, of G. Further, let T 1 and T 2 be triangles in ∆ which correspond to v 1 and v 2 on the path η −1 (e). Then T 1 and T 2 share an interior lattice point of P , and this is dual to C.

Lemma 3.
Let v ∈ V be some lattice point in P with two incident triangles T 1 = conv{z, a 1 , b 1 } and T 2 = conv{z, a 2 , b 2 } both of which are in ∆. Further, let L i be the line spanned by a i and b i , for i = 1, 2. Suppose that L 1 and L 2 meet in some point, say w, such that a i is closer to w than b i , for i = 1, 2. Then the interior of the quadrangle conv{z, a 1 , w, a 2 } does not contain a point in V , unless a 1 = a 2 = w.
With this we can make a small first step to our main results. Proof. From Lemma 1 the cut edges e 1 and e 2 yields two split lines, S 1 and S 2 , which may not be unique. Unless S 1 and S 2 are compatible, they must meet in a point of the point configuration V , which does not lie on ∂P , i.e., an interior lattice point of P . Yet, by Lemma 1 the line S 1 (and similarly S 2 ) contains precisely two points of V , and neither lies in the interior.
The connection to tropical geometry comes about as follows. See the book by Maclagan and Sturmfels [11] for the general context and [2] for a detailed analysis of the case which is of interest here. Let Φ be a bivariate tropical polynomial. Its vanishing locus, the tropical hypersurface T = T (Φ), is a tropical plane curve in R 2 . The Newton polygon of Φ is a lattice polygon, and the terms of Φ correspond to lattice points in P . Moreover, the coefficients of Φ induce a regular subdivision, ∆, on those lattice points which are dual to T , i.e., T essentially agrees with Γ(∆). The tropical plane curve T is smooth if its dual subdivision ∆ is unimodular. Conversely, each unimodular regular triangulations of P arises in this way. In that case the edges of G (with their lengths, which we do not need here) describe T as a point in the moduli space M planar g of tropical plane curves of genus g. Skeleta of unimodular regular triangulations of lattice polygons are tropically planar or "troplanar" graphs in [5].  Table 4]. See Section 5 for a more comprehensive discussion of anti-honeycomb polygons, their triangulations and the notation (1).
We now sketch the forbidden patterns which are known already. A node in G is sprawling if its removal leaves three connected components; cf. Figure 1 (left). This obstruction to tropical planarity was identified in [3, Proposition 4.1] and [2,Proposition 8.3]. Note that graphs with a sprawling node were called "sprawling" in [2] and [5]. A planar embedding of a graph G is called crowded if either: there exist two bounded regions sharing at least two edges; or, there exists a bounded region sharing an edge with itself. If all planar embeddings of G are crowded, then G itself is said to be crowded.
In [12,Lemma 3.5], it is shown that crowded graphs can never be tropically planar. Additionally, [12, Corollary 3.7] describes a family of crowded graphs, and this is depicted in Figure 1 (center). A graph is a TIE-fighter if it looks like the one in Figure 1 (right). TIE-fighter graphs can never be tropically planar was shown in [5, Theorem 3.4].

Heavy Cycles and Sprawling Triangles
Let P be a lattice polygon with precisely g interior lattice points. Moreover, let ∆ be a unimodular triangulation of P with dual graph Γ = Γ(∆) and skeleton G(∆) = G. The contraction map η sends nonredundant edges of Γ to edges of G. The g interior lattice points of P bijectively correspond to the bounded regions of the planar graph G. By Euler's formula we have g = m − n + 1, where m and n are the numbers of edges and nodes of G, respectively. Our arguments in this section do not require ∆ to be regular. That is, our results extend to a class of planar graphs, which is slightly more general than tropical plane curves. Lemma 6. Assume that P contains a unimodular triangle with vertices a, b, z such that neither a nor b is a vertex of P , and z is an interior lattice point. If a and b lie on ∂P then either a and b lie on a common edge of P or the lattice point a + b − z is contained in P .
Proof. Up to unimodular equivalence we may assume a = (1, 0), b = (0, 1) and z = (0, 0). Let L and M be the lines spanned by the edges containing a and b, respectively. Suppose that (1, 1) is not contained in P . Since P is a lattice polygon, then the intersection of P with the positive orthant agrees with the unit triangle abz. This entails that L and M agree, which means that L = M defines an edge of P , and this contains a as well as b. Figure 3. Graph with the heavy cycle C Definition 7. We say that the planar graph G has a heavy cycle if it looks like the graph in Figure 3. That is, G has a cycle, C, with two vertices, v 1 and v 2 , such that v i is incident with a cut edge e i connecting v i with subgraphs G i of positive genus; a third subgraph, G 3 , also of positive genus, which shares at least one vertex with the cycle C.
In particular, a graph with a heavy cycle has genus at least four. From the classification of hyperelliptic graphs in [12], we infer that a graph with a heavy cycle is not hyperelliptic. Moreover, it follows from Lemma 1 that there are split lines, S 1 and S 2 , dual to the edges e 1 and e 2 of G. While the split lines may not be unique we just pick some. By Lemma 4 the splits S 1 and S 2 are compatible, and thus P decomposes into a union of three lattice polygons P 1 , P 2 and P ′ such that ∆ induces triangulations of all three. In this way, we get triangulations ∆ 1 , ∆ 2 and ∆ ′ such that the cycle G i is the skeleton of ∆ i for i = 1, 2, and G 3 ∪ C is the skeleton of ∆ ′ . The triangles in ∆ ′ which are dual to v 1 and v 2 are denoted T 1 and T 2 , respectively. We refer to the polygon P ′ as the heavy component of P , and likewise ∆ ′ is the heavy component of ∆. Expressed in the language of [5], the heavy component ∆ ′ arises from ∆ via "bridge-reduction".
The following lemma is the technical core of this paper. Its proof is a bit cumbersome, with several cases to distinguish. However, it is rather powerful as it delineates a fine border between trivalent planar graphs which are realizable as skeleta of tropically planar curves and those which are not.
Lemma 8. Suppose that G has a heavy cycle with cut edges e 1 and e 2 as in Figure 3. Then the triangles T 1 and T 2 in ∆ share an edge [z, w], where z is the interior lattice point dual to C, and the split lines S 1 and S 2 intersect in w, which is a vertex of P ′ , and which lies in the boundary of P .
Proof. By Lemma 2 the triangles T 1 and T 2 share a vertex z, which is the interior lattice point in P dual to the heavy cycle C. Up to a unimodular transformation we may assume that z = (0, 0), and T 1 = conv{z, (1, 0), (0, 1)}.
We distinguish two cases, depending on whether the two split lines are parallel or not. First, suppose the split line S 2 is parallel to S 1 . As T 2 is a unimodular triangle S 2 is the line through (−1, 0) and (0, −1). Then, S 2 and α, β, γ, δ ≤ 0 forces α = δ = −1 and β = γ = 0. We consider the lattice points q = (−1, 1) and r = (1, −1). As P ′ has positive genus, it follows from Lemma 6 that either q or r is an interior lattice point of P ′ . By symmetry we may assume r ∈ int P ′ . Then, the point q can either lie on the boundary of P or is not in P . Now the interval [(1, 1), (1, −1)] is the convex hull of two interior lattice points of P , but it contains the boundary point p 1 in its relative interior. Hence we obtain a contradiction.
We conclude that the split lines S 1 and S 2 intersect at some point w = (ψ, ω), as depicted in the Figure 4. Exploiting the symmetry along the line x = y, we may assume that ω ≤ 0. We use the labels from Figure 4. and intend to show that w lies in ∂P . Suppose the contrary. Then, due to Lemma 4, the point w must lie outside P . From our choice of w we see that point p 1 := (1, 0) is closer to w than (0, 1). From (2) it follows that p 2 := (γ, δ) is closer to w than (α, β). Applying Lemma 3 we infer that the interior of the quadrangle conv{z, p 1 , w, p 2 } does not contain an interior lattice point of P . We realize that since the triangles T 1 and T 2 do not share an edge, and since p 1 , p 2 ∈ ∂P therefore in this case [p 1 , p 2 ] is a boundary edge and the triangle conv{z, p 1 , p 2 } belongs to ∆.
We now look at possible coordinates of the point p 2 = (γ, δ). Since the triangle conv{z, p 1 , p 2 } is unimodular and as δ ≤ 0 we have δ = −1. Therefore, finally, we conclude that there is no lattice point p 2 = (γ, δ) such that the triangle conv{z, p 1 , p 2 } belongs to ∆. Hence, our initial assumption was wrong, and the triangles T 1 and T 2 do share the edge [z, w], where w = p 1 = p 2 is the intersection of S 1 and S 2 . Consequently, s 1 and s 2 are two distinct edges of P ′ , and so w is a vertex of P ′ . Definition 9. We say that the planar graph G has a sprawling triangle if it is of the form in Figure 5, where the edges e 1 , e 2 and e 3 represent cut edges and the shaded regions represent subgraphs with positive genus.
Note that a graph with a sprawling triangle is a special case of a graph with a heavy cycle; cf. Definition 7. An example arises from the antihoneycomb triangulation of genus 4 in Example 5 and Figure 2, and this is characterized by the following. Proof. We use the notation from Figure 5. Let z be the central interior lattice point corresponding to the sprawling triangle. Let s 1 , s 2 and s 3 be the splits corresponding to the cut edges e 1 , e 2 and e 3 . Using Lemma 8 on these three splits, we obtain that each pair of them intersects at a point on ∂P . Therefore, the dual of the point z in ∆ is a triangle, too, whose vertices we call a, b, c. It follows that the lattice triangle abc has normalized area three, and its unimodular triangulation into abz, acz, and bcz forms an induced subcomplex of ∆.
As in the proof of Lemma 6 we may assume that a = (1, 0), b = (0, 1) and z = (0, 0). It then follows that c = (−1, −1). Directly from Definition 9 it follows that the line ab is dual to a cut edge in G and thus induces a split of ∆. All the interior lattice points corresponding to a region of the subgraph G 1 must lie in those halfspace defined by ab which does not contain z. Since z is the only interior lattice point of P which does not correspond to a region of G 1 ∪ G 2 ∪ G 3 , the lattice points a, b and c must lie in the boundary of P . Now Definition 9 requires the subgraphs G 1 , G 2 and G 3 to have positive genus. This excludes the possibility that one of the three points a, b, c is a vertex of P . Thus they must lie on pairwise distinct edges of P . Now we can apply Lemma 6 three times to learn that the three lattice points is not contained in P either. Yet this implies that the intersection of the lines L and M lies strictly between the two parallel line (1, 0) + λ(1, 1) and (0, 1) + λ(1, 1). Thus there is at least one vertex of P , which must be a lattice point, on the line (0, 0) + λ(1, 1). This forces that (2, 2) lies in P .
Again the situation is symmetric, which is why also (−2, 0) and (0, −2) lie in P . The single choice left is that (2, 2) as well as (−2, 0) and (0, −2) are vertices of P , and L, M, N are the only facets. This establishes P = conv{(−2, 0), (0, −2), (2, 2)}. The claim about ∆ follows as there is only one way to extend the triangulation from the triangle abc to all of P .
Remark 11. Contracting a sprawling triangle in a planar graph yields a graph with a sprawling node, and this cannot be tropically planar. Therefore, Example 5 shows that the class of tropically planar graphs is not minor closed.

Graph with a heavy cycle and two loops
Graphs with a sprawling triangle form special cases of graphs with a heavy cycle. Thus our main technical result, Lemma 8, allowed us to derive decisive structural constraints for triangulations whose skeleton has a sprawling triangle in Theorem 10. Now we are looking into another special class of groups with a heavy cycle, aiming for a second structural result on unimodular triangulations of lattice polygons.
Definition 12. We say that a connected trivalent planar graph G has a heavy cycle with two loops if it has the form as described in Figure 6, where the shaded region represents a subgraph of positive genus. Example 13. The quadrangle Q 4 of genus 5, cf. [2, Figure 22], admits a (regular) unimodular triangulation whose skeleton features a heavy cycle with two loops, cf. [2, Figure 23].
Our aim in this section is to establish the following, Theorem 14. Suppose G is a graph with a heavy cycle C and two loops with cut edges, e 1 and e 2 , as in Figure 6. Then the heavy component P ′ can have at most three interior lattice points, and these lie on the line spanned by the edge [z, w] ∈ ∆, where z is the interior lattice point dual to C, and w is the intersection point of the split edges s 1 and s 2 . In particular, P ′ is hyperelliptic and g ≤ 5.
Proof. It follows from Lemma 8 that the two triangles share the edge [z, w], where w ∈ ∂P is the point where the two split edges meet. We will first show that the interior lattice points of P ′ lie on the line spanned by [z, w].
As previously, we fix T 1 = conv{(0, 0), (0, 1), (1, 0)} and use the labels from Figure 7; in particular, we may assume that w = (0, 1). Using Lemma 8 and the unimodularity of T 2 , we realize that p 2 must lie on the line x = −1; so let p 2 = (−1, −k) for some integer k. We use Lemma 6 on the points z = (0, 0), w = (0, 1) and p 1 = (1, 0) and infer that the point r := (1, 1) is Figure 7. This illustrates Theorem 14: general sketch (left) and the case when g(P ′ ) ≥ 4 (right), which is impossible an interior lattice point of P ; moreover it is the unique interior lattice point of P 1 . Similarly, we infer that the point r ′ := (−1, −k + 1) is the unique interior lattice point of P 2 . By considering the lines connecting the interior point r with the boundary points p 1 and w, respectively, we see that k ≥ 0. The same argument shows that the entire polygon P ′ is squeezed between the lines x = 1 and x = −1. In particular, the interior lattice points of P ′ lie on x = 0, which is the line spanned by z and w. Next we will show that g(P ′ ) ≤ 3. We assume the contrary, i.e., g(P ′ ) ≥ 4 Then, since all interior lattice points of P ′ lie on the line x = 0, the point (0, −3) must be an interior lattice point of P .
The vertical line x = 1 contains the point p 1 , which is a boundary point, and r, which is an interior lattice point. It follows that there is a lattice point (1, λ) in the boundary of P 1 for λ > 1; in particular, either (1, 2) ∈ P 1 or the boundary edge at w passes through a point in the open interval ((1, 2), (1, 1)). Also, as (0, −3) is an interior point, no point in ∂P 1 is present on the line y = 3x − 3. We realize that in this case P 1 is contained in the triangle conv{p 1 , w, (1, 3)}. However, this triangle has no valid lattice point which could be a vertex of P 1 ; recall that (1, 3) has been excluded; see Figure 7. This provides the desired contradiction, and thus g(P ′ ) ≤ 3.
The above result is sharp as Example 13 shows. The following summarizes the known obstructions to tropical planarity together with our new results.
Theorem 15. A trivalent planar graph of genus g ≥ 3 is not tropically planar if one of the following holds: (1) it contains a sprawling node, or (2) it contains a sprawling triangle and g ≥ 5, or (3) it is crowded, or (4) it is a TIE-fighter, or (5) it has a heavy cycle with two loops such that the interior lattice points of the heavy component do not align with the intersection of the two split lines. For instance, the preceding result excludes the graphs in Figure 8.

Anti-honeycombs
The purpose of this section is to define a class of lattice polygons each of which admits a special triangulation. This is motivated by Theorem 10, which characterizes one of these triangulations. We believe that the entire family deserves some attention. Consider three families of parallel lines: where k, ℓ, m ∈ Z. By picking a sixtuple π = (k, k ′ ; ℓ, ℓ ′ ; m, m ′ ), with k < k ′ , ℓ < ℓ ′ , and m < m ′ we obtain a polygon A π which is defined by three pairs of inequalities, where each pair comes from one of the parallel families (3). The polygon A π has integral vertices, and we call it the anti-honeycomb polygon of type π. The name comes about from the connection to the "honeycomb polygons" studied in [2, pp. 10ff]. Note that not all of the six inequalities need to be facet defining, whence A π is a hexagon, a pentagon, a quadrangle or a triangle. For instance, A (0,k;0,k;0,k) = conv{(−k, −k), (0, k), (k, 0)} is a triangle, and its genus equals g(A (0,k;0,k;0,k) ) = 3k 2 − 3k + 2 2 . We fix a type π = (k, k ′ ; ℓ, ℓ ′ ; m, m ′ ), and we let V = A π ∩ Z 2 be the set of lattice points in A π . Intersecting with the shifted lattice of index 3 we obtain a subset V ′ = V ∩ L of the lattice points in A π . The families of lines (3) yield weakly compatible splits of the point configuration V ′ , and they induce a triangulation ∆ ′ π of the lattice points in V ′ . Notice that all the points in V \ V ′ lie in the interior int A π . Since none of these lattice points lies on any of the lines (3) it follows that each of them is contained in the interior of a unique triangle of ∆ ′ π . Employing stellar subdivisions at the points in V \ V ′ this yields a triangulation ∆ π of V , which we call the anti-honeycomb triangulation of type π. Figure 2, Example 5, and Theorem 10 are concerned with the case π = (−2, 0; −2, 0; −2, 0) of genus 4. Figure 9 shows a genus 19 anti-honeycomb triangulation along with its corresponding skeleton for π = (0, 4; 0, 4; 0, 4).
Up to an affine transformation, the three families of lines in (3) form a Coxeter hyperplane arrangement of typeÃ 2 . This generalizes to arbitrary dimensions, and so does the construction of the anti-honeycomb triangulations. The resulting anti-honeycomb polytopes are affine images of the "alcoved polytopes" of Lam and Postnikov [10]. Further details will be explored elsewhere.

Conclusion
The classication of the tropically planar graphs of genus g ≤ 5 was obtained in [2]. Theorem 15 now allows for a combinatorial characterization: Corollary 16. A trivalent planar graph of genus g ≤ 5 is tropically planar if and only if none of the obstructions in Theorem 15 occurs.
Proof. The trivalent graphs of low genus have been classified in [1]. For g = 3 there are five such graphs, one of which has a sprawling node; the other four are tropically planar [2, Theorem 5.1]. For g = 4 there are 17 graphs: one is non-planar, three have a sprawling node, the remaining 13 are tropically planar [2, Theorem 7.1]. This was known before.
There are exactly 71 trivalent graphs of genus 5. Among them only 52 are planar without a sprawling node [2]. Of these 14 were ruled out by explicit computations, which leaves 38 tropically planar graphs of genus 5. One of the key contributions of [5, Figure 8] is to obtain general obstructions to tropical planarity, which rules out another ten, which are crowded or TIE-fighters. a b c d Figure 10. The four genus 5 graphs that are not ruled out by any prior known criteria As our new contribution we can now discuss the remaining four graphs, which are shown in Figure 10. Firstly, we observe that all of these exhibit a heavy cycle. The graph labeled "a" has a heavy cycle with two loops, but the component away from the two loops is not hyperelliptic; i.e., it is ruled out by Theorem 14. The second graph, labeled "b" also has a heavy cycle with two loops, the component (of genus 3) away from the two loops is even hyperelliptic. However, the three interior lattice points of the genus 3 component do not lie on a common line with the point of intersection of the two splits. Thus "b" is ruled out by Theorem 14, too. The graphs labeled "c" and "d" feature sprawling triangles, whence they are ruled out by Theorem 10. This completes our combinatorial characterization of the tropically planar graphs of genus at most five.
For genus 6, there are 388 trivalent graphs altogether, 354 of which are planar [1]. In [5] it was shown that 152 tropically planar graphs of genus 6 a b c d e f g h Figure 11. The eight trivalent planar graphs of genus 6, which are not tropically planar [5], but which are not covered by Theorem 15.
remain. There are 28 graphs which are non-realizable and could not be ruled out using any prior known criteria; cf. [5, Figure 17]. Out of these 28 graphs, 19 have a heavy cycle with two loops and can be ruled out using Theorem 14 because the genus is too high. One of the remaining graphs has a sprawling triangle, and thus excluded by Theorem 10. We are left with eight graphs of genus 6, which are shown in Figure 11; for these we are not aware of any a priori obstruction. It would be interesting to know how the tropically plane curves of a fixed genus fit into the moduli space of all tropical curves. For genus 3 this was recently answered in terms of modifications by Hahn et al. [8].